Verified Solution • 2025–26 Session moderate

QJEEM22Phf7f3: JEE Main 2022 Physics - Class 11 JEEM22Phf7f3 Solution: A ball is spun with angular acceleration = 6t 2 2t where t is…

A ball is spun with angular acceleration α\alpha = 6t2 −- 2t where t is in second and α\alpha is in rads −- 2. At t = 0, the ball has angular velocity of 10 rads −- 1 and angular position of 4 rad. The most appropriate expression for the angular position of the ball is :

Multiple Choice Options

  • A 32t4−t2+10t{3 \over 2}{t^4} - {t^2} + 10t
  • B t42−t33+10t+4{{{t^4}} \over 2} - {{{t^3}} \over 3} + 10t + 4
  • C 2t43−t36+10t+12{{2{t^4}} \over 3} - {{{t^3}} \over 6} + 10t + 12
  • D 2t4−t32+5t+42{t^4} - {{{t^3}} \over 2} + 5t + 4

Step-by-Step Verified Solution

α=dωdt=6t2−2t\alpha = {{d\omega } \over {dt}} = 6{t^2} - 2t

∫0ωdω=∫0t(6t2−2t)dt\int_0^\omega {d\omega = \int_0^t {(6{t^2} - 2t)dt} }

so ω=2t3−t2+10\omega = 2{t^3} - {t^2} + 10

and dθdt=2t3−t2+10{{d\theta } \over {dt}} = 2{t^3} - {t^2} + 10

so ∫4θdθ=∫0t(2t3−t2+10)dt\int_4^\theta {d\theta = \int_0^t {(2{t^3} - {t^2} + 10)dt} }

θ=t42−t33+10t+4\theta = {{{t^4}} \over 2} - {{{t^3}} \over 3} + 10t + 4

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