A meter bridge setup is shown in the figure. It is used to determine an unknown resistance R using a given resistor of 15 Ω. The galvanometer (G) shows null deflection when tapping key is at 43 cm mark from end A. If the end correction for end A is 2 cm, then the determined value of R will be ____________ Ω.

Solution

Physics sir · 3 steps

Add the 2 cm end correction to the 43 cm side before you use the balance ratio.

1. Read the setup. The 15 Ω resistor sits in the gap on the A side and R sits in the gap on the B side. At null deflection, the bridge is balanced.
2. Correct the lengths. End A behaves like 2 cm of extra wire, so the effective length on the A side is 43 + 2 = 45 cm. The B side is 100 - 43 = 57 cm, and no correction is given for end B.

   15/(l_A + e_A) = R/(100 - l_A) ⇒ 15/45 = R/57

3. So. R = (15 × 57)/45 = 19 Ω, so the answer is 19.

The common slip: 20. Ignoring the end correction gives R = (15 × 57)/43 ≈ 19.9, which rounds to 20 instead of 19.

The idea to keep. An end correction is extra wire hidden at that end. Add it only to the length measured from that end, then use the usual metre bridge balance ratio.

Answer from the NTA final answer key. Solution by PrepWiser.