Jharkhand Board JAC Class 9 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.5 Textbook Exercise Questions and Answers.

## JAC Board Class 9th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.5

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Question 1.

A matchbox measures 4 cm × 2.5 cm × 1.5 cm. What will be the volume of a packet containing 12 such boxes?

Ans. Dimension of matchbox = 4 cm × 2.5 cm × 1.5 cm

l = 4 cm, b = 2.5 cm and h = 1.5 cm

Volume of one matchbox

= (l × b × h) = (4 × 2.5 × 1.5) cm^{3} = 15 cm^{3}

Volume of a packet containing 12 such boxes = (12 × 15) cm^{3}= 180 cm^{3}

Question 2.

A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold? (1 m^{3} = 10001)

Answer:

Dimensions of water tank = 6 m × 5 m × 4.5 m

l = 6 m , b = 5 m and h = 4.5 m

Therefore, Volume of the tank

= lbh m^{3} = (6 × 5 × 4.5) m^{3} =135 m^{3}

Therefore , the tank can hold = 135 × 1000 litres

[Since, 1 m^{3} = 1000 litres]

= 135000 litres of water.

Question 3.

A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic metres of a liquid?

Answer:

Length = 10 m

Breadth = 8 m and

Volume = 380 m^{3}

Volume of cuboid = Length × Breadth × Height

⇒ Height = \(\frac{Volume of cuboid}{(Length × Breadth)}\)

= \(\frac{380}{10 × 8}\) m = 4.75 m

Question 4.

Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at the rate of ₹ 30 per m^{3}.

Answer:

l = 8 m, b = 6 m and h = 3 m

Volume of the pit = lbh m^{3} = (8 × 6 × 3) m^{3} = 144 m^{3}

Rate of digging = ₹ 30 per m^{3}

Total cost of digging the pit = ₹ (144 × 30) = ₹ 4320

Question 5.

The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its length and depth are respec-tively 2.5 m and 10 m.

Answer:

Length = 2.5 m,

depth = 10 m and

volume = 50000 litres

1 m^{3} = 1000 litres

, ,

∴ 50000 litres = \(\frac{50000}{1000}\) m^{3} = 50 m^{3}

Breadth = \(\frac{Volume of cuboid }{(Length × Depth)}\)

= \(\frac{50}{(2.5 × 10)}\) = 2 m

Question 6.

A village, having a population of 4000, requires 150 litres of water per head per day. It has a tank measuring 20 m × 15 m × 6 m. For how many days will the water of this tank last?

Answer:

Dimensions of tank = 20 m × 15m × 6m

l = 20 m , b = 15 m and h = 6 m

Capacity of the tank = l × b × h = (20 × 15 × 6) m^{3}

= 1800 m^{3}

Water requirement per person per day =150 litres

Water required for 4000 person per day = (4000 × 150) litres

= \(\frac{(4000 × 150)}{1000}\) = 600 m^{3}

Number of days the water will last = Capacity of tank + Total water required per day

= \(\frac{1800}{600}\) = 3

The water will last for 3 days.

Question 7.

A godown measures 40 m × 25 m × 15 m. Find the maximum number of wooden crates each measuring 1.5 m × 1.25 m × 0.5 m that can be stored in the godown.

Answer:

Dimensions of godown = 40 m × 25 m × 15 m

Volume of the godown = (40 × 25 × 15) m^{3} = 15000 m^{3}

Dimensions of crates = 1.5 m × 1.25 m × 0.5 m

Volume of 1 crate = (1.5 × 1.25 × 0.5) m^{3} = 0.9375 m^{3}

Number of crates that can be stored = \(\frac{Volume of the godown}{Volume of 1 crate}\)

= \(\frac{15000}{0.9375}\) = 16000

Question 8.

A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the new cube? Also, find the ratio between their surface areas.

Answer:

Edge of the cube = 12 cm.

Volume of the cube

= (edge)^{3} cm^{3} = (12 × 12 × 12) cm^{3} = 1728 cm^{3}

Number of smaller cubes = 8

Volume of the 1 smaller cubes = \(\frac{1728}{8}\) cm^{3} = 216 cm^{3}

Let side of the smaller cube = a cm

∴ a^{3} =216

⇒ a = 6 cm

Surface area of the cube = 6 (side)^{2}

Ratio of their surface area = \(\frac{(6×12×12)}{(6×6×6)}\)

= \(\frac{4}{1}\) = 4 : 1

Question 9.

A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute?

Answer:

Depth of river (h) = 3 m

Width of river (b) = 40 m

Rate of flow of water (l) = 2 km per hour

= \(\frac{2000}{60}\) m per minute

= \(\frac{100}{3}\) m per minute

Volume of water flowing into the sea in a minute = lbh m^{3} = (\(\frac{100}{3}\) × 40 × 3) m ^{3}

= 4000 m^{3}