A block of mass 'm' is placed on a rough inclined plane of inclination theta. If the block is just on the verge of sliding down, the coefficient of static friction (mu) is equal to: MCQ with Answer and Explanation
A block of mass 'm' is placed on a rough inclined plane of inclination theta. If the block is just on the verge of sliding down, the coefficient of static friction (mu) is equal to:
A. tan(theta)
B. sin(theta)
C. cos(theta)
D. cot(theta)
Answer: Option A
Solution (By JKExamLibrary)
When the block is on the verge of sliding (limiting friction), the downward component of weight along the plane equals the maximum static friction force. mgsin(theta) = mu * N. The normal reaction N is balanced by the perpendicular component of weight: N = mgcos(theta). Dividing the two equations gives mgsin(theta) / mgcos(theta) = mu, so mu = tan(theta). This angle is called the angle of repose.
Explanation:
Beats (periodic amplitude variation from interference of close frequencies) are used to tune instruments: adjust until beats disappear (frequencies match). Also, in Kundt's tube or other methods, beat frequency helps measure sound speed by comparing known and unknown frequencies. Memory tip: 'Beats: tuning (zero beat = matched frequency); speed measurement (beat frequency = |f₁-f₂|)'. This application question tests wave phenomena uses, frequently appearing in competitive exams. Always link physical phenomena to practical applications; competitive exams emphasize real-world relevance of physics concepts.
Explanation:
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).
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