1

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

A uniform wire of length 1 and radius r has a resistance of 100 $\Omega$. It is recast into a wire of radius ${r \over 2}.$ The resistance of new wire will be :
A
1600 $\Omega$
B
400 $\Omega$
C
200 $\Omega$
D
100 $\Omega$

## Explanation

Resistance of a wire of length l and radius r is given by

R = ${{\rho l} \over A}$ = ${{\rho l} \over A} \times {A \over A} = {{\rho V} \over {{A^2}}} = {{\rho V} \over {{\pi ^2}{r^4}}}$

$\Rightarrow$ R $\propto$ ${1 \over {{r^4}}}$

$\therefore$ ${{{R_1}} \over {{R_2}}} = {\left( {{{{r_2}} \over {{r_1}}}} \right)^4}$

Given, R1 = 100 $\Omega$, r1 = r, r2 = ${r \over 2}$ , R2 = ?

$\therefore$ R2 = R1${\left( {{{{r_1}} \over {{r_2}}}} \right)^4}$ = 16R1 = 1600 $\Omega$
2

### 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
3

### 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
${{20} \over 3}\,\Omega$
4

### 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 $\Omega$