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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Straight Line
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An elevator car whose floor to ceiling distance is 2.7 m starts ascending with constant acceleration of $1.2 m/s^2$ after 2 seconds of the start a bolt begins falling from ceiling of elevator. What is the time after which the bolt hits the floor of the elevator.

\[(a)\;0.8 s\quad (b)\;0.7 s \quad (c)\;0.6s \quad (d)\;0.5 s\]

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1 Answer

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Considering elevator at rest the relative acceleration of bolt is
$a_r=9.8+1.2=11m/s^2$
Now after 2 seconds after the car starts moving with acceleration $1.2 m/s^2$ its velocity will be
$v_{\large car}$$=at=(1.2) \times 2=2.4 m/s$
But w.r.t elevator the bolt has zero velocity , and acceleration $=11 m/s^2$
Therefore $t=\sqrt {\large\frac {2s}{a}}$
$\qquad\qquad=\sqrt {\large\frac{2 \times 2.7}{11}}$$=0.7 s$
Hence b is the correct answer
answered Jun 27, 2013 by meena.p
edited May 24, 2014 by lmohan717
 

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