A resistance of 15 Ω is connected in series with an unknown resistanc…

The correct answer is Option (2) → \( 30\,\Omega \)

A meter bridge works on the Wheatstone bridge principle. If the resistance in the left gap is \( P \), the one in the right gap is \( Q \) and the balance point is at a distance \( l \) (in cm) from the left end,

    \( \displaystyle \frac{P}{Q} = \frac{l}{100 - l} \)

First balance, with \( P = 15 + R_1 \), \( Q = R_2 \) and \( l = 50\,\mathrm{cm} \):

    \( \displaystyle \frac{15 + R_1}{R_2} = \frac{50}{50} = 1 \;\Rightarrow\; R_2 = 15 + R_1 \)

When the \( 15\,\Omega \) resistance is removed, the left gap has less resistance, so the balance point moves towards the left end, by 10 cm to \( l = 40\,\mathrm{cm} \):

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

Putting \( R_2 = 15 + R_1 \):

    \( 3R_1 = 2(15 + R_1) \)
    \( R_1 = 30\,\Omega \)

Note: This question is from the pre-rationalised NCERT book (2022-23). The topic was removed from the latest NCERT book (2026-27), and it also appears to be out of the CUET 2026 syllabus.

NCERT (Pre-rationalised edition, 2022-23):
Meter Bridge – “… Let the distance of the jockey from the end A at the balance point be l = l₁. The four resistances of the bridge at the balance point then are R, S, R_cm l₁ and R_cm(100–l₁). The balance condition, Eq. [3.83(a)] gives

    \( \displaystyle \frac{R}{S} = \frac{R_{cm}\,l_1}{R_{cm}(100 - l_1)} = \frac{l_1}{100 - l_1} \qquad (3.85) \)”