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31. If the diameter of the base of a closed right circular cylinder is equal to its height h, then its whole surface area is :
  A.  
2 / 3
πh2
  B.  
3 / 2
πh2
  C.  
2 / 5
πh2
  D.  
3 / 5
πh2
     
   
View Answer

Shortcut:
If the radius of the base of a cylinder is 'r' units and its height (or length) is 'h' units, then the total surface area of the cylinder is 2πrh + 2πr2 sq units.
or Total Surface Area = 2πr(h + r) sq units or Circumference x (height + radius)
Here, r =
1 / 2
x h
∴ Whole S.A = 2πrh + 2πr2
= 2π
(
1 / 2
x h
)
h + 2π
(
1 / 2
h
)
2
=
3 / 2
πh2

32. A spherical shell of metal has an outer radius of 9 cm and inner radius of 8cm. If the metal costs RS 1.80 per cu cm. find the cost of shell.
  A.  Rs 1636.80
  B.  Rs 1640.40
  C.  Rs 1638.20
  D.  Rs 1635.50
     
   
View Answer

Shortcut:
If the radius of a sphere is 'r' units, then volume of the sphere is
4 / 3
πr3 cubic units. If diameter is given, then volume of sphere becomes
1 / 6
πD3 cubic units
where, D = Diameter
Here, Vol. of metal =
4 / 3
π(9)3 −
4 / 3
π(8)3
=
4 / 3
π x (729 − 512)
=
4 / 3
π x 217
=
2728 / 3
cubic cm.
Cost of metal =
2728 / 3
x 1.80 = Rs 1636.80

33. A metal sphere of diameter 42 cm is dropped into a cylinderical vessel, which is partly filled with water.The diameter of the vessel is 1.68 m. If the sphere is completely submereged, find by how much the surface of will rise.
  A.  1.68 cm
  B.  1.80 cm
  C.  1.60 cm
  D.  1.75 cm
     
   
View Answer

Shortcut:
If the radius of a sphere is 'r' units, then volume of the sphere is
4 / 3
πr3 cubic units. If diameter is given, then volume of sphere becomes
1 / 6
πD3 cubic units
where, D = Diameter
Here, r = 21
Vol. of sphere =
4 / 3
x π x (21)3
=
4 / 3
x
22 / 7
x 21 x 21 x 21 = 38808 cubic cm
Let the water rise by 'h' cm.
Then,
22 / 7
x 84 x 84 x h = 38808
or, h =
38808 x 7 / 22 x 84 x 84
=
7 / 4
= 1.75
Hence, the surface will rise by 1.75 cm.

34. Find the weight of an iron shell, the external and internal diameters of which are 13 cm and 10 cm repectively,if 1 cu cm of iron weights 8 gms.
  A.  5.5 kg
  B.  4.5 kg
  C.  5.016 kg
  D.  5.020 kg
     
   
View Answer

Shortcut:
If the radius of a sphere is 'r' units, then volume of the sphere is
4 / 3
πr3 cubic units. If diameter is given, then volume of sphere becomes
1 / 6
πD3 cubic units
where, D = Diameter
Here, Vol. of iron =
4 / 3
π
{(
13 / 2
)
3 − (5)3
}

=
4 / 3
x
22 / 7
x
1197 / 8
= 627 cu cm
∴ Weight of iron =
627 x 8 / 1000
= 5.016
Hence, the weight of an iron shell is 5.016 kg.

35. A spherical ball of radius 3 cm is melted an recast into three spherical balls.The radii of two of these balls are 1.5 cm and 2 cm. Find the radius of the third ball.
  A.  3.5 cm
  B.  2.5 cm
  C.  3 cm
  D.  2 cm
     
   
View Answer

Shortcut:
If the radius of a sphere is 'r' units, then volume of the sphere is
4 / 3
πr3 cubic units. If diameter is given, then volume of sphere becomes
1 / 6
πD3 cubic units
where, D = Diameter
Here, Vol. of 3rd ball =
4 / 3
π(3)3 −  
{
4 / 3
π
(
3 / 2
)
3 +
4 / 3
π(2)3
}
=
125 / 6
π
∴
4 / 3
x π x (r)3 =
125 / 6
π

⇒ r =
5 / 2
= 2.5
Hence, the radius of 3rd ball is 2.5 cm.

