SATHEE: Current Electricity 4 Question 7

Current Electricity 4 Question 7

7. In the experimental set up of meter bridge shown in the figure, the null point is obtained at a distance of $40 c m$ from $A$ . If a $10 Ω$ resistor is connected in series with $R_{1}$ , the null point shifts by $10 c m$ .

The resistance that should be connected in parallel with $(R_{1} + 10) Ω$ such that the null point shifts back to its initial position is

(Main 2019, 11 Jan II)

(a) $60 Ω$

(b) $20 Ω$

(d) $40 Ω$

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Solution:

For meter bridge, if balancing length is $l c m$ , then in first case, $\frac{R_{1}}{R_{2}} = \frac{l}{(100 - l)}$

It is given that, $l = 40 c m$ ,

$\begin{aligned} So, & \frac{R_{1}}{40} & = \frac{R_{2}}{100 - 40} \\ or & \frac{R_{1}}{R_{2}} & = \frac{2}{3} \end{aligned}$

In second case, $R_{1}^{'} = R_{1} + 10$ , and balancing length is now $50 c m$ then

$\begin{matrix} \frac{R_{1} + 10}{50} = \frac{R_{2}}{(100 - 50)} \\ R_{1} + 10 = R_{2} \end{matrix}$

Substituting value of $R_{2}$ from (ii) to (i) we get,

$\begin{aligned} or & \frac{R_{1}}{10 + R_{1}} & = \frac{2}{3} \\ \Rightarrow & 3 R_{1} & = 20 + 2 R_{1} \\ or & R_{1} & = 20 Ω \\ \Rightarrow & R_{2} & = 30 Ω \end{aligned}$

Let us assume the parallel connected resistance is $x$ .

Then equivalent resistance is $\frac{x (R_{1} + 10)}{x + R_{1} + 100}$

So, this combination should be again equal to $R_{1}$ .

$\begin{aligned} \frac{(R_{1} + 10) x}{R_{1} + 10 + x} & = R_{1} \\ \Rightarrow & \frac{30 x}{30 + x} & = 20 \\ or & 30 x & = 600 + 20 x \\ or & x & = 60 Ω \end{aligned}$

Current Electricity 4 Question 7

7. In the experimental set up of meter bridge shown in the figure, the null point is obtained at a distance of $40 c m$ from $A$ . If a $10 Ω$ resistor is connected in series with $R_{1}$ , the null point shifts by $10 c m$ .

Solution:

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Current Electricity 4 Question 7

7. In the experimental set up of meter bridge shown in the figure, the null point is obtained at a distance of 40cm from A. If a 10Ω resistor is connected in series with R1, the null point shifts by 10cm.

Solution:

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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7. In the experimental set up of meter bridge shown in the figure, the null point is obtained at a distance of $40 c m$ from $A$ . If a $10 Ω$ resistor is connected in series with $R_{1}$ , the null point shifts by $10 c m$ .