Bhagya

Students should go through these JAC Class 9 Maths Notes Chapter 7 Triangles will seemingly help to get a clear insight into all the important concepts.

JAC Board Class 9 Maths Notes Chapter 7 Triangles

Triangle
A plane figure bounded by three lines in a plane is called a triangle. Every triangle has three sides and three angels. If ABC is any triangle then AB, BC and CA are three sides and ∠A, ∠B and ∠C are three angles.

Types of Triangles
→ On the basis of sides we have three types of triangles:

• Scalene triangle – A triangle whose no two sides are equal is called a scalene triangle.
• Isosceles triangle – A triangle having two sides equal is called an isosceles triangle.
• Equilateral triangle – A triangle in which all sides are equal is called an equilateral triangle.

→ On the basis of angles we have three types of triangles:

• Right triangle – A triangle in which any one angle is a right angle (= 90°) is called right triangle.
• Acute triangle – A triangle in which all angles are acute (0° >angle >90°) is called an acute triangle.
• Obtuse (90° < angle < 180° ) triangle – A triangle in which any one angle is obtuse is called an obtuse triangle.

Congruent Figures
The figures are called congruent if they have same shape and same size. In other words, two figures are called congruent if they are having equal length, width and height.

In the above figures {Fig. (i) and Fig. (ii)} both are equal in length, width and height, so these are congruent figures.

Congruent Triangles
Two triangles are congruent if and only if one of them can be made to superimposed on the other, so as to cover it exactly.

If two triangles ΔABC and ΔDEF are congruent then there exist a one to one correspondence between their vertices and sides. i.e. we get following six equalities.
∠A = ∠D, ∠B = ∠E, ∠C = ∠F and AB = DE, BC = EF, AC = DF.
If ΔABC and ΔDEF are congruent under one to one correspondence A ↔ D, B ↔ E, C ↔ F then we write ΔABC ≅ ΔDEF We cannot write it as ΔABC ≅ ΔDFE or ΔABC ≅ ΔEDF or in other forms because ΔABC ≅ ΔDFE have following one-one correspondence A ↔ D, B ↔ F, C ↔ E.
Hence, we can say that ‘two triangles are congruent if and only if there exists a oneone correspondence between their vertices such that the corresponding sides and the corresponding angles of the two triangles are equal.

Sufficient Conditions for Congruence of two Triangles
→ SAS Congruence Criterion:

Two triangles are congruent if two sides and the included angle of one triangle are equal to the corresponding sides and the included angle of the other triangle.

→ ASA Congruence Criterion:

Two triangles are congruent if two angles and the included side of one triangle are equal to the corresponding two angles and the included side of the other triangle.

→ AAS Congruence Criterion:
If any two angles and a non included side of one triangle are equal to the corresponding angles and side of another triangle, then the two triangles are congruent.

→ SSS Congruence Criterion:
Two triangles are congruent if the three sides of one triangle are equal to the corresponding three sides of the other triangle.

→ RHS Congruence Criterion:
Two right angled triangles are congruent if the hypotenuse and one side of one triangle are respectively equal to the hypotenuse and one side of the other triangle.

→ Congruence Relation in the Set of all Triangles:
By the definition of congruence of two triangles, we have following results.

• Every triangle is congruent to itself i.e., ΔABC ≅ ΔABC
• If ΔABC ≅ ΔDEF then ΔDEF ≅ ΔABC
• If ΔABC ≅ ΔDEF and ΔDEF ≅ ΔΡQR then ΔΑΒC ≅ ΔΡQR

NOTE: If two triangles are congruent then their corresponding sides and angles are also congruent by CPCT (corresponding parts of congruent triangles are also congruent).

Theorem 1.
Angles opposite to equal sides of an isosceles triangle are equal.

Given:
ΔABC in which AB = AC
To Prove: ∠B = ∠C
Construction: We draw the bisector AD of ∠A which meets BC in D.
Proof: In ΔABD and ΔACD, we have
AB = AC [Given]
∠BAD = ∠CAD [∵ AD is bisector of ∠A]
And, AD = AD [Common side]
∴ By SAS criterion of congruence, we have
ΔΑΒD ≅ ΔΑCD
⇒ ∠B = ∠C [by CPCT]
Hence, proved.

Theorem 2.
If two angles of a triangle are equal, then sides opposite to them are also equal.
Given: ΔABC in which ∠B = ∠C
To Prove: AB = AC
Construction: We draw the bisector of ∠A

which meets BC in D.
Proof: In ΔABD and ΔACD, we have
∠B = ∠C [Given]
∠BAD = ∠CAD [∵ AD is bisector of ∠A]
AD = AD [Common side]
∴ By AAS criterion of congruence, we get
ΔΑΒD ≅ ΔΑCD
⇒ AB = AC [By CPCT] Hence, proved.

Theorem 3.
If the bisector of the vertical angle bisects the base of the triangle, then the triangle is isosceles.

Given: ΔABC in which AD is the bisector of ∠A meeting BC in D such that BD = CD
To Prove: ΔABC is an isosceles triangle.
Construction: We produce AD to E such that AD = DE and join EC
Proof: In ΔADB and ΔEDC, we have
AD = DE [By construction]
∠ADB = ∠CDE [Vertically opposite angles]
BD = DC [Given]
∴ By SAS criterion of congruence, we get
ΔADR ≅ ΔEDC ⇒ AB = EC ……(i)
And, ∠BAD = ∠CED [By CPCT]
But, ∠BAD = ∠CAD
∴ ∠CAD = ∠CED
⇒ AC = EC [Sides opposite to equal angles are equal]
⇒ AC = AB [By eq. (i)] Hence, proved

Some Inequality Relations In A Triangle
→ If two sides of a triangle are unequal, then the longer side has greater angle opposite to it, i.e., if in any ΔABC, AB > AC then ∠C > ∠B.
→ In a triangle the greater angle has the longer side opposite to it, ie, if in any ΔABC, ∠A > ∠B then BC > AC.
→ The sum of any two sides of a triangle is always greater than the third side, i.e., in any ΔABC, AB + BC > AC, BC + CA > AB and AC + AB > BC.
→ Of all the line segments that can be drawn to a given line, from a point, not lying on it, the perpendicular line segment is the shortest.

P is any point not lying on line l, PM ⊥ l then PM < PN.
→ The difference of any two sides of a triangle is less than the third side, i.e., in any ΔABC, AB – BC < AC, BC – CA < AB and AC – AB < BC.

Students should go through these JAC Class 9 Maths Notes Chapter 3 Coordinate Geometry will seemingly help to get a clear insight into all the important concepts.

JAC Board Class 9 Maths Notes Chapter 3 Coordinate Geometry

Co-Ordinate System:
In two dimensional coordinate geometry, we generally use two types of coordinate systems.

• Cartesian or Rectangular coordinate system.
• Polar coordinate system.

In cartesian coordinate system we represent any point by ordered pair (x, y) where x and y are called x and y coordinate of that point respectively.
In polar coordinate system we represent any point by ordered pair (r, θ) where is called radius vector and ‘θ’ is called vectorial angle of that point, which will be studied in higher classes.

Cartesian Coordinate System:
→ Rectangular Coordinate Axes:
Let XX’ and YY’ are two lines such that XX’ is horizontal and YY’ is vertical lines in the same plane and they intersect each other at O. This intersecting point is called origin Now choose a convenient unit of length and starting from origin as zero, mark off a number scale on the horizontal line XX’, positive to the right of origin O and negative to the left of origin O. Also mark off the same scale on the vertical line YY’, positive upwards and negative downwards of the origin. The line XX’ is called X-axis and the line YY’ is known as Y-axis and the two lines taken together are called the coordinate axes.

→ Quadrants:
The coordinates axes XX’ and YY’ divide the plane of graph paper into four parts XY, X’Y, X’Y’ and XY’. These four parts are called the quadrants. The parts XY, X’Y, X’Y’ and XY’ are known as the first second, third and fourth quadrants respectively.

→ Cartesian Coordinates of a Point:
Let -axis and y-axis be the coordinate axes and P be any point in the plane. To find the position of P with respect of x-axis and y-axis, we draw two perpendicular line segment from P on both coordinate axes.

Let PM and PN be the perpendiculars on x-axis and y-axis resepectively. The length of the line segment OM is called the x-coordinate or abscissa of point P. Similarly the length of line segment ON is called the y-coordinate or ordinate of point P.

Let OM = x and ON = y. The position of the point P in the plane with respect to the coordinate axes is represented by the ordered pair (x, y). The ordered pair (x, y) is called the coordinates of point P. “Thus, for a given point, the abscissa and ordinate are the distances of the given point from y-axis and x-axis respectively”.

The above system of coordinating of ordered pair (x, y) with every point in plane is called the Rectangular or Cartesian coordinate system.

Cartesian coordinate system

→ Convention of Signs:
As discussed earlier that regions XOY, Χ’ΟΥ, Χ’ΟΥ’ and ΧΟΥ’ are known as the first second, third and fourth quadrants respectively. The ray OX is taken as positive X-axis, OX’ as negative x-axis, OY as positive y-axis and OY as negative y-axis. Thus we have.
In first quadrant: x > 0, y > 0
In second quadrant: x < 0, y > 0
In third quadrant: x < 0, y < 0
In fourth quadrant: x > 0, y < 0

→ Points on Axis:
If point P lies on x-axis then clearly its distance from x-axis will be zero, therefore we can say that its ordinate will be zero. In general, if any point lies on x-axis then its y-coordinate will be zero. Similarly if any point Q lies on y-axis, then its distance from y-axis will be zero therefore we can say its x-coordinate will be zero. In general, if any point lies on y-axis then its x-coordinate will be zero.

→ Plotting of Points:
In order to plot the points in a plane, we may use the following algorithm.
Step I: Draw two mutually perpendicular lines on the graph paper, one horizontal and other vertical.
Step II: Mark their intersection point as O (origin).
Step III: Choose a suitable scale on X-axis and Y-axis and mark the points on both the axes.
Step IV: Obtain the coordinates of the point which is to be plotted. Let the point be P(a, b). To plot this point start from the origin and |a| units move along OX, OX’ according as ‘a’ is positive or negative respectively. Suppose we arrive at point M. From point M move vertically upward or downward |b| units according as ‘b’ is positive or negative respectively The point where we arrive finally is the required point P(a, b).

Distance Between Two Points:
→ If there are two points A (x₁, y₁) and B(x₂, y₂) on the XY plane, the distance between them is given by
AB = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)

Students should go through these JAC Class 9 Maths Notes Chapter 14 Statistics will seemingly help to get a clear insight into all the important concepts.

JAC Board Class 9 Maths Notes Chapter 14 Statistics

Introduction:
The branch of science known as Statistics has been used in India from ancient times. Statistics deals with collection of numerical facts ie, data, their classification and tabulation and their interpretation. In statistics we shall try to study, in detail about collection, classification and tabulation of such data.
→ Importance of Data: Expressing facts with the helps of data is of great importance in our day-to-day life. For example, instead of saying that India has a large population it is more appropriate to say that the population of India, based on the census of 2001 is more than one billion.

→ Collection of Data: On the basis of methods of collection, data can be divided into two categories:
(i) Primary data: Data which are collected for the first time by the statistical investigator or with help of his workers is called primary data. For example if an investigator wants to study the condition of the workers working in a factory then for this he collects some data like their monthly income, expenditure, number of brothers, sisters, etc.

(ii) Secondary data: Data already collected by a person or a society and may be available in published or unpublished form is known as secondary data. Secondary should be carefully used. Such data is generally obtained from the following two sources.

• Published sources
• Unpublished sources

→ Classification of Data: When the data is compiled the same form and order in which it is collected, it is known as Raw Data. It is also called Crude Data. For example, the marks obtained by 20 students of class X in English out of 10 marks are as follow:
7. 4, 9, 5, 8, 9, 6, 7, 9, 2, 0 3, 7, 6, 2, 1, 9, 8, 3, 8
(i) Geographical basis: Here, the data is classified on the basis of place or region. For example, the production of food grains of different states is shown in the following table:

(ii) Chronological classification data’s classification is based on hour, day, week, month or year, then it is called chronological classification. For example, the population of India in different years is shown in following table:

(iii) Qualitative basis: When the data is classified into different groups on the basis of their descriptive qualities and properties such a classification is known as descriptive or qualitative classification. Since the attributes cannot be measured directly they are counted on the basis of presence or absence of qualities. For example intelligence, literacy, unemployment, honesty etc. The following table shows classification on the basis of sex and employment.

(iv) Quantitative basis: If facts are such that they can be measured physically e.g, marks obtained, height, weight, age, income, expenditure etc, such facts are known as variable values. If such facts are kept into classes then it is called classification according to quantitative or class intervals.

Definitions:
→ Variate: The numerical quantity whose value varies is called a variate, generally a variate is represented by x. There are two types of variate.
(i) Discrete variate: Its magnitude is fixed. For example, the number of teachers in different branches of a institute are 30, 35, 40 etc.
(ii) Continuous variate: Its magnitude is not fixed. It is expressed in groups like 10 – 20, 20 – 30, … etc.
→ Range: The difference between the maximum and the minimum values of the variable x is called range.
→ Class frequency: in each class, the number of times a data is repeated is known as its class frequency.
→ Class limits: The lowest and the highest value of the class are known as lower and upper limits respectively of that class.
→ Classmark: The average of the lower and the upper limits of a class is called the mid value or the class mark of that class. It is generally denoted by x.
If x is the mid value and his the class size, then the class limits are \(\left(x-\frac{h}{2}, x+\frac{h}{2}\right)\).

Example:
The mid values of a distribution are 54, 64, 74, 84 and 94. Find the class size and class limits.
Solution:
The class size is the difference of two consecutive class marks, therefore class size (h) = 64 – 54 = 10.
Here the mid values are given and the class size is 10. So, class limits are:

Therefore, class limits are 49 – 59, 59 – 69, 69 – 79, 79 – 89, and 89 – 99.

Frequency Distribution:
The marks scored by 30 students of IX class of a school in the first test of Mathematics out of 50 marks are as follows:

The number of times a mark is repeated is called its frequency. It is denoted by f.

Above type of frequency distribution is called ungrouped frequency distribution. Although this representation of data is shorter than representation of raw data, but from the angle of comparison and analysis it is quite bit. So to reduce the frequency distribution, it can be classified into groups in following ways and it is called grouped frequency distribution.

(a) Kinds of Frequency Distribution: Statistical methods like comparison, decision taken etc. depends on frequency distribution. Frequency distribution are of three types.
(i) Individual frequency distribution: Here each item or original price of unit is written separately. In this category, frequency of each variable is one.

Example:
Total marks obtained by 10 students in a class.

(ii) Discrete frequency distribution: When number of terms is large and variable are discrete, i.e., variate can accept some particular values only under finite limits and is repeated then it is called discrete frequency distribution. For example, the wages of employees and their numbers is shown in following table.

The above table shows ungrouped frequency distribution and same facts can be written in grouped frequency as follows:

Note: If variable is repeated in individual distribution then it can be converted into discrete frequency distribution.
(iii) Continuous frequency distribution:
When number of terms is large and variate is continuous, i.e. variate can accept all values under finite limits and they are repeated then it is called continuous frequency distribution. For example age of students in a school is shown in the following table:

Classes can be made mainly by two methods:
(i) Exclusive series: In this method upper limit of the previous class and lower limit of the next class is same. In this method the value of upper limit in a class is not considered in the same class, it is considered in the next class.

(ii) Inclusive series: In this method value of upper and lower limit are both contained in same class. In this method the upper limit of class and lower limit of next class are not same. Some time the value is not a whole number, it is a fraction or in decimals and lies in between the two intervals then in such situation the class interval can be constructed as follows:

Cumulative Frequency:
→ Discrete frequency distribution:
From the table of discrete frequency distribution, it can be identified that number of employees whose monthly income is 4000 or how many employees of monthly income 11000 are there. But if we want to know how many employees whose monthly income is upto 11000, then we should add 10, 8, 5 and 7 i.e., number of employees whose monthly income is upto 11000 is 10 + 8 + 5 + 7 = 30. Here we add all previous frequency and get cumulative frequency. It will be more clear from the following table:

→ Continuous frequency distribution: In (a) part, we obtained cumulative frequency for discrete series. Similarly, cumulative frequency table can be made from continuous frequency distribution also.
For example, for table:

Graphical Representation Of Data:
(i) Bar graphs
(ii) Histograms
(iii) Frequency polygons
(iv) Frequency curves
(v) Cumulative frequency curves or Ogives
(vi) Pie Diagrams

(i) Bar Graphs. A bar graph is a graph that present categorical data with rectangular bars with heights or lengths proportional to the values that they represent.

Example:
A family with monthly income of ₹ 20,000 had planned the following expenditure per month under various heads: Draw bar graph for the data giyen below:

Solution:

To draw a bar graph, class intervals are marked along x-axis on a suitable scale. Frequencies are marked along y-axis on a suitable scale, such that the areas of drawn rectangles are proportional to corresponding frequencies.

(ii) Histogram: Histogram is rectangular representation of grouped and continuous frequency distribution in which class intervals are taken as base and height of rectangles are proportional to corresponding frequencies.

Now we shall study construction of histo grams related with four different kinds of frequency distributions.

• When frequency distribution is grouped and continuous and class intervals are also equal.
• When frequency distribution is grouped and continuous but class interval are not equal.
• When frequency distribution is grouped but not continuous.
• When frequency distribution is ungrouped and middle points of the distribution are given.

Now we try to make the above facts clear with some examples.

Example:
Draw a histogram of the following frequency distribution.

Solution:
Here frequency distribution is grouped and continuous and class intervals are also equal. So mark the class intervals on the x-axis i.e., age in year (scale 1 cm = 5 year). Mark frequency i.e., number of students (scale 1 cm = 25 students) on the y-axis.

Example:
The weekly wages of workers of a factory are given in the following table. Draw histogram for it.

Solution:
Here frequency distribution is grouped and continuous but class intervals are not same. Under such circumstances the following method is used to find heights of rectangle so that heights are proportional to frequencies the least.
(i) Write the least class size (h), here h = 500.
(ii) Redefine the frequencies of classes by using following formula.
Redefined frequency of class = \(\frac{\mathrm{h}}{\text { class size }}\) × frequency of class interval.
So, here the redefined frequency table is obtained as follows:

Now mark class interval on x-axis (scale 1 cm = 500) and no of workers on y-axis (scale 1 cm = 5). On the basis of redefined frequency distribution, construct rectangles A, B, C, D and E.

This is the required histogram of the given frequency distribution.

(a) Difference between Bar Graph and Histogram

• In histogram there is no gap in between consecutive rectangles as in bar graph.
• The width of the bar is significant in histogram. In bar graph, width is not important at all.
• In histogram the areas of rectangles are proportional to the frequency, however if the class size of the classes are equal then heights of the rectangle are proportional to the frequencies.

(iii) Frequency polygon: A frequency polygon is also a form of a graphical representation of frequency distribution Frequency polygon can be constructed in two ways:

• With the help of histogram.
• Without the help of histogram.

Following procedure is useful to draw a frequency polygon with the help of histogram.

