Question:
An elastic string has length $ \beta $ when subjected to 5 N tension. Its length is a when tension 4 N. When subjected to a tension of 9 N, its length will become
An elastic string has length $ \beta $ when subjected to 5 N tension. Its length is a when tension 4 N. When subjected to a tension of 9 N, its length will become
Updated On: Jun 14, 2022
- $ 9(\beta -\alpha ) $
- $ 5\alpha -4\beta $
- $ 5\beta -4\alpha $
- $ \beta -\alpha $
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The Correct Option is C
Solution and Explanation
: $ k= $ force constant of spring = force/length. Increase in length due to force = force/ $ k=\frac{F}{k} $ Original length $ =L $ $ \therefore $ Final length $ =L+\frac{F}{k} $ $ \therefore $ $ \beta =L+\frac{5}{k} $ $ \alpha =L+\frac{4}{k} $ $ \therefore $ $ \beta -\alpha =\frac{I}{k} $ ?..(i) $ \therefore $ $ \beta =L+5(\beta -\alpha ) $ or $ \beta =L+5\beta -5\alpha $ or $ 5\alpha -4\beta =L $ When force $ =9\text{ }N, $ final length be $ l $ . $ l=L+\frac{9}{k}=(5\alpha -4\beta )+9(\beta -\alpha ) $ $ =5\alpha -4\beta +9\beta -9\alpha =5\beta -4\alpha $
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Concepts Used:
Hooke’s Law
Hooke’s Law states that for small deformities, the stress and strain are proportional to each other. Thus,
Stress ∝ Strain
Stress = k × Strain … where k is the Modulus of Elasticity.
When a limited amount of Force or deformation is involved then concept of Hooke’s Law is only applicable . If we consider the fact, then we can deviate from Hooke's Law. This is because of their extreme Elastic limits.