R.d Sharma 2022 _mcqs for Class 9 Maths Chapter 20 - Surface Area And Volume Of A Right Circular Cone

Answer:

A cone has 2 surfaces:

One is a curved surface and the other one is a circular face as a base.

Hence, the correct answer is option (b).

Answer:

The curved surface area of radius R and slant height L is given by $\mathrm{\pi }RL$.

Given: Radius (R) = 2r and slant height (L) = $\frac{l}{2}$

$\begin{array}{rcl}\mathrm{Curved} \mathrm{surface} \mathrm{area} \mathrm{of} \mathrm{the} \mathrm{cone}& =& \mathrm{\pi }RL\\ & =& \mathrm{\pi }\times 2r\mathit{\times }\frac{l}{2}\\ & \mathit{=}& \mathrm{\pi }rl\end{array}$

Hence, the correct answer is option (a).

Answer:

$\mathrm{Radius} \mathrm{of} \mathrm{cone}\left(R\right)=\frac{r}{2}, \mathrm{Slant} \mathrm{height}\left(L\right)=2l.$
$\begin{array}{rcl}\mathrm{Total} \mathrm{Surface} \mathrm{Area}& =& \mathrm{\pi }RL+\mathrm{\pi }{R}^2\\ & =& \mathrm{\pi }\times \left(\frac{r}{2}\right)\left(2l\right)+\mathrm{\pi }\times {\left(\frac{r}{2}\right)}^2\\ & =& \pi rl+\frac{\pi r^2}{4}\\ & =& \pi r\left(l+\frac{r}{4}\right)\end{array}$

Hence, the correct answer is option (b).

Answer:

Let the height of the cylinder be H and the height of the cone be h and radius of both be r.

A solid cylinder is melted and cast into a cone of same radius. Then the volume of both the cylinder and the cone should be equal.

$$\begin{gathered} \Rightarrow \mathrm{\pi }{r}^{2}H=\frac{1}{3}\mathrm{\pi }{r}^{2}h \\ \Rightarrow H=\frac{h}{3} \\ \Rightarrow \frac{h}{H}=\frac{3}{1} \\ \Rightarrow h:H=3:1 \end{gathered}$$

Hence, the correct answer is option (c).

Answer:

The volume of a right circular cone of base radius R and height H is $\frac{1}{3}\mathrm{\pi }{R}^{2}H$.

If the radius of the base of a right circular cone is 3r and its height is equal to the radius of the base, then its volume is given by

$\begin{array}{rcl}Volume& =& \frac{1}{3}\times \mathrm{\pi }\times {\left(3r\right)}^2\times 3r\\ & =& \frac{1}{3}\times \mathrm{\pi }\times 9\times 3r^3\\ & =& 9\mathrm{\pi }{r}^3\end{array}$

Hence, the correct answer is option (d).

Answer:

Let the volumes of both the cones be V₁ and V₂, the radius of both the cones be r₁ and r₂ and their heights be h₁ and h₂ respectively.

Given: $\frac{V_1}{V_2}=\frac{1}{4}$ and $\frac{d_1}{d_2}=\frac{4}{5}$
$$\begin{gathered} \Rightarrow \frac{2r_1}{2r_2}=\frac{4}{5} \\ \Rightarrow \frac{r_1}{r_2}=\frac{4}{5} \end{gathered}$$

The volume of a cone of radius R and height H is given by $\frac{1}{3}\mathrm{\pi }{R}^{2}H$.

$$\begin{gathered} \Rightarrow \frac{V_1}{V_2}=\frac{{\displaystyle \frac{1}{3}}\mathrm{\pi }{r_1}^2{h}_1}{{\displaystyle \frac{1}{3}}\mathrm{\pi }{r_2}^2{h}_2} \\ \Rightarrow \frac{1}{4}={\left(\frac{{\displaystyle r_1}}{{\displaystyle r_2}}\right)}^2\times \frac{{\displaystyle h_1}}{{\displaystyle h_2}} \\ \Rightarrow \frac{1}{4}={\left(\frac{{\displaystyle 4}}{{\displaystyle 5}}\right)}^2\times \frac{{\displaystyle h_1}}{{\displaystyle h_2}} \\ \Rightarrow \frac{1}{4}=\frac{{\displaystyle 16}}{{\displaystyle 25}}\times \frac{{\displaystyle h_1}}{{\displaystyle h_2}} \\ \Rightarrow \frac{{\displaystyle h_1}}{{\displaystyle h_2}}=\frac{25}{64} \\ \Rightarrow h_1:h_2=25:64 \end{gathered}$$

Hence, the correct answer is option (d).

Answer:

Let the volumes of both the cones be V₁ and V₂, the radius of both the cones be r₁ and r₂ and their heights be h₁ and h₂ respectively.

