question_answer

A metre rod of mass 1-2 kg is pivoted at one end and hangs vertically. It is displaced through an angle of \[60{}^\circ \]. The increase in potential energy is........ Given : \[g=10\,m{{s}^{-2}}\]

A)  1 Joule           

B)         \[1.5\] Joule

C)  3 Joule             

D)         6 Joule

Correct Answer: C

Solution :

 The mass of the rod is distributed all over its length. In such a situation, the entire mass can be considered to be concentrated at its\[C.G\]. Now in the given problem to calculate the increase in potential energy, we must calculate the change in height of the\[C.G\]. If in general, the length of the rod is\[l\], then the initial position of the \[C.G.\] is \[\frac{l}{2}\] from the top end, as shown in the figure below: When the rod is displaced through \[\theta \] from the vertical, the vertical distance of the new position of \[C.G.\] from the top end is given by distance \[OC\] in the figure shown below: In triangle\[OCB\],          \[\frac{OC}{OB}=\cos \theta \] or,          base\[(OC)=\]hypotanuse\[(OB)\cos \theta \]                                   \[=\frac{l}{2}\cos \theta \] Thus, the displacement in the\[C.G.=OA-OC\]                                  \[=\frac{l}{2}-\frac{l}{2}\cos \theta =\frac{l}{2}(1-\cos \theta )\] Substituting\[l=1\,\,m\] and \[\theta ={{60}^{o}}\] we get displacement in\[C.G.\Delta h=\frac{1}{2}(1-\frac{1}{2})\]                                        \[=\frac{1}{4}=0.25m\] Now,   increase in\[PE=mg\Delta h\]                                        \[=1.2\times 9.8\times 0.8=3\,\,J\]