A potentiometer is connected across A and B and a balance is obtained at 64.0 cm. When the potentiometer lead at B is moved to C, a balance is found at 8.0 cm. If the potentiometer is now connected across B and C, a balance will be found at
A
8.0 cm
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B
56.0 cm
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C
64.0 cm
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D
72.0 cm
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Solution
The correct option is B 56.0 cm Potential difference V=kl where k= potential gradient. Here, V1=kl1=64k and V1−V2=8k 64k−kl2=8k 64−l2=8⇒l2=64−8=56cm