In a meter bridge, null point is found at a distance of 20 cm from th…

The correct answer is Option (2) → 33.3 cm

At the null point of a meter bridge the two resistances are in the same ratio as the two lengths of the wire (the Wheatstone bridge balance condition). The 10 Ω resistance is in the gap on the side of end A, and \( S \) is the resistance in the other gap.

    \( \displaystyle \frac{R}{S} = \frac{l}{100 - l} \)

First balance, \( R = 10\,\Omega \), \( l = 20\,\mathrm{cm} \):

    \( \displaystyle \frac{10}{S} = \frac{20}{80} \;\Rightarrow\; S = 40\,\Omega \)

Second balance, \( R = 20\,\Omega \), \( S = 40\,\Omega \):

    \( \displaystyle \frac{l}{100 - l} = \frac{20}{40} = \frac{1}{2} \)

    \( \displaystyle 2l = 100 - l \;\Rightarrow\; l = \frac{100}{3} = 33.3\,\mathrm{cm} \)

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_cml₁ and R_cm(100–l₁). The balance condition, Eq. [3.83(a)] gives

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