Question

Sixteen beads in a string are placed on a smooth inclined plane to inclination $\theta =si{n}^{-1}\left(\frac{1}{3}\right)$, such that some of them lie along the incline, whereas the rest hang over the top of the plane. If $N$ beads lie along inclined plane then the acceleration of the bead (of the vertical portion) just below vertex is $\frac{g}{2}$ downwards. Value of $N$ is

Answer 1

If $n$ beads each of mass $m$ are hanging vertically, then
$nmg-T=\left(nm\right)a\Rightarrow nmg-T=\left(nm\right)\frac{g}{2}$
$\therefore T=n\left(\frac{mg}{2}\right).....\left(1\right)$
Also, $T-(16-n)mg\sin \theta =(16-n)m\left(\frac{g}{2}\right)$
$\Rightarrow n\left(\frac{mg}{2}\right)-(16-n)mg\sin \theta =(16-n)\left(\frac{mg}{2}\right)$
$\Rightarrow \left(\frac{n}{2}\right)-(16-n)\left(\frac{1}{3}\right)=\left(\frac{16-n}{2}\right)$
$\Rightarrow \frac{n}{2}=(16-n)\frac{5}{6}$
$\Rightarrow n=10$
So, no. of beads that lie along inclined plane
$N=16-n=16-10=6$