Given: $\frac{h_1}{h_2}=\frac{1}{4}$ and $\frac{r_1}{r_2}=\frac{4}{1}$

The volume of a cone of radius R and height H is given by $\frac{1}{3}\mathrm{\pi }{R}^{2}H$.

$$\begin{gathered} \Rightarrow \frac{V_1}{V_2}=\frac{{\displaystyle \frac{1}{3}}\mathrm{\pi }{r_1}^2{h}_1}{{\displaystyle \frac{1}{3}}\mathrm{\pi }{r_2}^2{h}_2} \\ \Rightarrow \frac{V_1}{V_2}={\left(\frac{{\displaystyle r_1}}{{\displaystyle r_2}}\right)}^2\times \frac{{\displaystyle h_1}}{{\displaystyle h_2}} \\ \Rightarrow \frac{V_1}{V_2}={\left(\frac{{\displaystyle 4}}{{\displaystyle 1}}\right)}^2\times \frac{{\displaystyle 1}}{{\displaystyle 4}} \\ \Rightarrow \frac{V_1}{V_2}=\frac{{\displaystyle 16}}{{\displaystyle 1}}\times \frac{{\displaystyle 1}}{{\displaystyle 4}} \\ \Rightarrow \frac{V_1}{V_2}=\frac{4}{1} \\ \Rightarrow V_1:V_2=4:1 \end{gathered}$$

Hence, the correct answer is option (d).

Answer:

The volume of a cone of radius R and height H is given by $V=\frac{1}{3}\mathrm{\pi }{R}^{2}H$.

If the height and radius of a cone of volume V are doubled, then the volume of the cone, is given by

$$\begin{gathered} V\text{'}=\frac{1}{3}\mathrm{\pi }{\left(2R\right)}^2\left(2\mathrm{H}\right) \\ \Rightarrow V\text{'}=\frac{1}{3}\mathrm{\pi }\times 4R^2\times 2H \\ \Rightarrow V\text{'}=8\times \frac{1}{3}\mathrm{\pi }{R}^{2}H \\ \Rightarrow V\text{'}=8V \end{gathered}$$

Hence, the correct answer is option (d).

Answer:

Let the height and radius of the cylinder and cone be h and r respectively and the volume of a right circular cylinder be V₁ and the volume of a right circular cone be V₂.

$$\begin{gathered} \therefore V_1=\mathrm{\pi }{r}^{2}h \\ \mathrm{And} V_2=\frac{1}{3}\mathrm{\pi }{r}^{2}h \\ \therefore \frac{V_1}{V_2}=\frac{\mathrm{\pi }{r}^{2}h}{{\displaystyle \frac{1}{3}}\mathrm{\pi }{r}^{2}h} \\ \Rightarrow \frac{V_1}{V_2}=\frac{3\mathrm{\pi }{r}^{2}h}{\mathrm{\pi }{r}^{2}h} \\ \Rightarrow \frac{V_1}{V_2}=\frac{3}{1} \\ \Rightarrow V_1:V_2=3:1 \end{gathered}$$

Hence, the correct answer is option (b).

Answer:

The volume of a cone of radius R and height H is given by $\frac{1}{3}\mathrm{\pi }{R}^{2}H$ and the volume of a right circular cylinder of radius R and height H is given by $\mathrm{\pi }{R}^{2}H$.

Let the radius of both be r and the heights of the right circular cylinder and a right circular cone be h₁ and h₂ respectively.

A right circular cylinder and a right circular cone have the same volume.

$$\begin{gathered} \Rightarrow \mathrm{\pi }{r}^2{h}_1=\frac{1}{3}\mathrm{\pi }{r}^2{h}_2 \\ \Rightarrow h_1=\frac{1}{3}{h}_2 \\ \Rightarrow \frac{h_1}{h_2}=\frac{1}{3} \\ \Rightarrow h_1:h_2=1:3 \end{gathered}$$

Hence, the correct answer is option (d).

Answer:

If the diameters of two cones are equal then their radius should also be equal.

The curved surface area of the cone of radius r and slant height l is given by πrl.

If the radius of cones is equal then the ratio of their curved surface areas is equal to the ratio of their slant heights.

So, the ratio of their curved surface areas is 5 : 4.

Hence, the correct answer is option (d).

Answer:

Let the curved surface areas of both cones be S₁ and S₂, the slant heights be l₁ and l₂ and the radii are r₁ and r₂.

The curved surface area of one cone is twice that of the other while the slant height of the latter is twice that of the former.

Given: $\frac{S_1}{S_2}=\frac{2}{1}$ and $\frac{l_1}{l_2}=\frac{1}{2}$

The curved surface area of a cone of radius R and slant height L is given by πRL.

$$\begin{gathered} \therefore \frac{S_1}{S_2}=\frac{\mathrm{\pi }{r}_1{l}_1}{\mathrm{\pi }{r}_2{l}_2} \\ \Rightarrow \frac{2}{1}=\frac{r_1}{r_2}\times \frac{1}{2} \\ \Rightarrow \frac{r_1}{r_2}=\frac{4}{1} \\ \Rightarrow r_1:r_2=4:1 \end{gathered}$$

Hence, the correct answer is option (b).

Answer:

The formula of the curved surface area of a cone with base radius 'r' and slant height 'l' is given by

Curved surface area, S = πrl

Now, the slant height of a cone is increased by 10%.

