Question

Two resistances are connected in two gaps of a metre bridge. The balance point is 20 cm from the zero end. A resistance of $ 15\,\Omega $ is connected in series with the smaller of the two. The null point shift to 40 cm. The value of the smaller resistance in Ohm is

• A. 3
• B. 6
• C. 9
• D. 12

Answer 1

Answer: (C)

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) $ $ 4R=\frac{3}{2}(R+15) $ $ 8R-3R=45 $ $ 5R=45 $ $ R=9\Omega $ .