In this page we have *NCERT solutions for Class 10 Maths: Chapter 13 Surface Area and Volume* for
Exercise 13.3 on pages 251 and 252. Hope you like them and do not forget to like , social_share
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This exercise is based on the volume of an object formed by combining any two of the basic solids, namely, cuboid, cone, cylinder, sphere and hemisphere

**Formula to used**

The total Volume of the any new solid is the sum of the Volume of each of the individual parts.

Volume of new solid = Volume of One part + Volume of Second part + Volume of Third part

**Volume of Common Shapes**

Volume of cone

$=\frac {1}{3}\pi r^2h$

Volume of Sphere

$=\frac {4}{3} \pi r^3$

Volume of Cylinder

$=\pi r^2 h$

Unless stated otherwise, take $\pi = \frac {22}{7}$.

The total Volume of the any new solid is the sum of the Volume of each of the individual parts.

Volume of new solid = Volume of One part + Volume of Second part + Volume of Third part

Volume of cone

$=\frac {1}{3}\pi r^2h$

Volume of Sphere

$=\frac {4}{3} \pi r^3$

Volume of Cylinder

$=\pi r^2 h$

Unless stated otherwise, take $\pi = \frac {22}{7}$.

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.

Let h be the height then

Then , volume of cylinder =Volumne of Sphere

$\pi (6)^2 h =\frac {4}{3} \pi (4.2)^3$

then h =2.74 cm

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.

Let R be the radius

Now

Volume of resulting sphere = Volume of 6 cm sphere + Volume of 8 cm sphere + Volume of 10 cm sphere

$\frac {4}{3} \pi R^3 = \frac {4}{3}\pi 6^3 + \frac {4}{3}\pi 8^3 + \frac {4}{3}\pi 10^3$

or

$R^3= 6^3 + 8^3 + 10^3$

$R^3=1728$

R=12 cm

A 20 m deep well with diameter 7 m 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.

Let h be the height of platform

Now

Volume Of well = Volume of the platform formed

$\pi (3.5)^2 20 = 22 \times 14 \times h$

h=2.5 m

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.

Let h be the height of the embankment.

It is aparant from figue the embankment is a cylinderical right with inner radii= 1.5 m and outer radii = 5.5 m

Now

Volume Of well = Volume of the embankment formed

$ \pi \times (1.5)^2 \times 14 = \pi \times h \times (5.5^2 - 1.5^2)$

h=1.125 m

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 height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.

Radius of Cone(r) = 6/2 = 3 cm

Height of cone(h)= 12 cm

Radius of hemispherical shape on the top(r) = 3cm

Volume of One cone icecream = $ \frac {1}{3} \pi r^2 h + \frac {2}{3} \pi r^3 = \frac {1}{3} \pi ( 108 + 54)$

Let n be the number of icecream cones filled

Then

Volume of Cylinder = n × Volume of One cone icecream

$ \pi (6)^2 15 = \frac {n}{3} \pi ( 108 + 54)$

or n=10

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 x 10 cm x 3.5 cm?

Let n be the silver coins needed

n × Volume of one silver coin= volume of cuboid

$n \times \pi \times (.875)^2 \times .2 = 5.5 \times 10 \times 3.5$

n=400

A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This 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.

Let r be the radius of the conical heap

Now Volume of cylinderical bucket = Volume of conical heap

$ \pi \times (18)^2 \times 32 = \frac {1}{3} r^2 \times 24$

r=36 cm

Now slant height = $\sqrt {r^2 + h^2} = \sqrt {36^2 + 24^2} = 12 \sqrt {13}$ cm

Water in a canal, 6 m wide 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?

Cross-section of the Canal = $6 \times 1.5 = 9 m^2 $

Now Water speed is 10 km/hr = 1000/6 m/min

So volume of Water flowed in 1 min = $9 \times \frac {1000}{6}$

Therefore Volume of water in 30 min = $9 \times \frac {1000}{6} \times 30=45000 m^3$

Let A be the Area irrigated then

Then

$ A \times .08 = 45000 $

A=562500 m

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

Cross-section of the pipe = $\pi \times (.1)^2 = .01 \pi m^2 $

Speed of water = 3km/h = 3000/60 m/min = 50 m/min

Let t be the time time taken in min, then Volume of water out of pipe

$= .01 \pi \times 50 \times t = .5 t \pi m^2$

Now Volume of cylindrical tank= $ \pi (5)^2 2= 50 \pi m^2 $

Now Volume of water out of pipe=Volume of cylindrical tank

$.5 t \pi=50 \pi$

t= 100 min

- Class 10 Maths NCERT solutions for Surface Area and Volume Exercise 13.3 has been prepared by Expert with utmost care. If you find any mistake.Please do provide feedback on mail.You can download this as pdf
- This chapter 13 has total 5 Exercise 13.1 ,13.2,13.3 ,13.4 and 13.5. This is the Third exercise in the chapter.You can explore previous exercise of this chapter by clicking the link below

**Notes**-
**Assignments** -
**NCERT Solutions**- NCERT Solution Surface Area and Volume Class10 Exercise 13.1
- NCERT Solution Surface Area and Volume Class10 Exercise 13.2
- NCERT Solution Surface Area and Volume Class10 Exercise 13.3

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