\(100 \,mg\) ના એક વિદ્યુતભારિત કણને \(1 \times 10^{5} \,NC ^{-1}\) જેટલી તીવ્રતા ધરાવતા નિયમિત વિદ્યુતક્ષેત્રની વિરૂદધ દિશામાં ફેકવામાં આવે છે. જે કણ પરનો વિદ્યુતભાર \(40 \,\mu C\) અને પ્રારંભિક વેગ \(200 \,ms ^{-1}\) હોય તો તે ક્ષણિક વિરામસ્થિતિમાં આવતા પહેલા કેટલું અંતર (\(m\) માં) કાપ્શે?

A \(1\)
B \(5\)
C \(10\)
D \(0.5\)

Verified Solution

Answer & Solution

Correct Answer

(D) \(0.5\)

Step-by-step Solution

Detailed explanation

Distance travelled by particle before stopping \(\frac{ V ^{2}}{2 a }= S \Rightarrow \frac{ v ^{2} m }{2 qE }\) \(\Rightarrow \frac{(200)^{2} \times 100 \times 10^{-6}}{2 \times 40 \times 10^{-6} \times 10^{5}}=0.5 \,m\)

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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