A bullet is fired into a fixed target loses one third of its velocity after travelling 4 cm. It penetrates further D × 10⁻³ m before coming to rest. The value of D is: (a) 23 (b) 32 (c) 42 (d) 52

Answer: (b). Constant resistance: distance ∝ KE lost. v → 2v/3: KE goes K → 4K/9, so 5K/9 is lost in 4 cm. Remaining 4K/9 needs 4 × 4/5 = 3.2 cm = 32 × 10⁻³ m. So the answer is (b) D = 32.

A bullet is fired into a fixed target loses one third of its velocity… | JEE Main 2024 (27 Jan, Shift 2)

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Work-Energy Theorem

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

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Answer: (b)

Constant resistance: distance $\propto$ KE lost.
$v \to \frac{2v}{3}$ : KE goes $K \to \frac{4K}{9}$ , so $\frac{5K}{9}$ is lost in $4\text{ cm}$ .
Remaining $\frac{4K}{9}$ needs $4 \times \frac{4}{5} = 3.2\text{ cm} = 32 \times 10^{-3}\text{ m}$ .
So the answer is (b) $D = 32$ .

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