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 :

Correct answer: A. 3 . 5 % Given that, R = (V)/(I) According to error analysis (dR)/(R) = (dV)/(V) + (dI)/(I) ⇒ (dR)/(R) = (5)/(200) + (0 . 2)/(20) ⇒ (dR)/(R) = (7)/(200) Thus, the percentage error can be calculated as follows: % error = (dR)/(R) × 100 = (7)/(200) × 100 = 3 . 5 %

JEE Main 202429 JanuaryMorning Shift

Exam: JEE Main
Year: 2024
Date: 29 January
Shift: Morning Shift
Subject: Physics
Chapter: Mathematics in Physics

JEE Main 2024 · 29 January · Morning Shift · Physics · Mathematics in Physics

The resistance R = (V)/(I) , where V = ( 200 ± 5 ) V and I = ( 20 ±...

The resistance

R = \dfrac{V}{I}

, where

V = ( 200 \pm 5 ) V

and

I = ( 20 \pm 0 . 2 ) A

, the percentage error in the measurement of

R

is :

Options

A $3 . 5 %$
B $7 %$
C $3 %$
D $5 . 5 %$

Correct answer

A. $3 . 5 %$

Step-by-step solution

Given that, $R = \dfrac{V}{I}$

According to error analysis

$\dfrac{dR}{R} = \dfrac{dV}{V} + \dfrac{dI}{I} ⇒ \dfrac{dR}{R} = \dfrac{5}{200} + \dfrac{0 . 2}{20} ⇒ \dfrac{dR}{R} = \dfrac{7}{200}$

Thus, the percentage error can be calculated as follows:

$% \text{error} = \dfrac{dR}{R} \times 100 = \dfrac{7}{200} \times 100 = 3 . 5 %$

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