Q8 / 122 · NTA Abhyas (JEE Main) 2020 · Physics · Rotational Motion

A uniform rod AB of mass 2 kg and length 1 m is placed on a sharp support O, such that AO=a=25 cm and OB=b=75 cm. A spring of force constant 600 N m-1 is attached to the end B, as shown. Initially, the spring is extended by 2 cm at equilibrium. As soon as the thread is burnt, find the normal force (in N) exerted by the support at O.

Figure
 

Correct answer: 8

Solution

I about O is mL212+mL216=7mL248

Spring force   ⇒KΔx=600×2100=12  N

Torque, just after burning the string,

τ=2×10×14+12×34

⇒ τ= 5 + 9 = 14 N m

∴α=14×487×2×1=48 rad s-2

∴ Acceleration of COM,

a=48×14=12 m s-2

Now, total vertical force

12 + 20 – N = 2 × 12

⇒N = 32 – 24 = 8 N

Hence, the correct answer is 8.

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