These NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.5 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.5

Question 1.

A matchbox measure 4 cm × 2.5 cm × 1.5 cm. What will be the volume of packets containing 12 such boxes?

Solution:

We have given the dimension of match box = 4 cm × 2.5 cm × 1.5 cm

Volume of matchbox = l × b × h

= 4 × 2.5 × 1.5

= 15 cm^{3}

Volume of 12 such matchbox = 12 × 15 = 180 cm^{3}

Question 2.

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

Solution:

We have given that

length of cuboidal tank = 6m

Breadth of cubodical tank = 5 m

and height of cubodical tank = 4.5 m

Volume of cubodical tank = l × b × h

= 6 × 5 × 4.5

= 135 m^{3}

Again, we have given that

1 m^{2} = 10001

135m^{3} = 135 × 10001 = 1350001

Therefore, the cuboidal tank holds 135000 liters 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 meters of a liquid?

Solution:

We have given that

length of cuboidal vessel = 10 m

Breadth of cuboidal vessel = 8 m

and volume of cuboidal vessel = 380 m^{3}

We know that,

Volume of cuboidal Vessel = l × b × h

⇒ 380 = 10 × 8 × h

⇒ h = 4.75 m

Therefore, the height of required cuboidal vessels = 4.75m.

Question 4.

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

Solution:

We have given that length of cuboidal pit = 8m

Breadth of cubodical pit = 6 m

and height of cubodical pit = 4 m

Volume of cubodical pit = l × b × h

= 8 × 6 × 3

= 144 m^{3}

Rate od digging = Rs. 30 per m^{3}

Total cost of digging tire cubodical pit = 30 × 144 = Rs. 4,320

Question 5.

The capacity of a cuboidal tank is 50,000 liters of water. Find the breadth of the tank, if its length and depth are respectively 2.5 m and 10 m.

Solution:

We know that

1000 liters = 1 m^{3}

50,000 liters = 50 m^{3}

Now, we have given that length of cuboidal tank = 2.5m

and height of cuboidal tank = 10m

and volume of cuboidal tank = 50 m^{3}

We know that

The volume of cuboidal tank = l × b × h

⇒ 50 = 2.5 × b × 10

⇒ b = 2 m

The breadth of the required tank = 2m.

Question 6.

A village, having a population of 4000, requires 1500 liters of water per head per day. It has a tank measuring 20m × 15m × 6m. For how many days will the water of this tank last.

Solution:

Volume of water tank = l × b × h

= 20 × 15 × 6

= 1800 m^{3}

We know that

1 m^{3} = 1000 litres

1800 m^{3} = 1800 × 1000 = 18,00,000 litres

Therefore, capacity of water tank = 18,00,000 litres.

Now, One person requires 150 litres per day

4000 person requires = 4000 × 150 = 6,00,000 litres per day

No. of days will the water of tank last = \(\frac{18,00,000}{6,00,000}\) = 3 days.

Question 7.

A godown measures 40 m × 25 m × 10 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.

Solution:

Volume of godown = 40 × 25 × 10 = 10,000 m^{3}

Therefore, total number of crates can be stored in the godown = \(\frac{10,000}{0.9375}\) = 10666.66 or 10,666.

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.

Solution:

Volume of larger cube = (side)^{3}

= (12)^{3}

= 1728 cm^{3}

Now, volume of each smaller cube = \(\frac {1}{8}\) × 1728

(side)^{3} = 216

side = \(\sqrt[3]{216}\) = 6 cm

The side of’new cube = 6 cm

Again, Surface area of larger cube = 6 × (side)^{2}

= 6 × (12)^{2}

= 864 cm^{2}

and surface area of smaller cube = 6 × (side)^{2}

= 6 × (6)^{2}

= 216 cm^{2}

Ratio between them is = 984 : 216 = 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 min?

Solution:

We have given that

Rate of flowing = 2 km/h 100 m

= \(\frac{2 \times 1000}{60}\) m/min

= \(\frac{100}{3}\) m/min.

∴ Volume of water will fall into the sea in one min is = 40 × 3 × \(\frac{100}{3}\) = 4,000 m^{3}.