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At `t = 0`, a force `F = at^(2)` is applied to s small body of mass m at an angle `alpha` resting on a smooth horizontal plane. A. Velocity of the body at the moment it breaks off the plane is `sqrt((mg^(3))/(9 a tan^(2) alpha sin alpha))`B. The distance travelled by the body before breaking off the plane is `(mg^(2))/(12a sin alpha tan alpha)`C. Its acceleration at the time of breaking off the plane is `g cot alpha`.D. Time at which it breaks off the plane is `sqrt((mg)/(a sin alpha))` |
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Answer» Correct Answer - A::B::C::D For breaking off the plane `F sin alpha = mg` `rArr a t_(0)^(2) sin alpha = mg " " rArr t_(0) = sqrt((mg)/(a sin alpha))` Speed at time of breaking off, `v = underset(0)overset(t_(0))int (at^(2) cos alpha)/(m) dt = (at_(0)^(2) cos alpha)/(3 m)` `= (a cos alpha)/(3m) . (mg)/(a sin alpha) sqrt((mg)/(a sin alpha))` `sqrt((mg^(3))/(9a tan^(2) alpha sin alpha))` `= (F cos alpha)/(m) = (at_(0)^(2) cos alpha)/(m) = (amg)/(a sin alpha) . (cos alpha)/(m) = g` `cot alpha` `s = int v dt = underset(0)overset(t_(0))int (at^(3))/(3m) cos alpha dt` `= (a)/(12m) t_(0)^(2) = (a)/(12m) (m^(2) g^(2) cos alpha)/(a^(2) sin^(2) alpha)` `= (mg^(2))/(12a tan alpha sin alpha)` |
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