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JEE Main 202622 JanuaryMorning Shift

Exam: JEE Main
Year: 2026
Date: 22 January
Shift: Morning Shift
Subject: Physics
Chapter: Current Electricity

JEE Main 2026 · 22 January · Morning Shift · Physics · Current Electricity

A meter bridge with two resistances R_ 1 and R_ 2 as shown in...

A meter bridge with two resistances

R_{1}

and

R_{2}

as shown in figure was balanced (null point) at 40 cm from the point

P

. The null point changed to 50 cm from the point

P

, when

16 \Omega

resistance is connected in parallel to

R_{2}

. The values of resistances

R_{1}

and

R_{2}

are

\_\_\_\_

.

A meter bridge with two resistances R_ 1 and R_ 2 as shown in figure was balanced (null point) at 40 cm from the point

Options

A $R_{2}=4 \Omega, R_{1}=\frac{4}{3} \Omega$
B $R_{2}=16 \Omega, R_{1}=\frac{16}{3} \Omega$
C $R_{2}=12 \Omega, R_{1}=\frac{12}{3} \Omega$
D $R_{2}=8 \Omega, R_{1}=\frac{16}{3} \Omega$

Correct answer

D. $R_{2}=8 \Omega, R_{1}=\frac{16}{3} \Omega$

Step-by-step solution

Initial balance at 40 cm:

\frac{R_1}{R_2} = \frac{40}{60} = \frac{2}{3}

...(i)
With

16\,\Omega

parallel to

R_2

, null point at 50 cm:

\frac{R_1}{\frac{16R_2}{16+R_2}} = \frac{50}{50} = 1 \Rightarrow R_1 = \frac{16R_2}{16+R_2}

...(ii)
From (i):

R_1 = \frac{2}{3}R_2

.
Substituting in (ii):

\frac{2}{3}R_2 = \frac{16R_2}{16+R_2}

.

\frac{2}{3} = \frac{16}{16+R_2} \Rightarrow 2(16+R_2) = 48 \Rightarrow R_2 = 8\,\Omega

.

R_1 = \frac{16}{3}\,\Omega

.

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