Question:medium

Edge of a variable cube increases at the rate of 5 cm/s. The rate at which the surface area of the cube increases when the edge is 2 cm long is :

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The rate of change of the surface area of a cube is $12x \frac{dx}{dt}$, where $x$ is the edge length and $\frac{dx}{dt}$ is the rate of change of the edge length.
Updated On: Jan 14, 2026
  • 24 cm\(^2\)/s
  • 120 cm\(^2\)/s
  • 12 cm\(^2\)/s
  • 5 cm\(^2\)/s
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The Correct Option is B

Solution and Explanation

The surface area $A$ of a cube is expressed by the formula $A = 6x^2$, where $x$ represents the edge length. The rate of change of the surface area with respect to time is: \[ \frac{dA}{dt} = 12x \frac{dx}{dt} \] With a given rate of change of edge length $\frac{dx}{dt} = 5$ cm/s and an edge length $x = 2$ cm, substitution yields: \[ \frac{dA}{dt} = 12(2)(5) = 120 \, \text{cm}^2/\text{s} \] Thus, option $(B)$ is the correct answer.
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