1

### JEE Main 2016 (Online) 9th April Morning Slot

Three capacitors each of 4 $\mu$F are to be connected in such a way that the effective capacitance is 6 $\mu$F. This can be done by connecting them :
A
all in series
B
two in series and one in parallel
C
all in parallel
D
two in parallel and one in series

## Explanation

(a)   ${1 \over {{C_{eq}}}} = {1 \over 4} + {1 \over 4} + {1 \over 4} = {3 \over 4}$

$\Rightarrow$   ${C_{eq}} = {4 \over 3}\,\mu F$

(b)   ${C_{eq}} = {{4 \times 4} \over {4 + 4}} + 4 = 6\mu F$

(c)   ${C_{eq}} = 4 + 4 + 4 = 12\,\mu F$

(d)   ${C_{eq}} = {{\left( {4 + 4} \right) \times 4} \over {\left( {4 + 4} \right) + 4}} = {8 \over 3}\mu F$
2

### JEE Main 2016 (Online) 9th April Morning Slot

The potential (in volts) of a charge distribution is given by.

V(z) = 30 $-$ 5x2 for $\left| z \right|$ $\le$ 1 m.
V(z) = 35 $-$ 10 $\left| z \right|$ for $\left| z \right|$ $\ge$1 m.

V(z) does not depend on x and y. If this potential is generated by a constant charge per unit volume ${\rho _0}$ (in units of ${\varepsilon _0}$) which is spread over a certain region, then choose the correct statement.
A
${\rho _0}$ = 10 ${\varepsilon _0}$ for $\left| z \right|$ $\le$ 1 m and ${\rho _0} = 0$ elsewhere
B
${\rho _0}$ = 20 ${\varepsilon _0}$ in the entire region
C
${\rho _0}$ = 40 ${\varepsilon _0}$ in the entire region
D
${\rho _0}$ = 20 ${\varepsilon _0}$ for $\left| z \right|$ $\le$ 1 m and ${\rho _0} = 0$ elsewhere

## Explanation

We know,

E(z) = $-$ ${{dv} \over {dz}}$

$\therefore$   E(z) = $-$ 10 z for $\left| z \right| \le 1$ m

and E(z) = 10 for $\left| z \right| \ge 1$ m

$\therefore$   The source is an infinity large non conducting thick of thickness z = 2 m.

$\therefore$   E = ${\sigma \over {2{\varepsilon _0}}}$ = ${{\rho \left( 2 \right)} \over {2{\varepsilon _0}}}$ = ${\rho \over {{\varepsilon _0}}}$

$\therefore$   ${\rho \over {{\varepsilon _0}}}$ = 10

$\Rightarrow$   $\rho$ = ${10\,{\varepsilon _0}}$
3

### 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) $\propto$ ${1 \over r}$
4

### 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