JEE Main 2019 — Physics solution

JEE Main2019Physics

A bullet of mass 20 g has an initial speed of 1 ms^–1, just before it starts penetrating a mud wall of thickness 20 cm. If the wall offers a mean resistance of 2.5 × 10^–2 N, the speed of the bullet after emerging from the other side of the wall is close to :

A
0.4 ms^–1
B
0.7 ms^–1
C
0.3 ms^–1
D
0.1 ms^–1

Solution

$v^{2} = u^{2} - 2 a S v^{2} = (1)^{2} - (2) [\frac{2.5 \times 10 ^{- 2}}{20 \times 10 ^{- 3}}] \frac{20}{100} v^{2} = 1 - \frac{1}{2} \Rightarrow v = \frac{1}{2} m/s = 0.7m/s$

← Back to full paper

Practice more Physics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →