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 ((1/ρ ) (d ρ /d t)) is constant. The velocity v of any point on the surface of the expanding sphere is proportional to

Question

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 ((1/ρ ) (d ρ /d t)) is constant. The velocity v of any point on the surface of the expanding sphere is proportional to (A) R3 (B) (1/R) (C) R (D) R(2/3)

Tardigrade Admin · Accepted Answer

Density of sphere is ρ=(m/v)=(3 m/4 π R3) ⇒ (1/ρ) (d ρ/d t)=-(3/R) (d R/d t) Since ⇒ (1/ρ) (d ρ/d t) is constant ∴ (d R/d t) propto R

$R^{3}$

$\frac{1}{R}$

$R$

$R^{\frac{2}{3}}$

Density of sphere is $ρ = \frac{m}{v} = \frac{3 m}{4 π R ^{3}}$
$\Rightarrow \frac{1}{ρ} \frac{d ρ}{d t} = - \frac{3}{R} \frac{d R}{d t}$
Since $\Rightarrow \frac{1}{ρ} \frac{d ρ}{d t}$ is constant
$∴ \frac{d R}{d t} \propto R$

R3

R1​

R

R32​

Density of sphere is ρ=vm​=4πR33m​ ⇒ρ1​dtdρ​=−R3​dtdR​ Since ⇒ρ1​dtdρ​ is constant ∴dtdR​∝R

$R^{3}$

$\frac{1}{R}$

$R$

$R^{\frac{2}{3}}$

Density of sphere is $ρ = \frac{m}{v} = \frac{3 m}{4 π R ^{3}}$
$\Rightarrow \frac{1}{ρ} \frac{d ρ}{d t} = - \frac{3}{R} \frac{d R}{d t}$
Since $\Rightarrow \frac{1}{ρ} \frac{d ρ}{d t}$ is constant
$∴ \frac{d R}{d t} \propto R$