A bullet is fired into a fixed target looses one third of its velocity after travelling 4 cm. It penetrates further $$D \times 10^{-3}$$ m before coming to rest. The value of $$D$$ is :

Let the initial velocity be $$u$$. After travelling 4 cm, velocity = $$\frac{2u}{3}$$.

Using $$v^2 = u^2 - 2as$$ (deceleration $$a$$):

$$\frac{4u^2}{9} = u^2 - 2a(0.04)$$

$$2a(0.04) = u^2 - \frac{4u^2}{9} = \frac{5u^2}{9}$$

$$a = \frac{5u^2}{9 \times 0.08} = \frac{5u^2}{0.72}$$

For the bullet to come to rest from $$v = \frac{2u}{3}$$:

$$0 = \frac{4u^2}{9} - 2a \cdot D'$$

$$D' = \frac{4u^2}{18a} = \frac{4u^2}{18 \times \frac{5u^2}{0.72}} = \frac{4 \times 0.72}{18 \times 5} = \frac{2.88}{90} = 0.032 \text{ m} = 32 \times 10^{-3} \text{ m}$$

So $$D = 32$$.

The answer is $$32$$, which corresponds to Option (1).