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. cot(theta)
C. cos(theta)
D. sin(theta)
Answer: Option A
Solution (By JKSSB Mock Tests)
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:
V_s/V_p = N_s/N_p ⇒ V_s = 100×(5/1) = 500V. Step-up transformer increases voltage. Memory aid: 'Turns ratio = voltage ratio; current ratio inverse'. Direct transformer formula application frequently tested in competitive exams with ratio calculations.
Explanation:
Fission: heavy nucleus (e.g., U-235) splits into smaller nuclei plus neutrons and energy. Fusion: light nuclei combine (e.g., hydrogen to helium). Both release energy. Fission used in reactors and bombs.
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