The ratio of the diameter and height of the right circular cylinder is 4 : 3. If the diameter of the cylinder is reduced by 25%, then its total surface area is reduced to $318.5π\, m^2$. What is the circumference of the base of the cylinder?

$28π\, m^2$

The correct answer is Option (4) → $28π\, m^2$

Let the original diameter be $d$ and height $h$.

Given: $\frac{d}{h} = \frac{4}{3} \Rightarrow h = \frac{3}{4}d$

Total Surface Area (TSA) of a cylinder:

$TSA = \pi d h + \pi \cdot \left(\frac{d}{2}\right)^2 \cdot 2 = \pi d h + \pi \cdot \frac{d^2}{2}$

$= \pi d h + \frac{\pi d^2}{2}$

Substitute $h = \frac{3}{4}d$:

$TSA = \pi d \cdot \frac{3}{4}d + \frac{\pi d^2}{2} = \pi d^2 \left(\frac{3}{4} + \frac{1}{2}\right) = \pi d^2 \cdot \frac{5}{4}$

New diameter after 25% reduction = $0.75d$

New height remains $h = \frac{3}{4}d$

New TSA = $\pi \cdot 0.75d \cdot \frac{3}{4}d + \frac{\pi}{2} \cdot (0.75d)^2$

$= \pi d^2 \cdot \left(\frac{9}{16} + \frac{9}{32}\right) = \pi d^2 \cdot \frac{27}{32}$

New TSA is given as $318.5\pi$:

$\pi d^2 \cdot \frac{27}{32} = 318.5\pi$

$d^2 = \frac{318.5 \cdot 32}{27} = \frac{10192}{27}$

$d = \sqrt{\frac{10192}{27}} = 28$

Circumference = $\pi d = \pi \cdot 28 = {28\pi}$