The extension in a spring when 5g wt was hung from it was 0.56cm. If Hooke's law is obeyed, What is the extension caused by a load of 20g wt?
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Correct Answer: Option E
Explanation:
Extension & Load
Extension = K constant
\(\frac{\text{load}}{0.56cm}\) = \(\frac{X}{20g}\)
X = 2.24cm
Extension & Load
Extension = K constant
\(\frac{\text{load}}{0.56cm}\) = \(\frac{X}{20g}\)
X = 2.24cm