IIT JEE , AIEEE Forum - Physics , Mechanics

rtiit

Asmooth right circular cone of semi vertical angle

=tan^-15/12 is at rest on a horizontal plane. A rubber ring of mass 2.5 kg which requires force of 15N for an extension of 10cm is placed on cone. Find increase in the radius of ring in equilibrium.

rtiit

experts plz help me 2 solve this

rtiit

wat's this rubbish? Can't any of the experts answer my question? I'm really disappointed

Ksudha_iit

hey pls tell me the source ..ill tell u the answer..

rtiit

Can u give the ans with full details?

Ksudha_iit

i can but pls tell..me where u got this question from!!!! plsssssss....ii need it liberately...

rtiit

I got it from Bansal notes

Ksudha_iit

can i post it tomorrow????? it will take time...to explain coz....its regarded ..as the most toughest problem of the sheet of 11 ...class... ex 2... :P

bhupesh

keep the rubber on the cone and consider infinitely small element dm

Force on dm are normal force tension on both of its ends it weight
    so for equilibrium net force on it has to be zero

15 = k X 0.1   k = 150 n/m

now splitting all the forces

N sin alpha = dm g

2 T sin dtheta /2 = n cos alpha ( T is the force on the rubber due to extension )

( For the left part of the equation you can understand that tension due to extension on the rubber will act on both the ends since dm is small length of this segment will be r d theta , d theta is the small angle made by this element at the center hence )

2 d theta /2 = dmg cos alpha / sin alpha

T d theta = dm g cot alpha

N = t dtheta / cos alpha

solving

T = mg / 2 pi ( cot alpha )

x = mg / 2 pi ( cot alpha ) /k   ( T = kx )

R = l/ 2 pi ( l = length of the conductor )

d R = x/2 pi ( x = elongation )

solving the answer is 1 cm

rtiit

Can u explain with diagram & well resolved forces plzzz?

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