Mechanical Properties Of Fluids

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Two bars A and B of circular cross-section and same volume made of same material are subjected under same tension. If the diameter of A is half that of B. The ratio of extension of A to B will be

16

2

$\frac{1}{4}$

4

Solution

$Y = \frac{stress}{strain} = \frac{\frac{F}{A}}{\frac{∆ l}{l}} \Rightarrow ∆ l = (\frac{F}{AY}) l But V = {πr}^{2} l so L = \frac{V}{{πr}^{2}} = \frac{FV}{{({πr}^{2})}^{2} Y} \propto \frac{1}{r^{4}} \frac{∆ l_{A}}{∆ I_{B}} = {(\frac{r_{B}}{r_{B}})}^{4} = {(\frac{r_{B}}{\frac{r_{B}}{2}})}^{4} = 16$

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Mechanical Properties Of Fluids

Two bars A and B of circular cross-section and same volume made of same material are subjected under same tension. If the diameter of A is half that of B. The ratio of extension of A to B will be

16

2

$\frac{1}{4}$

4

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Mechanical Properties Of Fluids

Two bars A and B of circular cross-section and same volume made of same material are subjected under same tension. If the diameter of A is half that of B. The ratio of extension of A to B will be 16 2 14 4

Some More Questions From Mechanical Properties of Fluids Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Two bars A and B of circular cross-section and same volume made of same material are subjected under same tension. If the diameter of A is half that of B. The ratio of extension of A to B will be

16

2

$\frac{1}{4}$

4