Practice MCQ Questions and Answer on Volume and Surface Area

91.

A spherical ball of lead, 3 cm in diameter is melted and recast into three spherical ball. The diameter of two of these are 1.5 cm and 2 cm respectively. The diameter of the third ball is :

• (A) 2.5 cm
• (B) 2.66 cm
• (C) 3 cm
• (D) 3.5 cm

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 Let the radius of the third ball be R cm Then, $$\frac{4}{3}\pi \times {\left( {\frac{3}{4}} \right)^3} + \frac{4}{3}\pi \times {\left( 1 \right)^3}$$   $$ + \frac{4}{3}\pi \times {R^3}$$   $$ = \frac{4}{3}\pi \times {\left( {\frac{3}{2}} \right)^3}$$ $$\eqalign{ & \Rightarrow \frac{{27}}{{64}} + 1 + {R^3} = \frac{{27}}{8} \cr & \Rightarrow {R^3} = \frac{{125}}{{64}} = \frac{{{{\left( 5 \right)}^3}}}{{{{\left( 4 \right)}^3}}} \cr & \Rightarrow R = \frac{5}{4} \cr} $$ ∴ Diameter of the third ball : $$ = 2R = \frac{5}{2}cm = 2.5\,cm$$

92.

If the radius of the base and height of a cylinder and cone are each equal to r, and the radius of a hemisphere is also equal to r, then the volumes of the cone, cylinder and hemisphere are in the ratio ?

• (A) 1 : 2 : 3
• (B) 1 : 3 : 2
• (C) 2 : 1 : 3
• (D) 3 : 2 : 1

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 Required ratio : = Volume of cone : Volume of cylinder : Volume of hemisphere $$\eqalign{ & = \frac{1}{3}\pi {r^2}:\pi {r^2}r:\frac{2}{3}\pi {r^3} \cr & = \frac{1}{3}:1:\frac{2}{3} \cr & = 1:3:2 \cr} $$

93.

Length of each edge of a regular tetrahedron is 1 cm. It volume is :

• (A) $$\frac{\sqrt{3}}{12}c{m}^3$$
• (B) $$\frac{1}{4}\sqrt{3}c{m}^3$$
• (C) $$\frac{\sqrt{2}}{6}c{m}^3$$
• (D) $$\frac{1}{12}\sqrt{2}c{m}^3$$

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 Length of each edge of a regular tetrahedron = 1 cm Volume of regular tetrahedron : $$\eqalign{ & = \frac{{{a^3}}}{{6\sqrt 2 }}{\text{ c}}{{\text{m}}^3} \cr & = \frac{1}{{6\sqrt 2 }} \cr & = \frac{{\sqrt 2 }}{{6\sqrt 2 \times \sqrt 2 }}{\text{ c}}{{\text{m}}^3} \cr & = \frac{{\sqrt 2 }}{{12}}{\text{ Or }}\frac{1}{{12}}\sqrt 2 {\text{ c}}{{\text{m}}^3} \cr} $$

94.

If the total length of diagonals of a cube is 12 cm, then what is the total length of the edges of the cube ?

• (A) 6$$\sqrt{3}$$ cm
• (B) 12 cm
• (C) 15 cm
• (D) 12$$\sqrt{3}$$ cm

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 Since a cube has 4 diagonals, we have : Length of a diagonal $$\eqalign{ & = \left( {\frac{{12}}{4}} \right)cm \cr & = 3\,cm \cr} $$ Let the length of each edge of the cube be a cm Then, $$\eqalign{ & \sqrt 3 a = 3 \cr & or,a = \sqrt 3 \cr} $$ ∴ Total length of the edges of the cube = $$12\sqrt 3\, $$ cm

95.

A copper wire of length 36 m and diameter 2 mm is melted to form a sphere. The radius of the sphere (in cm) is :

• (A) 2.5 cm
• (B) 3 cm
• (C) 3.5 cm
• (D) 4 cm

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 Let the radius of the sphere be r cm Then, $$\eqalign{ & \frac{4}{3}\pi {r^3} = \pi \times {\left( {0.1} \right)^2} \times 3600 \cr & \Leftrightarrow {r^3} = 36 \times \frac{3}{4} \cr & \Leftrightarrow {r^3} = 27 \cr & \Leftrightarrow r = 3\,cm \cr} $$

96.

