Two moving coil metres `M_(1) and M_(2)` have the following particular `R_(1) =10Omega, N_(1) =30, A_(1) = 3.6 xx 10^(-3) m^(2), B_(1) =0.25 T,`
`R_(2) = 14Omega , N_(2) =42, A_(2) = 1.8 xx 10^(-3) m^(2), B_(2) =0.50 T`
The spring constants are identical for the two metres. What is the ratio of current sensitivity and voltage sensitivity of `M_(2)" to" M_(1)?`

Options

A. 1.4, 1
B. 1.4, 0
C. 2.8, 2
D. 2.8, 0

Explanation

To find the ratio of current sensitivity and voltage sensitivity of meters \( M_2 \) to \( M_1 \), we can follow these steps:

### Step 1: Understand the formulas for current sensitivity and voltage sensitivity
The formulas for current sensitivity (\( I_s \)) and voltage sensitivity (\( V_s \)) are given by:
- Current sensitivity:
\[ I_s = \frac{N \cdot B \cdot A}{k} \]
- Voltage sensitivity:
\[ V_s = \frac{N \cdot B \cdot A}{k \cdot R} \]
Where:
- \( N \) = number of turns
- \( B \) = magnetic field strength
- \( A \) = area of the coil
- \( k \) = spring constant
- \( R \) = resistance

### Step 2: Calculate the current sensitivity for both meters
For meter \( M_1 \):
\[ I_{s1} = \frac{N_1 \cdot B_1 \cdot A_1}{k} = \frac{30 \cdot 0.25 \cdot 3.6 \times 10^{-3}}{k} \]

For meter \( M_2 \):
\[ I_{s2} = \frac{N_2 \cdot B_2 \cdot A_2}{k} = \frac{42 \cdot 0.50 \cdot 1.8 \times 10^{-3}}{k} \]

### Step 3: Find the ratio of current sensitivities
Now, we can find the ratio of current sensitivities:
\[ \frac{I_{s2}}{I_{s1}} = \frac{\frac{42 \cdot 0.50 \cdot 1.8 \times 10^{-3}}{k}}{\frac{30 \cdot 0.25 \cdot 3.6 \times 10^{-3}}{k}} = \frac{42 \cdot 0.50 \cdot 1.8}{30 \cdot 0.25 \cdot 3.6} \]
Calculating the above expression:
\[ = \frac{42 \cdot 0.50 \cdot 1.8}{30 \cdot 0.25 \cdot 3.6} = \frac{42 \cdot 0.9}{30 \cdot 0.9} = \frac{42}{30} = 1.4 \]

### Step 4: Calculate the voltage sensitivity for both meters
For meter \( M_1 \):
\[ V_{s1} = \frac{N_1 \cdot B_1 \cdot A_1}{k \cdot R_1} = \frac{30 \cdot 0.25 \cdot 3.6 \times 10^{-3}}{k \cdot 10} \]

For meter \( M_2 \):
\[ V_{s2} = \frac{N_2 \cdot B_2 \cdot A_2}{k \cdot R_2} = \frac{42 \cdot 0.50 \cdot 1.8 \times 10^{-3}}{k \cdot 14} \]

### Step 5: Find the ratio of voltage sensitivities
Now, we can find the ratio of voltage sensitivities:
\[ \frac{V_{s2}}{V_{s1}} = \frac{\frac{42 \cdot 0.50 \cdot 1.8}{k \cdot 14}}{\frac{30 \cdot 0.25 \cdot 3.6}{k \cdot 10}} = \frac{42 \cdot 0.50 \cdot 1.8 \cdot 10}{30 \cdot 0.25 \cdot 3.6 \cdot 14} \]
Notice that the \( k \) cancels out:
\[ = \frac{42 \cdot 0.50 \cdot 1.8 \cdot 10}{30 \cdot 0.25 \cdot 3.6 \cdot 14} = \frac{42 \cdot 0.50 \cdot 1.8 \cdot 10}{30 \cdot 0.25 \cdot 3.6 \cdot 14} = 1 \]

### Final Ratios
Thus, the final ratios are:
- Ratio of current sensitivity \( \frac{I_{s2}}{I_{s1}} = 1.4 \)
- Ratio of voltage sensitivity \( \frac{V_{s2}}{V_{s1}} = 1 \)