• Construct the histogram for the given frequency distribution.
• Find the middle point of each upper horizontal line of the rectangle.
• Join these middle points of the successive rectangles by straight lines.
• Join the middle point of the initial rectangle with the middle point of the previous expected class interval on the x-axis.
• Join the middle point of the last rectangle with the middle point of the next expected class interval on the x-axis.

Example:
For the following frequency distribution, draw a histogram and construct a frequency polygon with it.

Solution:
The given frequency distribution is grouped and continuous, so we construct a histogram by the method given earlier Join the middle points P, Q, R, S, T of upper horizontal line of each rectangles A, B, C, D, E by straight lines.

Example:
Draw a frequency polygon of the following frequency distribution table.

Solution:
Given frequency distribution is grouped and continuous. So we construct a histogram by using earlier method. Join the middle points P, Q, R, S, T, U, V, W, X, Y of upper horizontal lines of each rectangle A, B, C, D, E, F, G, H, I, J by straight line in successions.

Example:
Draw a frequency polygon of the following frequency distribution.

Solution:
Here frequency distribution is grouped and continuous so here we obtain following table on the basis of class.

Now taking suitable scale on graph mark the points (5, 15), (15, 12), (25, 10) (35, 4), (45, 11), (55, 14).

Measures Of Central Tendency:
The commonly used measure of central tendency are:
(a) Mean,
(b) Median,
(c) Mode

(a) Mean: The mean of a number of observations is the sum of the values of all the observations divided by the total number of observations. It is denoted by the symbol \(\overline{\mathrm{x}}\), read as x bar.
(i) Properties of mean:
→ If a constant real number ‘a’ is added to each of the observations, then new mean will be \(\overline{\mathrm{x}}\) + a.
→ If a constant real number ‘a’ is subtracted from each of the observations, then new mean will be \(\overline{\mathrm{x}}\) – a.
→ If a constant real number ‘a’ is multiplied with each of the observations, then new mean will be \(\overline{\mathrm{x}}\).
→ If each of the observation is divided by a constant no ‘a’ then new mean will be \(\frac{\overline{\mathrm{x}}}{\mathrm{a}}\).

(ii) Mean of ungrouped data: If x₁, x₂, x₃, ….., x_n are n values (or observations) then A.M. (Arithmetic mean) is

i.e. product of mean and no. of items gives sum of observations.

Example:
Find the mean of the factors of 10.
Solution:
Factors of 10 are 1, 2, 5 and 10.
\(\overline{\mathrm{x}}\) = \(\frac{1+2+5+10}{4}=\frac{18}{4}\) = 4.5

Example:
If the mean of 6, 4, 7, P and 10 is 8, find P.
Solution:
8 = \(\frac{6+4+7+P+10}{5}\)
⇒ P = 13

(iii) Method for Mean of ungrouped frequency distribution:

Then, mean \(\overline{\mathrm{x}}\) = \(\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\sum \mathrm{f}_{\mathrm{i}}}\)

(iv) Method for Mean of grouped frequency distribution:
Example:
Direct Method: For finding mean

\(\overline{\mathrm{x}}\) = \(\frac{\sum f_i x_i}{\sum f_i}=\frac{1430}{40}\) = 35.75

(v) Combined Mean:
\(\overline{\mathrm{x}}\) = \(\frac{\mathrm{n}_1 \overline{\mathrm{x}}_1+\mathrm{n}_2 \overline{\mathrm{x}}_2+\ldots \ldots .}{\mathrm{n}_1+\mathrm{n}_2+\ldots \ldots}\)

(vi) Uses of Arithmetic Mean

• It is used for calculating average marks obtained by a student.
• It is extensively used in practical statistics.
• It is used to obtain estimates.
• It is used by businessmen to find out profit per unit article, output per machine, average monthly income and expenditure etc.

(b) Median: Median of a distribution is the value of the variable which divides the distribution into two equal parts.
(i) Median of ungrouped data:

• Arrange the data in ascending or descending order.
  Count the no. of observations (Let there be ‘n’ observations)
  If n is odd then median = value of \(\left(\frac{\mathrm{n}+1}{2}\right)^{\mathrm{th}}\) observation.
  If n is even then median = value of mean of \(\left(\frac{n}{2}\right)^{\text {th }}\) observation and \(\left(\frac{\mathrm{n}}{2}+1\right)^{\mathrm{th}}\) observation.

Example:
Find the median of the following values:
37, 31, 42, 43, 16, 25, 39, 45, 32
Solution:
Arranging the data in ascending order, we have
25, 31, 32, 37, 39, 42, 43, 45, 46
Here the number of observations, n
= 9 (odd)
∴ Median
= Value of \(\left(\frac{9+1}{2}\right)^{\text {th }}\) observation
= Value of 5th observation = 39.

Example:
The median of the observation 11, 12, 14, 18, x + 2, x + 4, 30, 32, 35, 41 arranged in ascending order is 24. Find the value of x.
Solution:
Here, the number of observations, n = 10.
Since n is even, therefore
Median

(ii) Uses of Median:
(A) Median is the only average to be used while dealing with qualitative data which cannot be measured quantitatively but can be arranged in ascending or descending order of magnitude.
(B) It is used for determining the typical value in problems concerning wages, distribution of wealth etc.

(c) Mode:
(i) Mode of ungrouped data (By inspection only): Arrange the data in an array and then count the frequencies of each variate. The variate having maximum frequency is the mode.

Example: Find the mode of the following array of an individual series of scores 7, 10, 12, 12, 12, 11, 13, 13, 17.

∴ Mode is 12
(ii) Uses of Mode: Mode is the average to be used to find the ideal size, e.g., in business forecasting, in manufacture of ready-made garments, shoes etc.

Empirical Relation Between Mode, Median And Mean:
Mode = 3 Median – 2 Mean
Range:
The range is the difference between the highest and lowest scores of a distribution. It is the simplest measure of dispersion. It gives a rough idea of dispersion. This measure is useful for ungrouped data.
(a) Coefficient of the Range:
If R and h are the lowest and highest scores in a distribution then the coefficient of the Range = \(\frac{\mathrm{h}-\mathrm{R}}{\mathrm{h}+\mathrm{R}}\)

Example: Find the range of the following distribution: 1, 3, 4, 7, 9, 10, 12, 13, 14, 16 and 19.
Solution:
R = 1, h = 19
∴ Range = h – R = 19 – 1 = 18.

Students should go through these JAC Class 9 Maths Notes Chapter 9 Areas of Parallelograms and Triangles will seemingly help to get a clear insight into all the important concepts.

JAC Board Class 9 Maths Notes Chapter 9 Areas of Parallelograms and Triangles

Polygonal Region

Polygon region can be expressed as the union of a finite number of triangular regions in a plane such that if two of these intersect, their intersection is either a point or a line segment. It is the shaded portion including its sides as shown in the figure.

Parallelogram

(a) Base and Altitude of a Parallelogram:
→ Base: Any side of parallelogram can be called its base.
→ Altitude: The perpendicular to the base from the opposite side is called the altitude of the parallelogram corresponding to the given base.
In the given Figure
→ DL is the altitude of ||^gm ABCD corresponding to the base AB.
→ DM is the altitude of ||^gm ABCD, corresponding to the base BC.

Theorem 1.
A diagonal of parallelogram divides it into two triangles of equal area.

Proof:
Given: A parallelogram ABCD whose one of the diagonals is BD.
To prove: ar(ΔABD) = ar(ΔCDB).
Proof: In ΔABD and ΔCDB;
AB = DC [Opp sides of a ||^gm]
AD = BC [Opp. sides of a ||^gm]
BD = BD [Common side]
∴ ΔΑΒD ≅ ΔCDB [By SSS]
ar(ΔABD) = ar (ΔCDB) [Areas of two congruent triangles are equal]
Hence, proved.

Theorem 2.
Parallelograms on the same base or equal base and between the same parallels are equal in area.

Proof:
Given: Two Parallelograms ABCD and ABEF are on the same base AB and between the same parallels AB and FC.
To Prove: ar(||^gm ABCD) = ar(||^gm ABEF)
Proof: In ΔADF and ΔBCE, we have
AD = BC [Opposite sides of a ||^gm]
and AF = BE [Opposite sides of a ||^gm]
AD || BC (Opposite sides of a parallelogram)
⇒ ∠ADF = ∠BCE (Alternate interior angles)
∴ ΔADF ≅ ΔBCE [By AAS]
∴ ar(ΔADF) = ar(ΔBCE) …..(i)
[Congruent triangles have equal area]
∴ ar (||^gm ABCD) = ar(ABED) + ar(ΔBCE)
= ar (ABED) + ar(ΔADF) [Using (1)]
= ar(||^gm ABEF).
Hence, ar(||^gm ABCD) = ar(||^gm ABEF).
Hence, proved.

Theorem 3.
The area of parallelogram is the product of its base and the corresponding altitude.

Proof:
Given: A ||^gm ABCD in which AB is the base and AL is the corresponding height.
To prove: Area (||^gm ABCD) = AB × AL.
Construction: Draw BM ⊥ DC so that rectangle ABML is formed.
Proof: ||^gm ABCD and rectangle ABML are on the same base AB and between the same parallel lines AB and LC.
∴ ar(||^gm ABCD) = ar(rectangle ABML)
= AB × AL
∴ area of a ||^gm = base × height.
Hence, proved.

Area Of A Triangle
Theorem 4.
Two triangles on the same base (or equal bases) and between the same parallels are equal in area.

Proof:
Given: Two triangles ABC and PCB on the same base BC and between the same parallel lines BC and AP.
To prove: ar(ΔABC) = ar(ΔPBC)
Construction: Through B, draw BD || CA intersecting PA produced in D and through C, draw CQ || BP, intersecting line AP produced in Q.
Proof: We have, BD || CA (By construction) And, BC || DA [Given]
∴ Quad. BCAD is a parallelogram.
Similarly, Quad. BCQP is a parallelogram.
Now, parallelogram BOQP and BCAD are on the same base BC, and between the same parallels.
∴ ar (||^gm BCQP) = ar (||^gm BCAD)….(i)
We know that the diagonals of a parallelogram divides it into two triangles of equal area.
∴ ar(ΔPBC) = \(\frac{1}{2}\)(||^gm BCQP) …..(ii)
And ar (ΔABC) = \(\frac{1}{2}\)(||^gm BCAD)….(iii)
Now, ar (||^gm BCQP) = ar(||^gm BCAD) [From (i)]
\(\frac{1}{2}\)ar(||^gm BCAD) = \(\frac{1}{2}\)ar(||^gm BCQP)
Hence, \(\frac{1}{2}\)ar(ABC) = ar(APBC) (Using (ii) and (iii) Hence, proved.

Theorem 5.
The area of a trapezium is half the product of its height and the sum of the parallel sides.

Proof:
Given: Trapezium ABCD in which AB || DC, AL ⊥ DC, CN ⊥ AB and AL = CN = h (say). AB = a, DC = b.
To prove: ar(trap. ABCD) = \(\frac{1}{2}\)h × (a + b).
Construction: Join AC.
Proof: AC is a diagonal of quad. ABCD.
∴ ar(trap. ABCD) = ar(ΔABC) + ar(ΔACD)
= \(\frac{1}{2}\)h × a+\(\frac{1}{2}\)h × b= \(\frac{1}{2}\)h(a + b).
Hence, proved.

Theorem 6.
Triangles having equal areas and having one side of the triangle equal to corresponding side of the other, have their corresponding altitudes equal.

Proof:
Given: Two triangles ABC and PQR such that
(i) ar(ΔABC) = ar(ΔPQR) and
(ii) AB = PQ.
CN and RT and the altitudes corresponding to AB and PQ respectively of the two triangles.
To prove: CN = RT.
Proof: In ΔABC, CN is the altitude corresponding to the side AB
ar(ΔABC) = \(\frac{1}{2}\)AB × CN ……(i)
Similarly, ar(ΔPQR) = \(\frac{1}{2}\)PQ × RT ……(ii)
Since, ar(ΔABC) = ar(ΔPQR) [Given]
∴ \(\frac{1}{2}\)AB × CN = PQ × RT
Also, AB = PQ [Given]
∴ CN = RT Hence, proved.

Students should go through these JAC Class 9 Maths Notes Chapter 15 Probability will seemingly help to get a clear insight into all the important concepts.

JAC Board Class 9 Maths Notes Chapter 15 Probability

Probability:
Theory of probability deals with measurement of uncertainty of the occurrence of some event or incident in terms of percentage or ratio.
→ Sample Space: Set of all possible outcomes.
→ Trial: Trial is an action which results in one of several outcomes.
→ An experiment: An experiment is any kind the of activity such as throwing a die, tossing a coin, drawing a card. The different possibilities which can occur during an experiment. e.g. on throwing a dice, 1 dot, 2 dots, 3 dots, 4 dots, 5 dots, 6 dots can occur.
→ An event: Getting a ‘six’ in a throw of dice, getting a head, in a toss of a coin.
→ A random experiment: is an experiement that can be repeated numerous times under the same conditions.
→ Equally likely outcomes: The outcomes of a sample space are called equally likely if all of them have same chance of occurring.
→ Probability of an event A: Written as P(A) in a random experiment and is defined as –
P(A) = \(\frac{\text { Number of outcomes in favour of A }}{\text { Total number of possible outcomes }}\)

Important Properties:
(i) 0 ≤ P(A) ≤ 1
(ii) P (not happening of A) + P (happening of A) = 1
or, P(\(\bar{A}\)) + P(A) = 1
∴ P(\(\bar{A}\)) = 1 – P(A)
Probability of the happening of A = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}\)
= \(\frac{\mathrm{m}}{\mathrm{m}+\mathrm{n}}\)
Probability of not happening of A (failing of A) = \(\frac{\mathrm{n}}{\mathrm{m}+\mathrm{n}}\)
where event A can happen in m ways and fail in n ways. All these ways being equally likely to occur.

Problems of Die:
A die is thrown once. The probability of
→ Getting an even number in the throwing of a die: the total number of outcomes is 6. Let A be the event of getting an even number then there are three even numbers 2, 4, 6.
∴ Number of favourable outcomes = 3.
P(A) = \(\frac{\text { Number of favourable outcomes }}{\text { Total number of possible outcomes }}\)
= \(\frac{3}{6}=\frac{1}{2}\)

→ Getting an odd number: Total no. of outcomes = 6,
favourable outcomes = 3 i.e. {1, 3, 5}
∴ P(A) = \(\frac{3}{6}=\frac{1}{2}\)
→ Getting a natural number P(A) = \(\frac{6}{6}\) = 1
→ Getting a number which is multiple of 2 and 3 = \(\frac{1}{6}\)
→ Getting a number ≥3 i.e. {3, 4, 5, 6},
P(A) = \(\frac{4}{6}\) = \(\frac{2}{3}\)
→ Getting a number 5 or 6, P(A) = \(\frac{2}{6}\) = \(\frac{1}{3}\)
→ Getting a number ≤5 i.e. {1, 2, 3, 4, 5},
P(A) = \(\frac{5}{6}\)

Problems Concerning Drawing a Card:

• A pack of 52 cards
• Face cards (King, Queen, Jack)

JAC Board Class 9th Science Important Questions Chapter 11 Work and Energy

Multiple Choice Questions

Question 1.
The gravitational potential energy of an object is due to
(a) it’s mass
(b) its acceleration due to gravity
(c) its height above the earth’s surface
(d) all of the above
Answer:
(d) all of the above

Question 2.
The unit of work is the joule. The other physical quantity that has the same unit is
(a) power
(b) velocity
(c) energy
(d) force
Answer:
(c) energy

Question 3.
If the velocity of a body is doubled, its kinetic energy
(a) gets doubled
(b) becomes half
(c) does not change
(d) becomes 4 times
Answer:
(d) becomes 4 times

Question 4.
A student carries a bag weighing 5 kg from the ground floor to his class on the first floor that is 2 m high. The work done by the boy is
(a) 1J
(b) 10J
(c) 100J
(d) 1000J
Answer:
(c) 100J

Question 5.
How much time will be required to perform 520 J of work at the rate of 20 W?
(a) 24s
(b) 16s
(c) 20s
(d) 26s
Answer:
(d) 26s

Question 6.
A body of mass 2 kg is dropped from a height of lm. Its kinetic energy as it touches the ground is
(a) 19.6N
(b) 19.6J
(c) 9.8m
(d) 9.8J
Answer:
(b) 19.6J

Question 7.
The unit of power is
(a) watt per second
(b) joule
(c) kilojoule
(d) joule per second
Answer:
(d) joule per second

Question 8.
A coolie carries a load of 500 N to a distance of 100 m. The work done by him is
(a) 5 Nm
(b) 50,000 Nm
(c) 0 Nm
(d) 1/5 N m
Answer:
(c) 0 Nm

Question 9.
Power is a measure of the
(a) rate of change of momentum
(b) force which produces motion
(c) change of energy
(d) rate of change of energy
Answer:
(d) rate of change of energy

Question 10.
If the speed of an object is doubled, its kinetic energy is
(a) doubled
(b) quadrupled
(c) halved
(d) tripled
Answer:
(b) quadrupled

Question 11.
Which of the following is not correct?
(a) Energy is the ability of doing work
(b) Work can be expressed as F × s
(c) Unit of power is joule
(d) Power is the amount of work done per unit of time
Answer:
(d) Power is the amount of work done per unit of time

Question 12.
kW h is the unit of
(a) acceleration
(b) work
(c) power
(d) energy
Answer:
(c) power

Question 13.
Considering air resistance negligible, the sum of potential and kinetic energies of a free falling body would be
(a) zero
(b) increasing
(c) decreasing
(d) fixed
Answer:
(d) fixed

Question 14.
Two bodies of masses m] and m2 have equal kinetic energies. If P₁ and P₂ are their respective momenta, the ratio of P₁ to P₂ is
(a) m₁ : m₂
(b) m₂ : m₁
(c) \(\sqrt{\mathrm{m}_{1}} : \sqrt{\mathrm{m}_{2}}\)
(d) \(m_{1}^{2} : m_{2}^{2}\)
Answer:
(c) \(\sqrt{\mathrm{m}_{1}} : \sqrt{\mathrm{m}_{2}}\)

Question 15.
A light and a heavy body have equal momenta Which one has greater kinetic energy?
(a) The lighter body
(b) The heavier body
(c) Both have same KE
(d) None of these
Answer:
(b) The heavier body

Analysing & Evaluating Questions

Question 16.
A car is accelerated on a levelled road and attains a velocity four times of its initial velocity. In this process the potential energy of the car
(a) does not change
(b) becomes twice to that of initial
(c) becomes 4 times that of intial
(d) becomes 16 times that of initial
Answer:
(a) does not change

Question 17.
An iron sphere of mass 10 kg has the same diameter as an aluminium sphere of mass 3.5 kg. Both spheres are dropped simultaneously from a tower. When they are 10 m above the ground, they have the same
(a) acceleration
(b) momenta
(c) potential energy
(d) kinetic energy
Answer:
(a) acceleration

Question 18.
A girl is carrying a school bag of 3 kg mass on her back and moves 200 m on a levelled road. The work done against the gravitational force will be (g = 10 m^-2)
(a) 6 × 10³J
(b) 6J
(c) 0.6J
(d) zero
Answer:
(d) zero

Assertion Reason Questions

Directions: In the following questions, the Assertions and the Reasons have been put forward. Read the statements carefully and choose the correct alternative from the following:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.
(B) The assertion and the reason are correct but the reason is not the correct explanation of the assertion.
(C) The assertion is true but the reason is false.
(D) Both the statements are false.
1. Assertion: Work, the product of force and displacement, is a vector quantity.
Reason: Product of two vector quantities is always a vector quantity.
Answer:
(D) Both the statements are false.

