question_answer

A string is wound around a hollow cylinder of mass 5 kg and radius 0.5 m. If the string is now pulled with a horizontal force of 40 N, and the cylinder is rolling without slipping on a horizontal surface (see figure), then the angular acceleration of the cylinder will be (Neglect the mass and thickness of the string) [JEE  Main Online Paper (Held on 11-jan-2019 Evening)]

A) \[10\text{ }rad/{{s}^{2}}\]                               

B) \[\text{20 }rad/{{s}^{2}}\]

C) \[\text{12 }rad/{{s}^{2}}\]                   

D)   \[\text{16 }rad/{{s}^{2}}\]

Correct Answer: D

Solution :

\[40+f=ma=m(R\alpha )\]                                  ?(i) \[40\times R-f\times R=m{{R}^{2}}\alpha \]                               ?(ii)   From eqn (i) and (ii),\[80=2mR\alpha \] \[\alpha =\frac{40}{mR}=\frac{40}{5\times 0.5}\]\[=16rad/{{s}^{2}}\]