Question

The gravity controlled instrument has crowded scale because current is proportional to

• A. gravitational force
• B. balancing weight
• C. deflection angle
• D. sine of deflection angle

Hint:

In gravity-controlled instruments, the deflection is proportional to the sine of the deflection angle, which makes the scale crowded.

Answer 1

The Correct Option is D

Step 1: Understanding the principle.
In a gravity-controlled instrument, the deflection of the needle depends on the current. The deflection angle is proportional to the sine of the deflection angle, which causes the scale to be crowded.
Step 2: Conclusion.
Thus, the current is proportional to the sine of the deflection angle, corresponding to option (D).

Answer 2

In a gravity-controlled instrument, a small adjustable weight provides the controlling torque, while the current through the coil provides the deflecting torque. At equilibrium (steady deflection), these two torques balance. Assume the deflecting torque is directly proportional to the current, \( T_d \propto I \), while the controlling torque from gravity depends on the angle the weight arm makes, \( T_c \propto \sin\theta \) (since only the component of the weight's moment arm perpendicular to gravity provides a restoring torque). Setting them equal, \( I \propto \sin\theta \). Let's check each option against this torque-balance relationship.

1. Gravitational force: The gravitational force (the weight itself) is fixed and does not change with deflection — it's the sine of the angle that varies, not the force itself, so current is not directly proportional to this fixed force alone.
2. Balancing weight: Like gravitational force, the weight used for control is typically a fixed value on the instrument, not something that varies with deflection to track current.
3. Deflection angle: If current were directly proportional to \( \theta \) itself, the scale would be uniform (evenly spaced), not crowded, which contradicts the premise of the question.
4. Sine of deflection angle: This matches the torque-balance result \( I \propto \sin\theta \) directly, and since \( \sin\theta \) grows slowly at small angles and faster at larger ones, this naturally produces a scale that's crowded together near the low end.

Balancing the deflecting and gravity-controlling torques confirms the current is proportional to the sine of the deflection angle.

Therefore, the correct answer is sine of deflection angle.