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Mechanical Properties of Solids

Multiple Choice Questions

121.

A material has Poisson's ratio 0.50. If a uniform rod of it suffers a longitudinal strain of 2 x 10^-3 , then the percentage change in volume is

0.6
0.4
0.2
zero

122.

Let L be the length and d be the diameter of cross- section of a wire. Wires of the same material with different L and d are subjected to the same tension along the length of the wire. In which of the following cases, the extension of wire will be the maximum ?

L = 200 cm, d = 0.5 mm
L = 300 cm, d = 1.0 mm
L = 50 cm, d = 0.05 mm
L = 100 cm, d = 0.2 mm

123.

The following four wires are made of the same material. Which of these will have the largest extension when the same tension is applied ?

Length = 50 cm, diameter = 0.5 mm
Length = 100 cm, diameter = 1 mm
Length = 200 cm, diameter = 2 mm
Length= 300 cm, diameter= 3 mm

124.

A metal rod is fixed rigidly at two ends so as to prevent its thermal expansion. If L, α and Y respectively denote the length of the rod, coefficient of linear thermal expansion and Young's modulus of its material, then for an increase in temperature of the rod by ΔT, the longitudinal stress developed in the rod is

inversely proportional to α
inversely proportional to Y
directly proportional to ΔT/Y
independent of L

125.
The length of a metal wire is L₁ when the tension is T₁ and L₂ when the tension is T₂. The unstretched length of wire is

$\frac{L_{1} + L_{2}}{2}$

$\sqrt{L_{1} L_{2}}$

$\frac{T_{2} L_{1} - T_{1} L_{2}}{T_{2} - T_{1}}$

$\frac{T_{2} L_{1} + T_{1} L_{2}}{T_{2} + T_{1}}$

$\frac{T_{2} L_{1} - T_{1} L_{2}}{T_{2} - T_{1}}$

Let the initial length of the metal wire is L.

The strain at tension T₁ is ΔL₁ = L₁ − L

The strain at tension T₂ is ΔL₂ = L₂ − L

Suppose, the Young's modulus of the wire is Y, then

$\frac{\frac{T_{1}}{A}}{\frac{∆ L_{1}}{L}} = \frac{\frac{T_{2}}{A}}{\frac{∆ L_{2}}{L}}$

where, A is an cross-section of the wire, assume to be same at all the situations

$\Rightarrow \frac{T_{1}}{A} \times \frac{L}{∆ L_{1}} = \frac{T_{2}}{A} \times \frac{L}{∆ L_{2}} \Rightarrow \frac{T_{1}}{(L_{1} - L)} = \frac{T_{2}}{(L_{2} - L)} \Rightarrow T_{1} (L_{2} - L) = T_{2} (L_{1} - L) \Rightarrow L = \frac{T_{2} L_{1} - T_{1} L_{2}}{T_{2} - T_{1}}$

Previous Year Papers

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Multiple Choice Questions

125. The length of a metal wire is L1 when the tension is T1 and L2 when the tension is T2. The unstretched length of wire is L1 + L22 L1L2 T2L1 - T1L2T2 - T1 T2L1 + T1L2T2 + T1

125.
The length of a metal wire is L₁ when the tension is T₁ and L₂ when the tension is T₂. The unstretched length of wire is

$\frac{L_{1} + L_{2}}{2}$

$\sqrt{L_{1} L_{2}}$

$\frac{T_{2} L_{1} - T_{1} L_{2}}{T_{2} - T_{1}}$

$\frac{T_{2} L_{1} + T_{1} L_{2}}{T_{2} + T_{1}}$