A wire of 9.8 10^ - 3 kg m^ - 1 passes over a frictionless light … - Vidyadip

MCQ

A wire of $9.8 \times {10^{ - 3}}kg{m^{ - 1}}$ passes over a frictionless light pulley fixed on the top of a frictionless inclined plane which makes an angle of $30°$ with the horizontal. Masses $m$ and $M$ are tied at the two ends of wire such that $m$ rests on the plane and $M$ hangs freely vertically downwards. The entire system is in equilibrium and a transverse wave propagates along the wire with a velocity of $100 ms^{-1}$. Chose the correct option $m =$ ..... $kg$

✓
$20$
B
$5$
C
$ 2$
D
$ 7$

✓

Answer

Correct option: A.

$20$

a
(a) $v = \sqrt {\frac{T}{\mu }} $

For equilibrium $Mg = mg \,sin 30 = T$

==> $M = \frac{m}{2}$==> $100 = \sqrt {\frac{{Mg}}{{9.8 \times {{10}^{ - 3}}}}} = \sqrt {\frac{{M(9.8)}}{{9.8 \times {{10}^{ - 3}}}}} $

==> $100 = \sqrt {M(1000)} $ ==> $M = 10kg$ and $m = 20kg$

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