Question

A neutron star of enormous density is rotating at the rate of one rotation per second. If theradius of the star is $20km$, then the acceleration in $m/s^2$ units for any particle situated at the equator of the star will be $({\pi }^2\approx 10)$

• A. $8\times {10}^5$
• B. $20\times {10}^3$
• C. $12\times {10}^6$
• D. $4\times {10}^8$

Answer 1

The correct option is A $8\times {10}^5$

Given : Frequency of rotation $f=1$ rps $R=20km=20000$ m

Angular velocity $w=2\pi f=2\pi \times 1=2\pi$ rad/s

Centripetal acceleration of the particle at equator $a=R{w}^2=R\times 4{\pi }^2=20000\times 4\times 10=8\times {10}^5$ $m/s^2$