Physics Circle home Student Login

Home›Chapters›Rotational Motion PYQs›Rolling Motion›Question 7

JEE Main 2026 (5 Apr, Shift 2) · Rolling Motion

Rolling Motion question from JEE Main 2026 (5 Apr, Shift 2)

Q7JEE Main 2026 · 5 APR · SHIFT 2Medium

Rolling Motion

An object of uniform density rolls up the curved path with the initial velocity v0v_0 as shown in the figure. If the maximum height attained by an object is 7v0210g\frac{7v_0^{2}}{10g} ( g=g = acceleration due to gravity), the object is a ______.

Figure

Show answer and solution

Answer: (d)

  1. Rolling body with I=kmR2I = kmR^2: total KE =12mv02(1+k)= \frac{1}{2}mv_0^2(1+k).
  2. All KE becomes PE at the top: h=(1+k)v022gh = \dfrac{(1+k)v_0^2}{2g}.
  3. (1+k)2=710⇒k=25\dfrac{(1+k)}{2} = \dfrac{7}{10} \Rightarrow k = \dfrac{2}{5}, which is a solid sphere.
  4. So the answer is (d) solid sphere.
Report a mistake

More questions on Rolling Motion