JEE Advanced 2015 · Paper 2 · Physics Q13

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PhysicsMultiple correct+4 / -2Question 13 of 60

A spherical body of radius $R$ consists of a fluid of constant density and is in equilibrium under its own gravity. If $P(r)$ is the pressure at $r\,(r<R)$ , then the correct option(s) is(are)

P(r=0)=0

\dfrac{P(r=3R/4)}{P(r=2R/3)}=\dfrac{63}{80}

\dfrac{P(r=3R/5)}{P(r=2R/5)}=\dfrac{16}{21}

\dfrac{P(r=R/2)}{P(r=R/3)}=\dfrac{20}{27}

Answer and worked solution

Official answer: B, C

Hydrostatic equilibrium. Inside the sphere, $g(r)=\frac43\pi G\rho r$ and
$\frac{dP}{dr}=-\rho g(r).$
With $P(R)=0$ :
$P(r)=\frac23\pi G\rho^2(R^2-r^2).$

(A) $P(0)\ne0$ . False.
(B) $\dfrac{1-\frac9{16}}{1-\frac49}=\dfrac{7/16}{5/9}=\dfrac{63}{80}$ . True.
(C) $\dfrac{1-\frac9{25}}{1-\frac4{25}}=\dfrac{16}{21}$ . True.
(D) $\dfrac{3/4}{8/9}=\dfrac{27}{32}$ . False.

(B), (C)

Topics:Gravitational field and potential Pressure and Pascal's law

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JEE Advanced 2015 · Paper 2, Physics question 13