Verified Solution • 2025–26 Session moderate QJEEM24Ph7b2c: JEE Main 2024 Physics - Class 11 JEEM24Ph7b2c Solution: A bullet is fired into a fixed target looses one third of its… A bullet is fired into a fixed target looses one third of its velocity after travelling 4 cm4 \mathrm{~cm}4 cm . It penetrates further D×10−3 m\mathrm{D} \times 10^{-3} \mathrm{~m}D×10−3 m before coming to rest. The value of D\mathrm{D}D is : Multiple Choice Options A 23 B 32 C 42 D 52 Step-by-Step Verified Solution amp;v2−u2=2aSamp;(2u3)2=u2+2(−a)(4×10−2)amp;4u29=u2−2a(4×10−2)amp;−5u29=−2a(4×10−2)…(1)amp;0=(2u3)2+2(−a)(x)amp;−4u29=−2ax…(2)\begin{aligned}& v^2-u^2=2 a S \\& \left(\frac{2 u}{3}\right)^2=u^2+2(-a)\left(4 \times 10^{-2}\right) \\& \frac{4 u^2}{9}=u^2-2 a\left(4 \times 10^{-2}\right) \\& -\frac{5 u^2}{9}=-2 a\left(4 \times 10^{-2}\right) \ldots(1) \\& 0=\left(\frac{2 u}{3}\right)^2+2(-a)(x) \\& -\frac{4 u^2}{9}=-2 a x \ldots(2)\end{aligned}amp;v2−u2=2aSamp;(32u)2=u2+2(−a)(4×10−2)amp;94u2=u2−2a(4×10−2)amp;−95u2=−2a(4×10−2)…(1)amp;0=(32u)2+2(−a)(x)amp;−94u2=−2ax…(2) (1)/(2)(1)/(2)(1)/(2) amp;54=4×10−2xamp;x=165×10−2amp;x=3⋅2×10−2 mamp;x=32×10−3 m\begin{aligned}& \frac{5}{4}=\frac{4 \times 10^{-2}}{\mathrm{x}} \\& \mathrm{x}=\frac{16}{5} \times 10^{-2} \\& \mathrm{x}=3 \cdot 2 \times 10^{-2} \mathrm{~m} \\& \mathrm{x}=32 \times 10^{-3} \mathrm{~m}\end{aligned}amp;45=x4×10−2amp;x=516×10−2amp;x=3⋅2×10−2 mamp;x=32×10−3 m Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring