1
MCQ (Single Correct Answer)

### JEE Main 2019 (Online) 9th January Evening Slot

In the given circuit the the internal resistance of the 18 V cell is negligible. If R1 = 400 $\Omega$, R3 = 100 $\Omega$ and R4 = 500 $\Omega$ and the reading of an ideal voltmeter across R4 is 5V, then the value of R2 will be :

A
300 $\Omega$
B
450 $\Omega$
C
550 $\Omega$
D
230 $\Omega$

## Explanation

Voltage accross resistance R4 = 5 V

$\therefore$   IR4 = 5 V

$\Rightarrow$   500 $\times$ I = 5

$\Rightarrow$   I = ${1 \over {100}}$ A

$\therefore$   Voltage across resistor R3 = ${1 \over {100}}\left( {100} \right)$ = 1 A

$\therefore$   Total drop in resistance R3 and R4 = 5 + 1 = 6V

So, voltage accross R2 resistance is also 6V as R3, R4 and R2 are in parallel

$\therefore$   Voltage accross R1 resistor R1 resistor = 18 $-$ 6 = 12 V

$\therefore$   Current through R1 resistor = ${{12} \over {400}}$ = ${3 \over {100}}$ A

$\therefore$   Current through R2 resistor

= ${3 \over {100}} - {1 \over {100}}$

= ${2 \over {100}}$ A

$\therefore$   $\left( {{2 \over {100}}} \right)$ R2 = 6

$\Rightarrow$   R2 = 300 $\Omega$
2
MCQ (Single Correct Answer)

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

Two electric dipoles, A, B with respective dipole moments ${\overrightarrow d _A} = - 4qai$ and ${\overrightarrow d _B} = - 2qai$ are placed on the x-axis with a separation R, as shown in the figure. The distance from A at which both of them produce the same potential is -

A
${{\sqrt 2 R} \over {\sqrt 2 + 1}}$
B
${R \over {\sqrt 2 + 1}}$
C
${{\sqrt 2 R} \over {\sqrt 2 - 1}}$
D
${R \over {\sqrt 2 - 1}}$

## Explanation

V $= {{4qa} \over {\left( {R + x} \right)}} = {{2qa} \over {\left( {{x^2}} \right)}}$

$\sqrt 2 x = R + x$

$x = {R \over {\sqrt 2 - 1}}$

dist $= {R \over {\sqrt 2 - 1}} + R = {{\sqrt 2 R} \over {\sqrt 2 - 1}}$
3
MCQ (Single Correct Answer)

### 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 $= - {{K\overrightarrow P Q} \over {{{\left( {r/3} \right)}^3}}} = 27F.$
4
MCQ (Single Correct Answer)

### 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

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