A particle is rotating in a circular path and at any instant its motion can be described as θ = (5t^4)/(40) - (t^3)/(3) . The angular acceleration of the particle after 10 seconds is _______ rad/s ^2 .

Correct answer: C. 130 Given θ = (5t^4)/(40) - (t^3)/(3) = (t^4)/(8) - (t^3)/(3) Angular velocity ω = (dθ)/(dt) = (4t^3)/(8) - (3t^2)/(3) = (t^3)/(2) - t^2 Angular acceleration α = (dω)/(dt) = (3t^2)/(2) - 2t At t = 10 s, α = (3(10)^2)/(2) - 2(10) α = 150 - 20 = 130 rad/s ^2 Answer: 130

JEE Main 20262 AprilMorning Shift

Exam: JEE Main
Year: 2026
Date: 2 April
Shift: Morning Shift
Subject: Physics
Chapter: Rotational Motion

JEE Main 2026 · 2 April · Morning Shift · Physics · Rotational Motion

A particle is rotating in a circular path and at any instant its...

A particle is rotating in a circular path and at any instant its motion can be described as

\theta = \dfrac{5t^4}{40} - \dfrac{t^3}{3}

. The angular acceleration of the particle after

10

seconds is _______ rad/s

^2

Options

A $150$
B $120$
C $130$
D $170$

Correct answer

C. $130$

Step-by-step solution

Given

\theta = \dfrac{5t^4}{40} - \dfrac{t^3}{3} = \dfrac{t^4}{8} - \dfrac{t^3}{3}

Angular velocity

\omega = \dfrac{d\theta}{dt} = \dfrac{4t^3}{8} - \dfrac{3t^2}{3} = \dfrac{t^3}{2} - t^2

Angular acceleration

\alpha = \dfrac{d\omega}{dt} = \dfrac{3t^2}{2} - 2t

t = 10

\alpha = \dfrac{3(10)^2}{2} - 2(10)

\alpha = 150 - 20 = 130

rad/s

^2

Answer:

130

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