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Question: A bomb explodes at t=0 in a uniform, isotropic medium of density \[\rho \] and releases energy \(E\)...

A bomb explodes at t=0 in a uniform, isotropic medium of density ρ\rho and releases energy EE, generating a spherical blast wave. The radius rr of this blast wave varies with tt as:
A) tt
B) t2/5{t^{2/5}}
C) t1/4{t^{1/4}}
D) t3/2{t^{3/2}}

Explanation

Solution

The waves produced by the explosion are longitudinal in nature. Therefore consider a small spherical element of wave and then use the concept of velocity of a longitudinal wave in medium and compare it with the basic derivative definition of velocity i.e. v=drdtv = \dfrac{{dr}}{{dt}}, where rr is the radius of the wave from centre of the explosion. On integration, it will give the relation of the radius with time.

Complete step by step solution:
Step1: Since due to the explosion the waves so formed is longitudinal in nature hence the speed of longitudinal waves in a medium is given by-
v=γPρv = \sqrt {\dfrac{{\gamma P}}{\rho }} ………………(1)
Where γ\gamma is the ratio of two specific heats, PP is the pressure and ρ\rho is the density of the medium
Also from the Ideal gas equation, we have,
PV=γRTPV = \gamma RT
Where P = pressure, V= volume, γ\gamma = ratio of two specific heats, R= universal gas constant, T= temperature.
Therefore, P=γRTVP = \dfrac{{\gamma RT}}{V}
Put this in equation (1) we get,

v=γρ×γRTV v=γRTVρ  v = \sqrt {\dfrac{\gamma }{\rho } \times \dfrac{{\gamma RT}}{V}} \\\ v = \gamma \sqrt {\dfrac{{RT}}{{V\rho }}} \\\

For sphere V=43πr3V = \dfrac{4}{3}\pi {r^3}
As ρ\rho ,γ\gamma ,R and T all are constant let them together as another constant k , Therefore
v=γRT(43πr3)ρv = \gamma \sqrt {\dfrac{{RT}}{{\left( {\dfrac{4}{3}\pi {r^3}} \right)\rho }}}
v=kr3/2\Rightarrow v = \dfrac{k}{{{r^{3/2}}}}
Where,
k=3RT4πρk = \sqrt {\dfrac{{3RT}}{{4\pi \rho }}}
Step2: Also The velocity of propagation of this wave is given by –
v=drdtv = \dfrac{{dr}}{{dt}}
Combining both equations we get, drdt=kr3/2\dfrac{{dr}}{{dt}} = \dfrac{k}{{{r^{3/2}}}}
r3/2dr=kdt{r^{3/2}}dr = kdt
Now integrating both sides we get,
r3/2dr=kdt\int {{r^{3/2}}} dr = \int k dt
r5/25/2=kt\dfrac{r^{5/2}}{5/2}=kt
Neglecting the constants
rt2/5r \propto {t^{2/5}}

\therefore The correct option is (B).

Note:
As here we are finding a relation between two quantities so the constant of integration can be neglected. Also we need to remember the equation of velocity of a longitudinal waves in a medium i.e. given by v=γPρv = \sqrt {\dfrac{{\gamma P}}{\rho }}