Question

A charge particle entres at point ${}^{\prime }{A}^{\prime }$ and comes out from ${}^{\prime }{B}^{\prime }$. Its velocity makes angle ${}^{\prime }{\alpha }^{\prime }$ and ${}^{\prime }{\beta }^{\prime }$ with electric field at these two points. Ratio of kinetic energy of changed particle at these two points will be

• A. $\frac{{\sin }^2\alpha }{{\sin }^2\beta }$
• B. $\frac{{\sin }^2\beta }{{\sin }^2\alpha }$
• C. $\frac{{\cos }^2\alpha }{{\cos }^2\beta }$
• D. $\frac{{\cos }^2\beta }{{\cos }^2\alpha }$

Answer 1

The correct option is B $\frac{{\sin }^2\beta }{{\sin }^2\alpha }$

Let initial speed is $4$ ]

and final speed is $V$

$\begin{array}{c}\Rightarrow u\sin \alpha =v\sin \beta \\ \Rightarrow \frac{u}{v}=\frac{\sin \beta }{\sin \alpha }\\ \Rightarrow \frac{\frac{1}{2}m{u}^2}{\frac{1}{2}m{v}^2}=\frac{{\sin }^2\beta }{{\sin }^2\alpha }\\ hence,optionBiscorrectanswer.\end{array}$