JEE MAIN - Physics (2021 - 25th February Evening Shift - No. 2)
An electron with kinetic energy K1 enters between parallel plates of a capacitor at an angle '$$\alpha$$' with the plates. It leaves the plates at angle '$$\beta$$' with kinetic energy K2. Then the ratio of kinetic energies K1 : K2 will be :
$${{{{\cos }^2}\beta } \over {{{\cos }^2}\alpha }}$$
$${{\cos \beta } \over {\cos \alpha }}$$
$${{{{\sin }^2}\beta } \over {{{\cos }^2}\alpha }}$$
$${{\cos \beta } \over {\sin \alpha }}$$
Explanation
_25th_February_Evening_Shift_en_2_1.png)
$$ \because $$ $${v_1}\cos \alpha = {v_2}\cos \beta $$
$${{{v_1}} \over {{v_2}}} = {{\cos \beta } \over {\cos \alpha }}$$
Then the ratio of kinetic energies
$${{{k_1}} \over {{k_2}}} = {{{1 \over 2}m{v_1}^2} \over {{1 \over 2}m{v_2}^2}} = {\left( {{{{v_1}} \over {{v_2}}}} \right)^2} = {\left( {{{\cos \beta } \over {\cos \alpha }}} \right)^2}$$
$$ \Rightarrow $$ $${{{k_1}} \over {{k_2}}} = {{{{\cos }^2}\beta } \over {{{\cos }^2}\alpha }}$$
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