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Physics-

General

Easy

Question

At t=0, force F=ct is applied to a small body of mass m resting on a smooth horizontal plane (c is a constant). The force is at an angle $theta$ with the horizontal .The velocity of the body at the moment of its breaking off the plane and the distance travelled by the body up to this moment are

$(\frac{m g^{2} c o s θ}{2 c {s i n}^{2} θ}) m / s, (\frac{m g^{3} c o n θ}{6 c^{2} {s i n}^{3} θ}) m$
$(\frac{m g^{2} c o s θ}{2 c {s i n}^{2} θ}) m / s, (\frac{m^{2} g^{3} c o n θ}{6 c^{2} {s i n}^{3} θ}) m$
$(\frac{m g c o s θ}{2 c {s i n}^{2} θ}) m / s, (\frac{m^{2} g^{3} s i n θ}{6 c^{2} {c o s}^{3} θ}) m$
$(\frac{m g^{2} c o s θ}{2 c {s i n}^{2} θ}) m / s, (\frac{m^{2} g^{3} s i n θ}{6 c^{2} {c o s}^{3} θ}) m$

The correct answer is: $open parentheses fraction numerator m g to the power of 2 end exponent c o s invisible function application theta over denominator 2 c s i n to the power of 2 end exponent invisible function application theta end fraction close parentheses m divided by s comma open parentheses fraction numerator m g to the power of 3 end exponent c o n invisible function application theta over denominator 6 c to the power of 2 end exponent s i n to the power of 3 end exponent invisible function application theta end fraction close parentheses m$

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