2. Assertion: When a book is moved from a table to the top of an almirah, its potential energy increases.
Reason: Higher the height of a body from the ground level, higher is its potential energy.
Answer:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.

3. Assertion: A person carrying a suitcase on his head is not doing any work. Reason: The force applied by the person is acting in the downward direction.
Answer:
(C) The assertion is true but the reason is false.

4. Assertion: The work done by the force of gravity on the moon revolving around the earth is zero.
Reason: The gravitational force and the displacement of moon are at right angles to each other.
Answer:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.

5. Assertion: Work done on an object can be positive, negative or zero.
Reason: Work done is a scalar quantity.
Answer:
(B) The assertion and the reason are correct but the reason is not the correct explanation of the assertion.

Very Short Answer Type Questions

Question 1.
Define work.
Answer:
Work is said to be done when a force applied on a body produces a displacement of the body. It is given by W = F x s where ‘F’ is the force applied and ‘s’ is the displacement caused.

Question 2.
State reason why work is a scalar quantity.
Answer:
Work is the product of force (F) and displacement (s). Since both F and s are vector quantities and the dot product of vector quantities produces a scalar quantity, therefore, work is a scalar quantity.

Question 3.
Name two kinds of potential energies.
Answer:
Gravitational potential energy and elastic potential energy.

Question 4.
If the work done is 20J and displacement is 2m, find the force applied.
Answer:
Given, W = 20J and s = 2m.
W = F × s
20 = F × 2
F= 10 N

Question 5.
Name the energy stored in a rubber band when it is stretched?
Answer:
On stretching a rubber band, potential energy is stored in it.

Question 6.
State the law of conservation of energy.
Answer:
It states that energy can neither be created nor destroyed. It can only change its form.

Question 7.
When a book is lifted from a table, against which force is the work done?
Answer:
Work is done against the force of gravity.

Question 8.
Define the commercial unit of energy.
Answer:
The commercial unit of energy is kW h (kilowatt hour). 1 kW h is the energy used in one hour at the rate of 1000 J per second.
1 kWh = 1 kW × 1 h = 3.6 × 10⁶j

Question 9.
A light and a heavy body have equal kinetic energies. Which one is moving faster?
Answer:
The lighter body is moving faster because for the same kinetic energy, velocity is inversely proportional to the mass.
Analysing & Evaluating Question uestions

Question 10.
Two objects of same mass are placed at positions A and B as shown in the figure. Both of them are raised to the position C. In which case, the potential energy is more?
Answer:
The object at A will gain more potential energy than the object at B. But the final potential energy of both A and B will be equal when raised to position C.

Question 11.
A rocket is moving up with a velocity v. If the velocity of this rocket is suddenly tripled, what will be the ratio of the itwo kinetic energies?
Answer:
K.E. °c (velocity)²
When the velocity is tripled, K.E. increases by a factor of 9 ( = 3²)
Thus, the ratio of the two kinetic energies is 1 : 9.

Question 12.
Can any object have momentum even if its mechanical energy is zero? Explain.
Answer:
No. Zero mechanical energy means no potential energy and no kinetic energy. Zero kinetic energy means, zero velocity. As a result, momentum is also zero (as P = mv).

Short Answer Type Questions

Question 1.
Calculate the work done by a man in rotating a wheel of an amusement slide in a fair 40 times in 1 minute?
Answer:
The man is rotating the wheel of an amusement slide by just standing at a place. This concludes that the wheel is not undergoing any displacement. Since displacement is zero, therefore, work done is zero.

Question 2.
Define positive work done and negative work done.
Answer:

1. Positive work (done: Work done is positive when the displacement occurs in the direction of force.
2. Negative work done: Work done is negative when the displacement occurs opposite to the direction of force.

Question 3.
In which of the following cases, work is said to be done?
1. When we push a table and the table is displaced.
2. When a person holds a book in his hand and keeps it stationary.
3. When a wire is twisted.

Answer:

1. When we push a table and the force applied by us is large enough to move it from its original position, then work is said to be done.
2. When a person holds a book in his hand and keeps it stationary, there occurs no movement of the book. In this case, though a force is constantly being applied, there is no displacement and hence work done is zero.
3. When a wire is twisted, the shape of the wire changes which concludes that work is done as there occurred changes in the configuration of the wire.

Question 4.
What will be the nature of work done when the force acting on a body retards its motion? Justify your answer by quoting examples.
Answer:
When force retards the motion of a body, the motion is stopped, i.e., a force opposite to the direction of the motion is applied. Thus, a negative work is done by the force.
For example:

1. In tug of war, the work done by the losing team is negative.
2. When a ball is thrown up in the air, the gravitational force acting downwards upon the ball does negative work on the ball.

Question 5.
What is gravitational potential energy?
Answer:
The gravitational potential energy of an object at a point above the ground
is defined as the work done in raising an object from the ground to that point against gravity.
G_PE = mgh

Question 6.
Differentiate between potential energy and kinetic energy.
Answer:

Question 7.
State two situations where energy is supplied but no work is done.
Answer:
(a) A person pushing a heavy rock is using all the energy but if the rock does not move, no work is done, (b) A person standing with heavy load on his head is spending energy in doing this, but no work is done.

Question 8.
How are work and energy related to each other?
Answer:
An object having a capability to do work is said to possess energy. The object which does work loses energy and the object on which work is done, gains energy. The unit of both energy and work is joule.

Question 9.
On what factor does the gravitational potential energy depend?
Answer:
Gravitational potential energy depends on the height of object from the ground level zero level we choose. Example: A ball tossed from the second floor of a building will attain a height, say h, from its roof, but from the first floor its height will be h ‘where h > h’.
Hence, the potential energy of the ball on the first floor level is less as compared to that on the second floor.

Question 10.
Write the form of energy possessed by the body in the following situations:
(a) A coconut falling from tree
(b) An object raised to a certain height
(c) Blowing wind
(d) A child driving a bicycle on the road
Answer:
(a) Kinetic energy + Potential energy
(b) Potential energy
(c) Kinetic energy
(d) Kinetic energy

Question 11.
What is energy? Give the unit of energy. Name the different forms of energy.
Answer:
Energy of a body is defined as its capacity or ability to do work. When a body is capable of doing more work, it is said to possess more energy.
The SI unit of energy is joule (J).
Energy has many forms: potential energy, kinetic energy, heat energy, chemical energy, electrical energy, light energy, solar energy, etc.

Question 12.
Derive an equation for kinetic energy of an object?
Answer:
The kinetic energy of a body can be determined by calculating the amount of work required to set the body into motion with the velocity ‘v’ from its state of rest. Suppose,
m = mass of the body
u = 0 = initial velocity of the body
F = force applied on the body
a = acceleration produced in the body in
the direction of force
v = final velocity of the body
s = distance covered by the body
As v² – u² = 2as
= v² – 0² = 2as
a= \(\frac{v^{2}}{2 s}\)
As the force and displacement are in the same direction, the work done on the body is
W = Fs = mas = m \( \frac{\mathrm{v}^{2}}{2 \mathrm{~s}}\)s = \(\frac{1}{2}\) mv²
This work done appears as the kinetic energy of the body.
∴ KE = \(\frac{1}{2}\) mv²

Question 13.
Derive an equation for potential energy?
Answer:
Let the work done on the object against gravity be W.
Work done, W = force × displacement
Work done, W = mg × h
Work done, W = mgh
Since work done on the object is equal to mgh, an energy equal to mgh units is gained by the object. This is the potential energy (E_p) of the object.
E_p = mgh

Question 14.
An electric heater of 1000 W is used for 2 hours a day. What is the cost of using it for a month of 28 days, if 1 unit costs ₹ 3.00?
Answer:
Here, P = 1000W = lkW
Total time, t = 2 × 28 hours = 56 hours
Total energy consumed = P × t
= 1 kW × 56 h = 56 kW h
Cost of 1 kWh = ₹ 3.00
Cost of 56 kWh = 3 × 56 = ₹ 168.

Analysing & Evaluating Questions

Question 15.
Two identical pointed objects made from iron and wood are allowed to fall on a heap of sand from the same height. The iron object penetrates more in sand than the wooden object. Which of the objects has more potential energy?
Answer:
Of the two identical objects, the one made from iron will have greater mass. So when it falls from a height, it will possess greater kinetic energy as compared to the wooden object. As a result,
(a) the iron object will penetrate more in sand.
(b) the iron object will have more potential energy.

Question 16.
The weight of a person on a planet A is about half that on the earth. He can jump upto 0.4 m height on the surface of the earth. How high can he jump on the planet A?
Answer:
The weight of a person on planet A is about half that on the earth. This means, the acceleration due to gravity of the planet A is half that of the earth. So, Height of the jump on the surface of planet A = \(\frac{0.4 \mathrm{~m}}{1 / 2}\) = 0.8 m

Long Answer Type Questions

Question 1.
How is work done measured when a body moves in a direction inclined to the direction of the applied force?
Answer:
In the figure, a force F pulls a block making angle 0 with the horizontal surface. Under this force, suppose the block moves from position A to B after covering a distance ‘s’.

Let, F₁ = Component of force in the direction of displacement ‘s’
Then, \(\frac{F_{1}}{F}\) = cos θ or F₁ = F cos θ
Work done = Component of force in the direction of displacement × displacement
W = F₁ × s = F cos θ x s
W = Fs cos θ
Special cases:
(a) When θ = 0°,
cos θ = 1 and W = Fs
Thus, work done is maximum when the displacement of the body is along the direction of the force.

(b) When θ = 90°,
cos θ = 0 and W = 0
Thus, work done is zero when the displacement of the body is perpendicular to the direction of force.

(c) When θ = 180°,
cos θ = -1 and W = – Fs
Thus, work done is negative when displacement is opposite to the direction of force.

(d) When s = 0, W = 0 Thus, work done on a stationary body is zero.

Question 2.
What is meant by potential energy of a body? Give some examples.
Answer:
The energy possessed by a body by the virtue of its position, shape or configuration is called its potential energy.

1. Examples of P.E. due to position:
   • Water stored in a dam at a height has potential energy.
   • A stone lying on the roof of the building has potential energy.
2. Examples of P.E. due to shape:
   • In a toy car, the wound spring possesses potential energy. As the spring is released, its potential energy changes into kinetic energy which moves the toy car.
   • Energy possessed by a stretched rubber band is potential energy.
3. Example of P.E. due to configuration:
   • A stretched bow possesses potential energy. As soon as it is released, it shoots the arrow in the forward direction with a large velocity.
   • The potential energy of the stretched bow gets converted into the kinetic energy of the arrow.
     

Question 3.
A 100 W electric bulb is lighted for 2 hours every day and five 40 W tubes are lighted for 4 hours every day. Calculate:
(a) the energy consumed for 60 days and
(b) the cost of electricity consumed at the rate of ₹3 per kW h.
Answer:
(a) Energy consumed by a 100 W bulb each day = 100 W × 2 h
= 200 Wh = \(\frac{200}{1000}\)= 0.2 kW h
Energy consumed by five 40 W tubes each day = 5 × 40 W × 4h
= 800 Wh = \(\frac{800}{1000}\) = 0.8 kW h
Total energy consumed each day = 0.2 + 0.8 = 1.0 kWh
Total energy consumed in 60 days = 1.0 x 60 = 60 kWh

(b) Cost of 1 kW h = ₹ 3
Cost of 60 kW h = 3 × 60 = 180

Question 4.
Answer the following:
(a) List any three situations in your daily life where you can say that work has been done.
(b) How much work is done in increasing the velocity of a car from 15 km/h to 30 km/h if the mass of the car is 1000 kg?
Answer:
(a) Three situations where work is done are:

1. Pushing a pebble lying on the ground. The pebble moves through some distance. Here, we apply a force and the pebble gets displaced. So, we have done work on the pebble.
2. We apply a force to lift a book through a height. The book rises up. We have done work in moving up the book.
3. A bullock is pulling a cart and the cart moves. There is a force on the cart and the cart has moved. The bullock has done work on the cart.

(b) According to the question,
u = 15 km/h = 4 m/s
v = 30 km/h = 8 m/s
Mass = 1000 kg
W = ?
W = K.E. = \(\frac{1}{2}\) m(v² – u²)
\(\frac{1}{2}\) × 1000 × ((8)² – (4)²)
= \(\frac{1}{2}\) × 1000 × (64 – 16)
= \(\frac{1}{2}\) × 1000 × (675) = 24,000J
Hence, work done is 24,000J.

Analysing & Evaluating Questions

Question 5.
A boy is moving on a straight road against a frictional force of 5N. After traveling a distance of 1.5 km he forgot the correct path at a roundabout of radius 100 m. However, he moves on that circular path for one and half cycle and then he moves forward up to 2.0 km. Calculate the work done by him.
Answer:
Work done by the boy while moving on a straight road
W = F × s
W = 5N × 1.5 km
= 5 kg m s^-2 × 1500 m = 7500J
Work done during moving around circular path
= 5N × (2 × 100 m) = 1000J
Work done during moving further by
2.0 km = 5N × (2 × 1000 m) = 10,000J
Total work done by the boy = 7500J + 1000J + 10,000J
= 18500J

Activity 1

• Take a heavy ball. Drop it on a thick bed of sand. A wet bed of sand would be better. Drop the ball on the sand bed from a height of about 25 cm and observe the results.
• Repeat this activity from heights of 50 cm, lm and 1.5 m.
• Ensure that all the depressions are distinctly visible.
• Mark the depressions to indicate the height from which the ball was dropped.
• Compare their depths.

Observations

• The ball that falls from the height of 1.5 m creates the deepest depression.
• The ball that falls from the height of 50 cm creates the shallowest depression.
• Larger the height from which the ball is dropped, larger is the kinetic energy gained by the ball on reaching the ground and more is its capability of doing work.

Activity 2

1. Set up the apparatus as shown in the figure.
2. Place a wooden block of known mass in front of the trolley at a convenient fixed distance.
3. Place a known mass on the pan so that the trolley starts moving.
4. The trolley moves forward and hits the wooden block.
5. Fix a stop on the table in such a manner that the trolley stops after hitting the block. The block gets displaced.
6. Note down the displacement of the block.
7. Repeat this activity by increasing the mass on the pan. Observe, in which case is the displacement more.
   

Observations

1. The force of gravity pulls the mass in the pan in the downward direction. This force gets transferred to the trolley through the string.
2. The trolley moves and hits the block with a force.
3. The larger the mass in the pan, the larger is the force with which the trolley hits the block.
4. Consequently, larger will be the displacement and larger will be the work done. The moving trolley possess energy and hence does work on the block.

Activity 3

• Take a slinky as shown below.
• Ask a friend to hold one of its ends. You hold the other end and move away from your friend. Now, you release the slinky and observe.
  

Observations

• When released, the slinky regains its original length. The slinky has acquired potential energy due to the work done on it during stretching. On releasing, potential energy is converted into kinetic energy.
• The slinky will also acquire energy when it is compressed.

Value Based Questions

Question 1.
Apoorva saw few planter pots kept on the balcony sill of fourth floor in her building. She makes an effort and keeps all the planter pots down the sill.
1. What type of energy is present in the pot kept on the balcony sill of fourth floor?
2. If the pot falls from the fourth floor, what type of energy will be seen in the falling pot?
3. What value of Apoorva is reflected in the above act?
Answer:
1. The pot on the sill possesses potential energy.
2. The falling pot possesses both the kinetic energy and the potential energy.
3. Apporva showed the value of moral responsibility and awareness.

Question 2.
Siddharth saw a lady labourer who carried bricks on her head from one point of the construction site to the other end which was some 500 m away. He prepares a trolley for the labourer to carry the bricks to make her work easier.
1. In carrying the bricks from point A to point B on the head by the lady labourer in the construction site, is any work done by the labourer?
2. By pulling the trolley of bricks from point A to point B, is any work done?
3. What value of Siddharth is seen in the above act?
Answer:
1. In carrying the bricks from point A to point B on head by the lady, no work is said to be done.
2. By pulling the trolley of bricks, work is said to be done.
3. Siddharth showed kindness, general awareness and sympathy.

JAC Class 9 Science Important Questions

JAC Board Class 9th Science Important Questions Chapter 10 Gravitation

Multiple Choice Questions 

Question 1.
The mass of a body on moon is 40 kg. What is its weight on the earth?
(a) 240 kg
(b) 392 N
(c) 240 N
(c) 400 kg
Answer:
(b) 392 N

Question 2.
The gravitational force between two objects is
(a) attractive at large distance only
(b) attractive at small distance only
(c) attractive at all distances
(d) attractive at large distance but repulsive at small distance
Answer:
(c) attractive at all distances

Question 3.
A body of mass 1 kg is attracted by the earth with a force equal to
(a) 9.8 N
(b) 1 N
(c) 6.67 × 10¹¹ N
(d) 9.8 m/s
Answer:
(a) 9.8 N

Question 4.
The earth attracts a body with a force of 10 N. With what force does that the body attract the earth?
(a) 10N
(b) 1 N
(c) \(\frac{1}{10 \mathrm{~N}}\)
(d) 2 N
Answer:
(a) 10N

Question 5.
The SI unit of G is
(a) Nm² kg^-1
(b) Nm² kg^-2
(c) Nm² kg
(d) N^-1 m² kg^-2
Answer:
(b) Nm² kg^-2

Question 6.
The gravitational force of attraction between two bodies of 1 kg each and 1 m apart is
(a) 6.67 × 10^-11 N
(b) 6.67 × 10^-8 N
(c) 6.67 × 10¹¹ N
(d) 6.67 × 10⁸ N
Answer:
(a) 6.67 × 10^-11 N

Question 7.
The value of acceleration due to gravity near the earth’s surface is
(a) 8.9 m^-2
(b) 9.8 ms^-2
(c) 8.9 cms^-2
(d) 9.8 cms^-2
Answer:
(b) 9.8 ms^-2

Question 8.
Consider an elevator moving downwards with an acceleration a The force exerted by a passenger of mass m on the floor of the elevator is
(a) ma
(b) ma – mg
(c) mg – ma
(d) mg + ma
Answer:
(c) mg – ma

Question 9.
If the earth suddenly shrinks to half of its present size, the value of acceleration due to gravity will
(a) become twice
(b) remain unchanged
(c) become half
(d) become four times
Answer:
(d) become four times

Question 10.
The weight of a body would not be zero
(a) at the centre of the earth
(b) during a free fall of an elevator
(c) in interplanetary space
(d) on a frictionless surface
Answer:
(d) on a frictionless surface

Question 11.
Newton’s law of gravitation
(a) can be verified in a laboratory
(b) cannot be verified but is true
(c) is valid only on earth
(d) is valid only in the solar system
Answer:
(a) can be verified in a laboratory

Question 12.
10 kg weight is equal to
(a) 9.8 N
(b) 98 N
(c) 980 N
(d) \(\frac{1}{9.8}\) N
Answer:
(b) 98 N

Question 13.
The weight of a body cannot be expressed in
(a) kg wt
(b) N
(c) dyne
(d) kg
Answer:
(d) kg

Question 14.
A stone is dropped from a cliff. Its speed after it has fallen 100m is
(a) 9.8 m/s
(b) 44.2 m/s
(c) 19.8 m/s
(d) 98 m/s
Answer:
(b) 44.2 m/s

Question 15.
At which of the following locations, the value of g is the largest?
(a) On top of the Mount Everest
(b) On top of Qutub Minar
(c) At a place on the equator
(d) A camp site in Antarctica
Answer:
(d) A camp site in Antarctica

Analysing & Evaluating Questions

Question 16.
An apple falls from a tree because of gravitational attraction between the earth and apple. If Ft is the magnitude of force exerted by the earth on the apple and F2 is the magnitude of force exerted by the apple on the earth, then
(a) F₁ is very much greater than F₂
(b) F₂ is very much greater than F₁
(c) F₁ is only a little greater than F₂
(d) F₁ and F₂ are equal
Answer:
(d) F₁ and F₂ are equal

Question 17.
An object is put one by one in three liquids having differei dnsities. The object floats with\(\frac{1}{9}\), \(\frac{2}{11}\) and \(\frac{3}{7}\) part of its volume outside dit liquid surface in liquids of densities d₁, d₂ and d₃ respectively. Which of the following is correct?
(a) d₁ > d₂ > d₃
(b) d₁ >d₂ <d₃
(c) d₁ < d₂ > d₃
(d) d₁ < d₂ < d₃
Answer:
(d) d₁ < d₂ < d₃

Question 18.
An object weighs 10N in air. When immersed fully in water, it weighs only 8N. The weight of the liquid displaced by the object will be
(a) 2 N
(b) 8 N
(c) 10 N
(d) 12
Answer:
(a) 2 N

Assertion Reason Questions

Directions: In the following questions, the Assertions and the Reasons have been put forward. Read the statements carefully and choose the correct alternative from the following:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.
(B) The assertion and the reason are correct but the reason is not the correct explanation of the assertion.
(C) The assertion is true but the reason is false.
(D) Both the statements are false.
1. Assertion: The law of gravitation is not applicable on the two bodies lying on the surface of the earth.
Reason: Law of gravitation is applied to celestial bodies only.
Answer:
(D) Both the statements are false.

