A) Lighter sphere
B) Heavier sphere
C) Both will reach at the same time
D) None of the above
Correct Answer: C
Solution :
Time of descend on an inclined plane \[{{T}_{r}}=\frac{1}{\sin \theta }\sqrt{\frac{2h}{g}\left( 1+\frac{{{k}^{2}}}{{{R}^{2}}} \right)}\] which is independent of mass and radius of the two spheres. Hence, both the spheres will reach the ground at the same time. Alternative Acceleration of a body on an inclined plane \[a=\frac{g\,\,\sin \theta }{1+\frac{{{K}^{2}}}{R{{}^{2}}}}\] For both spheres, \[\frac{{{K}^{2}}}{{{R}^{2}}}=1\] Therefore, acceleration of both spheres are same, hence they will reach at the same time
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