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A block released from rest from the top of a smooth inclined plane of angle of inclinatione, reaches the bottom in time t_(1) The same block, released from rest from the top of another smooth inclined plane of angle of inclination theta_(2)reaches the bottom in time ty. If the two inclined planes have the same height, the relation between t_(1) and t_(2) is : |
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Answer» `(t_(2))/(t_(1))=1` cases is`l_(1)=(h)/(sin theta_(1))`and `l_(2) = (h)/(sintheta_(2)`and acceleration `a_(1) = g sin theta-(1)`, and `a_(2) = g sin theta_(2)` `l=ut+(1)/(2)at^(2)` `l_(1)=0+(1)/(2)a_(1)t_(1)^(2)...(i)` and`l_(2)=0+(1)/(2)a_(2)t_(2)^(2)...(ii)` DIVIDING (ii) by (i) `(l_(2))/(l_(1))=(a_(2))/(a_(1))xx(t_(2)^(2))/(t_(2)^(2))` `:.(t_(2)^(2))/(t_(1)^(2))=(l_(2))/(l_(1))xx(a_(20))/(a_(1))=(sin theta_(1))/(sin theta_(2))xx(g sin theta_(1))/(g sintheta_(2))` `:.(t_(2))/(t_(1))=((sin theta_(1))/(sin theta_(2)))` Hence CORRECT choice is (b) |
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