\(l\) લંબાઈની દોરી ધરાવતાં અને \(m\) દ્રવ્યમાન ગોલક ધરાવતા એક સાદા લોલને કોઇ એક નાના કોણ \(\theta_0\) થી છોડવામાં આવે છે. ખરબચડી સમક્ષિતિજ સપાટી પર મુકેલ \(M\) દ્રવ્યમાનના ચોસલાને તે તેના નિમ્ન બિંદુ પર સ્થિતિસ્થાપક રીતે અથડાય છે. તે પાછો ફેંકાય છે અને કોણ \(\theta_1\) સુધી પહોંચે છે, તો \(M\) દળ કેટલું હશે?

A \(\frac{m}{2}\left( {\frac{{{\theta _0} + {\theta _1}}}{{{\theta _0} - {\theta _1}}}} \right)\)
B \(m\left( {\frac{{{\theta _0} - {\theta _1}}}{{{\theta _0} + {\theta _1}}}} \right)\)
C \(m\left( {\frac{{{\theta _0} + {\theta _1}}}{{{\theta _0} - {\theta _1}}}} \right)\)
D \(\frac{m}{2}\left( {\frac{{{\theta _0} - {\theta _1}}}{{{\theta _0} + {\theta _1}}}} \right)\)

Verified Solution

Answer & Solution

Correct Answer

(C) \(m\left( {\frac{{{\theta _0} + {\theta _1}}}{{{\theta _0} - {\theta _1}}}} \right)\)

Step-by-step Solution

Detailed explanation

Just before collision speed of \(m\) \(v = \sqrt {2gL\left( {1 - \cos {\theta _ \circ }} \right)} \) Just after collision speed os \( M \) \({v_1} = \sqrt {2gL\left( {1 - \cos {\theta _1}} \right)} \)…

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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