JEE Main202226 Jul 2022Morning ShiftPhysicsLaws of MotionActual
Three masses M = 100 kg , m 1 = 10 kg and m 2 = 20 kg are arranged in a system as shown in figure. All the surfaces are frictionless and strings are inextensible and weightless. The pulleys are also weightless and frictionless. A force F is applied on the system so that the mass m 2 moves upward with an acceleration of 2 ms - 2 . The value of F is (Take g = 10 ms - 2 )
Options
- A3360   N
- B3380   N
- C3120 N
- D3240 N
Correct answer
A. 3360   N
Step-by-step solution
Let acceleration of 100 ~kg block =a₁ FBD of 100 ~kg block w.r.t ground F - T - N ₁=100 a ₁ ... (i) FBD of 20 block wrt 100 ~kg T -20 ~g =20(2) T =240 ... (ii) ~N ₁=20 a ₁ ... (iii) FBD of 10 ~kg block wrt 100 ~kg 10 a ₁-240=10(2) a ₁=26 ~m / s ^2 ~F -240-20(26)=100 26 F =3360 ~N