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JEE Main 2023 (8 Apr, Shift 2) · Rolling Motion

Rolling Motion question from JEE Main 2023 (8 Apr, Shift 2)

Q45JEE Main 2023 · 8 APR · SHIFT 2Medium

Rolling Motion

A hollow spherical ball of uniform density rolls up a curved surface with an initial velocity 3 m/s3\text{ m/s} (as shown in figure). Maximum height with respect to the initial position covered by it will be __________ cm. (Take g=10 m/s2g = 10\text{ m/s}^{2})

Figure

Numerical answer type. Work it out, then open the answer.

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Answer: 75

  1. Hollow sphere k=23k = \frac{2}{3}: KE =12mv2(1+23)=56mv2= \frac{1}{2}mv^2\left(1+\frac{2}{3}\right) = \frac{5}{6}mv^2.
  2. mgh=56mv2⇒h=5v26g=5×960=0.75mgh = \frac{5}{6}mv^2 \Rightarrow h = \dfrac{5v^2}{6g} = \dfrac{5 \times 9}{60} = 0.75 m.
  3. Maximum height =75= 75 cm.
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