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 ONLINE 09-04-2016]

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

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

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

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

Correct Answer: D

Solution :

[d] 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

\[\frac{W}{20}v=\frac{P}{2}=\frac{3W}{4}\]            \[v=15m/s\]