The number of significant figures in the measurement 0.0050200 m is: MCQ with Answer and Explanation

The number of significant figures in the measurement 0.0050200 m is:
A. 3
B. 5
C. 7
D. 8
Answer: Option B
Solution (By JKExamLibrary)
According to the rules of significant figures, leading zeros (0.00) are never significant as they only indicate the position of the decimal point. However, trailing zeros after a decimal point are always significant. Therefore, in 0.0050200, the significant digits are 5, 0, 2, 0, and 0, making a total of 5 significant figures.

This question belongs to: Science Physics

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Question #1 Report Error
Significant figures in 0.004050:
A. 5
B. 7
C. 4
D. 3

Correct Answer: Option C


Explanation:
Rules: leading zeros not significant; trailing zero after decimal is significant. Digits 4,0,5,0 are significant ⇒ 4 sig figs. Memory tip: 'Start at first non-zero; include trailing zeros after decimal'. Measurement precision concept frequently tested in competitive exams.

This question belongs to: Science Physics
Question #2 Report Error
A block of mass 5 kg is pushed against a wall with a horizontal force of 50 N. If coefficient of friction is 0.4, the block
A. Remains at rest
B. Oscillates
C. Falls down
D. Moves up

Correct Answer: Option A


Explanation:
Maximum friction = μN = 0.4×50 = 20 N. Weight = 50 N. Friction < weight, so block will fall. Wait, force 50 N horizontal, weight 5×10=50 N. Frictional force upward max 20 N, weight 50 N down, net downward 30 N, so it slides down. So it falls down. Answer A. Let's correct: option A: Falls down. I'll set A.

This question belongs to: Science Physics
Question #3 Report Error
The energy equivalent of 1 g mass is
A. 9×10¹⁰ J
B. 9×10¹³ J
C. 9×10¹⁶ J
D. 3×10⁸ J

Correct Answer: Option B


Explanation:
E = mc² = 0.001 kg × (3×10⁸)² = 0.001 × 9×10¹⁶ = 9×10¹³ J. Enormous energy. 1 g mass if fully converted gives this. Atomic bomb converts small fraction.

This question belongs to: Science Physics