Two springs of equal lengths and equal cross – sectional areas are made of material whose Young’s modulii are in the ratio of 2 : 3. They are suspended and loaded with the same mass. When stretched and released, they will oscillate with time periods in the ratio of

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Solution

The correct option is A
Young’s modulus $Y = \frac{F}{A} . \frac{L}{l}$
Force constant $k = \frac{F}{l} = \frac{Y A}{L}$
Where l is the extension in the spring of original length L and cross-sectional area A when a force F = Mg is applied. Now, the time period of vertical oscillations is given by
$T = 2 π \sqrt{\frac{M}{k}} = 2 π \sqrt{\frac{M L}{Y A}} ∴ \frac{T_{1}}{T_{2}} = \sqrt{\frac{Y_{2}}{Y_{1}}} = \sqrt{\frac{3}{2}}$

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Two springs of equal lengths and equal cross – sectional areas are made of material whose Young’s modulii are in the ratio of 2 : 3. They are suspended and loaded with the same mass. When stretched and released, they will oscillate with time periods in the ratio of

The correct option is A Young’s modulus Y=FA.Ll Force constant k=Fl=YAL Where l is the extension in the spring of original length L and cross-sectional area A when a force F = Mg is applied. Now, the time period of vertical oscillations is given by T=2π√Mk=2π√MLYA∴T1T2=√Y2Y1=√32