A tuning fork A produces 5 beats/second with tuning fork B of frequency 256 Hz. If a little wax is applied to fork A, the beat frequency becomes 2 beats/second. The original frequency of fork A was: MCQ with Answer and Explanation
A tuning fork A produces 5 beats/second with tuning fork B of frequency 256 Hz. If a little wax is applied to fork A, the beat frequency becomes 2 beats/second. The original frequency of fork A was:
A. 258 Hz
B. 251 Hz
C. 254 Hz
D. 261 Hz
Answer: Option D
Solution (By JKSSB Mock Tests)
The beat frequency is |fA - fB|. So fA is either 256+5=261 Hz or 256-5=251 Hz. Adding wax increases mass, which strictly decreases frequency. If fA was 261, decreasing it brings it closer to 256, reducing beats to 2 (e.g., 258-256=2). If it was 251, decreasing it would increase the beats (e.g., |248-256|=8). Thus, it was 261 Hz.
Explanation:
In Simple Harmonic Motion (like a swinging pendulum), velocity is maximum when the particle passes through its equilibrium or mean position. Since Kinetic Energy = ½mv², KE is maximum at the mean position. At the extreme positions, the velocity is briefly zero, meaning kinetic energy is zero and potential energy is maximum.
Explanation:
SI unit Newton (kg m/s²), CGS unit dyne (g cm/s²). 1 N = 10⁵ dyne. Conversion: 1 kg = 10³ g, 1 m = 100 cm, so 1 N = 10³ × 10² = 10⁵ dyne. Ratio N/dyne = 10⁵.
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