The rate of change of the surface area of a sphere of radius $r$ , when the radius is increasing at the rate of $2 c m / s$ is proportional to

\frac{1}{r}

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\frac{1}{r^{2}}

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r

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r^{2}

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Solution

The correct option is D $r$
Surface area of sphere,
$S = 4 π r^{2}$

and $\frac{d r}{d t} = 2$

$∴ \frac{d s}{d t} = 4 π \times 2 r \frac{d r}{d t} = 8 π r \times 2 = 16 π r$

$\Rightarrow \frac{d s}{d t} \propto r$

Hence, option C is correct.

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Consider an expanding sphere of instantaneous radius $R$ whose total mass remains constant. The expansion is such that the instantaneous density $ρ$ remains uniform throughout the volume. The rate of fractional change in density $(\frac{1}{ρ} \frac{d ρ}{d t})$ is constant. The velocity $v$ of any point on the surface of the expanding sphere is proportional to

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The rate of change of the surface area of a sphere of radius r, when the radius is increasing at the rate of 2 cm/s is proportional to

The correct option is D rSurface area of sphere,S=4πr2and drdt=2∴dsdt=4π×2rdrdt=8πr×2=16πr⇒dsdt∝rHence, option C is correct.

The rate of change of the surface area of a sphere of radius $r$ , when the radius is increasing at the rate of $2 c m / s$ is proportional to

The correct option is D $r$
Surface area of sphere,
$S = 4 π r^{2}$

and $\frac{d r}{d t} = 2$

$∴ \frac{d s}{d t} = 4 π \times 2 r \frac{d r}{d t} = 8 π r \times 2 = 16 π r$

$\Rightarrow \frac{d s}{d t} \propto r$

Hence, option C is correct.