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  4. Two resistances are connected in two gaps of a metre bridge. The balance point is 20 cm from the zero end. A resistance of 15Q is connected in series with the smaller of the two. The null point shifts to 40 cm. The value of the smaller resistance in ohm is

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Q. Two resistances are connected in two gaps of a metre bridge. The balance point is 20 cm from the zero end. A resistance of 15Q is connected in series with the smaller of the two. The null point shifts to 40 cm. The value of the smaller resistance in ohm is

CMC MedicalCMC Medical 2010

Solution:

Let S be the large and R be the smaller resistance. From formula for meter bridge $ S=\left( \frac{100-l}{l} \right)R $ $ =\left( \frac{100-20}{20} \right)R=4R $ Again, $ S=\left( \frac{100-l}{l} \right)(R+15) $ $ =\left( \frac{100-40}{40} \right)(R+15) $ $ =\frac{3}{2}(R+15) $ $ \therefore $ $ 4R=\frac{3}{2}(R+15) $ or $ 8R-3R=45 $ or $ 5R=45 $ $ \therefore $ $ R=9\,\Omega $