A particle moves along circle of radius $(\frac{20}{π}) m$ with constant tangential acceleration. If the speed of the particle is $80 m / s$ at the end of the second revolution after motion has begun, the tangential acceleration is

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

$40 m / s^{2}$

$640 π m / s^{2}$

$40 π m / s^{2}$

$160 π m / s^{2}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$ω^{2} - ω_{0}^{2} = 2 α θ, θ = 2 π N$

$ω^{2} = 2 α (2 π \times 2) (ω_{0} = 0, N = 2)$

$ω = \sqrt{8 π α} rad / s$

$V = r ω \Rightarrow 80 = \frac{20}{π} ω \Rightarrow ω = 4 π rad / s$

$16 π^{2} = 8 π α \Rightarrow α = 2 π rad / s^{2}$

$a_{t} = r α = (\frac{20}{π}) (2 π) = 40 m / s^{2}$

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE