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← JEE Advanced 2024 · Paper 1
PhysicsInteger+4Question 30 of 51

JEE Advanced 2024 · Paper 1, Physics question 13

A thin uniform rod of length LL and certain mass is kept on a frictionless horizontal table with a massless string of length LL fixed to one end (top view is shown in the figure). The other end of the string is pivoted to a point O. If a horizontal impulse PP is imparted to the rod at a distance x=L/nx = L/n from the mid-point of the rod (see figure), then the rod and string revolve together around the point O, with the rod remaining aligned with the string. In such a case, the value of nn is _____.

Top view inside a square frame. A thin rod of length L (thick black bar) lies along a horizontal line; its right end is tied to a string of length L (thin line) whose other end is pivoted at point O, so rod and string are collinear and the rod's outer (left) end is at distance 2L from O. A dashed circle centred at O passes through the rod's outer end. Dimension arrows above mark L for the rod and L for the string. A dashed line marks the rod's midpoint, at L/2 from the end where the string is attached. The impulse P is a red arrow perpendicular to the rod, applied at a point on the outer side of the midpoint (farther from O), at distance x from the midpoint.
Answer and worked solution

Official answer: 18

For rod and string to revolve together, right after the impulse the rod must move as if rotating rigidly about O with some angular velocity ω\omega.

  • The rod's centre is L+L2=3L2L + \frac{L}{2} = \frac{3L}{2} from O, so vcm=32Lωv_{cm} = \frac32 L\omega.
  • The rod itself turns at ω\omega.

The string can only pull along its length, perpendicular to these velocities, so the impulse PP supplies both:

  • Linear: P=Mvcm=32MLωP = Mv_{cm} = \frac32 ML\omega.
  • Angular about the centre: Px=Icm ω=ML212 ωPx = I_{cm}\,\omega = \dfrac{ML^2}{12}\,\omega.

Dividing: x=L2/123L/2=L18x = \dfrac{L^2/12}{3L/2} = \dfrac{L}{18}, so n=18n = 18.

(Check: the end tied to the string moves at vcm−ωL2=Lωv_{cm} - \omega\frac L2 = L\omega, matching rotation about O, so the string needs no impulsive tension.)