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Question

A ball of mass m=2 kg is projected at an angle θ=45 with initial velocity of u=10 m/s as shown in figure. Find the torque acting on the particle due to gravity about origin at it's maximum height.
(Take g=10 m/s2)


A
100^k
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B
10^k
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C
100^k
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D
10^k
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Solution

The correct option is C 100^k

We know,
Range of projectileR=u2sin(2θ)g=102×sin(2×45)10=10 m
Maximum heightH=u2sin2θ2g=102sin2452×10=2.5 m


When ball reaches to maximum height H
r=R2^i+H^j =(5^i+2.5^j) m
Force due to gravityF=mg=2×10 N=20^j N

Torque due to gravity about originτ=r×F=(5^i+2.5^j)×20^j =100^k Nm

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