Resistances in the two gaps of a meter bridge are 10 ohm and 30 ohm, respectively. If the resistances are interchanged the balance point shifts by

Resistances in the two gaps of a meter bridge are 10 ohm and 30 ohm, respectively. If the resistances are interchanged the balance point shifts by

1. A

33.3 cm

2. B

66.67 cm

3. C

25 cm

4. D

50 cm

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Solution:

$\begin{array}{l}\mathrm{S}=\text{\hspace{0.17em}}\left(\frac{100-\mathrm{l}}{\mathrm{l}}\right)\text{\hspace{0.17em}\hspace{0.17em}}.\text{\hspace{0.17em}}\mathrm{R}\\ \mathrm{Initially},\text{\hspace{0.17em}\hspace{0.17em}}30\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\left(\frac{100-\mathrm{l}}{\mathrm{l}}\right)\text{\hspace{0.17em}}×\text{\hspace{0.17em}\hspace{0.17em}}10\text{\hspace{0.17em}\hspace{0.17em}}⇒\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{l}=25\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{cm}\\ \mathrm{Finally},\text{\hspace{0.17em}\hspace{0.17em}}10\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\left(\frac{100-\mathrm{l}}{\mathrm{l}}\right)\text{\hspace{0.17em}}×\text{\hspace{0.17em}\hspace{0.17em}}30\text{\hspace{0.17em}\hspace{0.17em}}⇒\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{l}=75\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{cm}\end{array}$

So, shift = 50 cm.