91Ó°ÊÓ

• Intensity of light,$I=6.0\text{″¾°Â}/{\text{m}}^{\text{2}}\text{´Ç°ù }6.0×{10}^{−3}{\text{W/m}}^{\text{2}}$
• The radius of sphere,$r=2.0\text{ }\mathrm{μ}\text{m´Ç°ù }2.0×{10}^{−6}\text{ }\text{m}$
• The density of sphere,$\mathrm{ÒÏ}=5.0×{10}^3\text{ }{\text{kg/m}}^{\text{3}}$

Here, we need to use the concept of radiation pressure and the equation of force on area A due to the total absorption of light. Then, to get the acceleration, we can use the equation of Newton’s second law of motion.

Formulae:

The force due to Newton’s second law,$F=maâ‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots \left(1\right)$

The force acting on a reflective surface due to intensity of the incident beam,

$F=\frac{IA}{c}â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots \left(2\right)$

Area of a circle,$A=\mathrm{Ï€}{r}^2â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots \left(3\right)$

Volume of a sphere,$V=\frac{4}{3}\mathrm{Ï€}{r}^3â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots \left(4\right)$

The mass of a body due to its density and volume,$m=\mathrm{ÒÏ}Vâ‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots \left(5\right)$

Equation equations (1) and (2), we can get the acceleration equation as follows:

$a=\frac{IA}{mc}â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots â‹\dots \left(6\right)$

Now, using equation (4) in equation (5), we can get the mass of the sphere as follows:

$m=\mathrm{ÒÏ}×\frac{4}{3}\mathrm{Ï€}{r}^3$

Substituting the above value and equation (3) in equation (6), we get the magnitude of acceleration of the sphere as follows:

$\begin{array}{c}a=\frac{3I}{4\mathrm{ÒÏ}cr}\\ =\frac{3×6.0×{10}^{−3}}{4×5.0×{10}^3×3.0×{10}^8×2.0×{10}^{−6}}\\ =1.5×{10}^{−9}\text{ }{\text{m/s}}^{\text{2}}\end{array}$

Hence, the value of acceleration is.$1.5×{10}^{−9}\text{ }{\text{m/s}}^{\text{2}}$