2. Assertion: Mass of a body remains same at all the places.
Reason: Mass of a body is independent of acceleration due to gravity.
Answer:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.

3. Assertion: Value of acceleration due to gravity is greater at the poles than at the equator.
Reason: Distance between pole and the centre of the earth is less than the distance between the equator and the centre of the earth.
Answer:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.

4. Assertion: One can jump higher on the surface of the moon than on the earth. Reason: The value of g (acceleration due to gravity) on the moon is greater than that on the earth.
Answer:
(C) The assertion is true but the reason is false.

5. Assertion: Fluids exert an upthrust on the objects immersed in them.
Reason: The upthrust is equal to the weight of the liquid displaced.
Answer:
(B) The assertion and the reason are correct but the reason is not the correct explanation of the assertion.

Very Short Answer Type Questions

Question 1.
Define the term ‘gravitation’.
Answer:
Every object in this universe attracts every other object with a force known as ‘force of gravitation’. Gravitation is the force of attraction between any two bodies in the universe.

Question 2.
Give the SI unit and value of G.
Answer:
SI unit of G = \(\frac{\mathrm{Nm}^{2}}{\mathrm{~kg} \cdot \mathrm{kg}}\) = N m² kg^-2
Its value = 6.67 × 10^-11 N m² kg^-2.

Question 3.
g = GM/R², what do the symbols in this formula denote?
Answer:
g = Acceleration due to gravity
G = Gravitational constant
M = Mass of the earth
R = Radius of the earth

Question 4.
Name the scientist who first determined the value of G experimentally.
Answer:
Henry Cavendish first determined the value of G experimentally, in the year 1778, using a sensitive balance.

Question 5.
Why is G called a universal gravitational constant’?
Answer:
The value of constant G is same for any pair of objects in the universe. Also, its value does not depend on the nature of the intervening medium. That is why constant G is called ‘universal’, whether the bodies are big or small, or whether they are celestial or terrestrial.

Question 6.
State any two properties of gravitational force.
Answer:

1. It is always attractive in nature.
2. It does not depend on the nature of the medium between the two bodies.

Question 7.
Which force is responsible for the stability of our universe?
Answer:
The force of gravitation.

Question 8.
How is ‘g’ on the surface of the earth related to ‘G’?
Answer:
g = \(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\) = \(\frac{1.67 \times\left(1.74 \times 10^{6}\right)^{2}}{6.67 \times 10^{-11}}\)

where M = mass of the earth,
R = radius of the earth.

Question 9.
How is weight of an object related to its mass?
Answer:
Weight (w) = mass (m) × acceleration due to gravity (g).

Question 10.
What is the SI unit of pressure?
Answer:
The SI unit of pressure = N/m² = Pascal.

Question 11.
Define thrust. What is the SI unit of thrust?
Answer:
The net force exerted by a body in a particular direction is called thrust. The SI unit of thrust is newton.

Question 12.
Why does a truck or a motorbike have much wider tyres?
Answer:
A truck or a motorbike has much wider types so that the pressure exerted by it can be distributed to more surface area of the road and avoid the wear and tear of tyres.

Question 13.
In what sense does the moon fall towards the earth? Why does it not actually fall on the earth’s surface?
Answer:
At each point of its orbit, the moon falls towards the earth instead of going straight. It does not fall on the earth because it is moving in a circular orbit. The moon is kept in its circular orbit by the centripetal force provided by the force of attraction of the earth.

Question 14.
The earth attracts an apple from the tree and it falls but the earth does not appear to move towards the apple. Why?
Answer:
The mass of the earth is extremely large as compared to that of the apple. So the acceleration produced is very small as compared to that in the apple. Hence, the motion of the earth towards the apple is not noticeable.

Question 15.
What do you mean by the term ‘free fall’?
Answer:
The motion of a body under the influence of the force of gravity alone is called a ‘free fall’.

Question 16.
Suppose a planet exists whose mass and radius, both are half that of the earth. Calculate the acceleration due to gravity on the surface of this planet.
Answer:
On the surface of the earth, g = \(\frac{\mathrm{GM}}{\mathrm{R}^{2}}\)
On the surface of the planet, \(g^{\prime}=\frac{G\left(\frac{M}{2}\right)}{\left(\frac{R}{2}\right)^{2}}=\frac{2 G M}{R^{2}}=2 g\)

Question 17.
Define mass of a body. What is its SI unit?
Answer:
The mass of a body represents the quantity of matter contained in the body. It gives a measure of inertia of the body. The greater the mass of a body, the greater is its inertia. The SI unit of mass is kilogram (kg).

Question 18.
Define one kilogram – weight. How many newtons are there in 1kg wt?
Answer:
One kilogram – weight (kg – wt)
= 1kg × 9.8 m/s² = 9.8 N.

Question 19.
State Archimedes’ principle.
Answer:
This principle states that when a body is immersed fully or partially in a fluid, it experiences an upward thrust equal to the weight of the fluid displaced by it.

Analysing & Evaluating Questions

Question 20.
A beam balance is used for measuring the mass of a body, whereas a spring balance gives the weight of a body. What is measured by a digital balance?
Answer:
Digital balance measures the weight of a body.

Question 21.
Under what conditions do the two hollow balls – one of aluminium and the other of iron, experience equal upthrust when placed in water?
Answer:
The two balls will experience equal upthrust in water when their volumes inside the water are equal.

Question 22.
The density of liquid B is greater than that of liquid A. A hydrometer is placed one by one in the two liquids. In which liquid will the hydrometer sink to the greater depth?
Answer:
The hydrometer will sink deeper in liquid A.

Short Answer Type Questions

Question 1.
During free – fall of an elevator, what is the weight of a body inside it? Give reasons for your answer.
Answer:
During free – fall of an elevator the apparent weight of the object becomes zero. This is because both the body and the elevator are in free fall, and have a downward acceleration of ‘g’. In this situation normal reaction on the body becomes zero. Since normal reaction is responsible for the sensation of weight in a body, the apparent weight of the body in this case becomes zero.

Question 2.
What is relative density?
Answer:
The relative density of a substance is the ratio of its density to that of the water at 4°C. Relative density Density of the substance Density of water at 4°C

Question 3.
Differentiate between ‘g’ and ‘G’ in tabular form.
Answer:

Question 4.
Why is it easier to swim in sea water than in river water?
Answer:
The density of sea water is high due to dissolved salts in it as compared to the density of river water. Hence, the buoyant force applied on the swimmer by the sea water is high which helps in floating and makes swimming simpler.

Question 5.
A ship made of iron does not sink but an iron nail sinks in water. Why?
Answer:
The iron nail sinks because of its high density and less buoyant force acting on it due to lesser surface area. Whereas, the surface area of a ship is greater and thus experiences a higher buoyant force. Due to this fact, a ship floats but an iron nail sinks.

Question 6.
A stone and feather are thrown from a tower. Both the objects should reach the ground at the same time but it does not happen. Give reasons.
Answer:
As per the motion of objects due to gravitational pull of the earth, both the bodies are acted upon by the same force of the earth but the stone will fall first and then the feather. Feather being lighter, experiences greater air resistance, so it will reach later.

Question 7.
The relative density of gold is 18.3. The density of water is 10³ kg/m³? What is the density of gold in S.I units?
Answer:
The relative density of gold is 18.3.
Relative density of gold = \(\frac{Density of gold}{Density of water}\)
That is, density of gold = Relative density of gold × Density of water = 18.3 × 10³ kg/m³ = 18300 kg/m³.

Question 8.
The acceleration due to gravity at the moon’s surface is 1.67 m/s². If the radius of the moon is 1.74 × 106m, calculate the mass of the moon.
Answer:
Here, G = 6.67 × 10^-11 Nm²/kg²
g = 1.67 m/s²
R= 1.74 × 10⁶ m.
Mass of the moon is
M = \(\frac{\mathrm{gR}^{2}}{\mathrm{G}}\) = \(\frac{1.67 \times\left(1.74 \times 10^{6}\right)^{2}}{6.67 \times 10^{-11}}\)
= 7.58 × 10²² kg.

Question 9.
Derive the formula for the gravitational force acting between two objects.
Answer:
Let two objects A and B of masses ‘M’ and’m’, lie at a distance d from each other as shown in the figure.

Let the force of attraction between two objects be F.
According to the universal law of gravitation, the force between two objects is directly proportional to the product of their masses.
F ∝ Mm … (1)
And the force between two objects is inversely proportional to the square of the distance between them,
F ∝ \(\frac{1}{\mathrm{~d}^{2}}\) …………(2)
F ∝ \(\frac{\mathrm{Mm}}{\mathrm{d}^{2}}\)
where G is universal gravitational constant

Question 10.
Find the value of ‘g’, acceleration due to gravity.
Answer:
g = \( \frac{\mathrm{GM}}{\mathrm{R}^{2}}\)
G = 6.67 × 10^-11 Nm²/kg² (constant)
M = 6 × 10²⁴ kg (mass of the earth)
= \( \frac{6.7 \times 10^{-11} \mathrm{Nm}^{2} / \mathrm{kg}^{2} \times 6 \times 10^{24} \mathrm{~kg}}{\left(6.4 \times 10^{6} \mathrm{~m}\right)^{2}}\)
= 9.8 m/s²

Question 11.
Camels can walk easily on desert sand but we are not comfortable walking on the sand. Why do you think so?
Answer:
The surface area of a camel’s feet is broad and large. Thus, the pressure exerted is low. However, when we walk, our legs sink because the pressure exerted by our body is not equally distributed but is directed towards the legs.

Question 12.
Define lactometer and hydrometer.
Answer:
Lactometer is a device used to find the purity of milk. Hydrometer is a device used to find the density of a liquid.

Question 13.
Two wood pieces of same size and mass are dipped in two beakers containing water and oil. One wood floats on water, but the other one sinks in oil. Why?
Answer:
The wood floats on water because the density of wood is lower than the density of water, and the other wood sinks in the oil because the density of wood is more than that of the oil.

Question 14.
What are fluids? Why is Archimedes’ principle applicable only for fluids? Give the applications of Archimedes’ principle.
Answer:
Liquids are the substances which can flow. Archimedes principle is based on the upward force exerted by fluids on any object immersed in them. Hence, it is applicable only for fluids. Applications of Archimedes’ principle are as follows:

1. It is used to design ships and submarines.
2. To determine the purity of milk using lactometers which are designed based on the Archimedes’ principle.
3. To make hydrometers which are used to determine the density of liquids.

Question 15.
The volume of 60g of a substance is 10cm². If the density of water is 1g/cm², find out whether the substance will float or sink in water.
Answer:
Mass = 60g
Volume = 10 cm³
The density of water is 1g/cm². As the density of the given substance is more than the density of water, the substance will get submerged in water.

Question 16.
What is meant by buoyancy?
Answer:
When a body is submerged in a fluid, the fluid exerts an upward force on the submerged body. This upward force is equal to the weight of the fluid displaced by the submerged body and is called the buoyant force. In other words, it is the force exerted by the fluid when an object is submerged in it.

Analysing & Evaluating Questions

Question 17.
Identical packets are dropped from two aeroplanes, one over the equator and the other over the north pole, both from the height h. Assuming that all conditions are identical, explain if those packets take same time to reach the surface of earth or not. Justify your answer.
Answer:
No. The two packets will take different time intervals to reach the earth’s surface. This is because the acceleration due to gravity (g) is greater at the poles than that at the equator. The packet will fall slowly at the equator than that at the north pole.

Question 18.
You must have seen two types of balances. A grocer has a balance with two platforms or pan, and a needle moving on a scale. Some junk dealers (kabadiwalas) may be using a spring balance to weigh used newspapers. Suppose, the two balances give the same measure for a given body on the earth. Now, you take the balances to the moon. Would the two balances give the same measure? Explain your answer.
Answer:
No. The two balances will not give the same measure on the moon. This is because, the spring balance measures the weight, and the grocer’s balance measures the mass. The weight of a body depends upon the acceleration due to gravity at the place of measurement. The acceleration due to gravity on the moon is not equal to that on the earth. Therefore, spring balance will give different readings relative to the grocer’s balance on the moon.

Long Answer Type Questions

Question 1.
What is centripetal force? From where does the moon get the centripetal force required for its motion around the earth?
Answer:

1. Centripetal force When a body moves along a circular path with a uniform speed, its direction of motion changes at every point. The change in direction involves change in velocity or acceleration.
2. The force that provides this acceleration and keeps the body moving along the circular path, acts towards the centre. This force is called centripetal (centre seeking) force.
3. Therefore, a force which is required to make a body move along a circular path with uniform speed is called centripetal force. Centripetal force always acts along the radius and towards the centre of the circular path.
   
4. The figure given here shows that it is the centripetal force which continuously deflects a particle from its straight line path to make it move along a circle.
5. Example: The moon needs a centripetal force for its circular motion around the earth. This centripetal force is provided by the gravitational attraction exerted by the earth on the moon.

Question 2.
Show that the weight of an object on the moon is 1/6,h its weight on the earth. Given that the mass of the earth is 100 times the mass of the moon and its radius is 4 times that of the moon.
Answer:
Suppose, mass of the object = m
mass of the earth = M_e
mass of the moon = M_m
radius of the earth = R_e
radius of the moon = R_m
Then, M_e = 100 Mm and Re = 4Rm
Weight of an object of mass m on the earth is
We = the force with which the earth
attracts the object = G \(\frac{\mathrm{M}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}_{\mathrm{e}}^{2}}\)
Weight of the object on the moon is
W_m = The force with which the moon
attracts the object = \( \mathrm{G} \frac{\mathrm{M}_{\mathrm{m}} \mathrm{m}}{\mathrm{R}_{\mathrm{m}}^{2}}\)
\(\frac{W_{m}}{W_{e}}\) = \(\frac{\left(G \frac{M_{m} m}{R_{m}^{2}}\right)}{\left(G \frac{M_{m} m}{R_{e}^{2}}\right)}\)
= \(\frac{M_{m}}{M_{e}} \times \frac{\left(R_{e}\right)^{2}}{\left(R_{m}\right)^{2}}\)
= \(\frac{\mathrm{M}_{\mathrm{m}}}{100 \mathrm{M}_{\mathrm{m}}} \times \frac{4 \mathrm{R}_{\mathrm{m}}^{2}}{\mathrm{R}_{\mathrm{m}}^{2}}\) = \(\frac{16}{100} \approx \frac{1}{6}\)
Thus, the weight of an object on the moon is about one – sixth of its weight on the earth.

Question 3.
Briefly explain how can Archimedes’ principle be verified experimentally.
Answer:
According to Archimedes’ principle, when a body is immersed in a liquid, completely or partially, it loses its weight. The loss in weight is equal to the weight of liquid displaced by the body. The loss in weight of a body is due to the presence of upthrust which is equal to the weight of liquid displaced.
Thus, a Loss in weight = Weight of body in air – Weight of body immersed in water =
W W₁ = upthrust in water on the body = Weight of liquid displaced.

As shown in figure, measure the weight of a solid, say a metallic ball, in air using spring balance. Weigh the empty beakers using spring balance. Set the spring balance, overflow can with tap water and beaker. Now, allow the bob to immerse completely in water in overflow can. Note down the new position of the pointer of the spring balance.

This will give you the weight of the brass bob in tap water. It is found to be less than the weight of bob in air. Weigh the beaker containing displaced water which is collected from the overflow can while immersing the bob in it completely. It is observed that the loss in weight of the bob is equal to the weight of the water collected in the beaker, i.e., weight of the water displaced.

Question 4.
With the help of an activity, prove that the force acting on a lesser area exerts a larger pressure.
Answer:
Consider a block of wood kept on a tabletop. The mass of the wooden block is 5kg. Its dimensions are 30 cm × 20 cm × 10 cm.
Now, we have to find the pressure applied by the wooden block on the tabletop by keeping it vertically and horizontally.

The mass of the wooden block = 5kg
Weight of the wooden block applies a thrust on the tabletop.
Thrust = F = mg = 5kg × 9.8 m/s² = 49 N
(Case a) : When the wooden box is kept vertically with sides 20 cm × 10 cm,
Area of a side = length × breadth = 20 cm × 10 cm = 200 cm² = 0.02m²
Pressure = \(\frac{Thrust}{Area}\) = \(\frac{49 \mathrm{~N}}{0.02 \mathrm{~m}^{2}}\)
= 2450 N/m²

(Case b): When the block is kept horizontally with side 30 cm × 20 cm, Area = length × breadth = 30 cm × 20 cm = 600 cm² = 0.06 m²

Pressure = \(\frac{Thrust}{Area}\) = [/latex] \frac{49 \mathrm{~N}}{0.02 \mathrm{~m}^{2}}[/latex]
= 816.7 N/m²
The pressure exerted by the box in case (a) in more as compared to the pressure exerted in case (b). The area is reduced and the pressure exerted is more.
This shows that pressure ∝= \( \frac{1}{\text { Area }}\)
Hence, pressure will be larger if the area is reduced.

