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 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

• 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, $S=\left(\frac{100-l}{l}\right)(R+15)$ $=\left(\frac{100-40}{40}\right)(R+15)$ $=\frac{3}{2}(R+15)$ $\therefore$ $4R=\frac{3}{2}(R+15)$ or $8R-3R=45$ or $5R=45$ $\therefore$ $R=9\Omega$