1

### JEE Main 2019 (Online) 11th January Morning Slot

The resistance of the meter bridge AB in given figure is 4 $\Omega$. With a cell of emf $\varepsilon$ = 0.5 V and rheostat resistance Rh = 2 $\Omega$ the null point is obtained at some point J. When the cell is replaced by another one of emf $\varepsilon$ = $\varepsilon$2 the same null point J is found for Rh = 6 $\Omega$. The emf $\varepsilon$2 is, : A
0.3 V
B
0.6 V
C
0.5 V
D
0.4 V

## Explanation

Potential gradient with Rh = 2$\Omega$

is $\left( {{6 \over {2 + 4}}} \right) \times {4 \over L} = {{dV} \over {dL}};$   L $=$ 100 cm

Let null point be at $\ell$ cm

thus $\varepsilon$1 $=$ 0.5V $=$ $\left( {{6 \over {2 + 4}}} \right) \times {4 \over L} \times \ell$      . . .(1)

Now with Rh $=$ 6$\Omega$ new potential gradient is

$\left( {{6 \over {4 + 6}}} \right) \times {4 \over L}$ and at null point

$\left( {{6 \over {4 + 6}}} \right)\left( {{4 \over L}} \right) \times \ell = {\varepsilon _2}$                          . . .(2)

dividing equation (1) by (2) we get

${{0.5} \over {{\varepsilon _2}}} = {{10} \over 6}$ thus ${\varepsilon _2} = 0.3$
2

### JEE Main 2019 (Online) 11th January Evening Slot

In the experimental set up of metre bridge shown in the figure, the null point is obtained at a distance of 40 cm from A. If a 10 $\Omega$ resistor is connected in series with R1, the null point shifts by 10 cm. The resistance that should be connected in parallel with (R1 + 10) $\Omega$ such that the null point shifts back to its initial position is : A
40 $\Omega$
B
30 $\Omega$
C
20 $\Omega$
D
60 $\Omega$

## Explanation

${{{R_1}} \over {{R_2}}} = {2 \over 3}\,\,$                                      . . .(i)

${{{R_1} + 10} \over {{R_2}}} = 1 \Rightarrow {R_1} + 10 = {R_2}$     . . .(ii)

${{2{R_2}} \over 3} + 10 = {R_2}$

$10 = {{{R_2}} \over 3} \Rightarrow {R_2} = 30\Omega$

& ${R_1} = 20\Omega$

${{{{30 \times R} \over {30 + R}}} \over {30}} = {2 \over 3}$

$R = 60\,\Omega$
3

### JEE Main 2019 (Online) 12th January Morning Slot

Determine the electric dipole moment of the system of the three charges, placed on the vertices of an equilateral triangle, as shown in the figure : A
$2q\ell \widehat j$
B
$\left( {q\ell } \right){{\widehat i + \widehat j} \over {\sqrt 2 }}$
C
$\sqrt 3 \,q\ell {{\widehat j - \widehat i} \over {\sqrt 2 }}$
D
$- \sqrt 3 \,q\ell \widehat j$

## Explanation $\left| {{P_1}} \right| =$ q(d)

$\left| {{P_2}} \right| =$ qd

|Resultant| $=$ 2 P cos30o

2 qd$\left( {{{\sqrt 3 } \over 2}} \right)$ = $\sqrt 3$ qd
4

### JEE Main 2019 (Online) 12th January Morning Slot

The galvanometer deflection, when key K1 is closed but K2 is open, equals $\theta$0 (see figure). On closing K2 also and adjusting R2 to 5$\Omega$, the deflection in galvanometer becomes ${{{\theta _0}} \over 5}.$ . The resistance of the galvanometer is, then, given by [Neglect the internal resistance of battery] : A
5 $\Omega$
B
25 $\Omega$
C
12 $\Omega$
D
22 $\Omega$

## Explanation

case I :

ig = ${E \over {220 + {R_g}}}$ = C$\theta$0     . . .(i)

Case II :

ig = $\left( {{E \over {220 + {{5{R_g}} \over {5 + {R_g}}}}}} \right)$ $\times$ ${5 \over {\left( {{R_g} + 5} \right)}}$ = ${{C{\theta _0}} \over 5}$       . . .(ii)

$\Rightarrow$  ${{5E} \over {225{R_g} + 1100}}$ = ${{C{\theta _0}} \over 5}$       . . .(ii)

${E \over {220 + {R_g}}} = C\theta$       . . .(i)

$\Rightarrow$  ${{225{R_g} + 1100} \over {1100 + 5{R_g}}} = 5$

$\Rightarrow$  5500 + 25Rg = 225Rg + 1100

200Rg = 4400

Rg = 22$\Omega$