AP EAMCET · PHYSICS · Laws of Motion
A bead of mass \(100 \mathrm{~g}\) is attached to one end of a spring of natural length \(L\) and spring constant \(k=\frac{(\sqrt{3}+1) m g}{L}\), where \(m\) is the mass of bead. The other end of the spring is fixed at point \(A\) on a smooth vertical ring of radius \(R\) as shown in the figure. The normal reaction at \(B\) just after it is released to move is (take, \(g=9.8 \mathrm{~ms}^{-2}\) )

- A \(1.73 \mathrm{~N}\)
- B ,
\(2.23 \mathrm{~N}\) - C \(2.44 \mathrm{~N}\)
- D \(2.55 \mathrm{~N}\)
Answer & Solution
Correct Answer
(D) \(2.55 \mathrm{~N}\)
Step-by-step Solution
Detailed explanation
Normal reaction \(=\) Resultant of spring force \(k x\) and weight \(\mathrm{mg}\). \[ =\sqrt{\left(k^2 x^2+m^2 g^2\right)}=2.55 \mathrm{~N} \]
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