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A rope AB of linear mass density lamda is placed on a quarter vertical fixed disc of radius R as shown in the figure. The surface between the disc and rope is rough such that the rope is just is equilibrium. Gravitational acceleration is g. Choose the correct option (s). |
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Answer» Coefficient of static friction between rope and disc is `mu=1` `lamda Rg int_(0)^((pi)/2) cos theta d theta =mu lamda Rg int_(0)^((pi)/2) SIN theta d theta` `:.mu=1` At the position of maximum tension in the rope `lamda R d THETAG cos theta=mu(lamda R d theta g sin theta)` `:. theta=45^(@)` At any `theta` `dT=lamda R d theta g cos theta -mu lamda d theta g sin theta` `int_(0)^(T_("max")) dT =lamda Rg int_(0)^((pi)/4) (cos theta-sintheta)d theta` `T_("max")=lamda Rg[sintheta+costheta]_(0)^((pi)/4)=lamda Rg[1/(sqrt(2))+1/(sqrt(2))-1]=lamdaRg(sqrt(2)-1)` 
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