A body is moving with uniform speed in a circular path. Which statement is correct? MCQ with Answer and Explanation

A body is moving with uniform speed in a circular path. Which statement is correct?
A. Velocity is constant
B. Acceleration is zero
C. Acceleration is directed towards the center
D. No force acts on the body
Answer: Option C
Solution (By JKSSB Mock Tests)
In uniform circular motion, speed is constant but velocity direction changes continuously, requiring centripetal acceleration directed toward the center (a = v²/r). By Newton's second law, a centripetal force must act. Velocity isn't constant (direction changes); acceleration isn't zero; force is required. Memory tip: 'Uniform circular motion: constant speed, changing velocity, center-seeking acceleration'. This conceptual question tests understanding of circular dynamics, crucial for mechanics in competitive exams. Always distinguish speed (scalar) from velocity (vector) in motion analysis.

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Practice More Physics Questions

Question #1
A wire of resistance 10 Ω is stretched to double its length. Its new resistance is
A. 10 Ω
B. 5 Ω
C. 40 Ω
D. 20 Ω

Correct Answer: Option C


Explanation:
Resistance R = ρL/A. Volume constant: A ∝ 1/L. Stretching length to 2L reduces area to A/2. New R' = ρ(2L)/(A/2) = 4ρL/A = 4R = 40 Ω. Resistance proportional to (length)² when volume constant.

This question belongs to: Science Physics
Question #2
The apparent weight of a person in an elevator moving downwards with an acceleration 'a' (where a < g) is:
A. Equal to true weight
B. Zero
C. Less than true weight
D. Greater than true weight

Correct Answer: Option C


Explanation:
When an elevator moves downwards with acceleration 'a', the net downward force on the person is mg - R = ma, where R is the normal reaction (apparent weight). Solving for R gives R = m(g - a). Since g - a is less than g, the apparent weight (R) is less than the true weight (mg).

This question belongs to: Science Physics
Question #3
Projectile peak: velocity and acceleration are:
A. Both max
B. v=0, a=g downward
C. Both zero
D. v max, a=0

Correct Answer: Option B


Explanation:
At maximum height, vertical velocity = 0 momentarily, but acceleration = g downward throughout motion. Memory tip: 'Peak: v=0, a=g; gravity never switches off'. Kinematics concept frequently tested to correct misconception that acceleration vanishes at peak.

This question belongs to: Science Physics