Figure shows an arrangement of a rod of length l and mass M and a… - Vidyadip

MCQ

Figure shows an arrangement of a rod of length $l$ and mass $M$ and a bead of mass $m$ attached to a weightless string passing over a frictionless pulley. At $t = 0$ , the bead is in level with the lower end of the rod. The bead slides down the string with considerable friction and is opposite to the other end of the rod after $T\ second$ . Assuming friction between the beadand the string to be constant all through, the frictional force is

✓
$\frac{{2Mm}}{{\left( {M - m} \right)}}\,\frac{l}{{{t^2}}}$
B
$\frac{{\left( {M - m} \right)}}{{2Mm}}\,\frac{l}{{{t^2}}}$
C
$\frac{{2Mm}}{{\left( {M + m} \right)}}\,\frac{l}{{{t^2}}}$
D
$\frac{{\left( {M + m} \right)}}{{2Mm}}\,\frac{l}{{{t^2}}}$

✓

Answer

Correct option: A.

$\frac{{2Mm}}{{\left( {M - m} \right)}}\,\frac{l}{{{t^2}}}$

a
Let $a_{1}$ and $a_{2}$ be the accelerations of $M$ and $m$ respectively. Then

$M g-F=M a_{1}$ $...(i)$

$m g-F=m a_{2}$ $...(ii)$

Now, $\quad 1+\frac{1}{2} a_{2} t^{2}=\frac{1}{2} a_{1} t^{2}$

or $a_{1}=\frac{2 l}{t^{2}}+a_{2}$ $...(iii)$

Solving equations $( i ), (ii)\, and \,(iii),$ we get;

$F=\frac{2 M m}{(M-m)} \frac{l}{t^{2}}$

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