question_answer

A car of weight W is on an inclined road that rises by 100 m over a distance of 1km and applies a constant frictional force \[\frac{W}{20}\] on the car. While moving uphill on the road at a speed of \[10m{{s}^{-1}},\]the car needs power \[\frac{P}{2}\]If it needs power while moving downhill at speed v then value of v is :   JEE Main Online Paper (Held On 09 April 2016)

A) \[5\,m{{s}^{-1}}\]                                           

B) \[20\,m{{s}^{-1}}\]                

C) \[10\,m{{s}^{-1}}\]                                         

D) \[15\,m{{s}^{-1}}\]

Correct Answer: D

Solution :

                While going up\[(\sin \theta =\tan \theta =\frac{1}{10})\] \[W\sin \theta 10+\frac{W}{20}10=P\] \[\frac{3W}{2}=P\] While going down \[\frac{W}{20}v=\frac{P}{2}=\frac{3W}{4}\]         \[v=15m/s\]