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Q.

A body is free to rotate about an axis passing through the y-axis. A force of F→=(3i^+2j^+6k^)N is acting on the body the position vector of whose point of application is r→=(2i^−3j^)m. The moment of inertia of body about y-axis is 10 kg m2. Then the angular acceleration of the body should be equal to:

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a

(−1.8i^−1.2j^+1.3k^)rad⁡s−2

b

−1.8i^rads−1

c

−1.2j^rads−2

d

1.3k^rads−2

answer is C.

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Detailed Solution

A body is free Torque about origin experienced by body isτ→=r→×F→=(2i^−3j^)×3i^+2j^+6k^ N-m=−18i^-12j^+13k^ But as the body is allowed to rotated only along y-axis, the y-component of torque only causes the angular acceleration of the body. Soα=τyI=−1210j^=−1.2j^rad/s2
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