Question

The radius of front wheel of arrangement shown in figure is $a$ and that of rear wheel is $b$. If a dust particle driven from a highest point of rear wheel alights on the highest point of front wheel, the velocity of arrangement is:

• A. ${\left[g\frac{(b-a)(c-a+b)}{(c+a-b)}\right]}^{1/2}$
• B. $\frac{g(c+a+b)}{4(b-a)}$
• C. ${\left[g\frac{(c+a-b)(c-a+b)}{4(b-a)}\right]}^{1/2}$
• D. ${\left[\frac{g(c-a+b)}{4(c-a)}\right]}^{1/2}$

Answer 1

The correct option is C ${\left[g\frac{(c+a-b)(c-a+b)}{4(b-a)}\right]}^{1/2}$

Let t be time of flight of the particle from P to Q. Since c is the distance between the centres of the two wheels, the horizontal distance between the wheels,
$c_h=\sqrt{c^2-(b-a{)}^2}$
If v is the velocity of the carriage, then
$c_h=vt$ (horizontal range)
or $vt=\sqrt{c^2-(b-a{)}^2}$ or $t=\frac{1}{v}\sqrt{c^2-(b-a{)}^2}$
In this time, the particle has covered the vertical distance $=2b-2a$

(The vertical distance between the highest points of both the wheels)
$\therefore 2b-2a=\frac{1}{2}g{t}^2$
or $2(b-a)=\frac{1}{2}g\frac{[c^2-(b-a{)}^2]}{v^2}$
or $v^2=\frac{g}{4}\frac{c^2-(b-a{)}^2}{(b-a)}$
or $v={\left[g\frac{(c+a-b)(c-a+b)}{4(b-a)}\right]}^{1/2}$