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The number of turns of the coil of a mov...

The number of turns of the coil of a moving coil galvanometer is increased in order to increase current sensitivity by `50%`.The percentage change in voltage sintivity of the galvanometer will be:

A

`50%`

B

`75%`

C

`0%`

D

`100%`

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The correct Answer is:
To solve the problem, we need to understand the relationship between current sensitivity and voltage sensitivity in a moving coil galvanometer. ### Step-by-Step Solution: 1. **Understanding Current Sensitivity**: - Current sensitivity (S_I) of a galvanometer is defined as the deflection per unit current. It is directly proportional to the number of turns (N) of the coil. - Mathematically, we can express this as: \[ S_I \propto N \] 2. **Increasing Current Sensitivity**: - According to the problem, the number of turns of the coil is increased to increase the current sensitivity by 50%. - If the initial current sensitivity is \( S_{I0} \), then the new current sensitivity \( S_{I1} \) can be expressed as: \[ S_{I1} = S_{I0} + 0.5 S_{I0} = 1.5 S_{I0} \] 3. **Relating Voltage Sensitivity**: - Voltage sensitivity (S_V) is defined as the deflection per unit voltage. It is related to current sensitivity by the formula: \[ S_V = S_I \cdot R \] - Where R is the resistance of the galvanometer. Assuming R remains constant, any change in current sensitivity will directly affect voltage sensitivity. 4. **Calculating the Change in Voltage Sensitivity**: - The initial voltage sensitivity \( S_{V0} \) can be expressed as: \[ S_{V0} = S_{I0} \cdot R \] - The new voltage sensitivity \( S_{V1} \) becomes: \[ S_{V1} = S_{I1} \cdot R = (1.5 S_{I0}) \cdot R = 1.5 S_{V0} \] 5. **Finding the Percentage Change**: - The percentage change in voltage sensitivity can be calculated as: \[ \text{Percentage Change} = \left( \frac{S_{V1} - S_{V0}}{S_{V0}} \right) \times 100\% \] - Substituting the values: \[ \text{Percentage Change} = \left( \frac{1.5 S_{V0} - S_{V0}}{S_{V0}} \right) \times 100\% = \left( \frac{0.5 S_{V0}}{S_{V0}} \right) \times 100\% = 50\% \] ### Final Answer: The percentage change in voltage sensitivity of the galvanometer is **50%**.
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