Figure shows a right angle rod (having one arm vertical and other horizontal). The acceleration of free end A just after the system is released is \frac{x \sqrt{2} g}{10} then x will be :

Engineering

Physics

Rolling

Question

Figure shows a right angle rod (having one arm vertical and other horizontal). The acceleration of free end A just after the system is released is $\frac{x \sqrt{2} g}{10}$ then x will be :

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

$τ = \frac{mg}{2} \frac{ℓ}{2} = (\frac{m}{2} \times \frac{ℓ^{2}}{3} + \frac{m}{2} \frac{ℓ^{2}}{12} + \frac{m}{2} \times (ℓ^{2} + \frac{ℓ^{2}}{4})) α$

$\frac{mgℓ}{4} = \frac{5}{6} m 12 α$

$α = \frac{3 g}{10 ℓ}$

$a_{A} = \sqrt{2} ℓα = \frac{3 \sqrt{2} g}{10}$

Please subscribe our Youtube channel to unlock this solution.