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Volume & Surface Areas Questions and Answers

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31.

If the ratio of volumes of two cones is 2:3 and the ratio of the radii of their bases is 1:2, then the ratio of their heights will be:

Exam Year(s)

DSSSB PGT - 2023,

Answer: B

The ratio of heights is 8 : 3. The volume of a cone is one third of pi r squared h, so h is proportional to V divided by r squared. With V in the ratio 2 : 3 and r in the ratio 1 : 2, h1 : h2 = (2/1) : (3/4) = 2 : 0.75 = 8 : 3. The ratio 2 : 1 ignores the radius, 3 : 8 is inverted and 3 : 2 uses the volumes only.

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32.

The volume of a sphere is 27 times that of another sphere. What is the ratio of their surface areas?

Exam Year(s)

DSSSB PGT - 2023,

Answer: A

The ratio of surface areas is 9 : 1. The volume of a sphere is proportional to the cube of its radius, so a volume ratio of 27 : 1 gives a radius ratio of 3 : 1. The surface area is proportional to the square of the radius, so the areas are in the ratio 9 : 1. 1 : 9 is inverted, 6 : 1 and 3 : 1 do not follow from squaring the radius ratio. The ratio of the radii is the cube root of 27 over the cube root of 1, which is 3 : 1. Surface area depends on the square of the radius, so the areas are in the ratio 9 : 1; the option 3 : 1 is the radii ratio, not the areas.

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33.

The maximum volume of a cone that can be carved out of a solid hemisphere of radius r is:

Exam Year(s)

DSSSB - 2023,

Answer: C

The maximum volume is \(\frac{1}{3}\pi r^{3}\).

The largest cone that fits in a solid hemisphere of radius r has its base on the flat face of radius r and its height equal to r.

\[ V=\frac{1}{3}\pi r^{2}\cdot r=\frac{1}{3}\pi r^{3} \]

\(\frac{2}{3}\pi r^{3}\) is the volume of the hemisphere itself, and \(\frac{4}{3}\pi r^{3}\) is a full sphere.

The cone with the greatest volume has its base on the flat face of the hemisphere and its apex at the top of the curved surface, which makes the height equal to the radius. A hemisphere has volume 2/3 pi r cubed, so the cone is exactly half of it: 1/3 pi r cubed.

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34.

Three spheres of radii 2cm, 3cm and 6cm are melted to form a new sphere. Find the radius of the new sphere, correct to two decimals.

Exam Year(s)

DSSSB - 2023,

Answer: A

The radius is 6.31 cm.

The volume of the new sphere equals the sum of the three volumes, so the cubes of the radii add up.

\[ R^{3}=2^{3}+3^{3}+6^{3}=8+27+216=251 \]

\[ R=\sqrt[3]{251}\approx6.31\text{ cm} \]

6.35, 6.55 and 6.15 cm would have cubes of about 256, 281 and 233.

You can check 6.31: its cube is 6.31 × 6.31 × 6.31, about 251.2, which is almost exactly 251. For 6.35 the cube is about 256, for 6.55 about 281 and for 6.15 about 233, which are all too far from 251.

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35.

The external and internal radii of an empty sphere are 10 cm and 3 cm, respectively. Find the volume of the metal which is required to make it.

Exam Year(s)

DSSSB - 2023,

Answer: C

The volume is \(4077.\overline{3}\ \text{cm}^{3}\).

The metal is the difference between the outer and inner spheres. Using 22/7 for pi:

\[ V=\frac{4}{3}\cdot\frac{22}{7}\,(10^{3}-3^{3}) \]

\[ V=\frac{88}{21}\times973=4077.\overline{3} \]

The other options differ in the decimal part and do not match.

The shell is hollow, so the metal fills the space between the radius 3 cm and radius 10 cm spheres. The difference of the cubes is 1000 − 27 = 973 and the volume is 4/3 × 22/7 × 973 = 4077.33 cm cubed, which is 4077 with the 3 repeating.

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36.

A swimming pool that is 24 m long 12 m wide and 3 m deep will have its bottom and its sides tiled. If the cost of tiling 1 m3 is Rs.750, what will be the total cost (in Rs.) incurred in tiling the pool?

Exam Year(s)

DSSSB - 2023,

Answer: A

The total cost is Rs. 3,78,000.

The bottom is 24 × 12 = 288 m2. The sides are the four walls: 2(24 + 12) × 3 = 216 m2. The tiled area is 504 m2. The printed rate is Rs. 750 per square metre (the question says m3, which is a slip).

\[ 504\times750=378000 \]

5,94,000 and 5,88,000 would include the top surface, and 3,96,000 does not come from these areas.

