The Young’s modulus of steel is twice that of brass. Two wires of same 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 w

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Solution

Q.
The Young’s modulus of steel is twice that of brass. Two wires of same 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 [2015]

1 4 : 1

2 1 : 1

3 1 : 2

4 2 : 1

Ans.
(4)

Let $L$ and $A$ be the length and area of cross-section of each wire. In order to have the lower ends of the wires to be at the same level (i.e. same elongation is produced in both wires), let weights $W_{s}$ and $W_{b}$ are added to steel and brass wires respectively. Then, by definition of Young’s modulus, the elongation produced in the steel wire is

$Δ L_{s} = \frac{W_{s} L}{Y_{s} A}$ $(as Y = \frac{W / A}{Δ L / L})$

and that in the brass wire is $Δ L_{b} = \frac{W_{b} L}{Y_{b} A}$

But $Δ L_{s} = Δ L_{b}$ (given)

$∴$ $\frac{W_{s} L}{Y_{s} A} = \frac{W_{b} L}{Y_{b} A} or \frac{W_{s}}{W_{b}} = \frac{Y_{s}}{Y_{b}}$

As $\frac{Y_{s}}{Y_{b}} = 2;$ $∴$ $\frac{W_{s}}{W_{b}} = \frac{2}{1}$

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