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The nuclear charge $(\mathrm{Ze})$ 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 rhe radial direction.
Figure:$Image$
$1.$ The electric field at $\mathrm{r}=\mathrm{R}$ is
$(A)$ independent of a
$(B)$ directly proportional to a
$(C)$ directly proportional to $\mathrm{a}^2$
$(D)$ inversely proportional to a
$2.$ For $a=0$, the value of $d$ (maximum value of $\rho$ as shown in the figure) is
$(A)$ $\frac{3 Z e}{4 \pi R^3}$ $(B)$ $\frac{3 Z e}{\pi R^3}$ $(C)$ $\frac{4 Z e}{3 \pi R^3}$ $(D)$ $\frac{\mathrm{Ze}}{3 \pi \mathrm{R}^3}$
$3.$ The electric field within the nucleus is generally observed to be linearly dependent on $\mathrm{r}$. This implies.
$(A)$ $a=0$ $(B)$ $\mathrm{a}=\frac{\mathrm{R}}{2}$ $(C)$ $a=R$ $(D)$ $a=\frac{2 R}{3}$
Give the answer question $1,2$ and $3.$

$(A,B,C)$
$(C,B,D)$
$(A,D,C)$
$(B,A,C)$
Solution
$2.$ $\mathrm{q}=\int_0^R \frac{d}{R}(R-x) 4 \pi x^2 d x=Z e $
$ d=\frac{3 Z e}{\pi R^3}$
$3.$ If within a sphere $\rho$ is constant $E \propto r$
Similar Questions
The electric field $E$ is measured at a point $P (0,0, d )$ generated due to various charge distributions and the dependence of $E$ on $d$ is found to be different for different charge distributions. List-$I$ contains different relations between $E$ and $d$. List-$II$ describes different electric charge distributions, along with their locations. Match the functions in List-$I$ with the related charge distributions in List-$II$.
List-$I$ | List-$II$ |
$E$ is independent of $d$ | A point charge $Q$ at the origin |
$E \propto \frac{1}{d}$ | A small dipole with point charges $Q$ at $(0,0, l)$ and $- Q$ at $(0,0,-l)$. Take $2 l \ll d$. |
$E \propto \frac{1}{d^2}$ | An infinite line charge coincident with the x-axis, with uniform linear charge density $\lambda$ |
$E \propto \frac{1}{d^3}$ | Two infinite wires carrying uniform linear charge density parallel to the $x$-axis. The one along ( $y=0$, $z =l$ ) has a charge density $+\lambda$ and the one along $( y =0, z =-l)$ has a charge density $-\lambda$. Take $2 l \ll d$ |
plane with uniform surface charge density |