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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Plane
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An aeroplane is moving on a circular path with a speed of $250 \;kmhr^{-1}$ The change in its velocity in half revolution is

$\begin{array}{1 1}(A)\;0 \\(B)\;250\;km\;hr^{-1} \\(C)\;500\;km\;hr^{-1}\\(D)\;250 \sqrt 2 \;km\;hr^{-1} \end{array} $

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1 Answer

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If $\overrightarrow{v_1} =(250 km hr^{-1} ) \hat i$
then $\overrightarrow{v_2} =-(2250\;hm\;hr^{-1} ) \hat i$
Hence $ \Delta \overrightarrow{v}= (\overrightarrow{v_2}-\overrightarrow{v_1})$
$\qquad= (-500 km hr^{-1} ) \hat i $
So, $ | \Delta \overrightarrow{v}|=500 \;km\;hr^{-1} $
Hence C is the correct answer.
answered Jul 16, 2014 by meena.p

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