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Question: A body of mass m starting from rest slides down a frictionless inclined surface of gradient αfixed o...

A body of mass m starting from rest slides down a frictionless inclined surface of gradient αfixed on the floor of a lift accelerating upward with acceleration a. Taking width of inclined plane as W, the time taken by body to slide from top to bottom of the plane is

A

(2W(g+a)sin⁡α)1/2\left( \frac { 2 W } { ( g + a ) \sin \alpha } \right) ^ { 1 / 2 }

B

(4W(g−a)sin⁡α)1/2\left( \frac { 4 W } { ( g - a ) \sin \alpha } \right) ^ { 1 / 2 }

C

(4W(g+a)sin⁡2α)1/2\left( \frac { 4 W } { ( g + a ) \sin 2 \alpha } \right) ^ { 1 / 2 }

D

(W(g+a)sin⁡2α)1/2\left( \frac { W } { ( g + a ) \sin 2 \alpha } \right) ^ { 1 / 2 }

Answer

(4W(g+a)sin⁡2α)1/2\left( \frac { 4 W } { ( g + a ) \sin 2 \alpha } \right) ^ { 1 / 2 }

Explanation

Solution

cos α = WOA\frac { \mathrm { W } } { \mathrm { OA } } or OA=Wcos⁡α\mathrm { OA } = \frac { \mathrm { W } } { \cos \alpha }

Also OA = 12\frac { 1 } { 2 } (g + a) sin αt2

(by using S = ut + 12\frac { 1 } { 2 } at2)

∴ Wcos⁡α=12( g+a)sin⁡αt2\frac { \mathrm { W } } { \cos \alpha } = \frac { 1 } { 2 } ( \mathrm {~g} + \mathrm { a } ) \sin \alpha \mathrm { t } ^ { 2 }

t = [2 Wcos⁡α(g+a)sin⁡α]1/2\left[ \frac { 2 \mathrm {~W} } { \cos \alpha ( \mathrm { g } + \mathrm { a } ) \sin \alpha } \right] ^ { 1 / 2 }

= [4W(2cos⁡αsin⁡α)(g+a)]1/2\left[ \frac { 4 W } { ( 2 \cos \alpha \sin \alpha ) ( g + a ) } \right] ^ { 1 / 2 }

= [4W(g+a)sin⁡2α]1/2\left[ \frac { 4 W } { ( g + a ) \sin 2 \alpha } \right] ^ { 1 / 2 }