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Q.

A particle of mass 1 kg is fired with velocity 50 m/s at an angle of 60° from horizontal. It is acted by viscous force of 0.2v during its journey. The horizontal distance travelled by it in first 10 seconds is :

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a

90 m

b

108 m

c

125 m

d

213 m

answer is B.

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Detailed Solution

dvxdt=−v5cosθ=−vx5⇒∫dvxvx=−∫15dt⇒Invxuxvx=−t50t⇒Invxux=−t5⇒vx=uxe−t/5=50cos60°e−t/5=25e−t/5Now dxdt=vx=25e−t/5⇒x=251−1/5e−t/5010=1251−e−2=1251−12.72=108m
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