Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
0 votes

A heavy particle hanging from fixed point by light inextensible string of length l is projected horizontally with speed $\sqrt {gl}$. Find speed of particle at the instant of the motion when tension in the string is equal to weight of particle

\[(a)\;v=\sqrt {\frac{g}{3l}} \quad (b)\;v=\sqrt {\frac{gl}{3}} \quad (c)\;v=\frac{gl}{3}\quad (d)\;v= \frac{g}{3l}\]

Can you answer this question?

1 Answer

0 votes
Let $T=mg$ when angle is $\theta$
$h=l(1-\cos \theta)$
Using concervation of energy at A and B
$\large\frac{1}{2} $$m(u^2-v^2)=mgh$
$u=gl,v=$speed of particle at B
$T-mg \cos \theta=\large\frac{mv^2}{l}$
$mg-mg \cos \theta=\large\frac{mv^2}{l}$
$v^2=gl(1-\cos \theta)$-----(1)
$gl(1-\cos \theta)=gl-2gl(1-\cos \theta)$
$\cos \theta= \large\frac{2}{3}$ , Substituting in (1)
Therefore $v=\sqrt {\large\frac{gl}{3}}$
answered Jul 18, 2013 by meena.p

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App