SATHEE: Current Electricity 4 Question 9

Current Electricity 4 Question 9

9. The resistance of the meter bridge $A B$ in given figure is $4 Ω$ . With a cell of emf $ε = 0.5 V$ and rheostat resistance $R_{h} = 2 Ω$ . The null point is obtained at some point $J$ . When the cell is replaced by another one of $emf ε = ε_{2}$ , the same null point $J$ is found for $R_{h} = 6 Ω$ . The emf $ε_{2}$ is

(2019 Main, 11 Jan I)

(a) $0.6 V$

(b) $0.3 V$

(d) $0.4 V$

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Solution:

Let length of null point ’ $J$ ’ be ’ $x$ ’ and length of the potentiometer wire be ’ $L$ ‘. In first case, current in the circuit

$I_{1} = \frac{6}{4 + 2} = 1 A$

$∴$ Potential gradient $= I \times R = \frac{1 \times 4}{L}$

$\Rightarrow$ Potential difference in part ’ $A J$ '

Given,

$\begin{aligned} = & \frac{1 \times 4}{L} \times x = ε_{1} \\ ε_{1} & = 0.5 = \frac{4 x}{L} or \frac{x}{L} = \frac{1}{8} \end{aligned}$

In second case, current in the circuit

$I_{2} = \frac{6}{4 + 6} = 0.6 A$

$∴$ Potential gradient $= \frac{0.6 \times 4}{L}$

$\Rightarrow$ Potential difference in part ’ $A J$ '

$\begin{aligned} = \frac{0.6 \times 4}{L} \times x = ε_{2} \\ \Rightarrow & ε_{2} & = \frac{0.6 \times 4}{L} \times \frac{L}{8} \\ \Rightarrow & ε_{2} & = 0.3 V \end{aligned}$

Current Electricity 4 Question 9

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Current Electricity 4 Question 9

9. The resistance of the meter bridge AB in given figure is 4Ω. With a cell of emf ε=0.5V and rheostat resistance Rh=2Ω. The null point is obtained at some point J. When the cell is replaced by another one of emfε=ε2, the same null point J is found for Rh=6Ω. The emf ε2 is

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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