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JEE Main 2019 (8 Apr, Shift 2) · Rolling Motion

Rolling Motion question from JEE Main 2019 (8 Apr, Shift 2)

Q69JEE Main 2019 · 8 APR · SHIFT 2Medium

Rolling Motion

A solid sphere and solid cylinder of identical radii approach an incline with the same linear velocity (see figure). Both roll without slipping all throughout. The two climb maximum heights hsph and hcyl on the incline. The ratio hsph/hcyl is given by :-

Figure

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Answer: (b)

  1. h=v22g(1+k)h = \dfrac{v^2}{2g}(1+k), so h∝(1+k)h \propto (1+k).
  2. hsphhcyl=1+251+12=7/53/2=1415\dfrac{h_{sph}}{h_{cyl}} = \dfrac{1 + \frac{2}{5}}{1+\frac{1}{2}} = \dfrac{7/5}{3/2} = \dfrac{14}{15}.
  3. So the answer is (b) 14/15.
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