1

### JEE Main 2019 (Online) 10th January Evening Slot

The Wheatstone bridge shown in figure, here, gets balanced when the carbon resistor used as R1 has the colour code (Orange, Red, Brown). The resistors R2 and R4 are 80$\Omega$ and 40$\Omega$, respectively. Assuming that the colour code for the carbon resistors gives their accurate values, the colour code for the carbon resistor, used as R3, would be - A
Brown, Blue, Brown
B
Grey, Black, Brown
C
Red, Green, Brown
D
Brown, Blue, Black

## Explanation

R1 = 32 $\times$ 10 = 320

for wheat stone bridge

$\Rightarrow$  ${{{R_1}} \over {{R_3}}} = {{{R_2}} \over {{R_4}}}$

${{320} \over {{R_3}}} = {{80} \over {40}}$

${R_3} = 160$

$\therefore$  Correct answer is Brown  Blue  Brown
2

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

Three charges Q, + q and + q are placed at the vertices of a right-angle isosceles triangle as shown below. The net electrostatic energy of the configuration is zero, if the value of Q is : A
${{ - q} \over {1 + \sqrt 2 }}$
B
+ q
C
$-$ 2q
D
${{ - \sqrt 2 q} \over {\sqrt 2 + 1}}$

## Explanation U = K$\left[ {{{{q^2}} \over a} + {{Qq} \over a} + {{Qq} \over {a\sqrt 2 }}} \right] = 0$

$\Rightarrow$  q = $-$ Q$\left[ {1 + {1 \over {\sqrt 2 }}} \right]$

$\Rightarrow$  Q = ${{ - q\sqrt 2 } \over {\sqrt 2 + 1}}$
3

### 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 ${\varepsilon _2} = 0.3$
4

### 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)

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

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

& ${R_1} = 20\Omega$

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

$R = 60\,\Omega$