Search any question & find its solution
Question:
Answered & Verified by Expert
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, $g=9.8 \mathrm{~ms}^{-2}$ )
Options:
Solution:
1301 Upvotes
Verified Answer
The correct answer is:
$2.55 \mathrm{~N}$
Normal reaction $=$ Resultant of spring force $k x$ and weight $\mathrm{mg}$.
$$
=\sqrt{\left(k^2 x^2+m^2 g^2\right)}=2.55 \mathrm{~N}
$$
$$
=\sqrt{\left(k^2 x^2+m^2 g^2\right)}=2.55 \mathrm{~N}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.