Analysing & Evaluating Questions

Question 5.
(a) A cube of side 5 cm is immersed in water and then in saturated salt solution. In which case it will experience a greater buoyant force. If each side of the cube is reduced to 4 cm and then immersed in water, what will be the effect on the buoyant force experienced by the cube as compared to the first case for water.

(b) A ball of weight 4 kg and density 4000 kg m³ is completely immersed in water of density 10³ kg m³. Find the force of buoyancy on it. (Given g = 10 ms^-2)
Answer:
(a) The cube will experience greater buoyant force in salt solution. When the side of the cube is reduced to 4 cm, it will experience lesser buoyant force as compared to the first case.
This is because the volume of water displaced by the smaller cube is lesser than that displaced by the original cube,

(b) Volume of water displaced by the ball = Volume of the ball = \(\frac{Mass}{Density}\)
= \(\frac{4 \mathrm{~kg}}{4000 \mathrm{~kg} \mathrm{~m}^{-3}}\) = 10^-3 m³
Weight of water displaced by the ball = Volume of water × Density × g
= 10^-3 m³ × 10³ kg m³ × 10 ms^-2 = 10 N
So, Buoyant force acting on the ball = Weight of water displaced = 10 N

Activity 1

• Take a piece of thread.
• Tie a small stone at one end. Hold the other end of the thread and whirl it round.
• Note the motion of the stone.
• Now release the thread.
• Again, note the direction of motion of the stone.
• Record your observations and draw conclusion about the direction of the stone.
  
• If thread is released when stone is here, stone goes straight towards A not towards B Top view A stone describing a circular path

Observations

• The motion gets accelerated and stone moves in a circular path.
• When the thread is released, the stone makes a tangent to the circle and falls down.

ACTIVITY 2

• Take a sheet of paper and a stone. Drop them simultaneously from the first floor of a building.
• Observe whether both of them reach the ground simultaneously or not.

Observations

• We see that the paper reaches the ground a little later than the stone. This happens because of air resistance. The air offers resistance due to friction to the motion of the falling objects. The resistance offered by air to the paper is more than the resistance offered to the stone.
• If we do the experiment in a glass jar from which air has been sucked out, the paper and the stone would fall at the same rate.

ACTIVITY 3

• Take an empty plastic bottle. Close the mouth of the bottle with an airtight stopper. Put it in a bucket filled with water and observe.
• Push the bottle into the water and observe again. Try to push it further down.
• Now, release the bottle. Record your observations.
• Does the force due to the gravitational attraction of the earth act on this bottle? If so, why doesn’t the bottle stay immersed in water after it is released? How can you immerse the bottle in water?

Gravity

Observations

• The plastic bottle floats on water. When it is pushed deeper, the upward force on the bottle increases.
• When released, the bottle bounces back to the surface because the upthrust on the bottle was larger than the downward pull of gravity
• Take a piece of stone and tie it to one end of a rubber string or a spring balance.

ACTIVITY-4

• Take a piece of stone and tie it to one end of a rubber string or a spring balance.
  Thread
  
• Suspend the stone by holding the balance or the string as shown in the figure (A).
• Note the elongation of the string or the reading on the spring balance due to the weight of the stone.
• Now, slowly dip the stone in the water in a container as shown in Figure (B).
• Observe what happens to the elongation of the string or the reading on the balance.

Observations

• In fig. A, the elongation of the string is more.
• In fig. B, when the stone is dipped in water, the length of the string is reduced.
• The length of the string in case (B) decreases due to the upward force exerted by water on the stone called buoyant force.

Value Based Questions

Question 1.
Priya had a bad experience during the take off of the plane when she boarded it for the first time. Her friend assisted and helped her during the landing of plane. She told her to fasten the seat belt and involved her in gossip. Priya faced less problems while landing of the plane.
1. Why do we tie our seat belts in moving cars or during take off of the plane?
2. What is free fall?
3. What value of Priya’s friend is seen in this set?
Answer:

1. We tie our seat belts to remain intact on the seat so that our body does not fall forward.
2. When an object falls towards the earth under the gravitational force alone, we say that the object is in free fall.
3. Priya’s friend showed the value of concerned, sympathetic, responsible and caring friend.

Question 2.
A goldsmith measured the purity of gold by using a special measuring device. He told the customer that there was impurity present in gold ornament that he wanted to buy and was not 22 carat but 18 carat jewellery.
1. How can we find the purity of gold?
2. What is the unit of relative density?
3. What value of goldsmith is reflected in this act?
Answer:

1. The purity of gold can be obtained by knowing the density of the gold.
2. Relative density does not have any unit.
3. Goldsmith showed the value of honesty and trustworthiness.

JAC Class 9 Science Important Questions

JAC Board Class 9th Science Important Questions Chapter 9 Force and Laws of Motion

Multiple Choice Questions

Question 1.
The SI unit of force is
(a) kg m/s
(b) newton
(c) dyne
(d) kg wt
Answer:
(b) newton

Question 2.
The combined effect of mass and velocity is taken into account by a physical quantity called
(a) torque
(b) momentum
(c) moment
(d) force
Answer:
(b) momentum

Question 3.
Quantitative definition of force is given by
(a) Newton’s first law of motion
(b) Newton’s second law of motion
(c) Newton’s third law of motion
(d) Newton’s law of gravitation
Answer:
(b) Newton’s second law of motion

Question 4.
Momentum gives a measure of
(a) mass
(b) weight
(c) velocity
(d) motion
Answer:
(d) motion

Question 5.
An athlete runs some distance before taking a long jump because
(a) he gains energy to take himself through long distance
(b) it helps him to apply a larger force
(c) by running, action and reaction forces increase
(d) by running, the athlete gives himself larger inertia of motion
Answer:
(d) by running, the athlete gives himself larger inertia of motion

Question 6.
Rocket works on the principle of conservation of
(a) velocity
(b) energy
(c) linear momentum
(d) mass
Answer:
(c) linear momentum

Question 7.
The rate of change of momentum of an object is proportional to the
(a) mass of the body
(b) velocity of the body
(c) net force applied on the body
(d) none of these
Answer:
(c) net force applied on the body

Question 8.
A force of 50 N moves a body. Which of the following is correct?
(a) Frictional force exerted on the body is less than 50N
(b) Frictional force exerted on the body is more than 50N
(c) None of these
(d) Both (a) and (b)
Answer:
(a) Frictional force exerted on the body is less than 50N

Question 9.
If a football and a stone have same mass, then both will have
(a) same inertia
(b) same momentum
(c) different inertia
(d) different momentum
Answer:
(a) same inertia

Question 10.
The rate of change of momentum w. r. t time is measured in
(a) kg m
(b) kg
(c) kg ms^-1
(d) kg ms^-1
Answer:
(d) kg ms^-1

Question 11.
A block of mass M is pulled with a force F along a smooth horizontal surface with a rope of mass m. The acceleration of the block will be
(a) F/M
(b) F/m
(c) F/(M + m)
(d)F/(M – m)
Answer:
(c) F/(M + m)

Question 12.
Action and reaction forces act on two
(a) different objects
(b) same objects
(c) either (a) or (b)
(d) none of these
Answer:
(a) different objects

Question 13.
A man is standing in a boat in still water. If he tries to walk towards the shore, the boat will
(a) move away from the shore
(b) remain stationary
(c) sink
(d) move towards the shore
Answer:
(a) move away from the shore

Question 14.
When an object undergoes acceleration
(a) its speed always increases
(b) its velocity always increases
(c) it always falls towards the earth
(d) a force always acts on it
Answer:
(d) a force always acts on it

Question 15.
On applying a constant force to a body, it moves with uniform
(a) momentum
(b) speed
(c) acceleration
(d) velocity
Answer:
(c) acceleration

Analysing & Evaluating Questions

Question 16.
A goalkeeper in the game of football pulls his hands backwards after holding the ball shot at the goal. This enables the goal keeper to
(a) exert larger force on the ball
(b) reduce the force exerted by the ball on hands
(c) increase the rate of change of momentum
(d) decrease the rate of change of momentum
Answer:
(b) reduce the force exerted by the ball on hands

Question 17.
A passenger in a moving train tosses a coin which falls behind him. It means that motion of the train is
(a) accelerated
(b) uniform
(c) retarded
(d) along circular tracks
Answer:
(a) accelerated

Question 18.
An object of mass 2kg is sliding with a constant velocity of 4 ms^-1 on a frictionless horizontal table. The force required to keep the object moving with the same velocity is
(a) 32 N
(b) 0 N
(c) 2 N
(d) 8 N
Answer:
(b) 0 N

Assertion Reason Questions

Directions: In the following questions, the Assertions and the Reasons have been put forward Read the statements carefully and choose the correct alternative from the following:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.
(B) The assertion and the reason are correct but the reason is not the correct explanation of the assertion.
(C) The assertion is true but the reason is false.
(D) Both statements are false.
1. Assertion: A brick has more inertia than a hollow wooden block of the same shape and size.
Reason: Heavier the body, more is the inertia.
Answer:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.

2. Assertion: A body falling towards the earth also pulls the earth towards itself.
Reason: The forces always occur in pairs.
Answer:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.

3. Assertion: When a bus starts, the person inside it falls forward.
Reason: The bus pushes the person forward.
Answer:
(D) Both statements are false.

4. Assertion: The third law of motion defines acceleration.
Reason: The third law of motion states that the net force experienced by a body is proportional to the rate of change in momentum of the body.
Answer:
(D) Both statements are false.

5. Assertion: Action and reaction forces are always equal and opposite.
Reason: Action and reaction forces cancel each other.
Answer:
(C) The assertion is true but the reason is false.

Very Short Answer Type Questions

Question 1.
Define force.
Answer:
Force may be defined as a push or a pull which changes or tends to change the state of rest or of uniform motion or direction of motion of a body.

Question 2.
Define one Newton.
Answer:
One newton is the force which produces an acceleration of m/s² in an object of mass 1kg.
1N = 1kg m/s²

Question 3.
State the various effects of force.
Answer:
A force applied on an object can do five things:
(a) It can change the speed of the object.
(b) It can change the direction of motion of the object.
(c) It can change the shape and size of the object.
(d) It can set an object at rest in motion.
(e) It can bring a moving object to rest.

Question 4.
Define resultant force.
Answer:
The resultant force or resultant of several forces acting simultaneously on a body is that single force which produces the same effect on a body as all these forces produce together.

Question 5.
What is frictional force?
Answer:
The force that always opposes the motion of an object is called force of friction.

Question 6.
What is inertia?
Answer:
The natural tendency of an object to resist a change in their state of rest or of uniform motion along a straight line is called inertia.

Question 7.
State Newton’s first law of motion.
Answer:
An object remains in its state of rest or of uniform motion in a straight line unless acted upon by an external unbalanced force.

Question 8.
Place a water – filled tumbler on a tray. Hold the tray and turn around as fast as you can. We observe that the water spills. Why?
Answer:
The water spills when the tray turns around fastly, because initially, the water filled in the tumbler placed on the tray was in a state of rest. While turning, we apply force on the tray which starts moving but the water remains at rest due to its inertia, and hence, it spills out of the tumbler.

Question 9.
State Newton’s second law of motion.
Answer:
The rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of the force.

Question 10.
State Newton’s third law of motion.
Answer:
To every action, there is an equal and opposite reaction and they act on two different bodies.

Question 11.
Why are shockers used in cars, scooters and motorcycles?
Answer:
Due to the shockers, the time interval of the jerk increases. As the rate of change of momentum will be smaller, comparatively less force acts on the passengers during the jerks.

Question 12.
Why is glass or chinaware packed with straw?
Answer:
The straw paper between the chinaware increases the time of experiencing the jerk during transportation. Hence, they strike against each other with a less force and are less likely to be damaged.

Question 13.
A bird hit the windscreen of a fast moving car and fell on the bonnet. Which of the two, the car or the bird, suffers greater change in momentum?
Answer:
By the law of conservation of momentum, both the car and the bird suffered equal and opposite change in momentum.

Question 14.
What decides the rate of change of momentum of an object?
Answer:
The rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of force.

Question 15.
The diagram shows a moving truck. Forces A, B, C and D are acting on the truck. Name the type of each of these forces acting on the truck.

Answer:
The forces A, B, C and D acting on the truck are:
A driving force
B normal reaction
C frictional force
D weight gravitational force.

Analysing & Evaluating Questions

Question 16.
Two identical bullets are fired, one by a light rifle and another by a heavy rifle, with the same force. Which rifle will hurt the shoulder more and why?
Answer:
The velocity of recoil will be higher for ligher rifle. So lighter rifle will hurt the shoulder more.

Question 17.
When a fast moving carrom striker hits the carrom coin at the bottom of a pile, ¡t moves out without disturbing
the pile. Which kind of inertia of the pile is responsible for it?
Answer:
Inertia of rest of the pile is responsible for it.

Question 18.
There are three solid balls made up of aluminium, steel and wood, of the same shape and same volume. Which of them would have the highest inertia?
Answer:
Steel ball will have the highest inertia This is because the density of steel is higher than aluminium and wood As a result, the steel ball will have the highest mass and hence highest inertia

Short Answer Type Questions

Question 1.
Differentiate between balanced and unbalanced forces.
Answer:

Question 2.
Give examples to show that balanced forces can change the shape of an object.
Answer:
Consider a spring attached to a rigid support at one of its ends, as shown in the figure.

If we pull the free ends of the spring, it gets elongated Thus, on applying force, a spring expands. Similarly, if we hold a rubber ball between our palms and push the two palms against each other, we find that the ball is no longer round but oblong. The force exerted on the ball changes its shape, as shown in the figure. In both cases, the two forces, being equal and opposite, balance each other but change the shape of the object.

Question 3.
A football player kicks the ball which travels in the air for a while and lands on the ground where the ball travels on the ground for a short distance and then stops. What may be the reason behind it?
Answer:
The reason is frictional force. There is always a contact between the ball and the ground The opposing force always acts against the motion of the ball, thereby stopping the ball.

Question 4.
A table is moved across the floor with a constant velocity where the horizontal force acting on it is 400N. What will be the frictional force that will be acting on the table?
Answer:
According to Newton’s third law of motion, for every action there will be an equal and opposite reaction. The net force acting on the table is zero because the table moves with constant velocity in forward direction. Hence, an equal amount of frictional force must act in opposite direction as that of the table, i.e., F = – 400N.

Question 5.
When the car, we are moving in, makes a sharp turn at a very high speed, we tend to get thrown to one side. Explain the statement.
Answer:
The reason behind this is the law of inertia While moving in a car along a straight line, we tend to maintain our state of motion. But when the car takes a sharp turn, our bodies resist the change in direction and are thrown to the opposite side due to inertia.

Question 6.
Athletes always have a special posture by resting their right foot on a solid support, why?
Answer:
During the race, athletes have to run the heats and they rest their right foot on the solid support before the start, so that this support can give them a lot of force during the start of the race. They push the support backwards and get an equal and opposite forward push to get a very good start.

Question 7.
The safety belts in the car help in preventing accidents. Justify the statement.
Yes, safety belts help in preventing accidents. When a car is moving with a high speed, our body tends to be in movement due to inertia of motion in the forward direction. So when there is a sudden  collision, serious injuries can happen. However, seat belts exert a force on our body to slow down the forward motion and hence prevent injuries.

Question 8.
A karate player breaks the pile of tiles with a single blow. Give reason.
Answer:
A karate player strikes the pile of tiles by applying a very large velocity. The overall momentum of fast moving hand is reduced to zero in a very short interval. So this increases the rate of change of momentum which increases the force on the tiles and they eventually break.

Question 9.
Athletes are made to fall on a sand bed while performing a high jump. Give reason.
Answer:
During the high jump event, an athlete is made to fall on the sand bed because it increases the time to attain the rest
position. We know that, F = \(\frac{m(v-u)}{t}\)
So, force and time are inversely proportional to each other. If there is an increase in time, there is a decrease in rate of change of momentum and therefore the force or the impact is reduced This prevents injuries to the feet of athlete.

Question 10.
Explain the law of conservation of momentum.
Answer:
According to the law of conservation of momentum, the total momentum of the objects before collision is equal to the total momentum after collision, provided there is no external unbalanced force acting on them.

Question 11.
A bullet of mass 20g moving with a speed of 500 ms^-2 strikes a wooden block of mass 1kg and gets embedded in it. Find the speed with which the block moves along with the bullet.
Answer:
Let the final velocity of the block along with the bullet embedded in it be v.
For bullet, m₁ = 20g = 0.02 kg,
U₁ = 500 m/s, v₁ = v
For block, m₁ = 1kg, u₂= 0, v₂ = v
According to the law of conservation of momentum,
Total momentum before collision = Total momentum after collision
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
m₁u₁ + m₂u₂= (m₁ + m₂) v
v = \(\frac{m_{1} u_{1}+m_{2} u_{2}}{m_{1}+m_{2}}\)
= \(\frac{(0.02 \times 500)+(1 \times 0)}{(0.02+1)}\)
=\(\frac{10}{1.02}\) = \(\frac{1000}{102}\)
= 9.8 m/s.

Analysing & Evaluating Questions

Question 12.
Two balls of the same size but of different materials, rubber and iron, are kept on the smooth floor of a moving train. The brakes are applied suddenly to stop the train. Will the balls start rolling? If so, in which direction? Will they move with the same speed?

Answer:
The balls will start rolling in the forward direction. No. The two balls will move with different speed This is because the inertia of the two balls is not the same.

Question 13.
Two forces F₁ = 40N and F₂ = 60N are acting on an object as shown in the figure.

1. What is the net force acting on the object?
2. What is the direction of the net force acting on the object?
3. How much extra force is acting on the object if the object is not moving due to the application of these two forces? Name that force. Where does that force act and what is the direction of that force?
Answer:

1. Net force on the object = 60N – 40N = 20N
2. Net force of 20N acts in the direction of F₂ (i.e., towards left)
3. A force equal to 20N is acting on the object.
   This force is the frictional force. This force acts in the direction of F₁

Long Answer Type Questions

Question 1.
Describe Galileo’s experiments with inclined planes and state their conclusion.
Answer:
By observing the motion of objects on inclined planes, Galileo deduced that objects move with constant speed when no force acts on them.

1. In the first experiment, Galileo studied the motion of marbles on an inclined plane. He observed that when a marble rolls down an inclined plane, its velocity increases, as shown in the figure A.
2. Here the marble falls under the unbalanced force of gravity. The velocity of the marble decreases when it rolls up the inclined plane (against the force of gravity), as shown in the figure B.
3. From these observations, Galileo argued that the velocity of a marble rolling on a flat horizontal surface should remain constant. To test his idea, Galileo used double inclined planes.
4. He observed that when the inclinations of the planes on both sides were equal, the marble rolled down from one plane from a certain height will climb up to the same height on the other plane.
   
   side plane were gradually decreased, the marble would travel larger and larger distances to reach the same height.
5. Ultimately, if the right side plane were made horizontal, the marble would continue to travel forever to reach the same height from which it was released No unbalanced force acts on the marble in this case.
6. The above experiment suggests that an unbalanced force (external force) is required to change the motion of an object while no unbalanced force is needed to keep an object moving with a constant velocity.
7. In actual practice, the bodies stop due to the force of friction which always acts opposite to the direction of motion.

