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### JEE Main 2017 (Online) 8th April Morning Slot

A potentiometer PQ is set up to compare two resistances as shown in the figure. The ammeter A in the circuit reads 1.0 A when two way key K3 is open. The balance point is at a length $\ell$1 cm from P when two way key K3 is plugged in between 2 and 1, while the balance point is at a length $\ell$2 cm from P when key K3 is plugged in between 3 and 1. The ratio of two resistances ${{{R_1}} \over {{R_2}}}$, is found to be :

A
${{{l_1}} \over {{l_1} + {l_2}}}$
B
${{{l_2}} \over {{l_2} - {l_1}}}$
C
${{{l_1}} \over {{l_1} - {l_2}}}$
D
${{{l_1}} \over {{l_2} - {l_1}}}$

## Explanation

When key K3 is plugged in between 1 and 2,

V1 = iR1 = x$l$1 . . . . . (1)

When key K3 is plugged in between 3 and 1,

V2 = i(R1 + R2) = x$l$2 . . . . (2)

On dividing (1) and (2),

${{{R_1}} \over {{R_1} + {R_2}}} = {{{l_1}} \over {{l_2}}}$

$\Rightarrow $$\,\,\, {{{R_1}} \over {\left( {R{}_1 + {R_2}} \right) - {R_1}}} = {{{l_1}} \over {{l_2} - {l_1}}} \Rightarrow$$\,\,\,$ ${{{R_1}} \over {{R_2}}}$ = ${{{l_1}} \over {{l_2} - {l_1}}}$
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### JEE Main 2017 (Online) 8th April Morning Slot

A 9 V battery with internal resistance of 0.5 $\Omega$ is connected across an infinite network as shown in the figure. All ammeters A1, A2, 3 and voltmeter V are ideal.
Choose correct statement.
A
Reading of A1 is 2 A
B
Reading of A1 is 18 A
C
Reading of   V   is  9 V
D
Reading of   V   is   7 V

## Explanation

Assume equivalent resistance of the infinite network is x.

The circuit can be redrawn like this,

Here 4 $\Omega$ and x $\Omega$ are parallel,

$\therefore\,\,\,$ ${1 \over {{x_{eq}}}}$ = ${1 \over 4}$ + ${1 \over x}$

$\Rightarrow $$\,\,\, xeq = {{4x} \over {4 + x}} \therefore\,\,\, Equivalent resistance between point A and B is , x = {{4x} \over {4 + x}} + 2 \Rightarrow$$\,\,\,$ x = ${{8 + 6x} \over {4 + x}}$

$\Rightarrow $$\,\,\, 4x + x2 = 8 + 6x \Rightarrow$$\,\,\,$ x2 $-$ 2x $-$ 8 = 0

$\Rightarrow$ $\,\,\,$ x = ${{2 \pm \sqrt {4 - 4\left( 1 \right)\left( { - 8} \right)} } \over 2}$

$\Rightarrow $$\,\,\,x = {{2 \pm 6} \over 2} \Rightarrow$$\,\,\,$ = 4 $\Omega$

(as negative value of x is not acceptable)

$\therefore\,\,\,$ Reading of Aonmeter, A1 = ${V \over {R + r}}$

A1 = ${9 \over {4\,\, + 0.5}}$ = 2 A
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### 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 $\Omega$
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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