Motion in a Plane - Result Question 50
«««< HEAD:content/english/neet-pyq-chapterwise/physics/motion-in-a-plane/motion-in-a-plane-result-question-50.md
53. A particle moves along a circle of radius $(\frac{20}{\pi}) m$ with constant tangential acceleration. If the velocity of the particle is $80 m / s$ at the end of the second revolution after motion has begun, the tangential acceleration is [2003]
======= ####53. A particle moves along a circle of radius $(\frac{20}{\pi}) m$ with constant tangential acceleration. If the velocity of the particle is $80 m / s$ at the end of the second revolution after motion has begun, the tangential acceleration is [2003]
3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed:content/english/neet-pyq-chapterwise/physics/motion-in-a-plane/motion-in-a-plane—result-question-50.md (a) $40 \pi m / s^{2}$
(b) $40 m / s^{2}$
(c) $640 \pi m / s^{2}$
(d) $160 \pi m / s^{2}$
Show Answer
Answer:
Correct Answer: 53. (b)
Solution:
- (b) Given, $r=\frac{20}{\pi} m$
$v_j=80 m / sec \Rightarrow w_f=\frac{8 \theta \pi}{2 \theta} \Rightarrow 4 \pi$
$\theta=2$ re $N=4 \pi$ radian
From equation,
$w_f^{2}=w_0^{2}+2 \alpha \theta \quad[\because w_0=0]$
$(4 \pi)^{2}=0+2 . \alpha .4 \pi$
$\alpha=2 \pi$
tangential acceleration
$ \begin{aligned} & a_t=\alpha \cdot r \\ & a t=2 \pi \cdot \frac{20}{\pi}=40 m / sec^{2} \end{aligned} $