1

### 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
2

### 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 $\Omega$
3

### 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.
4

### JEE Main 2018 (Online) 15th April Morning Slot In a meter bridge, as shown in the figure, it is given that resistance $Y = 12.5\,\,\Omega$ and that the balance is obtained at a distance $39.5$ $cm$ from end $A$ (by Jockey J). After interchanging the resistances $X$ and $Y$, a new balance point is found at a distance ${l_2}$ from end $A.$ What are the value of $X$ and ${l_2}$ ?
A
$8.16\,\,\Omega$ and $60.5$ $cm$
B
$19.15\,\,\Omega$ and $39.5$ $cm$
C
$8.16\,\,\Omega$ and $39.5$ $cm$
D
$19.15\,\,\Omega$ and $60.5$ $cm$

## Explanation

For balanced meter bridge,

${X \over {39.5}}$ = ${Y \over {\left( {100 - 39.5} \right)}}$

$\Rightarrow$$\,\,\,\,$ X = ${{12.5 \times 39.5} \over {60.5}}$

= 8.16 $\Omega$

Now, when we interchange X and Y then

${Y \over {{l_2}}} = {X \over {100 - {l_2}}}$

$\Rightarrow$ ${{12.5} \over {{l_2}}} = {{8.16} \over {100 - {l_2}}}$

$\Rightarrow$20.66 ${{l_2}}$ = 1250

$\Rightarrow$ ${{l_2}}$ = 60.5 $\Omega$