1

### JEE Main 2017 (Online) 9th April Morning Slot

In a meter bridge experiment resistances are connected as shown in the figure. Initially resistance P = 4 $\Omega$ and the neutral point N is at 60 cm from A. Now an unknown resistance R is connected in series to P and the new position of the neutral point is at 80 cm from A. The value of unknown resistance R is : A
${{33} \over 5}\,\Omega$
B
6 $\,\Omega$
C
7 $\,\Omega$
D
${{20} \over 3}\,\Omega$
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### JEE Main 2017 (Online) 9th April Morning Slot

The figure shows three circuits I, II and III which are connected to a 3V battery. If the powers dissipated by the configurations I, II and III are P1 , P2 and P3 respectively, then : A
P1 > P2 > P3
B
P1 > P3 > P2
C
P2 > P1 > P3
D
P3 > P2 > P1
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### JEE Main 2018 (Offline)

On interchanging the resistances, the balance point of a meter bridge shifts to the left by 10 cm. The resistance of their series combination is 1 k$\Omega$. How much was the resistance on the left slot before interchanging the resistances?
A
910 $\Omega$
B
990 $\Omega$
C
505 $\Omega$
D
550 $\Omega$

## Explanation ${X \over l}$ = ${{1000 - X} \over {100 - l}}$ . . . . . (1)

When X and Y are interchanged. ${{1000 - X} \over {l - 10}}$ = ${X \over {100 - l + 10}}$

$\Rightarrow $$\,\,\, {{1000 - X} \over {l - 10}} = {X \over {110 - l}} . . . . . (2) From (1) and (2) we get, {l \over {100 - l}} = {{110 - l} \over {l - 10}} \Rightarrow$$\,\,\,$ $l$2 $-$ 10 = 11000 $-$ 100$l$ $-$ 110$l$ + $l$2

$\Rightarrow $$\,\,\, 200l = 11000 l = 55 cm Putting this value of l, in equation (1) {X \over {55}} = {{1000 - X} \over {100 - 55}} \Rightarrow$$\,\,\,$ 45X = 55000 $-$ 55X

$\Rightarrow $$\,\,\, 100X = 55000 \Rightarrow$$\,\,\,$ X = 550 $\Omega$
4

### JEE Main 2018 (Offline)

Two batteries with e.m.f 12 V and 13 V are connected in parallel across a load resistor of 10 $\Omega$. The internal resistances of the two batteries are 1 $\Omega$ and 2 $\Omega$ respectively. The voltage across the load lies between :
A
11.7 V and 11.8 V
B
11.6 V and 11.7 V
C
11.5 V and 11.6 V
D
11.4 V and 11.5 V

## Explanation Let potential at S, T, R = V and potential at P, Q, U, = 0

Using kirchhoff's law at P :

Current at P is ,

${{V - 12} \over 1} + {{V - 13} \over 2} + {{V - 0} \over {10}} = 0$

$\Rightarrow $$\,\,\, {V \over 1} + {V \over 2} + {V \over 10} = 12 + {{13} \over 2} \Rightarrow$$\,\,\,$ ${{10V + 5V + V} \over {10}}$ = ${{37} \over 2}$

$\Rightarrow $$\,\,\, {{16V} \over {10}} = {{37} \over 2} \Rightarrow$$\,\,\,$ V = 11.56 Volt.