Question

The ratio between the radius of the base and the height of a cylindrical tank is 3 : 4. If its volume is $4851m^3,$ then the curved surface area of the cylinder is equal to

• A. $924m^2$
• B. $1617m^2$
• C. $900m^2$
• D. $425m^2$

Answer 1

The correct option is A $924m^2$

Let the radius of the base and the height of the cylinder be r and h, respectively.
∴ r : h = 3 : 4
$\Rightarrow h=\frac{4r}{3}$ ....(i)
$\text{Volume of cylinder}=4851m^3$
$\Rightarrow \pi r^{2}h=4851$
$\Rightarrow \pi \times (r{)}^2\times \frac{4r}{3}=4851$ [From (i)]
$\Rightarrow \pi r^3=\frac{4851\times 3}{4}$
$\Rightarrow r^3=\frac{4851\times 3\times 7}{4\times 22}$
$\Rightarrow r^3=1157.625$
$\Rightarrow r=10.5cm$ ....(ii)
$\therefore \text{CSA of cylinder}=2\pi rh$
$=2\pi \times 10.5\times \frac{4}{3}\times 10.5$ [From (i) and (ii)]
$=2\times \frac{22}{7}\times \frac{21}{2}\times \frac{4}{3}\times \frac{21}{2}$
$=2\times 22\times 21$
$=924m^2$
Hence, the correct answer is option (a).