RD Sharma 2020 solution class 9 chapter 19 Surface Area and Volume of a Right Circular Cylinder MCQS

MCQS

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Question 1:

Mark the correct alternative in each of the following:

In a cylinder, if radius is doubled and height is halved, curved surface area will be

(a) halved

(b) doubled

(c) same

(d) four times

Answer 1:

Let S₁ be the curved surface area of the cylinder with radius r₁ and height h₁, then

……. (1)

Now, let S₂ be the surface area after changing the dimension, then

So,

Hence, the correct option is (c)

Question 2:

Two cylindrical jars have their diameters in the ratio 3 : 1, but height 1 : 3. Then the ratio of their volumes is

(a) 1 : 4

(b) 1 : 3

(c) 3 : 1

(d) 2 : 5

Answer 2:

Let V₁ and V₂ be the volume of the two cylinders with radius r₁ and height h₁, and radius r₂ and height h₂, where

So,

……. (1)

Now,

…… (2)

From equation (1) and (2), we have

Hence, the correct option is (c)

Question 3:

The number of surfaces in right cylinder is

(a) 1

(b) 2

(c) 3

(d) 4

Answer 3:

In a cylinder there are three surfaces, one is curved surface area and other two are circular discs.

Hence, the correct option is (c)

Question 4:

Vertical cross-section of a right circular cylinder is always a

(a) square

(b) rectangle

(c) rhombus

(d) trapezium

Answer 4:

From the following figure of cylinder we observe that the vertical cross section is the rectangle PQRS.

Hence, the correct option is (b)

Question 5:

If r is the radius and h is height of the cylinder the volume will be

(a) $\frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h}$

(b) ${\mathrm{\pi r}}^2\mathrm{h}$

(c) 2$\mathrm{\pi r} (\mathrm{h} + \mathrm{r})$

(d) 2$\mathrm{\pi rh}$

Answer 5:

It is given that r is the radius and h is the height of the cylinder.

We know that the volume of cylinder is given by the formula

Hence, the correct option is (b)

Question 6:

The number of surfaces of a hollow cylindrical object is

(a) 1

(b) 2

(c) 3

(d) 4

Answer 6:

In a hollow cylinder, there are two curved surface areas: inner and outer and one circular base with inner and outer surface area.

Hence, the correct option is (d)

Question 7:

If the radius of a cylinder is doubled and the height remains same, the volume will be

(a) doubled

(b) halved

(c) same

(d) four times

Answer 7:

Let V₁ be the volume of the cylinder with radius r₁ and height h₁, then

……. (1)

Now, let V₂ be the volume after changing the dimension, then

So,

Hence, the correct option is (d)

Question 8:

If the height of a cylinder is doubled and radius remains the same, then volume will be

(a) doubled

(b) halved

(c) same

(d) four times

Answer 8:

Let V₁ be the volume of the cylinder with radius r₁ and height h₁, then

……. (1)

Now, let V₂ be the volume after changing the dimension, then

So,

Hence, the correct option is (a)

Question 9:

In a cylinder, if radius is halved and height is doubled, the volume will be

(a) same

(b) doubled

(c) halved

(d) four times

Answer 9:

Let V₁ be the volume of the cylinder with radius r₁ and height h₁, then

……. (1)

Now, let V₂ be the volume after changing the dimension, then

So,

Hence, the correct option is (c)

Question 10:

If the diameter of the base of a closed right circular cylinder be equal to its height h, then its whole surface area is

(a) $2{\mathrm{\pi h}}^2$

(b) $\frac{3}{2}{\mathrm{\pi h}}^2$

(c) $\frac{4}{3}{\mathrm{\pi h}}^2$

(d) ${\mathrm{\pi h}}^2$

Answer 10:

Let r be the radius of the cylinder and h be its height.

It is given that:

Therefore, total surface area S is:

Hence, the correct option is (b)

Question 11:

A right circular cylindrical tunnel of diameter 2 m and length 40 m is to be constructed from a sheet of iron. The area of the iron sheet required in m², is

(a) 40$\mathrm{\pi }$

(b) 80$\mathrm{\pi }$

(c) 160$\mathrm{\pi }$

(d) 200$\mathrm{\pi }$

Answer 11:

Let r be the radius of the cylinder and h be its height.

It is given that:

And,

Therefore, total surface area S of the sheet required is:

Hence, the correct option is (b)

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Question 12:

Two circular cylinders of equal volume have their heights in the ratio 1 : 2 Ratio of their radii is

(a) $1 : \sqrt{2}$

(b) $\sqrt{2}:1$

(c) 1 : 2

(d) 1 : 4

(d)

Answer 12:

Let V₁ and V₂ be the volume of the two cylinders with h₁ and h₂ as their heights:

Let r₁ and r₂ be their base radius.

