SATHEE: current-electricity Question 48

Current Electricity

current-electricity Question 48

Question: Q. 1. A resistance $R$ is to be measured using a meter bridge. Student chooses the standard resistance $S$ to be $100 Ω$ . He finds the null point at $l_{1} = 2.9 cm$ . He is told to attempt to improve the accuracy. Which of the following is a useful way?

(a) He should measure $l_{1}$ more accurately.

(b) He should change $S$ to $1000 Ω$ and repeat the experiment.

(d) He should give up hope of a more accurate measurement with a meter bridge.

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Solution:

Ans. Correct option : (c)

Explanation : To calculate resistance, $R$

$\begin{aligned} R & = S [\frac{l_{1}}{(100 - l_{1})}] & = 100 [\frac{2.9}{97.1}] & = 2.98 Ω \end{aligned}$

So to get balance point near to $50 cm$ (middle) we have to take $S = 3 Ω$ , as here $R : S = 2.9 : 97.1$ implies that $S$ is nearly 33 times to $R$ . In order to make ratio $R$ and $S = 1 : 1$ , we must take the resistance $S = 3 Ω$ .

Current Electricity

current-electricity Question 48

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Current Electricity

current-electricity Question 48

Table of Contents

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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