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Steel and copper wires of same length are stretched by the same weight one after the other. Young's modulus of steel and copper are $2 \times 10^{11} N / m^{2} a n d 1.2 \times 10^{11} N / m^{2} .$ The ratio of increase in length

$\frac{2}{5}$

$\frac{3}{5}$

$\frac{5}{4}$

$\frac{5}{2}$

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Explanation

(b)

$l = \frac{F L}{A Y} \Rightarrow \frac{l_{s}}{l_{e u}} = \frac{Y_{e u}}{Y_{s}} (F, L a n d Y a r e c o n s \tan t) ∴ \frac{l_{s}}{l_{e u}} = \frac{1.2 \times 10^{11}}{2 \times 10^{11}} = \frac{3}{5}$

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The temperature of a wire of length 1 metre and area of cross-section $1 c m^{2}$ is increased from 0°C to 100°C. If the rod is not allowed to increase in length, the force required will be $(α = 10^{- 5} / ° C a n d Y = 10^{11} N / m^{2})$

$10^{3} N$

$10^{4} N$

$10^{5} N$

$10^{9} N$

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Explanation

(b) F = force developed= YA $α (∆ θ)$

$= 10^{11} \times 10^{- 4} \times 10^{- 5} \times 100 = 10^{4} N$

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The force constant of a wire does not depend on

Nature of the material

Radius of the wire

Length of the wire

None of the above

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Explanation

(d) $K = \frac{Y A}{L} = \frac{Y \times {πr}^{2}}{L} \Rightarrow K \propto \frac{Y r^{2}}{L}$

i.e. force constant of a wire depends on young's modules (nature of the material), radius of the wire and length of the wire.

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The length of a wire is 1.0 m and the area of cross-section is $1.0 \times 10^{- 2} c m^{2}$ . If the work done for increase in length by 0.2 cm is 0.4 joule, then Young's modulus of the material of the wire is

$2.0 \times 10^{10} N / m^{2}$

$4 \times 10^{10} N / m^{2}$

$2.0 \times 10^{11} N / m^{2}$

$2 \times 10^{10} N / m^{2}$

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Explanation

(c) $W = \frac{1}{2} \frac{Y A δ^{2}}{L} \Rightarrow 0.4 = \frac{1}{2} \times \frac{Y \times 1^{- 6} \times {(0.2 \times 10^{- 2})}^{2}}{1} ∴ Y = 2 \times 10^{11} N / m^{2}$

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The quality of the material which opposes the change in shape, volume or length is called

Intermolecular repulsion

Intermolecular behaviour

Viscosity

Elasticity

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Explanation

(d)

Elasticity of a material resists change in shape, size, and volume of material.

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A wire of length L and radius r is rigidly fixed at one end. On stretching the other end of the wire with a force F, the increase in its length is l. If another wire of same material but of length 2L and radius 2r is stretched with a force of 2F, the increase in its length will be

$\frac{l}{2}$

$\frac{l}{4}$

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Explanation

(a)

$l = \frac{F L}{A Y} = \frac{F L}{{πr}^{2} Y} ∴ l \propto \frac{F L}{r^{2}} (Y = c o n s \tan t) ∴ \frac{l_{2}}{l_{1}} = \frac{F_{2}}{F_{1}} \times \frac{L_{2}}{L_{1}} {(\frac{r_{1}}{r_{2}})}^{2} = 2 \times 2 \times {(\frac{1}{2})}^{2} = 1$

$∴ l_{2} = l_{1}$ i.e. increment in its length will be l.

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In steel, the Young's modulus and the strain at the breaking point are $2 \times 10^{11} N m^{- 2}$ and 0.15 respectively. The stress at the breaking point for steel is therefore -

$1.33 \times 10^{11} N m^{- 2}$

$1.33 \times 10^{12} N m^{- 2}$

$7.5 \times 10^{- 13} N m^{- 2}$

$3 \times 10^{10} N m^{- 2}$

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Explanation

(d) Breaking stress = strain $\times$ Young's modulus

$= 0.15 \times 2 \times 10^{11} = 3 \times 10^{10} N m^{- 2}$

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Which one of the following substances possesses the highest elasticity

Rubber

Glass

Steel

Copper

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Explanation

(c)

Modulus of elasticity gives the idea of elasticity, and steel has highest modulus of elasticity.

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Which one of the following quantities does not have the unit of force per unit area

Stress

Strain

Young's modulus of elasticity

Pressure

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Explanation

(b) Because strain is a dimensionless and unitless quantity.

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A copper wire and a steel wire of the same diameter and length are connected end to end and a force is applied, which stretches their combined length by 1 cm. The two wires will have

Different stresses and strains

The same stress and strain

The same strain but different stresses

The same stress but different strains

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Explanation

(d) $S t r e s s = \frac{F o r c e}{a r e a}$

In the present case, force applied and area of cross-section of wires are same, therefore stress has to be the same.

$S t r a i n = \frac{S t r e s s}{Y}$

Since the Young’s modulus of steel wire is greater than the copper wire, therefore, strain in case of steel wire is less than that in case of copper wire.

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