JEE MAIN - Physics (2023 - 30th January Morning Shift - No. 3)
A small object at rest, absorbs a light pulse of power $$20 \mathrm{~mW}$$ and duration $$300 \mathrm{~ns}$$. Assuming speed of light as $$3 \times 10^{8} \mathrm{~m} / \mathrm{s}$$, the momentum of the object becomes equal to :
$$1 \times 10^{-17} \mathrm{~kg} \mathrm{~m} / \mathrm{s}$$
$$0.5 \times 10^{-17} \mathrm{~kg} \mathrm{~m} / \mathrm{s}$$
$$3 \times 10^{-17} \mathrm{~kg} \mathrm{~m} / \mathrm{s}$$
$$2 \times 10^{-17} \mathrm{~kg} \mathrm{~m} / \mathrm{s}$$
Explanation
Assuming the small object as photon.
Momentum $$(p)=\frac{E}{C}$$
$$=\frac{20\times10^{-3}\times300\times10^{-9}}{3\times10^8}$$
$$=2\times10^{-17}$$ kg m/s
Comments (0)
