A) \[x=1+xy\frac{dy}{dx}+\frac{{{(xy)}^{2}}}{2!}+{{\left( \frac{xy}{dx} \right)}^{2}}+\frac{{{(xy)}^{3}}}{3!}{{\left( \frac{dy}{dx} \right)}^{3}}\]
B) \[y={{\log }_{e}}(x)+C\]
C) \[y={{({{\log }_{e}}x)}^{2}}+C\]
D) \[y=\pm \sqrt{{{\log }_{e}}x{{)}^{2}}+2C}\]
Correct Answer: C
Solution :
\[\frac{l}{2}\]Linear acceleration for rolling, \[\frac{l}{6}\] For cylinder,\[\frac{4l}{3}\] \[\frac{{{\mu }_{0}}}{2\pi }\ln \left( \frac{d-a}{a} \right)\]\[\frac{{{\mu }_{0}}}{\pi }\ln \left( \frac{d-a}{a} \right)\] For rotation, the torque\[\frac{3{{\mu }_{0}}}{\pi }\ln \left( \frac{d-a}{a} \right)\](where, F = force of friction) But \[\frac{3{{\mu }_{0}}}{3\pi }\ln \left( \frac{d-a}{a} \right)\] \[2:\pi \] \[\pi :2\]\[\pi :4\]where N is normal reaction. \[4:\pi \] \[\frac{L}{R}\]For rolling without slipping of a roller down, the inclined plane, \[\frac{R}{L}\]You need to login to perform this action.
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