A balloon which always remains spherical, has a variable diameter $\frac{3}{2}(5x+7)$. Then the rate of change of its volume with respect to x is

$\frac{135\pi}{16}(5x+7)^2$

The correct answer is Option (4) → $\frac{135\pi}{16}(5x+7)^2$

Given: Diameter = $\frac{3}{2}(5x + 7)$

Radius $r = \frac{3}{4}(5x + 7)$

Volume $V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi \left( \frac{3}{4}(5x + 7) \right)^3$

Differentiate w.r.t. $x$:

$\frac{dV}{dx} = \frac{4}{3} \pi \cdot 3 \left( \frac{3}{4}(5x + 7) \right)^2 \cdot \frac{3}{4} \cdot 5$

Simplifying step-by-step:

$\frac{dV}{dx} = 4\pi \cdot \left( \frac{3}{4} \right)^2 (5x + 7)^2 \cdot \frac{15}{4}$

$= 4\pi \cdot \frac{9}{16} \cdot \frac{15}{4} \cdot (5x + 7)^2$

$= \pi \cdot \frac{135}{16} \cdot (5x + 7)^2$