કેપેસીટરની બે સમાંતર પ્લેટ વચ્ચે \('\alpha'\) કોણ રચાય તે પ્રમાણે \(K _{1}\) ગતિ ઊર્જા ધરાવતો ઈલેક્ટ્રોન બંને પ્લેટની વચ્ચે પ્રવેશે છે. તે પ્લેટોને \(K _{2}\) જેટલી ગતિ ઊર્જા સાથે \(' \beta '\) કોણે છોડે છે. તો ગતિ ઊર્જાનો ગુણોત્તર \(K _{1}: K _{2} ......\) થશે.

A \(\frac{\sin ^{2} \beta}{\cos ^{2} \alpha}\)
B \(\frac{\cos ^{2} \beta}{\cos ^{2} \alpha}\)
C \(\frac{\cos \beta}{\cos \alpha}\)
D \(\frac{\cos \beta}{\sin \alpha}\)

Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{\cos ^{2} \beta}{\cos ^{2} \alpha}\)

Step-by-step Solution

Detailed explanation

velocity along the plate will not change. \(\therefore v _{1} \cos \alpha= v _{2} \cos \beta\) \(\frac{ K _{1}}{ K _{2}} \Rightarrow \frac{ v _{1}^{2}}{ v _{2}^{2}}=\frac{\cos ^{2} \beta}{\cos ^{2} \alpha}\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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