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A solid cylinder of radius 5 cm and mass 300 g rolls down an inclined plane (1 in 20). The velocity of cylinder after 5s will be(a) 1 63. ms−1(b) 1 56. ms−1(c) 2 ms−1(d) 3 26. ms−1

Answer»

Answer is : (a) 1 63. ms−1

Acceleration with which the cylinder rolls down,

\(a=\cfrac{g\,sin\theta}{1+\frac{l}{MR^2}}=\cfrac{g\,sin\theta}{1+\frac{\frac{1}{2}MR^2}{MR^2}}\) \(=\frac{2}{3}g\,sin\theta\) \(=\frac{2}{3}\times 9.8\times \frac{1}{20}\) \(=0.326\,ms^{-2}\)

Using first equation of motion,

v = u + at \(= 0 + 0.326\times 5\) \(=1.63\,ms^{-1}\)


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