# Surface Area and Volume : Exercise - 13.3 (Mathematics NCERT Class 10th)

**Take , unless stated otherwise. **

**Q.1 A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder. **

**Sol.**

Volume of the sphere

If h is the height of a cylinder of radius 6 cm. Then its volume,

Since, the volume of metal in the form of sphere and cylinder remains the same , we have

h = 2.744

**Q.2 Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere. **

**Sol. **

Sum of the volumes of 3 gives spheres.

Let R be the radius of the new spheres whose volume is the sum of the volumes of 3 given spheres.

Therefore

R = 12

Hence, the radius of the resulting sphere is 12 cm.

Therefore 1728 = 2^{3} × 6^{3} = (2 × 6)^{3} = 12)^{3}

**Q.3 A 20 cm deep well with diameter 7 cm is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform. **

**Sol.**

Let h m be the required height of the platform.

The shape of the platform will be like the shape of a cuboid 22 m × 14 m × h with a hole in the shape of cylinder of radius 3.5 m and depth h m.

The volume of the platform will be equal to the volume of the earth dug out from the well.

Now, the volume of the earth = Volume of the cylindrical well

Also, the volume of the platform = 22 × 14 × h

But volume of the platform = Volume of the well

i.e., 22 × 14 × h = 770

Therefore Height of the platform = 2.5 m

**Q.4 A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment. **

**Sol. **

Let h be the required height of the embankment.

The shape of the embankment will be like the shape of a cylinder of internal radius 1.5 m and external radius (4 + 1.5)m = 5.5 m (see figure).

The volume of the embankment will be equal to the volume of earth dug out from the well.

Now, the volume of the earth = Volume of the cylindrical well

Also the volume of the embankment

Hence, we have

Hence, the required height of the embankment = 1.125 m

**Q.5 A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled into cones of height12 cm and diameter 6 cm, having hemispherical shape on the top. Find the number of such cones which can be filled with ice cream. **

**Sol. **

Volume of the cylinder

Volume of a cone having hemispherical shape on the top

Let the number of cone that can be filled with ice cream be h.

Then

Q.6 How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?

Sol. The shape of the coin will be like the shape of a cylinder of radius

= 0.875 cm and of height 2mm

Its volume

Volume of the cuboid

Number of coins required to form the cuboid

Hence, 400 coins must be melted to form a cuboid.

**Q.6 How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?
Sol.
**

The shape of the coin will be like the shape of a cylinder of radius

= 0.875 cm and of height 2mm

Its volume

Volume of the cuboid

Number of coins required to form the cuboid

Hence, 400 coins must be melted to form a cuboid.

**Q.7 A cylindrical bucket, 32 m high and with radius of base 18 cm, is filled with sand. The bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap. **

**Sol. **

Volume of the sand = Volume of the cylindrical bucket

Volume of the conical heap

, where r = ? , h = 24 cm

The volume of the conical heap will be equal to that of sand.

Therefore

r = 18 × 2 = 36

Here, slant height

Hence, the radius of the conical heap is 36 cm and its slant height is

**Q.8 Water in a canal 6 m wid and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed? **

**Sol. **

Width of the canal = 6 m

Depth of the canal = 1.5 m

Length of water column per hour = 10 km

Length of water column in 30 minutes or hour

Volume of water flown in 30 minutes

= 1.5 × 6 × 5000

Since

i.e., 0.08 m standing water is desired

Therefore Area irrigated in 30 minutes

**Q.9 A farmer connects a pipe of internal diameter 20 cm from a cannal into a cylindrical tank in his field, which is 10 m in diameter and 2m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled? **

**Sol. **

Diameter of the pipe = 20 cm

Radius of the pipe = 10 cm

Length of water column per hour

= 3 km = 3 × 1000 × 100 cm

Volume of water flown in one hour

Tank to be filled = Volume of cylinder (with r = 5m = 500 cm and h = 2m = 200 cm)

Time required to fill the tank

= 1 hour 40 minutes

= 60 + 40 minutes = 100 minutes.