JAMB Physics 2007 Objective — Question 30
Question 30 of 50 from the Joint Admissions and Matriculation Board (JAMB) Physics 2007 Objective paper, with the correct answer and a full explanation.
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A string of length 4m is extended by 0.02m when a load of 0.4kg is suspended at its end. What will be the length of the string when the applied force is 15N? [g=10ms-2]
- A. 10sqrt3 m
- B. 5.05m
- C. 6.08m
- D. 4.08mCorrect
Explanation
Load = 0.4 x 10 = 4N produces extension 0.02m, so K = 4/0.02 = 200N/m. For 15N: extension = 15/200 = 0.075m. New length = 4 + 0.075 = 4.08m.
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