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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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A body A of mass 1kg is resting on a table and connected to body B and C as shown. The surface are smooth and string and spring are massless. The string is inextensible. Find the extension in the spring if force constant of spring $=R=50 N/m$ take $(g=10 m/s^2)$

\[(a)\;0.2 m \quad (b)\; 0.1 m \quad (c)\;0.05 m \quad(d)\;0.25 m \]

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1 Answer

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Free body diagram of mass 3 kg in
For the system net pulling force is due to mass B and massC
$F=(2+3) g=5 \times 10=50 N$
Totla mass to be pulled $=(1+2+3)$
$\qquad=6 kg$
Therefore $ a=\large\frac{50}{6} $$m/s^2$
From free body diagram
$30 -kx=ma$
$30-kx=3 \bigg( \large\frac{50}{6} \bigg)$
$30-kx=25$
$-kx=25-30=-5$
$x=\large\frac{5}{R}$
$\quad= \large\frac{5}{50}$
$\quad=0.1$
Hence b is the correct answer.

 

answered Jul 8, 2013 by meena.p
edited Jan 26, 2014 by meena.p
 

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