Work Energy And Power Question 212
Question: Two identical beads of $ m=100 $ gram are connected by an inextensible massless string can slide along the two arms AC and BC of a rigid smooth wire frame in a vertical plane. If the system is released from rest, the kinetic energy of the first particle when they have moved by a distance of 0.1 m is $ x \times {10^{-3}}J $ . Find the value of $ x.( g=10m/s^{2} )~~~~~~ $
Options:
A) 18
B) 29
C) 36
D) 45
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Answer:
Correct Answer: C
Solution:
[c] let, the string make angle θ with the horizontal bar.
let, the first bead has velocity $=V _{1}$
Second bead velocity $=V _{2}$
now, equaliting the componets of velocity of the beads along the string,
$V _{1} cos \theta =V _{2}sin \theta$
$V _{1} \times \frac{0.9}{0.5}=V _{2} \times \frac{0.3}{0.5}$
$V _{2} =\frac{9}{3}V _{1}$
now, the second bead descend by 0.1 m so, from conservation of energy
mgh $=\frac{1}{2}m{V _1}^2 +\frac{1}{2}m{V _2}^2$
or $10 \times 0.1=\frac{1}{2}({V _1}^2+\frac{16}{9} {V _1}^2)$
or $2=\frac{25}{9}{V _1}^2$
or ${V _1}^2=\frac{18}{25}$
kinetic energy of the first bead
K1 $=\frac{1}{2} \times \frac{100}{1000} \times \frac{18}{25}$
∵$m=\frac{100}{1000}kq$
$=36 \times 10^-3$
$∵{V _1}^2=\frac{18}{25}$
The correct answer is 36
BETA