The length of the longest rod that can be placed in a room of dimensions 10 m × 10 m × 5 m is :

• (A) 15$$\sqrt{3}$$ m
• (B) 15 m
• (C) 10$$\sqrt{2}$$ m
• (D) 5$$\sqrt{3}$$ m

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 Required length : $$\eqalign{ & = \sqrt {{{\left( {10} \right)}^2} + {{\left( {10} \right)}^2} + {{\left( 5 \right)}^2}} \,m \cr & = \sqrt {225} \,m \cr & = 15\,m \cr} $$

97.

A right triangle with sides 3 cm, 4 cm and 5 cm is rotated the side of 3 cm to form a cone. The volume of the cone so formed is:

• (A) 12$$\pi$$ cm3
• (B) 15$$\pi$$ cm3
• (C) 16$$\pi$$ cm3
• (D) 20$$\pi$$ cm3

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 $$\eqalign{ & {\text{Clearly}}, \cr & {\text{We have }}r = 3\,{\text{cm}}\,{\text{and}}\,h = 4\,{\text{cm}} \cr & \therefore \,\,\,{\text{Volume}} \cr & = \frac{1}{3}\pi {r^2}h \cr & = \left( {\frac{1}{3} \times \pi \times {3^2} \times 4} \right){\text{c}}{{\text{m}}^3} \cr & = 12\pi \,{\text{c}}{{\text{m}}^3} \cr} $$

98.

In a right circular cone, the radius of its base is 7 cm and its height is 24 cm. A cross-section is made through the mid-point of the height parallel to the base. The volume of the upper portion is :

• (A) 154 cm3
• (B) 169 cm3
• (C) 800 cm3
• (D) 1078 cm3

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 $$\eqalign{ & r = 7{\text{ cm, }}h{\text{ = 24 cm}} \cr & {\text{Now, }}\vartriangle {\text{AOB}} \sim \vartriangle {\text{COD}} \cr & {\text{So, }}\frac{{OA}}{{OC}} = \frac{{AB}}{{CD}} \cr & \Rightarrow \frac{h}{{\frac{h}{2}}} = \frac{r}{{CD}} \cr & \Rightarrow CD = \frac{r}{2} \cr} $$ ∴ Volume of upper portion :$$\eqalign{ & = \frac{1}{3}\pi {\left( {\frac{r}{2}} \right)^2}\left( {\frac{h}{2}} \right) \cr & = \left( {\frac{1}{3} \times \frac{{22}}{7} \times \frac{7}{2} \times \frac{7}{2} \times 12} \right){\text{ c}}{{\text{m}}^3} \cr & = 154{\text{ c}}{{\text{m}}^3} \cr} $$

99.

The ratio of the surface area of a sphere and the curved surface area of the cylinder circumscribing the sphere is :

• (A) 1 : 1
• (B) 1 : 2
• (C) 2 : 1
• (D) 2 : 3

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 Let the radius of the sphere be r Then, radius of the cylinder = r Height of the cylinder = 2r Surface area of sphere = $$4\pi {{\text{r}}^2}$$ Surface area of the cylinder = $$2\pi {\text{r}}(2r) = 4\pi {{\text{r}}^2}$$ ∴ Required ratio : = $$4\pi {{\text{r}}^2}$$ : $$4\pi {{\text{r}}^2}$$ = 1 : 1

100.

The curved surface of a right circular cone of height 15 cm and base diameter 16 cm is :

• (A) 60π cm2
• (B) 68π cm2
• (C) 120π cm2
• (D) 136π cm2

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 h = 15 cm, r = 8 cm So, $$\eqalign{ & l = \sqrt {{r^2} + {h^2}} \cr & \,\,\,\,\,\, = \sqrt {{8^2} + {{\left( {15} \right)}^2}} \cr & \,\,\,\,\,\, = 17\,cm \cr} $$ ∴ Curved surface area : $$\eqalign{ & = \pi rl \cr & = \left( {\pi \times 8 \times 17} \right){\text{ c}}{{\text{m}}^2} \cr & = 136\pi {\text{ c}}{{\text{m}}^2} \cr} $$