1

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

Charges –q and +q located at A and B, respectively, constitude an electric dipole. Distance AB = 2a, O is the mid point of the dipole and OP is perpendicular to AB. A charge Q is placed at P where OP = y and y >> 2a. The charge Q experiences an electrostatic force F. If Q is now moved along the equatorial line to P' such that OP' = $\left( {{y \over 3}} \right)$, the force on Q will be close to - $\left( {{y \over 3} > > 2a} \right)$ A
9F
B
3F
C
F/3
D
27F

## Explanation

Electric field of equitorial plane of dipole

$= - {{K\overrightarrow P } \over {{r^3}}}$

$\therefore$  At P, F $= - {{K\overrightarrow P } \over {{r^3}}}$Q.

At   P1 , F1 $= - {{K\overrightarrow P Q} \over {{{\left( {r/3} \right)}^3}}} = 27F.$
2

### 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
3

### 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 = ${{ - q\sqrt 2 } \over {\sqrt 2 + 1}}$
4

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