A bullet of mass 20 g fired from a gun of mass 5 kg with velocity 200 m/s. Recoil velocity of gun is MCQ with Answer and Explanation

A bullet of mass 20 g fired from a gun of mass 5 kg with velocity 200 m/s. Recoil velocity of gun is
A. 1 m/s
B. 0.8 m/s
C. 2 m/s
D. 0.4 m/s
Answer: Option B
Solution (By JKSSB Mock Tests)
Conservation of momentum: 0 = (0.02×200) + (5×v) => v = -4/5 = -0.8 m/s (magnitude 0.8 m/s).

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Practice More Physics Questions

Question #1
The velocity-time graph of a particle moving in a straight line is shown. The displacement of the particle from t=0 to t=4s is: [Graph: triangle from (0,0) to (2,10) to (4,0)]
A. 30 m
B. 10 m
C. 40 m
D. 20 m

Correct Answer: Option D


Explanation:
Displacement = area under velocity-time graph. The graph forms a triangle with base 4 s and height 10 m/s. Area = ½ × base × height = ½ × 4 × 10 = 20 m. Since velocity is always positive, displacement equals distance travelled. Graphical analysis is powerful: area gives displacement, slope gives acceleration. This triangle represents motion with uniform acceleration followed by uniform retardation. Exam tip: For piecewise linear v-t graphs, calculate area of each geometric segment separately. Such graph-based questions assess conceptual clarity in motion analysis.

This question belongs to: Science Physics
Question #2
In a p-n junction diode, the depletion region is created primarily due to:
A. Recombination of minority carriers.
B. Diffusion of majority charge carriers across the junction.
C. Heating of the semiconductor material.
D. Drift of electrons under a strong electric field.

Correct Answer: Option B


Explanation:
When a p-type and an n-type semiconductor are joined, electrons from the n-side (majority carriers) diffuse across the junction to the p-side, and holes from the p-side diffuse to the n-side due to concentration gradients. They recombine near the junction, leaving behind immobile ionized atoms (positive donor ions on n-side, negative acceptor ions on p-side). This uncovered charge region is the depletion layer.

This question belongs to: Science Physics
Question #3
A tuning fork A produces 5 beats/second with tuning fork B of frequency 256 Hz. If a little wax is applied to fork A, the beat frequency becomes 2 beats/second. The original frequency of fork A was:
A. 254 Hz
B. 258 Hz
C. 261 Hz
D. 251 Hz

Correct Answer: Option C


Explanation:
The beat frequency is |fA - fB|. So fA is either 256+5=261 Hz or 256-5=251 Hz. Adding wax increases mass, which strictly decreases frequency. If fA was 261, decreasing it brings it closer to 256, reducing beats to 2 (e.g., 258-256=2). If it was 251, decreasing it would increase the beats (e.g., |248-256|=8). Thus, it was 261 Hz.

This question belongs to: Science Physics