In the given setup of meter bridge, the null point D is obtained at a…

The correct answer is Option (1) → 8 Ω and 12 Ω

At the null point of a meter bridge the ratio of the two resistances equals the ratio of the wire lengths on the two sides (balance condition of a Wheatstone bridge).

    \( \displaystyle \frac{R_1}{R_2} = \frac{l}{100 - l} \)

First balance, \( l = 40\,\mathrm{cm} \):

    \( \displaystyle \frac{R_1}{R_2} = \frac{40}{60} = \frac{2}{3} \;\Rightarrow\; R_1 = \frac{2}{3}R_2 \)

Adding \( 10\,\Omega \) to \( R_1 \) makes the left gap larger, so the null point moves towards C by 20 cm, to \( l = 60\,\mathrm{cm} \):

    \( \displaystyle \frac{R_1 + 10}{R_2} = \frac{60}{40} = \frac{3}{2} \)

Substituting \( R_1 = \tfrac{2}{3}R_2 \):

    \( \displaystyle \frac{2}{3}R_2 + 10 = \frac{3}{2}R_2 \;\Rightarrow\; 10 = \frac{5}{6}R_2 \;\Rightarrow\; R_2 = 12\,\Omega \)

    \( \displaystyle R_1 = \frac{2}{3} \times 12 = 8\,\Omega \)