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Standard XII

Physics

Thermal Stress

The Young's m...

Question

The Young's modulus of steel is twice that of brass. Two wires of sample length and of same area of cross section, one of steel and another of brass are suspended from the same roof. If we want the lower ends of the wires to be at the same level, then the weights added to the steel and brass wires must be in the ratio of:

1 : 1

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1 : 3

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2 : 1

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4 : 1

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Solution

The correct option is A $2 : 1$
Young's modulus , $Y = \frac{F l}{A Δ l}$
here A, l, and $Δ l$ is same for both material , so, $F \propto Y$
Thus, $\frac{F_{s}}{F_{b}} = \frac{Y_{s}}{Y_{b}} = \frac{2 Y_{b}}{Y_{b}} = 2$

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The correct option is A 2:1Young's modulus , Y=FlAΔlhere A, l, and Δl is same for both material , so, F∝YThus, FsFb=YsYb=2YbYb=2

The correct option is A $2 : 1$
Young's modulus , $Y = \frac{F l}{A Δ l}$
here A, l, and $Δ l$ is same for both material , so, $F \propto Y$
Thus, $\frac{F_{s}}{F_{b}} = \frac{Y_{s}}{Y_{b}} = \frac{2 Y_{b}}{Y_{b}} = 2$