\('l'\) લંબાઈના સળિયાને શિરોલંબ અક્ષ સાથે એક છેડાને જોડેલો છે,અક્ષ એ \(w\) કોણીય ઝડપથી કરવાનું શરૂ કરે ત્યારે સળિયા અક્ષ સાથે \(\theta\) ખૂણો બનાવે છે, દ્રવ્યમાન કેન્દ્રને અનુલક્ષીને સમક્ષિતિજ અને શિરોલંબ બળ \(F_{H}\) અને \(F_{V}\) દ્વારા મળતું ટોર્ક દ્વારા કોણીય વેગમાનનો ફેરફારનો સમયદર \(\frac{ m \ell^{2}}{12} \omega^{2} \sin \theta \cos \theta\) મળે છે.તો \(\theta\)નું મૂલ્ય ..... .

A \(\cos \theta=\frac{g}{2 \ell \omega^{2}}\)
B \(\cos \theta=\frac{3 g}{2 \ell \omega^{2}}\)
C \(\cos \theta=\frac{2 g}{3 \ell \omega^{2}}\)
D \(\cos \theta=\frac{g}{\ell \omega^{2}}\)

Verified Solution

Answer & Solution

Correct Answer

(B) \(\cos \theta=\frac{3 g}{2 \ell \omega^{2}}\)

Step-by-step Solution

Detailed explanation

\(F_{V}=m g\) \(F_{H}=m \omega^{2} \frac{\ell}{2} \sin \theta\) \(mg \frac{\ell}{2} \sin \theta- m \omega^{2} \frac{\ell}{2} \sin \theta \frac{\ell}{2} \cos \theta=\frac{ m \ell^{2}}{12} \omega^{2} \sin \theta \cos \theta\) \(\cos \theta=\frac{3}{2} \frac{g}{\omega^{2} \ell}\)…

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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