Question

An electron revolving in a circular path with initial angular velocity $6\text{rad/s}$ and constant angular acceleration of $4{\text{rad/s}}^2$. Find the ratio of ${\theta }_{5^{th}}:{\theta }_{7^{th}}:{\theta }_{9^{th}}$ (where ${\theta }_{n^{th}}$ is the angular displacement during $n^{th}$ second).

• A. $3:4:5$
• B. $1:2:3$
• C. $5:7:9$
• D. $7:9:11$

Answer 1

The correct option is A $3:4:5$

For the angular displacement for the $n^{th}$ second ${\theta }_{n^{th}}$ is given by
${\theta }_{n^{th}}={\omega }_0+\frac{\alpha }{2}(2n-1)$, where $\alpha$ is the constant angular acceleration and ${\omega }_0$ is the initial angular velocity

${\theta }_{n^{th}}={\omega }_0+\frac{\alpha }{2}(2n-1)$

hence the ratio of the angular displacement
${\theta }_{5^{th}}:{\theta }_{7^{th}}:{\theta }_{9^{th}}$, where initial angular velocity is $({\omega }_0=6\text{rad/s})$ and constant angular acceleration of $\left(4{\text{rad/s}}^2\right)$
$\Rightarrow 6+\frac{4}{2}\left(2\right(5)-1):6+\frac{4}{2}\left(2\right(7)-1):6+\frac{4}{2}\left(2\right(9)-1)$
$=24:32:40$
$=3:4:5$
Option D is the correct answer.