1

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

In a Wheatstone bridge(see fig.), Resistances P and Q are approximately equal. When R = 400 $\Omega$, the bridge is balanced. On interchanging P and Q, the value of R, for balance, is 405 $\Omega$. The value of X is close to :

A
402.5 ohm
B
401.5 ohm
C
403.5 ohm
D
404.5 ohm

## Explanation

For a balanced bridge,

${P \over Q} = {{400} \over X}$ ......(i)

After interchanging P and Q,

${Q \over P} = {{405} \over X}$ ......(ii)

Multiplying equation (i) and (ii), we get

X2 = 400$\times$405

$\Rightarrow$ X = $\sqrt {400 \times 405}$ = 402.5 $\Omega$
2

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

An electric field of 1000 V/m is applied to an electric dipole at angle of 45o. The value of electric dipole moment is 10–29 C.m. What is the potential energy of the electric dipole?
A
- 7 $\times$ 10–27 J
B
$-$ 9 $\times$ 10–20 J
C
$-$ 10 $\times$ 10–29 J
D
$-$ 20 $\times$ 10–18 J

## Explanation

U = $-$ $\overrightarrow P .\overrightarrow E$

= $-$ PE cos $\theta$

= $-$ (10$-$29) (103) cos 45o

= $-$ 0.707 $\times$ 10$-$26 J

= $-$ 7 $\times$ 10$-$27 J.
3

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

A galvanometer having a resistance of 20 $\Omega$ and 30 divisions on both sides has figure of merit 0.005 ampere/division. The resistance that should be connected in series such that it can be used as a voltmeter upto 15 volt, is:
A
120 $\Omega$
B
125 $\Omega$
C
80 $\Omega$
D
100 $\Omega$

## Explanation

Rg = 20$\Omega$

NL = NR = N = 30

FOM = ${1 \over \phi }$ = 0.005 A/Div.

Current sentivity = CS = $\left( {{1 \over {0.005}}} \right)$ = ${\phi \over {\rm I}}$

${\rm I}$gmax = 0.005 $\times$ 30

= 15 $\times$ 10$-$2 = 0.15

15 = 0.15 [20 + R]

100 = 20 + R

R = 80
4

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

A copper wire is wound on a wooden frame, whose shape is that of an equilateral triangle. If the linear dimension of each side of the frame is increased by a factor of 3, keeping the number of turns of the coil per unit length of the frame the same, then the self inductance of the coil:
A
decreases by a factor of $9\sqrt 3$
B
increases by a factor of 27
C
decreases by a factor of 9
D
increases by a factor of 3

## Explanation

Total length L will remain constant

L = (3a) N        (N = total turns)

and length of winding = (d) N

(d = diameter of wire)

self inductance = $\mu$0n2A$\ell$

= $\mu$0n2$\left( {{{\sqrt 3 {a^2}} \over 4}} \right)$ dN

$\propto$ a2 N $\propto$ a

So self inductance will become 3 times