Given that, R=IV
According to error analysis
RdR=VdV+IdI⇒RdR=2005+200.2⇒RdR=2007
Thus, the percentage error can be calculated as follows:
The resistance R= V I, where V=(200± 5)V and I=(20± 0.2)A, the percentage error in the measurement of R is :
JEE Main 2024 — Physics Mechanics
The resistance R=IV, where V=(200±5)V and I=(20±0.2)A, the percentage error in the measurement of R is :
Held on 29 Jan 2024 · Verified 1 Jul 2026.
Options
- A
3.5
- B
7
- C
3
- D
5.5
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.