question_answer

A stone tied to the end of string of 80 cm long, is whirled in a horizontal circle with a constant speed. If the stone makes 25 revolutions in 14 s. Then, magnitude of acceleration of the same will be:

A)  \[990\,cm/{{s}^{2}}\]

B)                        \[680\,cm/{{s}^{2}}\]                              

C)  \[750\,cm/{{s}^{2}}\]   

D)         \[650\,cm/{{s}^{2}}\]

Correct Answer: A

Solution :

                When stone is tied to a 80 cm long string and whirled in a circle the centripetal force is provided by the tension in the string created by drawing the string inwards. If angular velocity is \[{{V}_{0}}={{V}_{e}}\] and radius of circular path is r then                                    acceleration \[\mathbf{\vec{P}}=a\mathbf{\hat{i}}+3\mathbf{\hat{k}}\] where \[\vec{Q}=a\mathbf{\hat{i}}-2\mathbf{\hat{j}}-\mathbf{\hat{k}}\] \[4.24\times {{10}^{6}}m/s\] and \[5.25\times {{10}^{6}}m/s\] Hence, \[6.25\times {{10}^{6}}m/s\]