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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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A body is suspended by a string from the ceiling of an elevator. It is observed that the tension in the string is doubled when the elevator is accelerated. The acceleration will be

\[(a)\;4.9\; m/s^2 \quad (b)\;9.8\; m/s^2 \quad (c)\;19.6 \; m/s^2 \quad (d)\;2.45\; m/s^2\]

Can you answer this question?
 
 

1 Answer

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Let $a_0$ be acceleration of the elevator
$T=2 T_0$
Also $T=T_0\bigg(1+\large\frac{a_0}{g}\bigg)$
Therefore $ 2= 1+\large\frac{a_0}{g}$
=>$a_0=g=9.8m/s^2$
Hence b is the correct answer. 

 

answered Jul 14, 2013 by meena.p
edited Feb 10, 2014 by meena.p
 

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