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Mensuration - Volume and Surface Areas

Multiple Choice Questions

31.

If the height of a cylinder is 4 times its circumference, the volume of the cylinder in terms of its circumference c, is

$fraction numerator 2 straight c cubed over denominator straight pi end fraction$
$4 πc cubed$
$straight c cubed over straight pi$
$straight c cubed over straight pi$

455 Views

32.

If the surface area of sphere is 346.5 cm², then its radius [ taking π = 22/7 ] is

7 cm
3.25 cm
5.25 cm
9 cm

262 Views

33.

If each side of a cube is increased by 10% the volume of the cube will increase by

30%
10%
33.1%
33.1%

190 Views

34.

A conical cup is filled with ice cream. The ice-cream forms a hemispherical shape on its open top. The height of the hemispherical part is 7 cm. The radius of the hemispherical part equals the height of the cone. Then the volume of the ice-cream is $open square brackets straight pi space equals space 22 over 7 close square brackets$

1078 cubic cm
1708 cubic cm
7108 cubic cm
7108 cubic cm

323 Views

35.

A large solid sphere is melted and moulded to form identical right circular cones with base radius and height same as the radius of the sphere. One of these cones is melted and moulded to form a smaller solid sphere. Then the ratio of the surface area of the smaller to the surface area of the larger sphere is

$1 space colon space 3 to the power of 4 over 3 end exponent$
$1 space colon space 2 to the power of 3 over 2 end exponent$
$1 space colon space 3 to the power of 2 over 3 end exponent$
$1 space colon space 3 to the power of 2 over 3 end exponent$

463 Views

36.
A ball of lead 4 cm in diameter is covered with gold. If the volume of the gold and lead are equal, then the thickness of gold [given $cube root of 2 space equals space 1.259$ ] is approximately.

5.038 m

5.190 cm

1.038 cm

1.038 cm

1.038 cm

Diameter of lead = 4 cm
Radius of lead = 2 cm
Volume of lead = $4 over 3 πr cubed space space equals space 4 over 3 straight pi cross times 2 cubed$
Let the thickness of gold be x cm, then
Volume of gold = $4 over 3 straight pi open parentheses left parenthesis 2 plus straight x right parenthesis cubed minus 2 cubed close parentheses space cu. space cm$
   $therefore space space 4 over 3 straight pi open parentheses open parentheses 2 plus straight x close parentheses cubed minus 2 cubed close parentheses space equals space 4 over 3 straight pi space cross times space 2 cubed$
$rightwards double arrow space space left parenthesis 2 plus straight x right parenthesis cubed space minus space 2 cubed space equals space 2 cubed rightwards double arrow space space left parenthesis 2 plus straight x right parenthesis cubed space space equals space 8 space plus space 8 space equals space 16 rightwards double arrow space space left parenthesis 2 plus straight x right parenthesis cubed space equals space 2 cubed. space 2$
   $rightwards double arrow space space 2 space plus space straight x space equals space 2 space cross times space cube root of 2 rightwards double arrow space space 2 space space plus space straight x space equals space 2 space cross times space 1.259 space equals space 2.518 therefore space space space space straight x space equals space 2.518 space minus space 2 space equals space 0.518 space cm$

551 Views

37.

The base of a right prism is an equilateral triangle. If the lateral surface area and volume is 120 cm², 40√3 cm³ respectively then the side of base of the prism is

4 cm
5 cm
7 cm
40 cm

849 Views

38.

The base of a right prism is a right angle triangle with two sides 5cm and 12 cm. The height of the prism is 10 cm. The height of the prism is 10 cm. The total surface area of the prism is

360 sq cm
300 sq cm
330 sq cm
325 sq cm

510 Views

39.

The height of the right pyramid whose area of the base is 30m² and volume is 500 m³, is

50 m
60 m
40 m
20 m

238 Views

40.

The height of a cone is 30 cm. A small cone is cut off at the top by a plane parallel to the base. If its volume be 1/27 th of the volume of the given cone, at what height above the base is the section made?

19 cm
20 cm
12 cm
15 cm

241 Views

Previous Year Papers

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Multiple Choice Questions

36. A ball of lead 4 cm in diameter is covered with gold. If the volume of the gold and lead are equal, then the thickness of gold [given ] is approximately. 5.038 m 5.190 cm 1.038 cm 1.038 cm

36.
A ball of lead 4 cm in diameter is covered with gold. If the volume of the gold and lead are equal, then the thickness of gold [given $cube root of 2 space equals space 1.259$ ] is approximately.

5.038 m

5.190 cm

1.038 cm

1.038 cm