આકૃતિમાં દર્શાવ્યા અનુસાર, સમક્ષિતિજ સપાટી ઉપર રહેલ \(10\,kg\) ના દળને સમક્ષિતિજ સાથે \(30^{\circ}\) ના કોણે \(F\) બળથી ખેંચવામાં આવે છે.\(\mu_{ s }=0.25\) માટે,બળ \(F\) ના \(........\,N\) મહતમ મૂલ્ય સુધી બ્લોક સ્થિર રહેશે.[\(g=10\,ms^{-2}\) આપેલ છે.]

A \(33.3\)
B \(25.2\)
C \(20\)
D \(35.7\)

Verified Solution

Answer & Solution

Correct Answer

(B) \(25.2\)

Step-by-step Solution

Detailed explanation

\(N = Mg - F Sin 30^{\circ}\) \(= mg -\frac{ F }{2}=100-\frac{ F }{2}=\frac{200- F }{2}\) \(F Cos 30^{\circ}=\mu N\) \(\sqrt{3} \frac{ F }{2}=0.25 \times\left(\frac{200- F }{2}\right)\) \(4 \sqrt{3} F =200- F\) \(F =\frac{200}{4 \sqrt{3}+1}=25.22\)

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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