The current sensitivity of moving coil galvanometer is increased by 25 %. This increase is achieved only by changing in the number of turns of coils and area of cross section of the wire while keeping the resistance of galvanometer coil constant. The percentage change in the voltage sensitivity will be:

Solution

Physics sir · 4 steps

Voltage sensitivity is just current sensitivity divided by R, and here R does not change.

1. Current sensitivity. Deflection per unit current is Iₛ = θ/I = NAB/k, where N is the number of turns, A the coil area, B the field and k the torsion constant.
2. Voltage sensitivity. Deflection per unit voltage uses V = IR, so it equals the current sensitivity divided by the coil resistance.

   Vₛ = θ/V = θ/IR = Iₛ/R

3. Why R stays fixed. More turns would normally mean a longer wire and a larger R. Here the wire is made thicker by just the right amount, so R stays the same. With R constant, Vₛ is directly proportional to Iₛ.
4. So. A 25% rise in Iₛ gives the same 25% rise in Vₛ. The change is +25%, option (a).

The idea to keep. Always link the two with Vₛ = Iₛ/R. If R rises along with N, Vₛ can stay unchanged. If R is held fixed, Vₛ changes by exactly the same percentage as Iₛ.

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