It is given that

Hence, the correct option is (b)

Question 13:

The radius of a wire is decreased to one-third. If volume remains the same, the length will become

(a) 3 times

(b) 6 times

(c) 9 times

(d) 27 times

Answer 13:

Let V₁ and V₂ be the volume of the two cylinders with h₁ and h₂ as their heights:

Let r₁ and r₂ be their base radius.

It is given that

Hence, the length will become 9 times.

Hence, the correct option is (c)

Question 14:

If the height of a cylinder is doubled, by what number must the radius of the base be multiplied so that the resulting cylinder has the same volume as the original cylinder?

(a) 4

(b) $\frac{1}{\sqrt{2}}$

(c) 2

(d) $\frac{1}{2}$

Answer 14:

Let V₁ be the volume of the cylinder with radius r₁ and height h₁, then

……. (1)

Now, let V₂ be the volume after changing the dimension, then

So,

It is given that V₁ =V₂Therefpre,

Hence, the correct option is (b)

Question 15:

The volume of a cylinder of radius r is 1/4 of the volume of a rectangular box with a square base of side length x. If the cylinder and the box have equal heights, what is r in terms of x?

(a) $\frac{x^2}{2\mathrm{\pi }}$

(b) $\frac{x}{2\sqrt{\mathrm{\pi }}}$

(c) $\frac{\sqrt{2x}}{\mathrm{\pi }}$

(d) $\frac{\mathrm{\pi }}{2\sqrt{x}}$

Answer 15:

Let V₁ be the volume of the cylinder with radius r and height h, then

……. (1)

Now, let V₂ be the volume of the box, then

It is given that V₁ =1/4 V₂Therefpre,

Hence, the correct option is (b)

Question 16:

The height h of a cylinder equals the circumference of the cylinder. In terms of h what is the volume of the cylinder?

Answer 16:

Let h be the height of the cylinder with radius r.

It is given that:

Therefore, the volume of the cylinder is

Hence, the correct option is (a)

Question 17:

A cylinder with radius r and height h  is closed on the top and bottom. Which of the following expressions represents the total surface area of this cylinder?

(a) $2\mathrm{\pi r} (\mathrm{r}+\mathrm{h})$

(b) $\mathrm{\pi r} (\mathrm{r}+2\mathrm{h})$

(c) $\mathrm{\pi r}+(2\mathrm{r}+\mathrm{h})$

(d) $2{\mathrm{\pi r}}^2+\mathrm{h}$

Answer 17:

Let S be the total surface area of the closed cylinder with radius r and height h, then

Hence, the correct option is (a)

Question 18:

The height of sand in a cylindrical shaped can drops 3 inches when 1 cubic foot of sand is poured out. What is the diameter, in inches, of the cylinder?

(a) $\frac{24}{\mathrm{\pi }}$

(b) $\frac{48}{\mathrm{\pi }}$

(c) $\frac{32}{\mathrm{\pi }}$

(d) $\frac{48}{\mathrm{\pi }}$

Answer 18:

Let r be the radius of the cylinder. It is given that the height drops 3 inches, when 1 cubic foot of sand is poured out and 1 foot = 12 inch

So,

Hence, the correct option is (b)

Question 19:

Two steel sheets each of length a₁ and breadth a₂ are used to prepare the surfaces of two right circular cylinders ⇀ one having volume v₁ and height a₂ and other having volume v₂ and height a₁. Then,

(a) v₁ = v₂

(b) a₁v₁ = a₂v₂

(c) a₂v₁ = a₁v₂

(d) $\frac{v_1}{a_1}=\frac{v_2}{a_2}$

Answer 19:

Let the radius of the base of the cylinders be r and R.
Now, let sheet with length a₁ be used to form a cylinder with volume v₁

So,
 $$\begin{gathered} 2\mathrm{\pi r}={\mathrm{a}}_1 \\ \Rightarrow \mathrm{r}=\frac{{\mathrm{a}}_1}{2\mathrm{\pi }} \end{gathered}$$
Volume $v_1={\mathrm{\pi r}}^2\mathrm{h}=\mathrm{\pi }{\left(\frac{{\mathrm{a}}_1}{2\mathrm{\pi }}\right)}^2{\mathrm{a}}_2=\frac{{{\mathrm{a}}_1}^2{\mathrm{a}}_2}{4\mathrm{\pi }}$
Similarly, let sheet with length a2 be used to form a cylinder with volume v2
So,
$$\begin{gathered} 2\pi R={\mathrm{a}}_2 \\ \Rightarrow R=\frac{{\mathrm{a}}_2}{2\mathrm{\pi }} \end{gathered}$$
Volume $v_2=\pi R^2\mathrm{h}=\mathrm{\pi }{\left(\frac{{\mathrm{a}}_2}{2\mathrm{\pi }}\right)}^2{\mathrm{a}}_1=\frac{{{\mathrm{a}}_2}^2{\mathrm{a}}_1}{4\mathrm{\pi }}$
Now,
 $$\begin{gathered} \frac{v_1}{v_2}=\frac{{a_1}^2{a}_2}{{a_2}^2{a}_1}=\frac{a_1}{a_2} \\ \Rightarrow v_1{a}_2=v_2{a}_1 \end{gathered}$$
Hence, the correct option is (c).

