Q151 / 160 · AP EAMCET 2018 (22 Apr Shift 1) · Physics

Two moving coil galvanometer, $X$ and $Y$ have coils with resistance $10 \Omega$ and $14 \Omega$ cross-sectional areas $4.8 \times 10^{-3} \mathrm{~m}^2$ and $2.4 \times 10^{-3} \mathrm{~m}^2$, number of turns 30 and 45 respectively. They are placed in magnetic field of $0.25 \mathrm{~T}$ and $0.50 \mathrm{~T}$ respectively. Then, the ratio of their current sensitivities and the ratio of their voltage sensitivities are respectively

  • Correct answer
    A
    2 : 3, 14 : 15
  • B
    5 : 7, 2 : 1
  • C
    2 :13, 1 : 2
  • D
    14 : 15, 2 : 9

Solution

There is no information regarding spring constant. So, let's assume their spring constants are same. Current sensitivity, $I_S=\frac{N B A}{k}$ So, $$ \begin{aligned} \frac{I_{S_1}}{I_{S_2}} & =\frac{N_1 B_1 A_1 k_2}{N_2 B_2 A_2 k_1} \\ & =\frac{30 \times 0.25 \times 4.8 \times 10^{-3}}{45 \times 0.5 \times 24 \times 10^{-3}} \times\left(\frac{1}{1}\right) \\ \frac{I_{S_1}}{I_{S_2}} & =2 / 3 \end{aligned} $$ Voltage sensitivity, $V_S=N B A / R k$ So, $$ \begin{aligned} \frac{V_{S_1}}{V_{S_2}} & =\frac{N_1 B_1 A_1 k_2 R_2}{N_2 B_2 A_2 k_1 R_1} \\ & =\frac{2}{3} \times\left(\frac{R_2}{R_1}\right)=\frac{2}{3} \times \frac{14}{10}=\frac{14}{15} \end{aligned} $$

Hence, the correct answer is option: 'A'.

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