Question

Consider the arrangement shown in figure. Disc is uniform of mass $m$ and of radius $R$. Intensity of parallel beam falling on disc is uniform. Disc is in equilibrium due to radiation falling on it. String is ideal and coefficient of absorption for disc is $\frac{1}{3}$. Choose the correct option(s) (Plane of disc is horizontal). (Take, $I$ as intenisty of light and $c$ as the speed of light)

• A. Tension in the string is $T=\frac{mg\tan \theta }{5}$
• B. Tension in the string is $T=\frac{2mg\tan \theta }{5}$
• C. Tension in the string is$T=\frac{I\pi R^2\cos \theta \sin \theta \left(\frac{1}{3}\right)}{c}$
• D. Tension in the string is $T=\frac{I\pi R^2{\cos }^2\theta (1+\frac{2}{3})}{c}$

Answer 1

Answer: (A), (C)

The correct options are
A Tension in the string is $T=\frac{mg\tan \theta }{5}$
C Tension in the string is$T=\frac{I\pi R^2\cos \theta \sin \theta \left(\frac{1}{3}\right)}{c}$

Force imparted by light of intenisty $I$ getting obsorbed by a plate of area $A$ is $\frac{IA\cos \theta }{c}$, where $c$ is the speed of light, and $\theta$ is the angle between the area and the direction of propagation of light.

Forces due to absorbtion (Only $\frac{1}{3}$rd):
$F_a=\frac{1}{3}\frac{IA\cos \theta }{c}$ along the direction of propagation.

Forces due to reflection (Only $\frac{2}{3}$rd):
Due to reflection, the force will only be in the direction normal to the area.
$F_b=\frac{2}{3}\frac{2IA{\cos }^2\theta }{c}$

The resultant components of these forces are shown in the figure.

$mg=\frac{I\pi R^2{\cos }^2\theta (1+\frac{2}{3})}{c}$
$T=\frac{I\pi R^2\cos \theta \sin \theta \left(\frac{1}{3}\right)}{c}$

$\frac{T}{mg}=\left[\frac{\frac{1}{3}}{1+\frac{2}{3}}\right]\tan \theta =\frac{1}{5}\tan \theta$
$T=\frac{mg.\tan \theta }{5}$