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. Continue reading Work Done in Hanging a Chain

Centripetal & Centrifugal Forces

In circular motion, why don’t centripetal and centrifugal forces cancel each other? Well, they don’t cancel each other because they are not present simultaneously for the same observer. Only one is present at a time for one observer. If you take ground observer – inertial observer to be more precise – then only centripetal force is applicable and of course it is applicable towards the centre. There is no centrifugal force present here. Since both are not present simultaneously, they don’t cancel each other. In case of non-inertial or accelerated observer, only centrifugal force is present – it acts away from the centre. There is no centripetal force applicable. Since both are not present simultaneously, they can’t cancel each other.

Prove that,

${\sin ^3}30^\circ – {\sin ^3}18^\circ = {\sin ^2}18^\circ $

We have, ${\sin ^3}18^\circ + {\sin ^2}18^\circ = {\sin ^2}18^\circ (\sin 18^\circ + 1)$

$ = {\left( {\frac{{\sqrt 5 – 1}}{4}} \right)^2}\left( {\frac{{\sqrt 5 – 1}}{4} + 1} \right)$

$ = \frac{{6 – 2\sqrt 5 }}{{16}} \times \frac{{\sqrt 5 + 3}}{4}$

$ = \frac{{3 – \sqrt 5 }}{8} \times \frac{{3 + \sqrt 5 }}{4}$

$ = \frac{{9 – 5}}{{8 \times 4}} = \frac{1}{8} = {\sin ^3}30^\circ $

$ \Rightarrow {\sin ^2}18^\circ = {\sin ^3}30^\circ – {\sin ^3}18^\circ $

$3x+4y=10$

$x>0,y>0$

$(x^2y^3)_{max}=?$

For positive numbers, A.M. $\geq $ G.M.

So, $\frac{{\frac{{3x}}{2} + \frac{{3x}}{2} + \frac{{4y}}{3} + \frac{{4y}}{3} + \frac{{4y}}{3}}}{5} \ge {\left[ {{{\left( {\frac{{3x}}{2}} \right)}^2}{{\left( {\frac{{4y}}{3}} \right)}^3}} \right]^{1/5}}$

$ \Rightarrow \frac{{3x + 4y}}{5} \ge {\left( {\frac{9}{4}{x^2} \times \frac{{4 \times 16}}{{9 \times 3}}{y^3}} \right)^{1/5}}$

$ \Rightarrow \frac{{10}}{5} \ge {\left( {\frac{{16}}{3}{x^2}{y^3}} \right)^{1/5}}$

$ \Rightarrow {2^5} \ge \frac{{16}}{3}{x^2}{y^3}$

$ \Rightarrow {x^2}{y^3} \le 6$

$ \Rightarrow {\left( {{x^2}{y^3}} \right)_{\max }} = 6$

Find G.P. such that

$\sum\limits_{r = 1}^3 {{t_r}} = 42$

$\sum\limits_{r = 1}^3 {{{({t_r})}^2}} = 1092$

Using summation formula of G.P. having 1st term as a and common ratio as r we have,

$\frac{{a({r^3} – 1)}}{{(r – 1)}} = 42$ ……..(A)

Squaring the terms of the G.P. yields another G.P. having 1st term as $a^2$ and common ratio as $r^2 $. Thus,

$\frac{{{a^2}({r^6} – 1)}}{{({r^2} – 1)}} = 1092$

$ \Rightarrow \frac{{{a^2}({r^3} – 1)({r^3} + 1)}}{{(r – 1)(r + 1)}} = 1092$ Continue reading Find G.P. such that

$\sum\limits_{r = 1}^3 {{t_r}} = 42$

$\sum\limits_{r = 1}^3 {{{({t_r})}^2}} = 1092$

Solve for x,

${4^x} – {3^{x – \frac{1}{2}}} = {3^{x + \frac{1}{2}}} – {2^{2\left( {x – \frac{1}{2}} \right)}}$

We have, ${2^{2x}} + \frac{{{2^{2x}}}}{2} = {3^x}\sqrt 3 + \frac{{{3^x}}}{{\sqrt 3 }}$

$ \Rightarrow \frac{3}{2}{.2^{2x}} = \frac{4}{{\sqrt 3 }}{.3^x}$

$ \Rightarrow \frac{{{2^{2x}}}}{8} = \frac{{{3^x}}}{{3\sqrt 3 }}$

$ \Rightarrow {2^{2x – 3}} = {3^{x – \frac{3}{2}}}$

$ \Rightarrow {2^{2\left( {x – \frac{3}{2}} \right)}} = {3^{x – \frac{3}{2}}}$

$ \Rightarrow {4^{x – \frac{3}{2}}} = {3^{x – \frac{3}{2}}}$

The above is only possible if $x – \frac{3}{2} = 0$ or $x=\frac {3}{2}$.

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