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MHT CET · Physics · Current Electricity

Moving coil galvanometers \(\mathrm{M}_{1}\) and \(\mathrm{M}_{2}\) have resistance, number of turns, area of coil and magnetic field as follows.
\(\mathrm{R}_{1}=10 \Omega, \mathrm{R}_{2}=14 \Omega, \mathrm{N}_{1}=30, \mathrm{~N}_{2}=42\)
\(\mathrm{A}_{1}=3 \cdot 6 \times 10^{-3} \mathrm{~m}^{2}, \mathrm{~A}_{2}=1 \cdot 8 \times 10^{-2} \mathrm{~m}^{2},\) \(\mathrm{~B}_{1}=0 \cdot 25 \mathrm{~T}, \mathrm{~B}_{2}=0 \cdot 50 \mathrm{~T}\)
(Spring constants are same for both materials)
The ratio of (i) current sensitivity and (ii) voltage sensitivity for galvanometer
\(\left(\mathrm{M}_{2}\right.\) to \(\left.\mathrm{M}_{1}\right)\) is respectively

  1. A \(1: 1, \quad 1 \cdot 4: 1\)
  2. B \(1: 1 \cdot 4\), \(1: 1\)
  3. C \(4: 1\) \(1: 1\)
  4. D \(1 \cdot 4: 1\) \(1: 1\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(1 \cdot 4: 1\) \(1: 1\)

Step-by-step Solution

Detailed explanation

(D)
We know, \(\mathrm{L}=\mu_{0} \mathrm{~N}^{2} \frac{\mathrm{A}}{\ell}\)
If \(x\) is the length of the wire and \(a\) is the area of cross section
\(\mathrm{R}=\frac{\rho \mathrm{x}}{\mathrm{a}} \quad \mathrm{m}=\mathrm{axD}\)
\(\mathrm{Rm}=\frac{\rho \mathrm{x}}{\mathrm{a}} \times \mathrm{axD} \quad \therefore \quad \mathrm{x}=\sqrt{\frac{\mathrm{Rm}}{\rho \mathrm{D}}}\)
Also,\(x=2 \pi r N\therefore \quad \mathrm{N}=\frac{\mathrm{x}}{2 \pi \mathrm{r}}\)
\(=\mu_{0}\left(\frac{\mathrm{x}}{2 \pi \mathrm{r}}\right)^{2} \frac{\pi \mathrm{r}^{2}}{\ell}=\frac{\mu_{0}}{4 \pi \ell} \frac{\mathrm{Rm}}{\rho \mathrm{D}}\)