Explanation
Given:
From the equation of motion,
$$v^2 – u^2 = 2 a S$$
$$\Rightarrow0 - 200^2 = 2 a S$$
$$\Rightarrow a = - \dfrac{2}{3}\times 10^6$$ (Here a is the deceleration)
Hence the force exerted by the sand on the bullet is
$$F = m (-a)$$
$$\Rightarrow F = 0.02 \times \dfrac{2}{3} \times 10^6$$
$$\Rightarrow F = 13.3 \times 10^3 N$$
Note: $$1 N = 10^5 dyne$$
$$\Rightarrow F = 13.3 \times 10^8 dyne$$
Hence option C is correct.
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