IIT-JEE 2008 · Paper 2 · Physics Q37

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Passage

Paragraph for Question Nos. 36 to 38

The nuclear charge ( $Ze$ ) is non-uniformly distributed within a nucleus of radius $R$ . The charge density $\rho(r)$ [charge per unit volume] is dependent only on the radial distance $r$ from the centre of the nucleus as shown in figure. The electric field is only along the radial direction.

ρ(r) against r: constant at d from r = 0 to r = a, then falling linearly to zero at r = R.

For $a=0$ , the value of $d$ (maximum value of $\rho$ as shown in the figure) is

\dfrac{3Ze}{4\pi R^3}

\dfrac{3Ze}{\pi R^3}

\dfrac{4Ze}{3\pi R^3}

\dfrac{Ze}{3\pi R^3}

Answer and worked solution

Official answer: B

With a = 0, $\rho(r)=d\left(1-\frac rR\right)$ . Total charge:
$Ze=\int_0^Rd\left(1-\frac rR\right)4\pi r^2dr=4\pi d\left(\frac{R^3}3-\frac{R^3}4\right)=\frac{\pi dR^3}3.$
So $d=\dfrac{3Ze}{\pi R^3}$ .

(B)

Topics:Electric flux and Gauss's law

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IIT-JEE 2008 · Paper 2, Physics question 37