JEE PYQ: Experimental Physics Question 4
Question 4 - 2021 (25 Feb 2021 Shift 1)
In the given circuit of potentiometer, the potential difference E across $AB$ (10 m length) is larger than $E_1$ and $E_2$ as well. For key $K_1$ (closed), the jockey is adjusted to touch the wire at point $J_1$ so that there is no deflection in the galvanometer. Now the first battery ($E_1$) is replaced by second battery ($E_2$) for working by making $K_1$ open and $E_2$ closed. The galvanometer gives then null deflection at $J_2$. The value of $\frac{E_1}{E_2}$ is $\frac{a}{b}$, where $a =$
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Answer: 1
Solution
$\frac{E_1}{E_2} = \frac{l_1}{l_2}$. From the figure, $l_1 = 380$ cm and $l_2 = 760$ cm. $\frac{380}{760} = \frac{1}{2}$, thus $a = 1$.
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