The volume of a cube is increasing at the rate 9 cm3/s. How fast is t…

The correct answer is Option (1) → 3.6 cm²/s

Let the edge be x cm. Then \( V=x^3 \) and \( S=6x^2 \). By the chain rule,

    \( \displaystyle \frac{dV}{dt}=3x^2\frac{dx}{dt} \)

    \( \displaystyle 9=3(10)^2\frac{dx}{dt} \)

    \( \displaystyle \frac{dx}{dt}=\frac{9}{300}=0.03\ \text{cm/s} \)

    \( \displaystyle \frac{dS}{dt}=12x\frac{dx}{dt} \)

    \( \displaystyle \frac{dS}{dt}=12(10)(0.03)=3.6\ \text{cm}^2\text{/s} \)

Shortcut: Divide the two rates directly: \( \frac{dS}{dV}=\frac{12x}{3x^2}=\frac{4}{x} \), so \( \frac{dS}{dt}=\frac{4}{10}\times 9=3.6 \) cm²/s.

Why the other options are incorrect

• Option (2) 36 cm²/s — Incorrect. 36 is ten times too large; it comes from \( \frac{dx}{dt}=0.3 \) instead of 0.03.

• Option (3) 120 cm²/s — Incorrect. 120 is \( 12x \) at x = 10, the value of \( \frac{dS}{dx} \); it still has to be multiplied by \( \frac{dx}{dt}=0.03 \).

• Option (4) 2.4 cm²/s — Incorrect. With \( \frac{dx}{dt}=0.03 \), the rate \( 12x\frac{dx}{dt} \) is 3.6, not 2.4.

NCERT (2026-27 edition):
Rate of Change of Quantities