1

### JEE Main 2016 (Online) 10th April Morning Slot

Within a spherical charge distribution of charge density $\rho$(r), N equipotential surfaces of potential V0, V0 + $\Delta$V, V0 + 2$\Delta$V, .......... V0 + N$\Delta$V ($\Delta$ V > 0), are drawn and have increasing radii r0, r1, r2,..........rN, respectively. If the difference in the radii of the surfaces is constant for all values of V0 and $\Delta$V then :
A
$\rho$ (r) $\alpha$ r
B
$\rho$ (r) = constant
C
$\rho$ (r) $\alpha$ ${1 \over r}$
D
$\rho$ (r) $\alpha$ ${1 \over {{r^2}}}$

$\Delta$

## Explanation

Here, $\Delta$v and $\Delta$r are same for any pair of surface.

we know,

Electric field, E = $-$ ${{dv} \over {dr}}$

$\therefore$   E = constant [As dv and dr are constant]

Electric field inside the spherical charge distribution.

E = ${{\rho r} \over {3{\varepsilon _0}}}$

Now,   as E = constant

$\therefore$   $\rho$ r = constant

$\Rightarrow$   $\rho$ (r) $\propto$ ${1 \over r}$
2

### JEE Main 2016 (Online) 10th April Morning Slot

Figure shows a network of capacitors where the numbers indicates capacitances in micro Farad. The value of capacitance C if the equivalent capacitance between point A and B is to be 1 $\mu$F is :

A
${{31} \over {23}}\,\mu F$
B
${{32} \over {23}}\,\mu F$
C
${{33} \over {23}}\,\mu F$
D
${{34} \over {23}}\,\mu F$

## Explanation

Equivalent capacitance of 6 $\mu$F and 12 $\mu$F is = ${{6 \times 12} \over {12 + 6}}$ = 4 $\mu$F

Equivalent capacitance of 4$\mu$F and 4$\mu$F

is = 4 + 4 = 8 $\mu$F

New circuit is $\to$

equivalent capacitance of 1 $\mu$F and 8 $\mu$F is

= ${{1 \times 8} \over {8 + 1}}$ = ${8 \over 9}$ $\mu$F

Equivalent capacitance of 8 $\mu$F and 4 $\mu$F is

= ${{8 \times 4} \over {12}}$ = ${8 \over 3}$ $\mu$F

New circuit is $\to$

Equation capacitance of ${8 \over 9}$ $\mu$F and ${8 \over 3}$ $\mu$F is

= ${8 \over 9}$ + ${8 \over 3}$ = ${32 \over 9}$ $\mu$F

$\therefore$   Equivalent capacitance of AB is

CAB = ${{C \times {{32} \over 9}} \over {C + {{32} \over 9}}}$

Given that,

CAB = 1 $\mu$F

$\therefore$   1 = ${{C \times {{32} \over 9}} \over {C + {{32} \over 9}}}$

$\Rightarrow$   C + ${{{32} \over 9}}$ = ${{32C} \over 9}$

$\Rightarrow$   ${{23C} \over 9} = {{32} \over 9}$

$\Rightarrow$   C = ${{32} \over {23}}$ $\mu$F
3

### JEE Main 2017 (Offline)

A capacitance of 2 $\mu$F is required in an electrical circuit across a potential difference of 1.0 kV. A large number of 1 $\mu$F capacitors are available which can withstand a potential difference of not more than 300 V. The minimum number of capacitors required to achieve this is:
A
2
B
16
C
32
D
24

## Explanation

To get a capacitance of 2 μF arrangement of capacitors of capacitance 1μF as shown in figure 8 capacitors of 1μF in parallel with four such branches in series i.e., 32 such capacitors are required.

${1 \over {{C_{eq}}}} = {1 \over 8} + {1 \over 8} + {1 \over 8} + {1 \over 8}$

$\Rightarrow$ ${1 \over {{C_{eq}}}} = {1 \over 2}$

$\Rightarrow$ ${{C_{eq}} = 2}$
4

### JEE Main 2017 (Offline)

An electric dipole has a fixed dipole moment $\overrightarrow p$, which makes angle $\theta$ with respect to x-axis. When subjected to an electric field $\mathop {{E_1}}\limits^ \to = E\widehat i$ , it experiences a torque $\overrightarrow {{T_1}} = \tau \widehat k$ . When subjected to another electric field $\mathop {{E_2}}\limits^ \to = \sqrt 3 {E_1}\widehat j$ it experiences a torque $\mathop {{T_2}}\limits^ \to = \mathop { - {T_1}}\limits^ \to$ . The angle $\theta$ is:
A
90o
B
45o
C
30o
D
60o

## Explanation

Torque experienced by the dipole in an electric field,

$T$ = pE sin$\theta$

$\overrightarrow T = \overrightarrow p \times \overrightarrow E$

$\overrightarrow p = p\cos \theta \widehat i + p\sin \theta \widehat j$

$\mathop {{E_1}}\limits^ \to = E\widehat i$

$\overrightarrow {{T _1}} = \overrightarrow P \times {\overrightarrow E _1}$

= ($p\cos \theta \widehat i + p\sin \theta \widehat j$) $\times$ $E\left( {\widehat i} \right)$

= pE sin$\theta $$\left( { - \widehat k} \right) \mathop {{E_2}}\limits^ \to = \sqrt 3 {E_1}\widehat j \overrightarrow {{T _2}} = (p\cos \theta \widehat i + p\sin \theta \widehat j) \times \sqrt 3 {E_1}\widehat j = \sqrt 3 pE\cos \theta \left( {\widehat k} \right) Now given, \overrightarrow {{T _2}} = -\overrightarrow {{T _1}} \Rightarrow \sqrt 3 pE\cos \theta \left( {\widehat k} \right) = -pE sin\theta$$\left( { - \widehat k} \right)$

$\Rightarrow$ $\tan \theta = \sqrt 3$

$\Rightarrow$ $\theta$ = 60o