એક કણ \(R\) ત્રિજ્યાના વર્તુળમાં સમાન ઝડપથી ગતિ કરે છે અને એક પરિક્રમણ પૂર્ણ કરવા માટે \(\mathrm{T}\) સમય લે છે. જો આ કણને તે જ ઝડપથી ક્ષિતિજ સાથે \(\theta\) ખૂણે પ્રક્ષિપ્ત કરવામાં આવે, તો તેના દ્વારા પ્રાપ્ત મહત્તમ ઊંચાઈ \(4 R\) જેટલી છે. તો પ્રક્ષેપણ કોણ \(\theta\) _________ દ્વારા અપાય છે.

A \(\sin ^{-1}\left[\frac{2 g T^2}{\pi^2 R}\right]^{\frac{1}{2}}\)
B \(\sin ^{-1}\left[\frac{\pi^2 R}{2 \mathrm{gT}^2}\right]^{\frac{1}{2}}\)
C \(\cos ^{-1}\left[\frac{2 \mathrm{gT}^2}{\pi^2 \mathrm{R}}\right]^{\frac{1}{2}}\)
D \(\cos ^{-1}\left[\frac{\pi R}{2 g T^2}\right]^{\frac{1}{2}}\)

Verified Solution

Answer & Solution

Correct Answer

(A) \(\sin ^{-1}\left[\frac{2 g T^2}{\pi^2 R}\right]^{\frac{1}{2}}\)

Step-by-step Solution

Detailed explanation

\(\frac{2 \pi R}{T}=V\) મહત્તમ ઊંચાઈ \(H=\frac{v^2 \sin ^2 \theta}{2 g}\) \(4 R=\frac{4 \pi^2 R^2}{T^2 2 g} \sin ^2 \theta\) \(\sin \theta=\sqrt{\frac{2 g T^2}{\pi^2 R}}\) \(\theta=\sin ^{-1}\left(\frac{2 g T^2}{\pi^2 R}\right)^{\frac{1}{2}}\)

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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