36. How many bullets can be made out of a cube of lead whose egde measures 22 cm,each bullet being 2 cm in diameter?
  A.  2531
  B.  2545
  C.  2541
  D.  2527
     
   
View Answer

Shortcut:
If the radius of a sphere is 'r' units, then volume of the sphere is
4 / 3
πr3 cubic units. If diameter is given, then volume of sphere becomes
1 / 6
πD3 cubic units
where, D = Diameter
Here, Vol. of cube = 22 x 22 x 22 cu cm
Vol. of 1 bullet =
4 / 3
x
22 / 7
x (1)3 cu cm
∴ No. of bullets =
22 x 22 x 22 x 3 x 7 / 4 x 22
= 2541
Hence, 2541 bullets can be made.

37. A cylinderical vessel 60 cm in diameter is partially filled with water.A sphere,30 cm in diameter is gently dropped into the vessel.To what further height will water in the cylinder rise?
  A.   45cm
  B.  42cm
  C.  38cm
  D.  40cm
     
   
View Answer

Shortcut:
If the radius of a sphere is 'r' units, then volume of the sphere is
4 / 3
πr3 cubic units. If diameter is given, then volume of sphere becomes
1 / 6
πD3 cubic units
where, D = Diameter
Here, Let H and h be the heights of water level before and after dropping the sphere into it.
Then, [π x (30)2 x H] − [π x (30)2 x h] =
4 / 3
x π x (30)3
π x 900 x (H − h) =
4 / 3
π x 27000
or, (H − h) = 40
Hence, the water will rise by 40 cm in the cylinder.

38. Find the surface area of a sphere whose volume is 310464 cu cm.
  A.  22180 sq cm
  B.  22176 sq cm
  C.  22175 sq cm
  D.  22182 sq cm
     
   
View Answer

Shortcut:
If the radius of a sphere is 'r' units, then the surface area of the sphere is 4πr2 sq units. If in place of radius, diameter of the sphere is given, then the formula becomes as follows,
S.A of sphere = 4πr2 = 4&pi
(
D / 2
)
2 = πD2 sq units.
Here, r = ?, Vol. = 310464
4 / 3
x π x (r)3 = 310464
(r)3 =
310464 x 3 x 7 / 4 x 22
= 74088
∴ r = 42
∴ S.A = 4πr2 = 4 x
22 / 7
x 42 x 42 = 22176
Hence, the surface area of a sphere is 22176 sq cm.

39. If a solid sphere of radius 10 cm is moulded into 8 spherical solid balls of equal radius, then surface area of each ball ( in cm2 ) is :
  A.  100 π
  B.  105 π
  C.  110 π
  D.  95 π
     
   
View Answer

Shortcut:
If the radius of a sphere is 'r' units, then the surface area of the sphere is 4πr2 sq units. If in place of radius, diameter of the sphere is given, then the formula becomes as follows, S.A of sphere = 4πr2 = 4&pi
(
D / 2
)
2 = πD2 sq units.
Here, Let 'r' be the radius of each moulded sphere. Then, 8 x
4 / 3
x π x (r)3 =
4 / 3
x π x (10)3
⇒ (2r)3 = (10)3
or, 2r = 10
or, r = 5 cm
So, the surface area of each ball = 4π x (25) = 100π sq cm

40. If the volume of surface area of a sphere are numerically the same, then its radius is :
  A.  5 units
  B.  2 units
  C.  3 units
  D.  6 units
     
   
View Answer

Shortcut:
If the radius of a sphere is 'r' units, then the surface area of the sphere is 4πr2 sq units. If in place of radius, diameter of the sphere is given, then the formula becomes as follows, S.A of sphere = 4πr2 = 4&pi
(
D / 2
)
2 = πD2 sq units.
Here, Let 'r' be the radius of each sphere.
Then,
4 / 3
x π x (r)3 =
4 / 3
x π x (r)2
⇒ r = 3
Hence, the radius is 3 units.

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