 

Question 20:

The altitude of a circular cylinder is increased six times and the base area is decreased one-ninth of its value. The factor by which the lateral surface of the cylinder increases, is

(a) $\frac{2}{3}$

(b) $\frac{1}{2}$

(c) $\frac{3}{2}$

(d) 2

Answer 20:

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{original} \mathrm{radius} \mathrm{of} \mathrm{the} \mathrm{base} \mathrm{of} \mathrm{cylinder} = \mathrm{r} \\ \mathrm{Let} \mathrm{the} \mathrm{original} \mathrm{height} \mathrm{of} \mathrm{cylinder} = \mathrm{h} \\ \mathrm{Now}, \mathrm{original} \mathrm{base} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{cylinder}, \mathrm{S} = {\mathrm{\pi r}}^2 \\ \mathrm{Now}, \mathrm{original} \mathrm{LSA} \mathrm{of} \mathrm{cylinder}, \mathrm{A} = 2\mathrm{\pi rh} ............\left(1\right) \\ \mathrm{When} \mathrm{the} \mathrm{altitude} \mathrm{is} \mathrm{increased} \mathrm{to} 6 \mathrm{times} \mathrm{of} \mathrm{its} \mathrm{initial} \mathrm{value} \mathrm{and} \mathrm{base} \mathrm{area} \mathrm{is} \mathrm{decreased} \mathrm{one}-\mathrm{ninth} \mathrm{of} \mathrm{its} \mathrm{initial} \mathrm{value}: \\ \mathrm{Let} \mathrm{the} \mathrm{new} \mathrm{height} \mathrm{of} \mathrm{the} \mathrm{cylinder} = \mathrm{h}\text{'} \\ \mathrm{Let} \mathrm{the} \mathrm{new} \mathrm{radius} \mathrm{of} \mathrm{base} \mathrm{of} \mathrm{cylinder} = \mathrm{r}\text{'} \\ \mathrm{Now}, \mathrm{new} \mathrm{base} \mathrm{area} \mathrm{of} \mathrm{cylinder}, \mathrm{S}\text{'} = \mathrm{\pi }{\left(\mathrm{r}\text{'}\right)}^2 \\ \mathrm{Now}, \mathrm{it} \mathrm{is} \mathrm{given} \mathrm{that}, \\ \mathrm{new} \mathrm{height} \mathrm{of} \mathrm{cylinder} = 6\times \mathrm{original} \mathrm{height} \mathrm{of} \mathrm{cylinder} \\ \Rightarrow \boxed{\mathrm{h}\text{'} = 6\mathrm{h}} \\ \mathrm{Also}, \mathrm{new} \mathrm{base} \mathrm{area} \mathrm{of} \mathrm{cylinder} = \frac{1}{9}\left(\mathrm{original} \mathrm{base} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{cylinder}\right) \\ \Rightarrow \mathrm{S}\text{'} = \frac{1}{9}\mathrm{S} \\ \Rightarrow \mathrm{\pi }{\left(\mathrm{r}\text{'}\right)}^2 = \frac{1}{9}\left({\mathrm{\pi r}}^2\right) \\ \Rightarrow {\left(\mathrm{r}\text{'}\right)}^2 = {\left(\frac{\mathrm{r}}{3}\right)}^2 \\ \Rightarrow \boxed{\mathrm{r}\text{'} = \frac{\mathrm{r}}{3}} \\ \mathrm{Now}, \mathrm{new} \mathrm{LSA} \mathrm{of} \mathrm{cylinder}, \mathrm{A}\text{'} = 2\mathrm{\pi r}\text{'}\mathrm{h}\text{'} \\ \Rightarrow \mathrm{A}\text{'} = 2\mathrm{\pi } \times \left(\frac{\mathrm{r}}{3}\right) \times \left(6\mathrm{h}\right) = 2\times \left(2\mathrm{\pi rh}\right) \\ \Rightarrow \mathrm{A}\text{'} = 2 \mathrm{A} \left[\mathrm{Using} \left(1\right)\right] \\ \mathrm{Hence}, \mathrm{lateral} \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{cylinder} \mathrm{becomes} \mathrm{twice} \mathrm{of} \mathrm{the} \mathrm{original}. \\ \mathrm{So}, \mathrm{option} \left(\mathrm{d}\right) \mathrm{is} \mathrm{correct}. \end{gathered}$$