Question 2.
Mathematically explain the concept of conservation of momentum during collision between two bodies.
Answer:
Let us consider two bodies, viz., A and B, moving with initial velocities u_A and u_B respectively, such that u_A > u_B.
Let mass of body A be m_A and that of B be m_B.
Before collision
For body A, momentum = m_Au_A
For bodv B. momentum = m_Bu_B

After collision, the bodies A and B move with velocities, say, v_A and v_B respectively. Let the collision occur for a time, say, t.
After collision
For body A, momentum = m_Av_A
For body B, momentum = m_Bv_B
The rate of change of momentum for the body A = \(\frac{m_{A}\left(v_{A}-u_{A}\right)}{t}\)
The rate of change of momentum for the body A = \(\frac{m_{B}\left(v_{B}-u_{B}\right)}{t}\)

According to Newton’s third law of motion, The force F_AB exerted by ball A on ball B and the force F_BA exerted by the ball B on ball A must be equal and opposite to each other. Therefore
\(\left\{\frac{m_{A}\left(v_{A}-u_{A}\right)}{t}=-\left(\frac{m_{B}\left(v_{B}-u_{B}\right)}{t}\right)\right\}\)
m_Au_A + m_Bu_B= m_Av_A + m_Bv_B
Here, (m_Au_A + m_Bu_A) is the total momentum of the two bodies, A and B, before collision and (m_Av_A+ m_Bv_B) is the total momentum of the two bodies, A and B, after collision. So, the total momentum of the two bodies remains constant during collision, when no external force acts on them.

Question 3.
Give some examples of Newton’s third law of motion.
Answer:
Newton’s third law of motion states that: For every action, there is always an equal and opposite reaction acting on two different bodies.
Examples:

1. The person who is swimming, pushes the water backward (action), while water pushes the person forward (reaction).
2. When a bullet is fired from a gun, the gun exerts a force on the bullet (action), the bullet exerts an equal and opposite force on the gun (reaction),
3. When a rower jumps out of the boat in the forward direction (action), the boat moves backward (reaction).

Question 4.
Define the term ‘inertia’. Explain its types and state some of its examples in daily life.
Answer:

1. The tendency of a body to resist change in its state of rest or of uniform motion is called inertia.
   • Inertia of rest: The ability of a body to resist any change in its state of rest.
   • Inertia of motion: The ability of a body to resist any change in its state of uniform motion.
   • Inertia of direction: The ability of a body to resist any change in its direction of motion.
2. Examples of application of inertia in our day to day life are as follows:
   • When a carpet is beaten with a stick, the dust falls down.
   • When the car applies a sudden brake, we tend to fall forward.
   • When a bus suddenly takes a turn, we fall in the direction opposite to the turn.

Question 5.
Mathematically explain the concept of Newton’s second law of motion and also explain its importance in sports.
Answer:
Newton’s second law of motion: The rate of change of momentum of an object is directly proportional to the applied unbalanced force and it acts in the direction of force.

Example: Let us consider an object of mass (m), moving along a straight line with an initial velocity (u). Suppose, it attains a final velocity (v) in time (t).
Initial momentum of the object
= p₁ = mu
Final momentum of the object
= p₂ = mv
The change in momentum
= p₂ – p₁ = (mv) – (mu) = m (v – u)
The rate of change in momentum
= \(\frac{\mathrm{m}(\mathrm{v}-\mathrm{u})}{\mathrm{t}}\)
Therefore,
The applied force, F is
F ∝ \(\frac{\mathrm{m}(\mathrm{v}-\mathrm{u})}{\mathrm{t}}\)
F = \(\frac{\mathrm{mk}(\mathrm{v}-\mathrm{u})}{\mathrm{t}}\) …………..(i)
(where k is proportionality constant)
From the first equation of motion,
v = u + at
Substituting (ii) in (i),
So, F = kma
The unit of force is so chosen that the value of k becomes one.
Hence, F = kma or
F = ma
The second law is related to sports in the following examples:
(a) In a cricket match, a fielder who is trying to attempt a catch, generally pulls his hands along the moving ball because it consumes time to bring the momentum of the ball to zero. This decreases the rate change of mometum, and hence the force. So, when the time is increased, the overall force reduces, thus saving the hand from getting injured.

(b) In a high jump athletic event, the athletes are made to fall on a sand bed to increase the time of the athlete’s fall to stop after making the jump.
This decreases the rate of change of momentum and hence the force,

(c) A karate expert breaks a slab of ice with a single blow as he moves his hand very fast and then stops it in a very short time. Since the rate of change of momentum increases, the force also increases and the ice slab breaks.

Analysing & Evaluating Questions

Question 6.
A test tube containing little water is corked and suspended from two strings in a slanting position. Afterwards, the test tube is heated Answer the following:

1. What happens to the cork?
2. In which direction does the test tube move?
3. Compare the velocities of the cork and of the recoiling test tube.
Answer:

1. On heating, the water inside the test tube vaporises and the cork is blown out.
2. The test tube moves in the direction opposite to the direction in which the cork has blown out.
3. Cork blows out with greater velocity than the recoiling test tube.

Activity 1

• Make a pile of similar carom coins on a table, as shown in the figure.
• Attempt a sharp and strong horizontal hit at the bottom of the pile using another carom coin or striker.

Observations

• If the hit is strong enough the bottom coin moves out quickly.
• Once the lowermost coin is removed, the inertia of the other coins makes them fall vertically on the table.

Activity 2

• Set a five – rupee coin on a stiff card covering an empty glass tumbler standing on a table as shown in the figure.
• Give the card a sharp horizontal flick with a finger.

Observations

• If we do it fast, the card shoots away, allowing the coin to fall vertically into the glass tumbler due to its inertia.
• The inertia of the coin tries to maintain its state of rest even when the card is flown away.
• The force applied on the card due to flicking changes the inertia of the card but the coin resists this change and stays at rest, i.e, due to the inertia and gravity, the coin falls down into the tumbler.

Activity 3

• Request two children to stand on two separate carts as shown in the figure.
  
• Give them a bag full of sand or some other heavy object.
• Ask them to play a game of catch with the bag and observe.
• You can paint a white line on cartwheels to observe the motion of the two carts when the children throw the bag towards each other.

Observations

• In this case, each of them receives an instantaneous reaction as a result of throwing the sand bag.
• This activity explains Newton’s third law of motion, i.e., a force is exerted in forward direction in throwing the bag full of sand and the person who is throwing it, gets pushed backward.

Activity 4

• Take a big rubber balloon and inflate it fully.
• Tie its neck using a thread Also, using adhesive tape, fix a straw on the surface of this balloon.
• Pass a thread through the straw and hold one end of the thread in your hand or fix it on the wall.
• Ask your friend to hold the other end of the thread or fix it on a wall at some distance.
• The arrangement is shown in the figure.
• Now, remove the thread tied on the neck of the balloon and let the air escape from the mouth of the balloon.
• Observe the direction in which the straw moves.

Observations.

• When the air escapes out from the balloon, the straw moves in the direction opposite to the direction in which the air moved out of the balloon. This activity explains the law of conservation of momentum.
• This activity is also based on Newton’s third law of motion, i.e., for every action, there is always an equal and opposite reaction between two bodies.

Value Based Questions

Question 1.
Class IX students were playing cricket with cork ball in the school campus. Abhishek, senior student, told them about the accidents that can occur due to cork ball in the campus and also advised them to bring a soft cosco ball to play the game.
1. Why is it safe to play with a soft ball and not with a hard cork ball?
2. A player pulls his hands backwards while catching the ball shot at high speed Why?
3. What value of Abhishek is seen in this act?
Answer:

1. The soft ball has less inertia as compared to the heavy ball and it would not hurt the players.
2. By pulling the hand backwards, it reduces the force exerted by the ball
   on the hand.
3. Abhishek showed the value of being responsible and helpful by nature.

Question 2.
Rahul saw his karate expert friend breaking a brick. He tried to break the brick but his friend stopped him from doing so and told him that it would hurt him as one needs a lot of practice in doing so.
1. How can a karate expert break the brick without any injury to his hand?
2. What is Newton’s third law of motion?
3. What value of Rahul’s friend is seen in the above case?
Answer:

1. A karate expert applies the force with a large velocity in a very short interval of time on the brick, therefore, a large force is exerted on the brick and it breaks.
2. To every action there is an equal and opposite reaction, both act on different bodies.
3. Rahul’s friend showed the value of being responsible and caring friend.

JAC Class 9 Science Important Questions

JAC Board Class 9th Science Important Questions Chapter 8 Motion

Multiple Choice Questions

Question 1.
Which of the following is correct for a car which travels a distance of 100 km in 2 hours?
(a) Its average speed is 50 km/h
(b) The car did not travel at 50 km/h all the time
(c) The car travelled at 50 km/h all the time
(d) All of the above
Answer:
(a) Its average speed is 50 km/h

Question 2.
The distance-time graph of an object shown in the given figure represents that the object is
(a) at rest position
(b) moving with constant speed
(c) moving with constant velocity
(d) moving with constant acceleration
Answer:
(a) at rest position

Question 3.
The SI unit of velocity is
(a) ms^-1
(b) ms^-2
(c) ms^-3
(d) Nn^-1
Answer:
(a) ms^-1

Question 4.
Deceleration of a body is expressed in
(a) m
(b) ms^-1
(c) ms^-2
(d) ms^-3
Answer:
(c) ms^-2

Question 5.
The initial velocity of a body is ‘u’. It is under uniform acceleration ‘a’. Its velocity ‘v’ at any time ‘t’ is given by
(a) v = u + at²
(b) v = u + \(\frac{1}{2}\) at²
(c) v = u + at
(d) v = u
Answer:
(c) v = u + at

Question 6.
A wooden slab starting from rest, slides down a 10m long inclined plane with an acceleration of 5 ms^-2 What would be its speed at the bottom of the inclined plane?
(a) 10 ms^-1
(b) 12ms^-1
(c) 10 cm^-1
(d) 12 cm^-1
Answer:
(a) 10 ms^-1

Question 7.
The velocity of a particle increases from ‘u’ to ‘v’ in time’t’ during which it covers a distance ‘s’. If the particle has a uniform acceleration, which of the following equations does not apply to the motion?
(a) 2s = (v + u) t
(b) v² = u² – 2as
(c) a = v – u/t
(d) s = (u + \(\frac{1}{2}\) at) t
Answer:
(b) v² = u² – 2as

Question 8.
In 12 minutes, a car whose speed is 35 km/h travels a distance of
(a) 7 km
(b) 3.5 km
(c) 2.4 km
(d) none of these
Answer:
(a) 7 km

Question 9.
The area under the distance – time graph gives
(a) uniform speed
(b) non – uniform speed
(c) velocity and speed
(d) both (a) and (b)
Answer:
(d) both (a) and (b)

Question 10.
The odometer of a car measures
(a) speed
(b) velocity
(d) acceleration
(d) distance
Answer:
(d) distance

Question 11.
If a particle moves with a constant speed, the distance time graph is a
(a) straight line
(b) curved line
(c) straight line parallel to time axis
(d) straight line parallel to velocity axis
Answer:
(a) straight line

Question 12.
An object moving with uniform circular motion shows
(a) constant acceleration in speed
(b) constant velocity
(c) constant acceleration in direction
(d) constant change in type of motion
Answer:
(c) constant acceleration in direction

Question 13.
The slope of speed – time graph gives
(a) speed
(b) velocity
(c) acceleration
(d) momentum
Answer:
(c) acceleration

Question 14.
The distance travelled by a freely falling body is proportional to the
(a) mass of body
(b) square of the time of fall
(c) square of the acceleration due to
Answer:
(b) square of the time of fall

Question 15.
The figure shows the displacement time graph of a body moving in a straight line. The velocity of the body during the

(a) 2 ms^-2
(b) zero
(c) 3 ms^-1
(d) 2.5 ms^-1
Answer:
(b) zero

Analysing & Evaluating Questions

Question 16.
In which of the following cases of motion, the distance moved and the magnitude of displacement are equal?
(a) If the car is moving on a straight road
(b) If the car is moving in a circular path
(c) The pendulum is moving to and fro
(d) The earth is revolving around the Sun
Answer:
(a) If the car is moving on a straight road

Question 17.
Four cars A, B, C and D are moving on a levelled road. Their distance versus time graphs are shown here. Choose the correct statement.

(a) Car A is faster than car D
(b) Car B is the slowest
(c) Car D is faster than car C
(d) Car C is the slowest
Answer:
(b) Car B is the slowest

Question 18.
Suppose a boy is enjoying a ride on a merry – go – round which is moving with a constant speed of 10 ms^-1. It implies that the boy is
(a) at rest
(b) moving with no acceleration
(c) in accelerated motion
(d) moving with uniform velocity
Answer:
(c) in accelerated motion

Assertion Reason Questions

Directions: In the following questions, the Assertions and the Reasons have been put forward. Read the statements carefully and choose the correct alternative from the following:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.
(B) The assertion and the reason are correct but the reason is not the correct explanation of the assertion.
(C) The assertion is true but the reason is false.
(D) Both the statements are false.

1. Assertion: Velocity is a scalar quantity.
Reason: Velocity cannot be zero.
Answer:
(D) Both the statements are false.

2. Assertion: A body moving in a circular path is in non – uniform motion.
Reason: The direction of a body moving in a circular path changes at every point.
Answer:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.

3. Assertion: A stone thrown vertically upwards has negative acceleration.
Reason: The acceleration of the stone thrown upward is in the direction opposite to the direction of its motion.
Answer:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.

4. Assertion: A freely falling body is in uniform motion.
Reason: A freely falling body covers equal distances in equal intervals of time.
Answer:
(D) Both the statements are false.

5. Assertion: An object under acceleration can have a uniform speed.
Reason: Both speed and acceleration are scalar quantities.
Answer:
(C) The assertion is true but the reason is false.

Very Short Answer Type Questions

Question 1.
Define uniform motion.
Answer:
If an object covers equal distances in equal intervals of time, however small the time intervals may be, the motion of the object is said to be uniform motion.

Question 2.
Define non – uniform motion.
Answer:
If an object covers unequal distances in equal intervals of time, it is said to be in ‘non – uniform motion’.

Question 3.
Define speed.
Answer:
It is the distance travelled by a body per unit time.
Speed. = \(\vec{a}\)

Question 4.
Define average speed.
Answer:
The total distance travelled by an object divided by the total time taken is called its average speed.

Question 5.
Define acceleration.
Answer:
It is defined as the rate of change of velocity with time.
Acceleration = \(\frac{Change in velocity}{Time Taken}\)
or a = \(\frac{v-\mathbf{u}}{t}\)
The SI unit of acceleration is m/s².

Question 6.
Draw a distance – time graph that represents uniform speed.
Answer:

Question 7.
With the help of a distance – time graph show that the object is stationary.
Answer:

Question 8.
Define uniform circular motion.
Answer:
When a body moves in a circular path with uniform speed, its motion is called uniform circular motion.

Question 9.
Define scalar quantities. Give examples.
Answer:
The physical quantities which require only magnitude, and not the direction, for their complete description are called ‘scalars’ or ‘scalar quantities’. Distance, speed, time, area, etc., are all scalar quantities.

Question 10.
Define vector quantities. Give examples.
Answer:
The physical quantities which need both magnitude and direction for their complete description are called ‘vectors’ or vector quantities. Displacement, velocity, force, etc., are all vector quantities.

Question 11.
What is the SI unit of displacement?
Answer:
Metre (m).

Question 12.
A particle moves over three – quarters of a circle of radius. What is the magnitude of its displacement?
Answer:
When the particle covers three quarters of a circle, the magnitude of its displacement is

AB = \(\sqrt{\mathrm{OA}^{2}+\mathrm{OB}^{2}}\) = \(\sqrt{r^{2}+r^{2}}\)
= \(\sqrt{2 r}\) = √2 r

Question 13.
Can the average speed of a moving body ever be zero?
Answer:
Speed is a scalar quantity and is always positive. So the average speed of a moving body can never be zero.

Question 14.
What is the relationship between the distance travelled and the time elapsed for motion with uniform velocity?
Answer:
Distance is directly proportional to the time elapsed. In fact,
Distance travelled = Uniform velocity × Time elapsed

Question 15.
What is the SI unit of acceleration?
Answer:
The SI unit of acceleration is m/s².

Question 16.
What is acceleration of a body moving with uniform velocity?
Answer:
The acceleration of a body moving with uniform velocity is zero.

Question 17.
State a relationship connecting u, v, a and t for an accelerated motion. Give an example of motion in which acceleration is uniform.
Answer:
The relationship between u, v, a and t is v = u + at A body falling freely towards the earth has a uniform acceleration of 9.8 ms^-2.

Question 18.
Express the velocity of a body in uniform circular motion in terms of its time period T.
Answer:
Suppose a body of mass m rotates in a circle of radiusr with velocity v. It completes one revolution in time T. Then,
Velocity = \(\frac{Distance}{Time}\)[/latex] = \(\frac{Circumference}{Time period }\)[/latex]
or v = \(\frac{2 \pi \mathrm{r}}{\mathrm{T}}\)

Analysing & Evaluating Questions

Question 19.
The cars A, B, C and D are moving on a levelled road. Their distance versus time graphs are shown in the figure. Which car is the slowest?
Answer:
Speed = Slope of distance – time graph. The smaller the slope, the smaller is the speed

From the given figure, slope is minimum for car D. So, D is the slowest car.

Question 20.
A rubber ball dropped from a certain height bounce to certain height. This height keeps decreasing in subsequent bounces. What type of motion does the ball exhibit?
Answer:
The ball exhibits non – uniform motion.

Question 21.
A vehicle is moving at a constant velocity of 10 km/h in the north – south direction. It turns and starts moving towards east. What is the velocity of the vehicle towards east before taking the turn?
Answer:
Zero. The component of velocity at right angle to the original direction of motion is zero.

Short Answer Type Questions

Question 1.
Define the term rest and motion.
Answer:

• Rest: If a body does not change its position with respect to its surroundings, the body is said to be at rest, e.g., a table lying in a room is at rest with respect to the walls of the room.
• Motion: A body is said to be in motion if it changes its position with respect to its surroundings, e.g., a car running on the road is in motion with respect to the lamp posts, trees or bus stop on the roadside.

Question 2.
What is meant by a point object? Give some examples.
Answer:
Whenever the size of the object is much smaller than the distance it moves in a given time interval, the size of the object can be neglected. The object can be regarded as a point object in such cases.
Examples:

1. A car covering a distance of 10km can be treated as a point object.
2. Earth can be regarded as a point object for studying its motion around the sun.

Question 3.
Give some examples of straight line motion.
Answer:
Examples:

1. A bus moving on a straight road
2. A train moving on a straight track
3. A runner running along a straight track
4. A ball moving along a straight path
5. An object falling vertically downwards towards the surface of the earth.

Question 4.
How can we specify the position of an object?
Answer:
1. The position of an object can be specified by choosing:

• a fixed point called ‘origin’ or reference point, and
• a fixed line passing through the origin, called reference axis.

2. So the position of an object can be fully described by knowing:

• its distance from origin O and
• the angle 0 which the line joining the origin ‘O’ and the object makes with the reference axis.

