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Physics: Post your doubts here!

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Well Guys, kitkat <3 :P mehria
Mohammed salik


I am sorry for yesterday night.. I got it, it was v.v.v.v.v.v.v.v.v.v.v.v.v.v.v.v.v.v.v.v.v.v.v. SIMPLE I made it complicated, Abb shamajh aya hadu ne a ko 1 q liya tha... :)
I was so stupid, sorry for wasting your imp hours on this topic. :cry:
its ok... :) m happy that finally u got it... n see i told u that u need rest :p
 
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Well I am sure of Q31 - greater the charge the more the electric field. since both Experiences a same force F it should be F/q. or u could say like F=E*charge now since F is constant if u increase the electric field strength u decrease the charge

Can you please elaborate it a bit more,
also can you please tell why ans is not B
 
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yound modulus=stress/strain
E=Fl/Ae
(2.1x10^11)= (70 x 9.81) x (20) / ((3.2 x 10^6)x(200) x e)
e= 13734 / (( 2.1 x 10^11) x ( 6.4 x 10^-4))
e= 1.02 x 10^-4 m
e= 0.1 mm
 
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