A bomb explodes at time $\mathrm{t}=0$ in a uniform, isotropic medium of density $\rho$ and releases energy E, generating a spherical blast wave. The radius $\mathrm{R}$ of this blast wave varies with time t as :

• A

  $\mathrm{t}$
• Correct answer
  B

  $t^{2 / 5}$
• C

  $t^{1 / 4}$
• D

  $t^{3 / 2}$

Solution

The energy will propagate in form of spherical blast wave which is longitudinal in nature. Hence velocity of propagate of disturbance
$$ \mathrm{v}=\sqrt{\frac{\gamma \mathrm{P}}{\mathrm{\rho}}} $$ Where $P=$ Pressure $$ \begin{array}{r} \because \quad \mathrm{PV}=\mathrm{nRT} \\ \mathrm{v}=\sqrt{\frac{\gamma \mathrm{RT}}{\mathrm{V} \rho}} \end{array} $$ $$ \begin{array}{l} \text { Where } V=\frac{4 \pi r^{3}}{3} \\ \& v=\text { velocity of propagation }=\frac{d r}{d t} \\ \Rightarrow \frac{d r}{d t}=\sqrt{\frac{\gamma R T}{\left(\frac{4}{3} \pi r^{3}\right) \rho}} \\ \Rightarrow \quad \frac{d r}{d t}=\frac{k}{r^{3 / 2}} \\ \quad k=c o n s t a n t=\sqrt{\frac{3 \gamma R T}{4 \pi \rho}} \\ \Rightarrow \quad r^{3 / 2} d r=k d t \\ \qquad r^{3 / 2} d r=\int k d t \\ r^{5 / 2}=k t \Rightarrow r=(k t)^{2 / 5} \\ \Rightarrow r \propto t^{2 / 5} \end{array} $$ Hence correct answer is (B) Alternate solution $\mathrm{R} \propto \mathrm{E}^{\mathrm{a}} \mathrm{P}^{\mathrm{b}} \mathrm{t}^{\mathrm{c}}$ Dimension of $R=L$ $\begin{array}{c} \mathrm{L} \propto\left(\mathrm{ML}^{2} \mathrm{~T}^{-2}\right)^{\mathrm{a}}\left(\mathrm{ML}^{-3}\right)^{\mathrm{b}}(\mathrm{T})^{\mathrm{c}} \\ 2 \mathrm{a}-3 \mathrm{~b}=1 \\ \mathrm{c}-2 \mathrm{a}=0 \\ \mathrm{a}+\mathrm{b}=0 \end{array}$ from (i), (ii) \& (iii) $\begin{array}{l} \mathrm{a}=\frac{1}{5} \\ \mathrm{c}=2 \mathrm{a} \\ \therefore \quad \mathrm{c}=\frac{2}{5} \\ \mathrm{R} \propto \mathrm{t}^{2 / 5} \end{array}$

Hence, the correct answer is option: 'B'.