The time period of a simple pendulum is 2 s at a place where g=9.8 m/s². Its length is about MCQ with Answer and Explanation

Question Report Error

The time period of a simple pendulum is 2 s at a place where g=9.8 m/s². Its length is about

A. 2 m

B. 0.5 m

C. 1 m

D. 0.25 m

Answer: Option C

Solution (By JKExamLibrary)

T=2π√(L/g) => L = gT²/(4π²) ≈ (9.8×4)/(4×9.87) ≈ 1 m.

This question belongs to: Science Physics

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Practice More Physics Questions

Question #1 Report Error

Which measuring instrument is most suitable for measuring the internal diameter of a narrow tube with high precision?

A. Measuring tape

B. Screw gauge

C. Meter scale

D. Vernier caliper

View Answer & Solution Discuss

Correct Answer: Option D

Explanation:
Vernier caliper is ideal for measuring internal diameters of narrow tubes due to its internal jaws designed for such measurements. Screw gauge measures external dimensions like wire diameter. Meter scale and measuring tape lack the precision (least count ~1 mm) needed for narrow tubes. Vernier calipers typically have least count 0.01 mm or 0.02 mm. Practical tip: Always use internal jaws for internal measurements, external jaws for outer dimensions. This application-based question tests knowledge of instrument selection in experimental physics, commonly featured in practical examination sections.

This question belongs to: Science Physics