JEE MAIN - Physics (2024 - 30th January Evening Shift - No. 18)
If the total energy transferred to a surface in time $$\mathrm{t}$$ is $$6.48 \times 10^5 \mathrm{~J}$$, then the magnitude of the total momentum delivered to this surface for complete absorption will be:
$$2.16 \times 10^{-3} \mathrm{~kg} \mathrm{~m} / \mathrm{s}$$
$$2.46 \times 10^{-3} \mathrm{~kg} \mathrm{~m} / \mathrm{s}$$
$$1.58 \times 10^{-3} \mathrm{~kg} \mathrm{~m} / \mathrm{s}$$
$$4.32 \times 10^{-3} \mathrm{~kg} \mathrm{~m} / \mathrm{s}$$
Explanation
$$\mathrm{p=\frac{E}{C}=\frac{6.48 \times 10^5}{3 \times 10^8}=2.16 \times 10^{-3}}$$
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