question_answer

A particle is released from the top of two inclined rough surfaces of height '\[h\]' each. The angle of inclination of the two planes are \[30{}^\circ \] and \[60{}^\circ \] respectively. All other factors (e.g. coefficient of friction, mass of block etc.) are same in both the cases. Let \[{{K}_{1}}\] and \[{{K}_{2}}\] be the kinetic energies of the particle at the bottom of the plane in two cases. Then

A)  \[{{K}_{1}}={{K}_{2}}\]                   

B)  \[{{K}_{1}}>{{K}_{2}}\]

C)  \[{{K}_{1}}<{{K}_{2}}\]       

D)  data insufficient

Correct Answer: C

Solution :

[c] Work done in friction is \[W=(\mu mgcos\theta )s\] \[=(\mu mgcos\theta )\left( \frac{h}{\sin \theta } \right)=\mu mgh\cot \theta \] Now \[\cot {{\theta }_{1}}=\cot 30{}^\circ =\sqrt{3}\] And \[\cot {{\theta }_{2}}=\cot 60{}^\circ =\frac{1}{\sqrt{3}}\] \[\therefore {{W}_{1}}>{{W}_{2}}\] i.e., kinetic energy in first case will be less. \[(K=mgh-W)\] Or \[{{K}_{1}}<{{K}_{2}}\]