IIT JEE , AIEEE Forum - Physics , Mechanics

ganesha1991

hcv page 79
Q 12

2Tcos

=mg
cos

= h/

d² +h²

2T h/

d² +h² = mg

or T = mg(

d² + h²) /2h

but then the answer given is
T = mg (

d² +4h² ) /4h

but how
explain
rates assured

anchitsaini

cos

= h/

d² +h²

change the above step to
--

cos

= h /

(d/2)² +h²

anandghegde

$\text{I think I have found a mistake.}\\ \\ \cos \theta = \frac{h}{\sqrt{\frac{d^2}{4}+h^2}}$

but you have taken it as d^2.

anandghegde

sorry again anchit...I was busy latexifying my post.......