Two resistances X and Y are in the left and right gaps of a meter bridge. The ba

Question

Two resistances X and Y are in the left and right gaps of a meter bridge. The balance point is 40 cm from left.Two resistances of

10 Ω

each are connected in series with X and Y separately. The balance point is 45 cm. The value of X and Y are

Options:

A . 12Ω,8Ω

B . 4Ω,6Ω

C . 8Ω,12Ω

D . 12Ω,16Ω

Answer: Option C
:
C

\frac{x}{y} = \frac{40}{100 - 40} = \frac{40}{60} = \frac{2}{3} \to (1) \frac{x + 10}{y + 10} = \frac{45}{55} = \frac{9}{11} \frac{\frac{2}{3} y + 10}{y + 10} = \frac{9}{11} \frac{2}{3} \times 11 y + 110 = 9 y + 90 20 = 9 y - \frac{22}{3} y 20 \times 3 = 5 y y = 12 Ω x = \frac{2}{3} \times 12 = 8 Ω

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