In the figure shown below, the position of an object located at point P is 6 cm from the origin and 30° north of east. v (Scale: 1cm = 1m)

Question 5.
A body moves in a circle of radius ‘2R’. What is the distance covered and displacement of the body after 2 complete rounds?
Answer:
Distance covered after 2 complete rounds = 2 × circumference
= 2 × 2π (2R) = 8πR
Displacement = zero, because initial and final positions of the body are the same and displacement is the shortest distance between initial and final positions.

Question 6.
Define the term velocity. What is its SI unit? Is it a scalar or a vector quantity?
Answer:
Velocity is a physical quantity that gives both speed and direction of motion of the body.
Definition: Velocity of a body is defined as the displacement produced per unit time. Velocity is also defined as the speed of a body in a givendirection. If ‘s’ is the distance travelled by a body in a given direction and ‘t’ is the time taken to travel that distance, then the velocity V is given by,
Velocity = \(\frac{Displacement}{Time}\)[/latex]
SI unit of velocity is m/s. Velocity is a vector quantity because it requires both magnitude and direction of a body.

Question 7.
Give the difference between distance and displacement.

Question 8.
Differentiate between speed and velocity.
Answer:

Question 9.
A cheetah is the fastest land animal and can achieve a peak velocity of 100 km/h upto distances less than 500 m. If a cheetah spots his prey at a distance of 100 m, what is the minimum time it will take to get its prey, if the average velocity attained by it is 90 km/h?
Answer:
Here v = 90Km/h = \(\frac{90 \times 1000 \mathrm{~m}}{3600 \mathrm{~s}}\)
= 25m/s
s = 100m
Minimum time, t = \(\frac{s}{v}\) = \(\frac{100}{25}\) = 4s

Question 10.
The Rajdhani Express travels a distance of 1384 km from Mumbai to Delhi. It starts from Mumbai at 4.00 p.m. and reaches Delhi at 9.00 a.m. the next day. What is its average speed?
Answer:
Total distance travelled = 1384 km
Total time taken = 17 hours
Average speed =\(\frac{Total distance travelled}{Total time taken}\)= \(\frac{1384}{17}\)
= 81.4 km/h

Question 11.
A body starts initially with a velocity ‘u’ and is accelerated at constant rate ‘a’. Find an expression for final velocity after time ‘t’
Answer:
First equation of motion : Let a body start with initial velocity ‘u’ and after time ‘t’, its velocity becomes ‘v’ due to uniform acceleration ‘a’. From the definition of acceleration,
Acceleration = \(\frac{Change in velocity }{Time taken}\)
= \(\frac{Final velocity – Initial velocity}{Time taken}\)
a = \(=\frac{\mathrm{V}-\mathrm{u}}{\mathrm{t}}\)
at = v – u
v = u + at.

Question 12.
Deduce the expression for the distance travelled by a body moving with uniform acceleration in a given time.
Answer:
Second equation of motion:
Suppose a body starts with initial velocity ‘u’ and due to uniform acceleration ‘a’ its final velocity becomes ‘v’ after time ‘t’ Then,
Average velocity = \(\frac{Initial velocity + Final velocity}{2}\)
=\(\frac{u+v}{2}\)
So, the distance covered by the body in time t is
s = Average velocity × Time =\(\frac{u+v}{2}\) × t = \(\frac{\mathrm{u}+(\mathrm{u}+\mathrm{at}) \times \mathrm{t}}{2}\)
= \(\frac{2 u t + a t^{2}}{2}\)
or s = ut + \(\frac{1}{2}\) at 2

Question 13.
Establish the relation v² – u² = 2as, where ‘u’ is the initial velocity, ‘v’ is the final velocity, ‘a’ is the uniform acceleration and ‘s’ is the distance covered by the body.
Answer:
Third equation of motion:
Let a body start with initial velocity ‘u’ and after covering distance ‘s’ under uniform acceleration ‘a’, its velocity becomes ‘v’ in ‘t’ seconds. Then
Average velocity = \(\frac{\mathrm{u}+\mathrm{v}}{2 \mathrm{}}\)
So the distance covered in time t is given by
s = Average velocity x Time
= \(\frac{u+v}{2}\) × t
or v + u = \(\frac{2 s}{t}\) …………..(1)

Using the first equation of motion: v = u + at
or v – u = at … (ii)

Multiplying equations (i) and (ii), we get
(v + u) (v – u) = \(2 \frac{s}{t}\) × at
or v² – u² = 2as

Question 14.
Show that the slope of distance – time graph gives velocity of the body.

Answer:
Given figure shows distance – time graph for a body moving with uniform velocity. Clearly, it covers distances s₁ and s₂ in times t₁ and t₂ respectively.
Slope of line AB = tan θ = \(\frac{\mathrm{BC}}{\mathrm{AC}}\)
= \(\frac{s_{2}-s_{1}}{t_{2}-t_{1}}\) = \(\frac{Displacement}{Time}\) = velocity
Hence, the slope of the distance-time graph gives velocity of the body.

Question 15.
Show that the slope of velocity time graph gives acceleration of the body.
Answer:
Given figure shows the velocity – time graph for a body in uniform acceleration. It is a straight line inclined to the time – axis. Body has velocities ‘u’ and ‘v’ at times ‘t₁’ and ‘t₂’ respectively
Img-1
Slope of line AE = tan θ = \( \frac{E D}{A D}\)
\(\frac{v-u}{t_{2}-t_{1}}\) = \( \frac{Change in velocity}{Time taken}\)
= Acceleration of the body
Hence, the slope of the velocity – time graph gives the acceleration of the body.

Question 16.
Draw velocity – time graph for a body moving with uniform velocity. Hence show that the area under the velocity time graph gives the distance travelled by the body in a given time interval.
Answer:
In the given figure, line PQ is the velocity – time graph of a body moving with a uniform velocity such that OP = v Area of rectangle ABCD = AD × AB
= OP × AB
= v × (t₂ – t₁)
= Velocity × Time
= Distance travelled in time interval (t₂ – t₁)

Hence, the area under the velocity – time graph gives the distance travelled by the body in the given time interval.

Question 17.
The odometer of a car reads 1800 km at the start of a trip and 2400 km at the end of the trip. If the trip took 10h, calculate the average speed of the car in km/h and m/s.
Answer:
Distance covered by the car (s)
= 2400 – 1800 = 600 km.
Trip time = 10h
Average speed = ?

(i) V_av = \(\frac{s}{t}\) = \(\frac{600}{10}\) km = 60 km/h

(ii) In m/s : (60 km/h) = \(\frac{60 \times 1000}{60 \times 60}\)
= 16.7 m/s
The average speed of the car in km/h is 60 km/h and in m/s is 16.7 m/s.

Question 18.
Draw velocity – time graphs to show the following:
(a) Uniform velocity
(b) Uniform acceleration
(c) Non – uniform acceleration
Answer:

Analysing & Evaluating Questions

Question 19.
The table given below shows distance (in km) travelled by bodies A, B and C. Read the data carefully and answer the following questions.

(a) Which of the bodies is moving with
(i) constant speed?
(ii) constant acceleration?
(b) Which of the bodies covers
(i) maximum distance in 3rd second?
(ii) minimum distance in 3rd second?
Answer:
(a) (i) Body A
(ii) Body C

(b) (i) Body C
(ii) Body A

Question 20.
An electron moving with a velocity of 5 × 10⁴ms^-1 enters a uniform electric field and acquires a uniform acceleration of 10⁴ms^-2 in the direction of its initial motion.
(a) Calculate the time in which the electron would acquire a velocity double of its initial velocity.
(b) How much distance the electron would cover in this time?
Answer:
(i) Acceleration = \(\frac{ Change in velocity}{Time taken}\)
or Time taken, t = \(\frac{5 \times 10^{4} \mathrm{~m} \mathrm{~s}^{-1}}{10^{4} \mathrm{~m} \mathrm{~s}^{-2}}\)
= 5 s

(ii) s = ut + \(\frac{1}{2}\) at²
= (5 × 10⁴ m s^-1 × 5 s) + \(\frac{1}{2}\) × 10⁴m s^-2 × (5s)²
= 37.5 × 104 m

Long Answer Type Questions

Question 1.
Describe the various types of motion observed in bodies.
Answer:
Various types of motion as observed in bodies are:

1. Translatory motion: When a body moves, as a whole, along a straight or curved path, it is said to be in translatory motion. Translatory motion is again of two types:
   • Rectilinear motion : Here a body moves as a whole along a straight path. For example, a train moving on a straight track has translatory rectilinear motion.
   • Curvilinear motion : In this case a body moves as a whole along a curved path. For example, motion of a bicycle taking a turn along a curved path.
2. Rotatory motion: When a body rotates about a fixed point or axis. it exhibits a rotatory motion. For example, motion of a flywheel about a shaft
3. Vibratory or oscillatory motion: When a body moves to and fro about a mean position, the motion is said to be vibratory or oscillatory motion. For example, the motion of the pendulum of a wall – clock.
4. Complex motion: When the motion of a body may be a combination of more than one type of motion, it is said to be a complex motion. For example, a ball rolling down an inclined plane has both translatory and rotatory motions.

Question 2.
Define average velocity when the velocity of a body changes at a non – uniform rate and a uniform rate.
Answer:
Average velocity : When the velocity of a body changes at a non – uniform rate, its average velocity is defined as the net displacement covered divided by the total time taken.
Average velocity = \(\frac{Net displacement}{Total time taken}\)
When the velocity of a body changes at a uniform rate, the average velocity is given by the arithmetic mean of initial velocity and final velocity for a given period of time.
Average velocity = \(\frac{Initial velocity + Final velocity}{2}\) :
If u is the initial velocity and v is the final velocity, the average velocity v_ay is given by,
v_ay = \(=\frac{u+v}{2}\)

Question 3.
Explain the difference regarding the nature of acceleration of the three moving bodies as expressed by the following velocity – time graphs:

(a) Uniform acceleration: A body increases velocity by equal amounts in equal intervals of time.
(b) Non – uniform acceleration: A body travels unequal distances in equal intervals of time.
(c) Uniform motion or zero acceleration: A body moves with constant

Analysing & Evaluating Questions

Question 4.
Two stones are thrown vertically upwards simultaneously with their initial velocities u₁ and u₂ respectively. Prove that the heights reached by them would be in the ratio of \(\mathbf{u}_{1}^{2}\) : \(\mathbf{u}_{2}^{2}\) (Assume upward acceleration is – g and downward acceleration to be + g).
Answer:
Stone 1
Initial velocity = u₁
Acceleration = – g
Height = h,
Final velocity, v = 0
Using the equation, v² – u² =2as
0 – \(\vec{a}\) = 2 (- g) × h₁
h₁ = \(\frac{-u_{1}^{2}}{-2 g}\) = \(\frac{u_{1}^{2}}{2 g}\) ……..(1)

Stone 2
Initial velocity = u₁
Acceleration = – g
Height = h2
Final velocity, v = 0
0² \(-u_{2}^{2}\) = 2 (- g) × h₂
h₂ = \(\frac{-u_{2}^{2}}{-2 g}\) =\(\frac{u_{2}^{2}}{2 g}\) ……..(ii)
This gives; h₁ : h₂ = \(\vec{a}\) : \(\vec{a}\)
= \(=\frac{u_{1}^{2}}{2 g}\) : \(\frac{\mathrm{u}_{2}^{2}}{2 \mathrm{~g}}\)
=\(u_{1}^{2}\) : \(\vec{a}\)

Activity 1

1. In your everyday life, you come across a range of motions in which:
   • acceleration is in the direction of motion
   • acceleration is against the direction of motion
   • acceleration is uniform
   • acceleration is non – uniform
2. Observe these motions carefully and identify one example each of the above type of motions.

Observations

• A car moving on a road.
• A ball thrown up.
• Fan blades rotating
• Windmill at time moves fast when the wind speed is more and becomes slow when the wind speed decreases.

Activity 2

• Take piece of thread and tie a small piece of stone at one of its end. Move the stone to describe a circular path with constant speed by holding the thread at the other end.
  
• Now, let the stone go by releasing the thread.
• Observe the direction in which the stone moves after it is released.
• By repeating the activity for a few more times and releasing the stone at different positions of the circular path, carefully observe whether the direction in which the stone moves remains the same or not.

Observations

• When the stone is released, it moves along a straight line, tangential to the circular path.
• By repeating the activity and releasing the stone at different positions of the circular path, the direction in which the stone moves does not remain the same. It changes every time.

Value Based Questions

Question 1.
The speed limits are prescribed for vehicles running on highways. Why is it essential to follow the speed limit rules?
Answer:
Speed limits are prescribed on the highways/expressways because:
1. When a vehicle is made to run at height speeds, its tyres get hot (due to friction). This makes them softer and the air inside the tubes hotter Both. these factors may lead to bursting of tyre/tube.

2. Reflex time varies from person to person. Any running vehicle needs a certain time period to stop after the brakes are applied. At higher speeds, this time period becomes shorter.

3. As a result, the chances of collision increase. Because of these reasons, one should strictly follow the speed limit regulations. Also, it is safer to drive in your lane depending upon the speed of your vehicle.

4. Moral: Drive below the speed limit and in your lane.

Question 2.
Most drivers involved in road accidents are found to be drunk. Give reason.
Answer:
The reflex time of a person increases when drunk, (average reflex time of a normal person is about 1/15s). Therefore, such a driver of vehicle will take more time in applying brakes. As a result, the vehicle may not stop well in time and cause an accident.

JAC Class 9 Science Important Questions

JAC Jharkhand Board Class 9th Sanskrit Solutions शेमुषी भाग 1

• Chapter 1 भारतीवसन्तगीतिः
• Chapter 2 स्वर्णकाकः
• Chapter 3 गोदोहनम्
• Chapter 4 कल्पतरुः
• Chapter 5 सूक्तिमौक्तिकम्
• Chapter 6 भ्रान्तो बालः
• Chapter 7 प्रत्यभिज्ञानम्
• Chapter 8 लौहतुला
• Chapter 9 सिकतासेतुः
• Chapter 10 जटायोः शौर्यम्
• Chapter 11 पर्यावरणम्
• Chapter 12 वाङ्मनः प्राणस्वरूपम्

JAC Class 9 Sanskrit अपठितावबोधनम्

• अपठित अनुच्छेदाः

JAC Class 9 Sanskrit व्याकरणम्

• उच्चारणस्थानानि
• सन्धि-प्रकरणम्
• समास प्रकरणम्
• कारक प्रकरणम्
• उपसर्ग प्रकरणम्
• प्रत्यय प्रकरणम्
• संज्ञा-शब्दरूप-प्रकरणम्
• सर्वनाम शब्दरूप प्रकरणम्
• संख्याज्ञानम्
• धातुरूप-प्रकरणम्

JAC Class 9 Sanskrit रचनात्मक कार्यम्

• पत्र-लेखनम्
• संकेताधारितलघुकथा, चित्रवर्णनम्
• संकेताधारित अनुच्छेदलेखनम्
• अनुवाद-प्रकरणम्

JAC Board Class 9th Science Important Questions Chapter 7 Diversity in Living Organisms

Multiple Choice Questions

Question 1.
Whittaker classified all organisms into
(a) five kingdoms
(b) four kingdoms
(c) three kingdoms
(d) two kingdoms
Answer:
(a) five kingdoms

Question 2.
A housefly belongs to which phylum?
(a) Nematoda
(b) Annelida
(c) Porifera
(d) Arthropoda
Answer:
(d) Arthropoda

Question 3.
The five kingdom classification is based on the complexity of
(a) mode of nutrition
(b) body organisation
(c) cell structure
(d) all of these
Answer:
(d) Arthropoda

Question 4.
Which one of the following belongs to coelenterata?
(a) Sycon
(b) Hydra
(c) Spongilla
(d) Planaria
Answer:
(b) Hydra

Question 5.
In which of the following are the reproductive organs hidden?
(a) Cryptogamae
(b) Phanerogamae
(c) Gymnosperms
(d) Angiosperms
Answer:
(a) Cryptogamae

Question 6.
Which phylum of animals is also called flatworms?
(a) Porifera
(b) Coelenterata
(c) Platyhelminthes
(d) Nematoda
Answer:
(c) Platyhelminthes

Question 7.
The excretory system in annelids consists of tubes called
(a) flame cells
(b) metanephridia
(c) nephridia
(d) protonephridia
Answer:
(c) nephridia

Question 8.
What is the phylum of octopus?
(a) Arthropoda
(b) Mollusca
(c) Annelida
(d) Cnidaria
Answer:
(b) Mollusca

Question 9.
What is the mode of nutrition in bacteria?
(a) Autotrophic
(b) Heterotrophic
(c) Both (a) and (b)
(d) None of these
Answer:
(c) Both (a) and (b)

Question 10.
In which organism do flame cells form the excretory system?
(a) Flatworm
(b) Earthworm
(c) Insect
(d) Crab
Answer:
(a) Flatworm

Question 11.
Which sub – group in plant kingdom produces flowers?
(a) Angiosperm
(b) Fungus
(c) Moss
(d) Fern
Answer:
(a) Angiosperm

Question 12.
The mode of nutrition in fungi is
(a) only saprotrophic
(b) saprotrophic or parasitic
(c) only parasitic
(d) none of the above
Answer:
(b) saprotrophic or parasitic

Question 13.
Which among the following is exclusively marine?
(a) Arthropoda
(b) Mollusca
(c) Echinodermata
(d) Coelenterata
Answer:
(c) Echinodermata

Analysing & Evaluating Questions

Question 14.
Meena and Hari observed an animal in their garden. Hari called it an insect, while Meena said it was an earthworm. Which of the following characteristics confirms that it was an insect?
(a) Bilaterally symmetrical body
(b) Body with jointed legs
(c) Cylindrical body
(d) Body with little segmentation
Answer:
(b) Body with jointed legs

Question 15.
After studying the characteristics of Agaric us, Madan noted them as follows:
I. It is fleshy.
II. It has an umbrella-like cap called pileus.
III. Spores are produced by the stipe.
IV. Its body is made of filaments.
Which observation is not correct?
(a) I
(b) II
(c) III
(d) IV
Answer:
(c) III

Question 16.
Preeti studied the following
features in a preserved specimen – bilateral symmetry, true segmentation, clitellum and mouth. It is a/an
(a) tapeworm
(b) cockroach
(c) earthworm
(d) roundworm
Answer:
(c) earthworm

Assertion Reason Questions

Directions: In the following questions, the Assertions and the Reasons have been put forward Read the statements carefully and choose the correct alternative from the following:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.
(B) The assertion and the reason are correct but the reason is not the correct explanation of the assertion.
(C) The assertion is true but the reason is false.
(D) Both the statements are false.

1. Assertion: Bacteria and blue – green algae belong to the same phylum, i.e. Monera.
Reason: Both the bacteria and blue – green algae are unicellular prokaryotic organisms.
Answer:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.

2. Assertion: Most of the amphibians lay their eggs in water.
Reason: Young ones of amphibians have gills at the initial stages of there lives.
Answer:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.

3. Assertion: Snake and turtle belong to the same group.
Reason: Both the snake and the turtle are cold – blooded and have four – chambered heart.
Answer:
(C) The assertion is true but the reason is false.

4. Assertion: Dolphins do not belong to pisces.
Reason: Dolphins respire through lungs and possess four – chambered heart.
Answer:
(A) Both the assertion and the reason are correct and the reason is the correct explanation of the assertion.

