Exam Preparation-Important Questions &
Solutions of Surface Area & Volume:Class-10
Q.1. How many spherical bullets can be made out of a solid cube of lead whose
edge measures 44cm, each bullet being 4cm in diameter?
Solution
Edge of the solid cube = 44 cm
Volume of the cube =L3= 443 = 85184 cm3
Diameter of the bullet = 4 cm
Radius = 2 cm
Volume of each bullet = 4/3πR3
= 33.5 cm3
Number of bullets = .5 = 2542.8=2542(approx)
Q.2. A right circular cone is 3.6cm and the radius of its base is 1.6cm. It is melted
and recast into a right circular cone with radius of its base as 1.2cm. Find its
height.
Solution:
For initial right circular core
h = 3.6 cm, r = 1.6 cm
Volume of the given cone =1/3πr2h=1/3π(1.6)2×3.6 cm2
For Final recast right circular cone
Radius of the second cone = 1.2 cm. Let its height be h
Volume of the second cone = 1/3πr2h=1/3π(1.2)2×h cm2 Now Volume of initial
right circular core= Volume of Final recast right circular cone
1/3pi(1.6)2×3.6=1/3pi(1.2)2×h
h=6.4 cm
Q.3. A conical vessel whose internal radius is 5cm and height 24cm is full of
water. The water is emptied into a cylindrical vessel with internal radius 10cm.
Find the height to which the water rises.
, Solution:
Volume of water in conical vessel= 13πr2h=13π(5)224
Let H be the height of water in cylinderical vessel.
Volume of water in cylinderical vessel= πr2h=π(10)2h
Now Volume of water in conical vessel=Volume of water in cylinderical vessel
13π(5)224=π(10)2h
h=2 cm
Q.4. A well, whose diameter is 7cm, has been dug 22.5m deep and the earth
dugout is used to form an embankment around it. If the height of the embankment
is 1.5m, find the width of the embankment.
Solution:
Let x be the width
Volume of well=πr2h=π×(3.5)2×22.5
Volume of embankment=πR2h−πr2h=π×(3.5+x)2×1.5−π×(3.5)2×1.5
Volume of well = Volume of embankment
So π×(3.5)2×22.5=π×(3.5+x)2×1.5−π×(3.5)2×1.5
On solving we get x = 10.5m
Q.5. The perimeters of ends of a frustum are 48cm & 36cm, if height of frustum be
11cm, find its volume.
Solution:
Volume of a frustum of a cone =13πh(r12+r22+r1r2)
Here h=11 cm , r2 =48/2π=24/π , r1= frac362π=18/π
Substituting these values
V=1554 cm3
Q.6. Water is being pumped out through a circular pipe whose internal diameter is
7cm. if the flow of water is 72cm per second, how many liters of water are being
pumped out in one hour?
Solution:
Radius of circular pipe (r) = 7/2 =3.5cm
Now Length of water per sec = 72cm
Therefore, volume of water per second = volume of cylinder per second
πr2h=227×(3.5)2×72=2772cm3
Solutions of Surface Area & Volume:Class-10
Q.1. How many spherical bullets can be made out of a solid cube of lead whose
edge measures 44cm, each bullet being 4cm in diameter?
Solution
Edge of the solid cube = 44 cm
Volume of the cube =L3= 443 = 85184 cm3
Diameter of the bullet = 4 cm
Radius = 2 cm
Volume of each bullet = 4/3πR3
= 33.5 cm3
Number of bullets = .5 = 2542.8=2542(approx)
Q.2. A right circular cone is 3.6cm and the radius of its base is 1.6cm. It is melted
and recast into a right circular cone with radius of its base as 1.2cm. Find its
height.
Solution:
For initial right circular core
h = 3.6 cm, r = 1.6 cm
Volume of the given cone =1/3πr2h=1/3π(1.6)2×3.6 cm2
For Final recast right circular cone
Radius of the second cone = 1.2 cm. Let its height be h
Volume of the second cone = 1/3πr2h=1/3π(1.2)2×h cm2 Now Volume of initial
right circular core= Volume of Final recast right circular cone
1/3pi(1.6)2×3.6=1/3pi(1.2)2×h
h=6.4 cm
Q.3. A conical vessel whose internal radius is 5cm and height 24cm is full of
water. The water is emptied into a cylindrical vessel with internal radius 10cm.
Find the height to which the water rises.
, Solution:
Volume of water in conical vessel= 13πr2h=13π(5)224
Let H be the height of water in cylinderical vessel.
Volume of water in cylinderical vessel= πr2h=π(10)2h
Now Volume of water in conical vessel=Volume of water in cylinderical vessel
13π(5)224=π(10)2h
h=2 cm
Q.4. A well, whose diameter is 7cm, has been dug 22.5m deep and the earth
dugout is used to form an embankment around it. If the height of the embankment
is 1.5m, find the width of the embankment.
Solution:
Let x be the width
Volume of well=πr2h=π×(3.5)2×22.5
Volume of embankment=πR2h−πr2h=π×(3.5+x)2×1.5−π×(3.5)2×1.5
Volume of well = Volume of embankment
So π×(3.5)2×22.5=π×(3.5+x)2×1.5−π×(3.5)2×1.5
On solving we get x = 10.5m
Q.5. The perimeters of ends of a frustum are 48cm & 36cm, if height of frustum be
11cm, find its volume.
Solution:
Volume of a frustum of a cone =13πh(r12+r22+r1r2)
Here h=11 cm , r2 =48/2π=24/π , r1= frac362π=18/π
Substituting these values
V=1554 cm3
Q.6. Water is being pumped out through a circular pipe whose internal diameter is
7cm. if the flow of water is 72cm per second, how many liters of water are being
pumped out in one hour?
Solution:
Radius of circular pipe (r) = 7/2 =3.5cm
Now Length of water per sec = 72cm
Therefore, volume of water per second = volume of cylinder per second
πr2h=227×(3.5)2×72=2772cm3
