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 uniform rod $AB$ of mass $m$ and length $l$ is hinged at its mid point $C$ . The left half $(AC)$ of the rod has linear charge density $-\lambda $ and the right half $(CB)$ has $+\lambda $  where $\lambda $ is constant . A large non conducting sheet of unirorm surface charge density $\sigma $ is also .present near the rod. Initially the rod is kept perpendicular to the sheet. The end $A$ of the rod is initially at a distance $d$ . Now the rod is rotated by a small angle in the plane of the paper and released. The time period of small angular oscillations is

According to Gauss’ Theorem, electric field of an infinitely long straight wire is proportional to

A hollow metal sphere of radius $R$ is uniformly charged. The electric field due to the sphere at a distance r from the centre

Let $E_1(r), E_2(r)$ and $E_3(r)$ be the respective electric fields at a distance $r$ from a point charge $Q$, an infinitely long wire with constant linear charge density $\lambda$, and an infinite plane with uniform surface charge density $\sigma$. if $E_1\left(r_0\right)=E_2\left(r_0\right)=E_3\left(r_0\right)$ at a given distance $r_0$, then