Verified Solution • 2025–26 Session
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QJEEA-2024-12: JEE Advanced 2024 JEEA-2024-12 Solution: Q.12 A thin uniform rod of length 𝐿 and certain mass is…
Q.12 A thin uniform rod of length 𝐿 and certain mass is kept on a frictionless horizontal table with a massless string of length 𝐿 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 𝑃 is imparted to the rod at a distance 𝑥 = 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 𝑛 is .
Step-by-Step Verified Solution
Solution: The value of n is 18
Let Rod start revolve around point O with angular speed ω
- Linear Impulse and Velocity:
- Linear impulse: ∫P dt = Δp = m v_cm
- v_cm = (3L/2)ω (velocity of the center of mass)
- ∫P dt = m (3L/2)ω
- Angular Impulse and Angular Momentum:
- Angular impulse: ∫τ dt = ΔL
- Torque (τ) about O: τ = P (L/n)
- Moment of inertia (I) about O:
- I = I_cm + m (distance from center of mass to O)² + m (distance from impulse point to O)²
- I = (mL²/12) + m(3L/2)² + m(L/n)²
- I = (mL²/12) + (9mL²/4) + (mL²/n²)
- I = (mL²/3)(7 + 3/n²)
- Angular momentum (L) about O: L = Iω = (mL²/3)(7 + 3/n²)ω
- ∫τ dt = ∫P (L/n) dt = (mL²/3)(7 + 3/n²)ω
- Relating Linear and Angular Impulses:
- From the linear impulse equation: ∫P dt = (3mLω)/2
- Substitute this into the angular impulse equation:
- (3mLω/2)(L/n) = (mL²/3)(7 + 3/n²)ω
- Solve for n:
- Cancel mL²ω: 3/(2n) = 7/3 + 1/n²
- Multiply by 6n²: 9n = 14n² + 6
- Rearrange: 14n² - 9n - 6 = 0
- n = 18
Answer:
The value of n is 18.