Name: Properties of Solid
Uploaded: 2019-07-19
Duration: 3 min 59 s
Description: Do you know which are the important mechanical properties of solids? In this video, we will discuss the solid state of matter and its different properties.

Question

An aluminium Wire and a steel wire of the same length and cross-section are joined end-to-end. The composite wire is hung from a rigid support and a load is suspended from the free end. If the increase in length of the composite wire is $1.35 mm$ , find the ratio of the (i) stress in the two wires and (ii) strain in the two wires.

$(Y_{A l} = 0.7 \times 10^{11} N m^{- 2}, Y_{s t e e l} = 2 \times 10^{11} N m^{- 2})$

Accepted Answer

The wires are loaded as shown in the figure given below.

Given,

$L_{A l} = L_{s t e e l} A_{A l} = A_{s t e e l} F_{A l} = F_{s t e e l} = m g$

$Y_{A l} = 0.7 \times 10^{11} N m^{- 2}$

$Y_{s t e e l} = 2 \times 10^{11} N m^{- 2}$

Also, it is given that the total extension of the composite wire is $1.35 mm$ .

$\Rightarrow ∆ L_{A l} + ∆ L_{s t e e l} = 1.35 mm$

We know that stress, $σ = \frac{F}{A}$

Therefore, $\frac{σ_{A l}}{σ_{s t e e l}} = \frac{\frac{F_{A l}}{A_{A l}}}{\frac{F_{s t e e l}}{A_{s t e e l}}} \Rightarrow \frac{σ_{A l}}{σ_{s t e e l}} = 1 \Rightarrow σ_{A l} : σ_{s t e e l} = 1 : 1$

Also, We know that Young's modulus $Y = \frac{Tensile stress (σ)}{Longitudinal strain (ε)}$

Therefore, $\frac{Y_{A l}}{Y_{s t e e l}} = \frac{\frac{σ_{A l}}{ε_{A l}}}{\frac{σ_{s t e e l}}{ε_{s t e e l}}} \Rightarrow \frac{0.7 \times 10^{11}}{2 \times 10^{11}} = \frac{ε_{s t e e l}}{ε_{A l}} \Rightarrow \frac{ε_{A l}}{ε_{s t e e l}} = \frac{20}{7} \Rightarrow ε_{A l} : ε_{s t e e l} = 20 : 7$

Important Questions on Mechanical Properties of Solids

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