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### Work Done in Hanging a Chain

A chain with increasing mass per unit length from one end as a function of distance x from lighter end as $\lambda (x) = kx$ lying horizontally is to be hung against a vertical wall in such a manner that the external agent has to do minimum work. The min. work done by external agent is given by,

(A) $\frac {1}{6} kgl^3$
(B) $\frac {1}{3} kgl^3$
(C) $\frac {1}{2} kgl^3$
(D) $\frac {2}{3} kgl^3$

Solution

A vertically hanging chain with heavy end upwards has much more potential energy in comparison to other situations. If the heavy end is downwards, the potential energy is minimum. Minimum work would be required if increase in potential energy is minimum.

If horizontal plane is taken as the reference level for potential energy, then the initial potential energy is 0.

Let us calculate the final potential energy of vertical chain with heavy end at the bottom just touching the horizontal plane.

Taking a small element dy of the chain at a height y above horizontal having mass dm.

$dU = dm.g.y$

$\Rightarrow dU = g.y.(\lambda dy)$

$\Rightarrow dU = g.y.[k(l-y)] dy$

$\Rightarrow U = gk\int\limits_0^l {y(l - y)dy}$

$\Rightarrow U = gk\left. {\left( {\frac{{{y^2}l}}{2} - \frac{{{y^3}}}{3}} \right)} \right|_0^l = gk\left( {\frac{{{l^3}}}{2} - \frac{{{l^3}}}{3}} \right) = \frac{1}{6}kg{l^3}$

Work done by external agent will cause this increase in potential energy.

Hence, Option (A).

### A man starts walking from the point P (-3, 4) ....

A man starts walking from the point P (-3, 4), touches the x-axis at R, and then turns to reach at the point Q (0, 2). The man is walking at a constant speed. If the man reaches the point Q in the minimum time, then $50 [(PR)^2 + (RQ)^2 ]$ is equal to _ _ _ _ . Solution For time to be minimum at constant speed, the directions must be symmetric. In other words, the angles made by PR and RQ with the vertical must be the same just like in the law of reflection in optics. $tan \theta = \frac {MP}{MR} = \frac {NQ}{NR}$ $\Rightarrow \frac {3-r}{4} = \frac {r}{2}$ $\Rightarrow r=1$ So, $R \equiv ( - 1,0)$ Now, $50(PR^2+RQ^2)=50[(4+16)+(1+4)]=1250$