एक कण \(\mathrm{R}\) त्रिज्या के एक वृत्त पर एक समान चाल से गति कर रहा है तथा एक चक्कर पूर्ण करने में \(\mathrm{T}\) समय लेता है। यदि इसको एक समान चाल से क्षैतिज से \(\theta\) कोण पर प्रक्षेपित किया जाता है तो इसके द्वारा तय की गई अधिकतम ऊँचाई \(4 \mathrm{R}\) है। तब प्रक्षेपण कोण \(\theta\)_______ होगा।

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

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

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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