The resistance of a wire is R. If its length is doubled and cross-sectional area halved, the new resistance is:
A. R/4
B. 2R
C. 4R
D. R
Answer: Option C
Solution (By JKSSB Mock Tests)
Resistance R = ρL/A, where ρ is resistivity (material property). New length L' = 2L, new area A' = A/2. Thus R' = ρ(2L)/(A/2) = ρ·2L·2/A = 4(ρL/A) = 4R. Resistance scales directly with length and inversely with area. Memory tip: 'Double length ⇒ double R; halve area ⇒ double R; combined ⇒ 4×'. This proportional reasoning problem is frequent in electricity sections of competitive exams. Always verify if resistivity changes (it doesn't here, same material).
Explanation:
Frequency determines pitch: higher frequency ⇒ higher pitch (e.g., whistle vs drum). Loudness depends on amplitude; quality (timbre) on waveform/harmonics; speed on medium properties (density, elasticity). This is a fundamental characteristic of sound perception. Memory aid: 'Frequency → Pitch; Amplitude → Loudness'. Competitive exams often test these distinctions to identify misconceptions. Note: Speed of sound is independent of frequency in a given medium (non-dispersive for audible sound in air). Always link physical quantities to perceptual attributes correctly.
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