5. Assertion: Roundworms lack a digestive tract.
Reason: Roundworms are endoparsites.
Answer:
(B) The assertion and the reason are correct but the reason is not the correct explanation of the assertion.

Very Short Answer Type Questions

Question 1.
What is evolution?
Answer:
Evolution is the gradual unfolding of organisms, from the pre – existing ones, through changes since the beginning of life.

Question 2.
Name the scientist who described the idea of organic evolution and the book in which he explained it.
Answer:
Charles Darwin first described the idea of evolution in his book ‘The Origin of Species’.

Question 3.
What is lichen?
Answer:
Lichens are a complex life form that is a symbiotic partnership of two separate organisms, i.e., a fungus and an alga The alga can be either a green alga or a blue-green alga (also known as cyanobacteria).

Question 4.
What is a symbiotic relationship or symbiosis?
Answer:
It is a relationship between two organisms in which both of them are benefited, e.g., in the symbiotic association of lichens, fungi get food from blue – green algae and in return, blue-green algae get shelter.

Question 5.
Who proposed the two kingdom classification?
Answer:
Carolus Linnaeus.

Question 6.
What is biodiversity?
Answer:
Different forms of living organisms found in a particular region is known as biodiversity of that region.

Question 7.
Define species.
Answer:
A species includes all organisms that are similar enough to interbreed and perpetuate naturally.

Question 8.
Define mycoplasma.
Answer:
Mycoplasma are the smallest and the simplest organisms. They are prokaryotes having a nucleoi(d) Their body can change shape easily. They are heterotrophs.

Question 9.
Define ‘taxonomy’.
Answer:
The branch of science that classifies living organisms among different categories or groups is called taxonomy. Taxonomy is the science of identifying and naming species and organising them into systems of classification.

Question 10.
Define Taxon.
Answer:
A taxon is a unit of classification of organisms which can be recognised to a definite category at any level of classification, for e.g., fishes, birds, insects, etc.

Question 11.
In how many kingdoms does Carolus Linnaeus divided living beings?
Answer:
Two kingdoms, viz., Plantae (plants) and Animalia (animals).

Question 12.
Name the division of plant kingdom which is also called the amphibians of plant kingdom.
Answer:
Bryophyta, e.g., Lunaria (moss).

Question 13.
What is meant by bilateral symmetry?
Answer:
When the left and right halves of the body have the same design, it is called bilateral symmetry.

Question 14.
What is thallus?
Answer:
Thallophytes have a simple plant body. The plant body is not differentiated into root, stem and leaves and is called thallus.

Question 15.
What is moss?
Answer:
Moss is a radially symmetrical leafy bryophyte having multicellular rhizoids, Funaria, Bryum, Sphagnum.

Question 16.
Why are bryophytes called the amphibians of the plant kingdom?
Answer:
Bryophytes are known as amphibians of the plant kingdom because these plants can live in soil but are dependent on water for sexual reproduction. Usually, they are found in humid and damp areas.

Analysing & Evaluating Questions

Question 17.
The external features of some living organisms are given below as I, II, III and IV. Read these and write the name of their respective animal phylum or class.
I. Streamlined body, covered with scales and divided externally into head, trunk and tail.
II. Skin is dry, covered with epidermal scales and bear two pairs of limbs each with five toes ending in horny claws.
III. Body is externally segmented and covered with a thick exoskeleton made up of chitin. The segments are grouped to form head, thorax and abdomen.
IV. Body is flattened, leaf-like or ribbon – like and bilaterally symmetrical.
Answer:
I – Phylum chordata (superclass pisces)
II – Phylum chordata (class reptilia)
III – Phylum arthropoda
IV – Phylum platyhelminthes

Question 18.
An animal is described as worm-like with unsegmented body and shows some features of  invertebrates and some of chordates. Identify the described animal.
Answer:
The animal is Balanoglossus.

Question 19.
Atrip to Himalayan foothills was organised by the school of Raman and Riya During the trip, they noticed tall trees having needle – like leaves and cones. Name the trees Raman and Riya saw there.
Answer:
They saw Pinus trees.

Short Answer Type Questions

Question 1.
What are the steps in classification of organisms?
Answer:
Organisms are classified into large groups first on the basis of independent characteristics. The characteristic in the next level would be dependent on the previous one and would decide the variety of organisms in the next level. Thus, one can build up a whole hierarchy of mutually related characteristics to be used for classification.

Question 2.
Write the hierarchy of classification proposed by Linnaeus.
Answer:
The hierarchical categories are as follows:

• Kingdom: Plant kingdom and Animal kingdom.
• Phylum (for animals)Division (for plants): A group of closely related classes having certain common characteristics.
• Class: A group of closely related orders having certain common characteristics.
• Order: A group of closely related families having certain common characteristics.
• Family: A group of closely related genera having certain common characteristics.
• Genus: A group of closely related species having certain common characteristics.
• Species: A group of organisms which are similar enough to breed and perpetuate.

Question 3.
What are the characteristics of kingdom Monera?
Answer:

1. These organisms do not have clearly defined nucleus, i.e., nucleus is not enclosed in a nuclear membrane.
2. Cell organelles are not covered with membranes.
3. The organisms do not show multicellular body design, i.e., they are unicellular.
4. Some organisms have cell wall, others do not have cell wall.
5. The mode of nutrition may be autotrophic or heterotrophic.

Question 4.
Why are blue green algae included under Monera and not under Plantae?
Answer:
Monera is a kingdom of prokaryotes while organisms of kingdom Plantae shows a definite nucleus, membrane- bound organelles and multicellular body design. Blue green algae are prokaryotes having nucleoid with naked DNA. The cell organelles are also not enclosed within a membrane. As they also do not possess multicellular body design, these characteristics bring them closer to Monera and exclude them from the kingdom Plantae.

Question 5.
Give four main features of organisms placed under Protista.
Answer:
The four main features of organisms placed under Protista are as follows:

1. This kingdom includes unicellular algae, diatoms and protozoans.
2. The organisms in this kingdom are unicellular, eukaryotic organisms (as well – defined nucleus and other membrane-bound cell organelles are present).
3. Mode of nutrition is either autotrophic (like in algae and diatoms) or heterotrophic (like in protozoans).
4. Some Protists bear hair – like cilia or whip – like flagella for movement. In some protists, like Amoeba, movement takes place by pseudopodia (false feet).

Question 6.
What are the characteristics of fungi? Give examples.
Answer:
Characteristics of fungi are as follows:

1. They are heterotrophic.
2. The organisms are saprophytic, i.e., they use decaying organic material as food or may be parasitic.
3. They have membrane bound nucleus.
4. They have cell walls made of chitin.
5. Some of them have the capacity to become multicellular organisms at certain stages in their lives.
6. Some of the fungal species have symbiotic relationship where they are mutually dependent on the blue – green algae. These are called lichens.
7. Examples: Rhizopus, Yeast, Agaricus (mushrooms), Penicillium.

Question 7.
On what basis are plants divided into two sub-kingdoms?
Answer:
Based on whether the reproductive organs are conspicuous (clearly visible) or not, plants are divided into two subkingdoms:

1. Cryptogamae: Non – flowering or seedless plants which include thallophytes, bryophytes and pteridophytes.
2. Phanerogamae: Flowering plants which include gymnosperms and angiosperms.

Question 8.
How are angiosperms further divided?
Answer:
Angiosperms are divided into two groups on the basis of the number of cotyledons present in the seeds.

1. Monocotyledonous (Monocots): These are the plants with seeds having a single cotyledon, e.g., maize, wheat, rice, etc.
2. Dicotyledonous (Dicots): These are the plants with seeds having two cotyledons, e.g., pea, gram, bean, etc.

Question 9.
List the similarities between plants and animals.
Answer:
In spite of certain external differences, plants and animals show a number of very obvious similarities which are as follows:

1. Plants and animals are made up of microscopic units called cell.
2. Both contain living substance called protoplasm.
3. Certain life processes, i.e., respiration, digestion, reproduction, assimilation, etc., take place in both the groups in an identical manner.
4. Both show response to external stimuli.
5. Both of them show growth.
6. Both reproduce and pass on their hereditary characters to their offsprings by the same mechanism.
7. Both are multicellular and eukaryotic organisms having division of labour in their cells.

Question 10.
Differentiate between algae and fungi.
Answer:

Question 11.
What are the conventions followed for writing the scientific names?
Answer:
The conventions followed while writing the scientific names are as follows:

1. The name of the genus begins with a capital letter.
2. The name of the species begins with a small letter.
3. When printed, the scientific name is given in italics.
4. If it is handwritten, the genus name and the species name have to be underlined separately.

Question 12.
Identify the different parts A, B, C and D of fern plant shown in the figure given below.

Answer:
(A) Leaflet
(B) Stipe
(C) Adventitious roots
(D) Rhizome

Question 13.
Define bryophytes. Give some examples.
Answer:

1. Bryophytes:
   • These are also called the amphibians of the plant kingdom.
   • Plant body may be thalloid or leafy.
   • True roots are absent, instead rhizoids develop.
   • No specialised conducting tissue. These grow on damp walls and on the bark of tree.
2. These include two groups:
   • Liverworts
   • Mosses
3. Examples: Marchantia, Riccia, Sphagnum, etc.

Question 14.
Draw a labelled diagram of Spirogyra cell.
Answer:

Question 15.
Define pteridophytes.
Answer:
Pteridophytes are vascular cryptogams and are the first vascular land plants. The main plant body is the saprophyte, which is differentiated into true roots, stems and leaves. Their xylem lacks companion cells. They have specialised tissue for the conduction of water and other substances from one part of the plant body to another. Examples: Mars ilea, ferns and horse tail.

Question 16.
Define coelom.
Answer:
Coelom is a cavity which lies in between the body wall and gut (alimentary canal) and is lined by mesodermal cells. It allows greater body flexibility. In it, the organs of the body can accommodate in a better way.

Question 17.
Differentiate between diploblastic and triploblastic animals.
Answer:

Question 18.
State the features of chordates.
Answer:
The main features of chordates are as follows:

1. They have bilaterally symmetrical body.
2. They have notochord and internal skeleton.
3. They have dorsal nerve cord.
4. They are triploblastic.
5. They are coelomate.
6. Respiration is mainly through lungs in land vertebrates and by gills in aquatic animals.

Question 19.
Discuss the characteristics of sponges.
Answer:
Phylum (Porifera) – Sponges:

1. Animals have pores (called ostia) all over the body.
2. Body is not well differentiated.
3. Non – motile animals, remain attached to solid support.
4. Body is covered with hard skeleton.
5. Reproduction is both asexual (by budding and gemmule formation) and sexual (through fertilisation).
6. Examples: Sycon, Spongilla and Euplectella.

Question 20.
Give specific features of coelenterata.
Answer:
Phylum (Cnidaria) – Coelenterata

1. All coelenterates are found in water.
2. Body is radially symmetrical.
3. These are the first of multicellular animals which possess tissue level of organisation with a distinct division of labour.
4. The body has a single, sac – like central cavity, called coelenteron or the gastrovascular cavity, with only one opening.
5. Some coelenterates live in colonies (Obelia) while others live solitary (Hydra).
6. Examples: Hydra, Aurelia, sea anemone and jelly fish.

Question 21.
Give the general characteristics of Platyhelminthes.
Answer:

1. Their body is dorsoventrally flattened and leaf-like or ribbon like.
2. Body symmetry is bilateral, i.e., left and right halves have the same design.
3. They are mostly hermaphrodite.
4. Body cavity or true coelom is absent.
5. There are three layers of cells from which differentiated tissues can be forme(d) So, animals are triploblastic.
6. They are either free living or parasitic.
7. Examples: Planaria, Fasciola hepatica (liver fluke) and Taenia solium (tape worm) are parasitic.

Question 22.
Give the characteristics of Arthropoda with two examples.
Answer:

1. This is the largest group of animals.
2. They possess jointed legs/ appendages.
3. They have bilaterally symmetrical and segmented body.
4. Body cavity is filled with blood.
5. There is an open circulatory system, i.e., the blood does not flow in definite blood vessels.
6. Examples: Apis (honey bee), Musca (housefly), Anopheles (mosquito), Palaemon (prawn), crabs, etc

Question 23.
Give the characteristics of Echinoder – mata.

1. They are free living, marine animals.
2. They are triploblastic and have a coelomic cavity.
3. They have peculiar water-driven tube system used for moving around.
4. They have a hard calcium carbonate structure that is used as skeleton.
5. Sexes are separate.
6. Examples: Asterias (starfish),

Question 24.
Echinus (sea urchin), Antedon (feather star), etc Give the characteristics of Nematoda.

1. Most of the Nematodes have small cylindrical or round body.
2. Body cavity is not a true coelom. A pseudocoelom is present.
3. Body is bilaterally symmetrical and triploblastic.
4. Sexes are separate.
5. Examples: Ascaris (round worm), Ancylostoma (hook worm) and Wuchereria (filarial worm).

Question 25.
Give the characteristics of phylum Annelida.

1. They have elongated and segmented body.
2. Body bears lateral appendages for locomotion in the form of chitinous setae or parapodia.
3. The body is bilaterally symmetrical and triploblastic.
4. Reproduction by sexual means. Sexes may be either separate or united.
5. They have true coelom (body cavity).
6. Examples: Pheretima (earthworm), Hirudinaria (blood sucking leech) and Nereis.

Question 26.
List the main features of phylum Mollusca.
Answer:

1. They have unsegmented or soft body with little segmentation.
2. The body is divided into three regions—head, dorsal visceral mass and ventral foot.
3. Body is bilaterally symmetrical.
4. The coelomic cavity is reduced.
5. They have open circulatory system and kidney-like organs for excretion.
6. Some molluscs have hard calcareous shell, an outer covering of the body.
7. Respiration is by gills called ctenidia.
8. Examples: Pila (snail), Unio (fresh water mussel) and Octopus.

Analysing & Evaluating Questions

Question 27.
You are given leech, Nereis, Scolopendra, prawn and scorpion and all have segmented body organisation. Will you classify them in one group? If no, give the important characteristics based on which you will separate these organisms into different groups.
Answer:
Leech, Nereis, Scolopendra, prawn and scorpion belong to different groups.

1. Leech and Nereis belong to phylum Annelida They have metameric segmentation, closed circulatory system and unjointed appendages.
2. Scolopendra, prawn and scorpion belong to phylum Arthropoda They have open circulatory system and jointed appendages.

Question 28.
Sakshi’s younger brother frequently suffered from stomach ache and vomiting. Her mother took him to a clinic for treatment. Doctor advised for a stool test before prescribing some medicines. The report diagnosed that the stool is infected with common roundworms. Answer the following questions asked by Sakshi to the doctor:
(a) What are common roundworms?
(b) Are there any other worms that live as parasites in our body and cause diseases?
(c) How do roundworms enter our body? What can be done to prevent such infections?
Answer:

1. Common roundworms are parasites that live inside the human digestive tract and feed on the food present inside the tract. During feeding, they damage the stomach and the intestines. The scientific name of roundworms is Ascaris.
2. Some other common parasitic worms that cause different diseases in our body are Wuchereria (filarial worm), Enterobius (pinworm) and Ancylostoma (hookworm).
3. Roundworms enter our body through contaminated food and water. Hence, one should always maintain hygienic food habits to avoid these parasites.

Long Answer Type Questions

Question 1.
Describe the three basic features for grouping all organisms into five major kingdoms.
Answer:
Basic criteria such as nature of cell, cellularity and mode of nutrition were used by Whittaker to classify the living organisms into five kingdoms,
(a) Nature of cell relates to the pres – ence or absence of membrane bound organelles in it. On the basis of this category, we can clas¬sify the living organisms in two broad categories prokaryotes and eukaryotes.

1. Prokaryotes, such as all bacteria, have poorly developed nuclear material called nucleoid and no membrane – bound organelles.
2. Eukaryotes such as protozoans, fungi, plants and animals have well developed genetic material called nucleus and all membrane – bound organelles.

(b) Cellularity refers to the number of cells present in an organism. Unicellular organisms, such as bacteria, and protozoans, are made up of single cell while multicellular organisms such as many fungi, plants and animals are made up of many cells.

(c) Mode of nutrition categorises all living organisms into autotrophs and heterotrophs. Euglena is a dual organism because of its mode of nutrition. It can act as both an autotroph (makes its own food) as it contains chlorophyll and heterotroph (feeds on other substances). Also, in presence of excess organic matter or darkness, it can act as a saprophytic organism. So, as it shows features of both plants and animals, it is a dual organism.

Question 2.
Give the outline classification of animal kingdom upto the level of phyla

Question 3.
What are the differences between plants and animals?
Answer:

Question 4.
Differentiate between bryophytes and pteridophytes.
Answer:

Question 5.
Give the classification of plant kingdom.

Question 6.
Write the characteristics of flat worms, round worms and segmented worms. Write their phylum also.
Answer:

Analysing & Evaluating Questions

Question 7.
Observe the plant given below and answer the questions that follow.
1. Identify the plant and name the phylum it belongs to.
2. What is its habitat?
3. List any two of its special features.
4. How does transport of substances take place in this plant?
Answer:

1. It is a Cycas plant. It belongs to the phylum gymmosperms.
2. It is found mainly in colder regions of the earth.
   
3. Plant body is sporophyte. Ovules are not enclosed in ovary, thus produce naked seeds.
4. Vascular tissues are present for the conduction of substances.

Activity 1

Find out the scientific names of the following animals and plants:
(a) Frog
(b) Peacock
(c) Human
(d) Neem
(e) Maize
(f) Honeybee

Observations
It is difficult to remember names of a species in different languages. There was a need for some system to create uniform naming convention.

Activity 2

• Soak seeds of green gram, wheat, maize, pea and tamarin(d) Once they become tender, try to split the seeds. Observe if the seeds break into two nearly equal halves.
• Now, observe the roots, leaves and flowers of these plants and record your observations.
  Observations
• All seeds do not break into two nearly equal halves.
• Out of the given seeds, gram, pea and tamarind are dicots while wheat and maize are monocots.
• Dicot plants have tap-roots and monocots have fibrous root system.
• Dicot plant leaves have reticulate venation, whereas monocot leaves have parallel venation.
• In monocot flowers, petals are three or in multiples of three (trimerous). In dicot flowers, petals are five or in multiples of five (pentamerous).

Value Based Questions

Question 1.
Many medicinal plants are getting extinct every year. A group of students who had gone for educational trip clicked photographs of endangered plants. These photographs were used by the school laboratory to study these plants.
1. Name two endangered plants.
2. Name any one medicinal plant and write its medicinal use.
3. What value of students is reflected in the above act?
Answer:

1. Two endangered plants are: Lotus comiculatus and Acacia planifrons.
2. Aloe – vera Juice of Aloe vera is used in case of indigestion, treating skin infections, etc.
3. Students are caring citizens, showing responsible behavior.

Question 2.
Due to global warming, coral is getting diminished in all the oceans/water bodies. People in the Lakshadweep island protect their corals by not allowing people/tourists to take few pieces away.
1. Name the phylum of coral.
2. What is coral made up of?
3. What values of people in Lakshadweep island are reflected?
Answer:

1. Phylum of coral is coelenterata.
2. Coral is made up of calcium carbonate.
3. People in Lakshadweep island reflect the values of being responsible citizens, respecting environment and nature.

JAC Class 9 Science Important Questions