If the rate of change of volume of a sphere is equal to the rate of change of its radius then its radius =

Question

If the rate of change of volume of a sphere is equal to the rate of change of its radius then its radius = (A) 1 (B) √ 2 π (C)  (1 /√ 2 π) (D) (1/2√π)

Tardigrade Admin · Accepted Answer

V=(4/3) π r3 (dV/dt)=(4/3) π⋅3r2 (dr/dt) =4 π r2 (dr/dt) Since (dV/dt)=(dr/dt) ∴ 1=4π r2 ⇒ r2=(1/4π) ⇒ r=(1/2√π)

Q. If the rate of change of volume of a sphere is equal to the rate of change of its radius then its radius =

1

$2 π$

$\frac{1}{2 π}$

$\frac{1}{2 π}$

$V = \frac{4}{3} π r^{3}$
$\frac{d V}{d t} = \frac{4}{3} π \cdot 3 r^{2} \frac{d r}{d t}$
$= 4 π r^{2} \frac{d r}{d t}$
Since $\frac{d V}{d t} = \frac{d r}{d t}$
$∴ 1 = 4 π r^{2}$
$\Rightarrow r^{2} = \frac{1}{4 π}$
$\Rightarrow r = \frac{1}{2 π}$

Q. If the rate of change of volume of a sphere is equal to the rate of change of its radius then its radius =

1

2π​

2π​1​

2π​1​

V=34​πr3 dtdV​=34​π⋅3r2dtdr​ =4πr2dtdr​ Since dtdV​=dtdr​ ∴1=4πr2 ⇒r2=4π1​ ⇒r=2π​1​

$2 π$

$\frac{1}{2 π}$

$\frac{1}{2 π}$

$V = \frac{4}{3} π r^{3}$
$\frac{d V}{d t} = \frac{4}{3} π \cdot 3 r^{2} \frac{d r}{d t}$
$= 4 π r^{2} \frac{d r}{d t}$
Since $\frac{d V}{d t} = \frac{d r}{d t}$
$∴ 1 = 4 π r^{2}$
$\Rightarrow r^{2} = \frac{1}{4 π}$
$\Rightarrow r = \frac{1}{2 π}$