Gujarati
Hindi
1. Electric Charges and Fields
normal

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

$(A,B,C)$

B

$(C,B,D)$

C

$(A,D,C)$

D

$(B,A,C)$

(IIT-2008)

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 $E \propto r$

Standard 12
Physics

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