The pool is open at the top, so only the bottom and four walls are tiled. The long walls are 24 × 3 × 2 = 144 and the short walls are 12 × 3 × 2 = 72, so the walls are 216 m squared. With the bottom, 288, the total area is 504 m squared.

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37.

The volume of a right circular cylinder is 554.4 cm3 and its height is 10 cm. What is the perimeter of the base of the cylinder? (Use \(\pi = \frac{22}{7}\))

Exam Year(s)

DSSSB TGT - 2023,

Answer: B

The perimeter of the base is 26.4 cm. The volume of a cylinder is \(\pi r^{2}h\), so \[ \pi r^{2} = \frac{554.4}{10} = 55.44 \] \[ \begin{aligned} r^{2} &= 55.44\times\frac{7}{22} \\ &= 17.64 \Rightarrow r \\ &= 4.2\text{ cm} \end{aligned} \] The perimeter (circumference) of the base is \[ \begin{aligned} 2\pi r &= 2\times\frac{22}{7}\times 4.2 \\ &= 26.4\text{ cm} \end{aligned} \] The base radius of 4.2 cm gives 26.4 cm, so 17.6 cm, 22 cm and 44 cm are not the circumference of that circle. Check: \(\frac{22}{7}\times 4.2^{2}\times 10 = 554.4\), which is the given volume.

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38.

The curved surface area of a right circular cone is 444 \(\pi\) cm3 and the diameter of its base is 24 cm. What is the volume of the cone (in cm3)?

Exam Year(s)

DSSSB TGT - 2023,

Answer: D

The volume of the cone is \(1680\pi\) cm3. The base radius is 12 cm. The curved surface area is \(\pi r l\), so \[ \begin{aligned} \pi\times 12\times l &= 444\pi \Rightarrow l \\ &= 37\text{ cm} \end{aligned} \] The height follows from the right triangle: \[ \begin{aligned} h &= \sqrt{37^{2} - 12^{2}} \\ &= \sqrt{1369 - 144} \\ &= \sqrt{1225} \\ &= 35 \end{aligned} \] \[ \begin{aligned} V &= \frac{1}{3}\pi r^{2}h \\ &= \frac{1}{3}\pi\times 144\times 35 \\ &= 1680\pi \end{aligned} \] The options 840, 1120 and 1400 times \(\pi\) correspond to heights of 17.5, 23.3 and 29.2, which do not fit a slant height of 37.

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39.

A square pyramid has a base perimeter x, and the slant height is half of the perimeter. What is the lateral surface area of the pyramid?

Exam Year(s)

GATE AG - 2016,

Answer: D

The lateral surface area is \(0.25\,x^{2}\).

The base is a square of perimeter \(x\), so each side is \(x/4\). The slant height is half the perimeter, \(x/2\). The lateral surface consists of four congruent triangles, each with base \(x/4\) and height equal to the slant height:

\[ \begin{aligned} A &= 4 \times \frac{1}{2} \times \frac{x}{4} \times \frac{x}{2} \\ &= \frac{x^{2}}{4} = 0.25\,x^{2} \end{aligned} \]

\(0.50\,x^{2}\) forgets the half in the triangle area, and \(x^{2}\) uses the full perimeter as the base of each face. \(0.75\,x^{2}\) matches no consistent working.

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40.

Arrange the following three-dimensional objects in the descending order of their volumes:

(i) A cuboid with dimensions 10 cm, 8 cm and 6 cm
(ii) A cube of side 8 cm
(iii) A cylinder with base radius 7 cm and height 7 cm
(iv) A sphere of radius 7 cm

Exam Year(s)

GATE Physics - 2018, GATE AG - 2018,

Answer: D

The order is (iv), (iii), (ii), (i).

  • (iv) sphere: \(\frac{4}{3}\pi(7)^{3} \approx 1437\ \text{cm}^{3}\)
  • (iii) cylinder: \(\pi(7)^{2}(7) \approx 1078\ \text{cm}^{3}\)
  • (ii) cube: \(8^{3} = 512\ \text{cm}^{3}\)
  • (i) cuboid: \(10 \times 8 \times 6 = 480\ \text{cm}^{3}\)

From largest to smallest this is sphere, cylinder, cube, cuboid. The other orders put the cuboid or cube ahead of the curved solids, whose radius 7 cm gives them much larger volumes.

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