So, the new slant height = 1.1l

New curved surface area, S ' = 1.1πrl = 1.1S

Thus, the curved surface area is increased by 10%.

Hence, the correct answer is option (a).

Answer:

Given: Height of the cone(h) = 12 cm

Area of circular base = 64π cm²

Let the radius of the circular base be r.

∴ πr² = 64π

⇒ r = 8 cm



Now, in △OCD and △OAB

∠O = ∠O  (Common)

∠OAB = ∠OCD   (Each 90°)

Therefore, △OCD ~ △OAB [By AA similarity criterion]

$$\begin{gathered} \therefore \frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{AB}}{\mathrm{CD}} \\ \Rightarrow \frac{12}{9}=\frac{8}{\mathrm{CD}} \\ \Rightarrow \mathrm{CD}=6 \mathrm{cm} \end{gathered}$$

Now the radius of the upper cone is 6 cm.

∴ Area ot the upper cone = πr² = π(6)² = 36 cm²

Hence, the correct answer is option (d).

Answer:

The formula of the volume of a cone with base radius 'r' and slant height 'l' is given by $V=\frac{1}{3}\mathrm{\pi }{r}^{2}h$.

Now, the base radius and height have increased by 20%.

So, the new base radius and new height are 1.2r and 1.2l respectively.

The new volume of the cone is given by

$$\begin{gathered} V\text{'}=\frac{1}{3}\mathrm{\pi }{\left(1.2r\right)}^2\left(1.2h\right) \\ \Rightarrow V\text{'}=1.728\times \frac{1}{3}\mathrm{\pi }{r}^{2}h \\ \Rightarrow V\text{'}=1.728V \\ \Rightarrow V\text{'}=V+0.73V \end{gathered}$$

Thus, the volume of the cone is increased by 73%.

Hence, the correct answer is option (c).

Answer:

If h, S and V denote respectively the height, curved surface area and volume of a right circular cone, then

$S=\mathrm{\pi }rl \mathrm{and} V=\frac{1}{3}\mathrm{\pi }{r}^{2}h, \mathrm{where} l=\sqrt{r^2+h^2}$

$\begin{array}{rcl}\therefore 3\mathrm{\pi }V{h}^3-S^2{h}^2+9V^2& =& 3\mathrm{\pi }\left(\frac{1}{3}\mathrm{\pi }{r}^{2}h\right)h^3-{\left(\mathrm{\pi }r\sqrt{r^2+h^2}\right)}^2{h}^2+9{\left(\frac{1}{3}\mathrm{\pi }{r}^{2}h\right)}^2\\ & =& {\mathrm{\pi }}^2{r}^2{h}^4-{\mathrm{\pi }}^2{r}^2\left(r^2+h^2\right)h^2+{\mathrm{\pi }}^2{r}^4{h}^2\\ & =& {\mathrm{\pi }}^2{r}^2{h}^4-{\mathrm{\pi }}^2{r}^4{h}^2-{\mathrm{\pi }}^2{r}^2{h}^4+{\mathrm{\pi }}^2{r}^4{h}^2\\ & \mathit{=}& 0\end{array}$

Hence, the correct answer is option (b).

Answer:

The formula of the volume of a cone with base radius 'r' and slant height 'l' is given by

Volume, $V=\frac{1}{3}\mathrm{\pi }{r}^{2}h$

It is given that a cone is cut into two parts by a horizontal plane passing through the mid-point of its axis.

Let the height of the cone be H and R respectively and the radius of the smaller cone be r.



Now, in △ABC and △ADE

∠A = ∠A  (Common)

∠ADE = ∠ABC   (Each 90°)

Therefore, △ABC ~ △ADE   [By AA similarity criterion]

$$\begin{gathered} \therefore \frac{\mathrm{AB}}{\mathrm{AD}}=\frac{\mathrm{DE}}{\mathrm{BC}} \\ \Rightarrow \frac{H}{{\displaystyle \frac{H}{2}}}=\frac{R}{r} \\ \Rightarrow R=2r \end{gathered}$$

Volume of the given cone = $\frac{1}{3}\mathrm{\pi }{R}^{2}H=\frac{1}{3}\mathrm{\pi }{\left(2r\right)}^{2}H=\frac{4}{3}\mathrm{\pi }{r}^{2}H$

Volume of the smaller cone = $\frac{1}{3}\mathrm{\pi }{r}^2\left(\frac{H}{2}\right)=\frac{1}{6}\mathrm{\pi }{r}^{2}H$

$$\begin{gathered} \therefore \frac{\mathrm{Volume} \mathrm{of} \mathrm{smaller} \mathrm{cone}}{\mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{cone}}=\frac{{\displaystyle \frac{1}{6}}\mathrm{\pi }{r}^{2}H}{{\displaystyle \frac{4}{3}}\mathrm{\pi }{r}^{2}H} \\ \Rightarrow \frac{\mathrm{Volume} \mathrm{of} \mathrm{smaller} \mathrm{cone}}{\mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{cone}}=\frac{1}{6}\times \frac{3}{4}=\frac{1}{8} \end{gathered}$$

Hence, the correct answer is option (d).