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Standard XII

Physics

Longitudinal Strain

A steel wire ...

Question

A steel wire of uniform area 2 m $m^{2}$ is heated upto $50^{\circ}$ C and is stretched by tying its ends rigidly. The tension when the temperature falls from $50^{\circ}$ C and $30^{\circ}$ C is (Take Y = 2 $\times$ $10^{11}$ N $m^{- 2}$ , $α$ = 1.1 $\times$ $10^{- 5}^{\circ}$ $C^{- 1}$ )

1.5 × 10¹¹N

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5 N

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88 N

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2.5 × 10¹¹N

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Solution

The correct option is C 88 N
$△$ = $α L (△ T)$ = $\frac{F}{A}$ $\frac{L}{Y}$
$∴$ F = $α (△ T) A Y$
= 1.1 $\times 10^{- 5} \times (50 - 30) \times 2 \times 10^{- 6} \times 2 \times 10^{11}$
= 88 N

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(coefficient of linear expansion α = 1.1 \times 10^{- 5} /^{\circ} C, Youngs modulus Y = 2.0 \times 10^{11} \frac{N}{m^{2}})

A steel wire of uniform cross - sectional area $2 m m^{2}$ is heated up to $50^{\circ}$ C and is stretched by clamping its two ends rigidly. The change in tension when the temperature falls from $50^{\circ}$ C to $30^{\circ}$ C is given by:
$α = 1.1 \times {10^{- 5}}^{\circ} C, Y = 2.0 \times 10^{11} \frac{N}{m^{2}}$

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A steel wire of uniform area 2 mm2 is heated upto 50∘C and is stretched by tying its ends rigidly. The tension when the temperature falls from 50∘C and 30∘C is (Take Y = 2 × 1011 Nm−2 , α = 1.1 × 10−5∘C−1)

The correct option is C 88 N △ = αL(△T) = FA LY ∴ F = α(△T)AY = 1.1 ×10−5×(50−30)×2×10−6×2×1011 = 88 N

A steel wire of uniform area 2 m $m^{2}$ is heated upto $50^{\circ}$ C and is stretched by tying its ends rigidly. The tension when the temperature falls from $50^{\circ}$ C and $30^{\circ}$ C is (Take Y = 2 $\times$ $10^{11}$ N $m^{- 2}$ , $α$ = 1.1 $\times$ $10^{- 5}^{\circ}$ $C^{- 1}$ )

The correct option is C 88 N
$△$ = $α L (△ T)$ = $\frac{F}{A}$ $\frac{L}{Y}$
$∴$ F = $α (△ T) A Y$
= 1.1 $\times 10^{- 5} \times (50 - 30) \times 2 \times 10^{- 6} \times 2 \times 10^{11}$
= 88 N