The resistance in the two arms of the meter bridge are 5Ω and RΩ respectively, balance point is at l1 from left end. When the resistance R is shunted with an equal resistance the new balance point is 1.6l1. The resistance R is
A
10Ω
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B
15Ω
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C
20Ω
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D
25Ω
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Solution
The correct option is B15Ω For first case: 5(100−l1) = Rl1 ⇒500−5l1 = Rl1…(1)
For second case:
Resistance in second arm will become R2
so, 5(100−1.6l1)= R2(1.6l1) ⇒500−8l1=0.8Rl1…(2)
Subtracting equations (1)−(2) −5l1+8l1=Rl1−0.8Rl1 3=0.2R ⇒R=15Ω