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A meter bridge is balanced by putting `20pi` resistance in the left gap and `40pi` in the right, gap, if `40pi` resistance is now shunted with `40pi` resistance the shift in the null point towards right is nearlyA. 16.67 cmB. 50 cmC. 25 cmD. 70.67 cm |
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Answer» Correct Answer - A For balanced bridge, `(X)/(R)=(l_(x))/(l_(r)) rArr (20)/(40)=(l_(x))/(l_(r))` `:.2l_(x)=l_(r)=(100-l_(x)) " " :.l_(x)=33.33 cm` Now, `40 Omega` . Thus, `R=(40xx40)/(40+40)=20Omega` Thus, to balanced bridge again, `(20)/(20)=(l_(x))/(l_(r))` `:.l_(x)=l_(r)=50cm` Thus, shift in null point is, `50-33.33=16.67cm` |
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