# Maths Exercise 13.3 Class 10 NCERT Solutions

Chapter 13 of Maths Class 10 is Surface Area and Volume . This chapter has 4 Exercise . Each exercise has questions related to surface are and volumes. Here we have provided the Solutions of Exercise 13.3 Class 10 Maths

Contents

## Class 10 Maths Chapter 13

Chapter 13 Surface Areas and Volume Exercise Solutions on this page are based on the latest curriculum and criteria as mentioned by CBSE. These are useful for students in preparation for their exams.

Class 10 Maths Exercise 13.1
Class 10 Maths Exercise 13.2
Class 10 Maths Exercise 13.3
Class 10 Maths Exercise 13.4

## Maths Exercise 13.3 Class 10 Solutions

Exercise 13.3 Class 10 Maths Question 1
Q1. 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.

Solution :
According to Question :
Radius of Metallic Sphere (r)= 4.2 cm
Radius of Cylinder (R)= 6 cm
Height of Cylinder (H) = ?
It is given that the metallic sphere is melted and recast into cylinder. So the Volume of two shapes is equal.
Therefore,
Volume of Metallic Sphere = Volume Cylinder
= $$\frac{4}{3}\ \pi\ r^3\ =\ \pi R^2H\$$
= $$\frac{4}{3}\ \pi\ \left(4.2\right)^{3\ }\ =\ \pi\left(6\right)^2H\$$
= H =2.74 cm

Exercise 13.3 Class 10 Maths Question 2
Q2. 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.

Solution :
According to Question ;
Radius of Metallic Sphere I (r1) = 6 cm
Radius of Metallic Sphere II ( r2) = 8 cm
Radius of Metallic Sphere III (r3) = 10 cm

Radius if Single Resulted Spheres ( R )= ?

It is given that Three metallic sphere are melted to form single sphere , So the volume of three metallic sphere is equal to the volume of single newly formed sphere.
Therefore,
Volume of Metallic Sphere I + Volume of Metallic Sphere II + Volume of Metallic Sphere III = Volume of newly formed Sphere
4/3 pi r3 = 4/3 pi 63 + 4/3 pi 83 + 4/3 pi 103
r3 = 216 +512 +1000
r = 12 cm

Exercise 13.3 Class 10 Maths Question 3
Q3.
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.

Solution :
According to Question ;
Diameter of the well = 7 m
Radius of well = 7/2 m = 3.5 m
Depth of earth Dug (h) = 20 m

The sand is then transferred in rectangular platform whose
length = 22 m
breadth = 14 m
Height (H)= ?

Volume of Earth Dug = Volume of the platform
pi r2 h = l * b * h
3.14 * (3.5)2 20 = 22 * 14 * H
H = 2.5 m

Exercise 13.3 Class 10 Maths Question 4
Q4.
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.

Solution :
According to questions,
The diameter of Well = 3m
Then radius is found to be = r = D/2 = 3/2 = 1.5m
Width of Embankment = 4 m
The radius of well with embankment R = r+4 = 1.5 +4 = 5.5
We have to find the height of embankment
Let us take height of embankment as h.
According to question ,
The volume of embankment = Volume of earth dug
pi[ R2 – r2 ] h = pi r2 h
[ (5.5)2 – (1.5)2 ] h = (1.5)2 *14
On Solving the value of h is found to be 1.125.
The height of embankment is 1.125 m

Exercise 13.3 Class 10 Maths Question 5
Q5.
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

Solution :
Radius of Cylinder full of ice cream = R = 6 cm
Height of Cylinder full of ice cream = H = 15 cm

Volume of Cylinder = pi R2H = pi (6)2 * (15)2 = 540 pi cm3
We are also provided with the dimensions of cone
Radius of cone = r=3 cm
height of ice cream cone = 12 cm
radius of hemispherical shape at top =r = 3 cm
Volume of cone = 1/3 pi r2 h
= 1/3 pi (3)2 * (12)2 = 36 pi cm3

Volume of Hemispherical shape over cone = 2/3 pi r3
2/3 pi (3)3 = 18 pi cm3

Total volume of ice cream in cone = Volume of cone +volume of hemispherical shape = 36 pi + 18 pi = 54 pi cm3

Number of ice cream cone required = (Volume of Cylinder / Volume of Ice cream in cone with hemispherical top) = (540 pi) /( 54 pi) = 10

Total 10 cones can be filled with the ice cream present in the given cylinder.

Exercise 13.3 Class 10 Maths Question 6
Q6.
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?

Solution :
Length of Cuboid (l) = 5.5 cm
breadth of Cuboid (b)= 10 cm
Height of Cuboid (h) = 3.5 cm

Volume of Cuboid = l * b * h
= 5.5 *10 *3.5 = 192.5 cm3

Now for coin ,
Radius of coin (r) = 1.75 /2 = 0.875 cm
Height of Coin (h’) = 2 mm = 0.2 cm

Volume of Silver Coin = pi r2 h’
= pi (0.875)2 (0.2) = 0.153 pi cm3

Total number of coins required to make the cuboid = (Volume of Cuboid ) /(Volume of Coin) = 192.5 / (0.153 pi) = 400

Total 400 silver coins will be required to make given cuboid

Exercise 13.3 Class 10 Maths Question 7
Q7.
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 .

Solution :
According to question
Radius of Cylindrical Basket = R = 18 cm
Height of Cylindrical Basket = H = 32 cm
Volume of Cylindrical Basket = pi R2 H
= pi (18)2 * (32) =10368 pi cm3

Height of Conical Heap of Sand = h = 24
Let the Radius of Conical Heap of Sand = r
Volume of conical heap of sand = 1/3 pi r2 h
= 1/3 pi (r)2 *( 24) = 8 pi r2 cm3

According to question,
volume of cylindrical basket = volume of conical heap of sand
10368 pi = 8 pi r2
On solving we get r =36
Let the slant height of the conical heap = l
To find l , we apply Pythagoras theorem formula
Slant Height (l) = Sqrt( r2 +h2 )
l = sqrt( 2 +2 )
slant height (l) = 12 sqrt(13) cm

Exercise 13.3 Class 10 Maths Question 8
Q8.
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?

Solution :
According to question,
Dimensions of Canal;
Height of canal = 1.5m
Width of canal =6 m
Speed of Water in Canal = 10km/h
So,
Water flows through canal in 1 hr = 10km
Water flows through canal in 30 minutes ( 1/2hrs) = 5 km = 5000 m
Now,
Volume of water flows in 30minutes = l * b * h = 5000 * 5 * 1.5 = 45000m3

Standing water required in field (h) = 8 cm = 0.08 m
Volume of water flowing through canal in 30 minutes = area of irrigated field * 0.08
= 45000 m3 =Area * 0.08 m
Area of irrigated field = 45000/0.08 = 562500 m2
= 56.25 hectare

Exercise 13.3 Class 10 Maths Question 9
Q9.
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?

Solution :
According to questions,
Dimensions of Cylindrical tank;
Radius of tank = r= 5 m
height of tank = h = 2 m
Volume of tank = pi r2 h
= pi *5*5*2 = 50pi

Speed of water = 3km/hr = 50m/minute
radius of pipe (r’) = 10 cm = 0.1 m
Volume of water flowed through the pipe in one minute = pi (r’)2 H
pi * 0.10 *0.10 *50 = 100
Time taken to get the tank completely filled = 100 minutes = 1 hour 40 minutes.