Question

If the volume of spherical ball is increasing at the rate of 4π cm³/s, then the rate of change of its surface area when the volume is 288π cm³, is ______.

विकल्प

• `4/3`π cm²/s

• `2/3`π cm²/s

• 4π cm²/s

• 2π cm²/s

Answer 1

If the volume of spherical ball is increasing at the rate of 4π cm³/s, then the rate of change of its surface area when the volume is 288π cm³, is `underlinebb(4/3π  cm^2//s)`.

Explanation:

Let r be the radius of spherical ball

∵ Volume of spherical ball V = `4/3 πr^3`  ...(i)

Now, 288π = `4/3πr^3`

`\implies` r³ = 72 × 3 = 8 × 27

`\implies` r = 6

After differentiating equation (i) w.r.t. 't', we get

`(dV)/dt = 4πr^2 (dr)/dt`

`\implies` 4π = `4πr^2 (dr)/dt`  `[∵ (dv)/dt = 4π  cm^3//s]`

`\implies` 1 = `(6)^2 (dr)/dt`

`\implies (dr)/dt = 1/36`

∵ Surface area of spherical ball, s = 4πr²

After differentiating on both sides, w.r.t. 't', we get

`(ds)/dt = 4 xx 2πr (dr)/dt`

`\implies (ds)/dt = 8 xx π xx 6 xx 1/36`

Hence, `(ds)/dt = (4π)/3` cm²/s