In the arrangement shown in the figure when the switch S2 is open, the galvanometer shows no deflection for l=L2 . When the switch S2 is closed, the galvanometer shows no deflection for l=512L. The internal resistance of 6V cell, and the emf E of the other battery are respectively (internal resistance of cell of emf E is negligible)

When S2 is open, l= L/2 ∴E=6VL/2L=12 Vwhen S2 is closed V10=12L×512L=5 VBut, V10=10×610+r10×610+r=5 ⇒ r=2 Ω

In the arrangement shown in the figure when the switch S2 is open, the galvanometer shows no deflection for l=L2 . When the switch S2 is closed, the galvanometer shows no deflection for l=512L. The internal resistance of 6V cell, and the emf E of the other battery are respectively (internal resistance of cell of emf E is negligible)

When S2 is open, l= L/2 ∴E=6VL/2L=12 Vwhen S2 is closed V10=12L×512L=5 VBut, V10=10×610+r10×610+r=5 ⇒ r=2 Ω

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In the arrangement shown in the figure when the switch $S_{2}$ is open, the galvanometer shows no deflection for $l = \frac{L}{2}$ . When the switch $S_{2}$ is closed, the galvanometer shows no deflection for $l = \frac{5}{12} L$ . The internal resistance of 6V cell, and the emf E of the other battery are respectively (internal resistance of cell of emf E is negligible)

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2 $Ω$ , 12V

2 $Ω$ , 24V

3 $Ω$ , 8V

3 $Ω$ , 12V

answer is B.

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Detailed Solution

When $S_{2}$ is open, l= L/2
$∴ E = (\frac{6 V}{L / 2}) L = 12 V$
when $S_{2}$ is closed
$V_{10} = (\frac{12}{L}) \times \frac{5}{12} L = 5 V$
$But, V_{10} = 10 \times (\frac{6}{10 + r})$
$10 \times (\frac{6}{10 + r}) = 5 \Rightarrow r = 2 Ω$

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