AP EAMCET · PHYSICS · Laws of Motion
A body is made to move up along an inclined plane of inclination \(30^{\circ}\) and the coefficient of friction is 0.5 , then its retardation is
( \(\mathrm{g}\) - acceleration due to gravity)
- A \(\left(\frac{2+\sqrt{3}}{4}\right) \mathrm{g}\)
- B \(\left(\frac{2-\sqrt{3}}{4}\right) \mathrm{g}\)
- C \(\left(\frac{2-\sqrt{3}}{2}\right) \mathrm{g}\)
- D \(\left(\frac{2+\sqrt{3}}{2}\right) \mathrm{g}\)
Answer & Solution
Correct Answer
(A) \(\left(\frac{2+\sqrt{3}}{4}\right) \mathrm{g}\)
Step-by-step Solution
Detailed explanation
Retardation, \(a=g(\sin \theta+\mu \cos \theta)\) \(=\mathrm{g}\left[\sin 30^{\circ}+0.5\left(\cos 30^{\circ}\right)\right]\)…
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