A wire of resistance R is stretched to double its length. Assuming volume remains constant, the new resistance is:
A. R
B. 2R
C. 4R
D. R/2
Answer: Option C
Solution (By JKSSB Mock Tests)
Resistance R = ρL/A. Volume V = A·L constant. If L' = 2L, then A' = V/L' = V/(2L) = A/2. Thus R' = ρ(2L)/(A/2) = 4(ρL/A) = 4R. Resistance increases by factor of 4. Memory tip: 'Stretching wire: R ∝ L² when volume constant'. This proportional reasoning problem tests resistance concepts, frequently appearing in competitive exams. Always verify volume conservation assumption; if area changed independently, result would differ. This problem assesses understanding of geometric effects on electrical properties.
Explanation:
T = 2π√(l/g): depends on l and g, independent of mass (gravitational and inertial mass cancel). Memory aid: 'Pendulum: T ∝ √(l/g); independent of mass for small angles'. Oscillations concept frequently tested in competitive exams.
Explanation:
At 0°C, speed ~ 331 m/s. At room temperature (20°C) ~ 343 m/s, often taken as 340. Increases with temperature v ∝ √T. For many problems 340 is used, but at 0°C it's 331. So option A.
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