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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Laws of Motion

A ball of mass 1 kg is at rest in position P by means of two light strings OP and RP. The string RP is now cut and the ball swing to position Q. If $\theta=45 ^{\circ}$ . Find the tension in the string OQ when RP is cut $(g=10 m/s^2)$

\[(a)\;10 \sqrt 2 N\quad (b)\; 5 \sqrt 2 N \quad (c)\;2 \sqrt 2 N \quad(d)\;5 N \]

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1 Answer

When the ball reaches Q the ball is temperorily at rest but begins to move perpendicular to OQ
Therefore Net force along OQ=0
$T_2= mg cos \theta$
$T_2=1 \times 10 \large\frac{1}{\sqrt 2}$
$\quad=5 \sqrt 2 N$
Hence b is the correct answer.

 

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

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