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A fluid container is containing a liquid of density `rho` is accelerating upward with acceleration a along the inclined plane of inclination `alpha` as shown in the figure. Then the angle of inclination `theta` of free surface isA. `tan^(-1)[(a)/(gcosalpha)]`B. `tan^(-1)[(a+gsinalpha)/(gcosalpha)]`C. `tan^(-1)[(a-gsinalpha)/(g(1+cosalpha))]`D. `tan^(-1)[(a-gsinalpha)/(g(1-cosalpha))]` |
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Answer» Correct Answer - B Apply pseudo force on a particle of mass `m`. Net force along the surface is zero. `macostheta=mgcos[90-(theta-alpha)]` `(a)/(g)costheta=gsin(theta-alpha)` `(a)/(g)=(sinthetacosalpha)/(costheta)-(costhetasinalpha)/(costheta)` `tantheta=((a)/(g)+sinalpha)/(cosalpha)=(a+gsinalpha)/(gcosalpha)` |
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