Question

The elongation in a metallic rod hinged at one end and rotating in a horizontal plane becomes four times of the initial value. The angular velocity of rotation becomes

• A.
  half of initial value
• B.
  two times the initial value
• C.
  one−third of initial value
• D.
  four times the initial value

Answer 1

The correct option is B

two times the initial value
Draw rough diagram for given situation


Find the angular velocity of rotation

$-{\int }_T^0\Delta T={\int }_0^l\frac{m}{l}dx{w}^{2}x$
$T=\frac{m}{l}{w}^2\frac{x^2}{2}$
$Y=\frac{Fl}{A\Delta l}$
$\Delta l=\frac{Fl}{AY}$
$\Delta l=\frac{\frac{m{w}^2{x}^2}{l2}dx}{AY}$
$\Delta l=\frac{m}{l}\frac{w^2{l}^3}{6AY}$
$\Delta l=\frac{\rho w^2{l}^3}{6Y}$
$\Delta l=\alpha w^2$

If the elongation is four times, then
$w_2=2w_1$