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Two identical potentiometer wires w_{1} and w_{2} of equal length l, connected to a battery of emf \varepsilon _{p} and internal resistance 1\; \Omega through two switches S_{1} and S_{2} . A battery of emf \varepsilon is balanced on these potentiometer wires one by one. If potentiometer wire w_{1} is of resistance 2\; \Omega and balancing length is  \frac{l}{2} on it, when only S_{1} is closed and S_{2} is open. On closing S_{2} and opening S_{1} the balancing length On w_{2} is found to be \frac{2l}{3} ,then the resistance (in \Omega) of potentiometer wire w_2 .

Option: 1

1\; \Omega


Option: 2

3\: \Omega


Option: 3

4\: \Omega


Option: 4

5\: \Omega


Answers (1)

best_answer

For \: \: w_1, \varepsilon=\frac{l}{2}\left[\left(\frac{\varepsilon_p}{1+2}\right) \frac{2}{1}\right] _______________(i)

For \: \: w_2, \varepsilon=\frac{2l}{3}\left[\left(\frac{\varepsilon_p}{1+R}\right) \frac{R}{1}\right] ___________(ii)

Dividing eq. (i) by (ii) and on solving, we get, Resistance of wire w_2=1 \Omega

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SANGALDEEP SINGH

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