The momentum of a body in two perpendicular directions at any instant of time 't' is given by
$P_{x} = 2 t^{2} + 6$ and $p_{y} = \frac{3 t^{2}}{2} + 3$ . The force acting on the body at t=2 sec is:

5 u n i t s

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2 u n i t s

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10 u n i t s

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7 u n i t s

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Solution

The correct option is C $10 u n i t s$
$F_{x} = \frac{d P_{x}}{d t} = 4 t ⟹ F_{x} = 4 (2) = 8 F_{y} = \frac{d P_{y}}{d t} = 3 t ⟹ F_{y} = 3 (2) = 6 F = \sqrt{{F_{x}}^{2} + {F_{y}}^{2}} = \sqrt{8^{2} + 6^{2}} = \sqrt{64 + 36} = \sqrt{100} = 10$

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Similar questions

P_{x} = 2 t^{2} + 6

P_{y} = \frac{3 t^{2}}{2} + 3

t = 2

P_{x} = A s i n (t)

P_{y} = A c o s (t)

^{'} A^{'}

^{'} t^{'}

v = \frac{t}{3} + \frac{t^{2}}{2} + 1

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\to L = (4 t^{2} + 6)^k

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