Question

The ratio of base radius and height of a cone is $3:4$. If the cost of smoothing the curved surface area at $5$ rupees per sq. cm is $Rs.$ $11550$, then the volume of liquid in it is:

• A. $12936{cm}^3$
• B. $12693{cm}^3$
• C. $12653{cm}^3$
• D. $12542{cm}^3$

Answer 1

Answer: (A)

Given, ratio of radius and height $=3:4$.

$\Rightarrow \frac{r}{h}=\frac{3}{4}$

$\Rightarrow 4r=3h$

$\Rightarrow r=\frac{3h}{3}$.

Also, total cost for smoothing $=Rs.$ $11550$.

and rate per sq. cm $=Rs.$ $5$.

Then, the curved surface area $=\frac{11550}{5}=2310c{m}^2$.

We know, the curved surface area of cone $=\pi rl=\pi r\sqrt{r^2+h^2}$

$\Rightarrow \pi r\sqrt{r^2+{\left(\frac{4r}{3}\right)}^2}=2310$

$\Rightarrow \pi r\sqrt{\frac{25r^2}{9}}=2310$

$\Rightarrow \pi r\times \frac{5r}{3}=2310$

$\Rightarrow r^2=\frac{2310\times 3\times 7}{22\times 5}$

$\Rightarrow r^2=441$

$\Rightarrow r=21cm$.

So $h=\frac{4r}{3}=\frac{4\times 21}{3}=28cm$.

Then, volume of liquid,

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

$\Rightarrow V=\frac{1}{3}\times \frac{22}{7}\times 28\times 21\times 21$

$\Rightarrow V=12936c{m}^3$.

Therefore, option $A$ is correct.