Initially at null deflection R2R1=32…(i) Finally at null deflection, when null point is shifted R2R1+10=1⇒R1+10=R2...(ii) Solving equations (i) and (ii) we get 32R2+10=R2 10=3R2⇒R2=30Ω &R1=20Ω Now if required resistance is R then 3030+R30×R=32 R=60Ω
In the experimental set up of metre bridge shown in the figure, the null point is obtaine data distance of 40 ~cm from A. If a 10 Ω resistor is…
JEE Main 2019 — Physics Mechanics
In the experimental set up of metre bridge shown in the figure, the null point is obtaine data distance of 40 cm from A. If a 10Ω resistor is connected in series with R1, the null point shifts by 10 cm. The resistance that should be connected in parallel with (R1+10)Ω such that the null point shifts back to its initial position is 
Held on 11 Jan 2019 · Verified 6 Jul 2026.
Options
- A
20Ω
- B
40Ω
- C
60Ω
- D
30Ω
Did you get this right?
Sign in to track your attempts and accuracy.
Your note
Sign in to keep a private note on this question. Nothing you write is ever public.
JEE Main Mechanics in other years
More JEE Main Mechanics questions
If $\epsilon, E$ and $t$ represent the free space permittivity, electric field and time respectively, then the unit of $\frac{\epsilon E}{t}$ will be :
When a part of a straight capillary tube is placed vertically in a liquid, the liquid raises upto certain height $h$. If the inner radius of the capillary tube, density of the liquid and surface tension of the liquid decrease by $1 \%$ each, then the height of the liquid in the tube will change by $\%$.
The velocity of a particle is given as $\vec{v} = -x\hat{i} + 2y\hat{j} - z\hat{k}$ m/s. The magnitude of acceleration at point $(1, 2, 4)$ is _______ m/s$^2$.
Suppose there is a uniform circular disc of mass $M \mathrm{~kg}$ and radius $r \mathrm{~m}$ shown in figure. The shaded regions are cut out from the disc. The moment of inertia of the remainder about the axis $A$ of the disc is given by $\frac{x}{256} M r^{2}$. The value of $x$ is $\_\_\_\_$. 
A soap bubble of surface tension $0.04 \mathrm{~N} / \mathrm{m}$ is blown to a diameter of 7 cm. If $(15000-x) \mu \mathrm{J}$ of work is done in blowing it further to make its diameter 14 cm, then the value of $x$ is $\_\_\_\_$. ($\pi=22 / 7$)
Other Physics topics for JEE Main
Keep practicing Mechanics
Work through every JEE Main Mechanics PYQ, year by year.