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. 254 Hz
C. 261 Hz
D. 251 Hz
Answer: Option C
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 series circuits, current is the same through all components because there's only one path for charge flow. Voltage divides across components proportional to resistance (V = IR). Resistance and power vary per component. Memory aid: 'Series: same current; Parallel: same voltage'. This fundamental circuit property is frequently tested in competitive exams to assess basic electronics understanding. Always apply Kirchhoff's current law: current entering a junction equals current leaving; in series, no junctions, so current constant.
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