Post-UTME Physics 2007 Objective — Question 8
Question 8 of 10 from the Post-UTME Screening (Post-UTME) Physics 2007 Objective paper, with the correct answer and a full explanation.
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The period of a S.H.M is 1.20 sec when the length of the pendulum is 40cm, find the period when the length is Increased to 90cm
- A. 1.40s
- B. 1.60s
- C. 1.80sCorrect
- D. 2.00s
Explanation
Period is proportional to the square root of length: T2 = 1.20 x sqrt(90/40) = 1.20 x 1.5 = 1.80s.
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