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Byju's Answer

Standard XII

Physics

Thermal Expansion

Two wires P a...

Question

Two wires P and Q of same cross-sectional area $A = 10 m m^{2}$ and the same length $l_{0} = 1$ m but made of different materials are joined together and their ends are rigidly clamp between two walls with negligible tension. Their respective Young's modulus and the coefficients of linear expansion are $Y_{1}, α_{1}$ and $Y_{2}, α_{2}$ respectively.
Given $Y_{1} = 2 \times 10^{9} N / m^{2}, α_{1} = 3 \times 10^{- 4} /^{\circ} C, Y_{2} = 10^{9} N / m^{2}, α_{2} = 6 \times 10^{- 4} /^{\circ} C$ ,

If the temperature of system is reduced by $20^{\circ} C$ the tension developed in wire P is

60 N

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120 N

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40 N

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180 N

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Solution

The correct option is B
120 N

If F is the horizontal force on the wire, then

$(\frac{E l_{0}}{A Y_{1}} + \frac{F l_{0}}{F Y_{2}}) = α_{1} Δ T + α_{2} Δ T$

Putting values gives F = 120 N.

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Similar questions

Two wires P and Q of same cross-sectional area $A = 10 m m^{2}$ and the same length $l_{0} = 1$ m but made of different materials are joined together and their ends are rigidly clamp between two walls with negligible tension. Their respective Young's modulus and the coefficients of linear expansion are as $Y_{1}, α_{1}$ and $Y_{2}, α_{2}$ respectively.
Given $Y_{1} = 2 \times 10^{9} N / m^{2}, α_{1} = 3 \times 10^{- 4} /^{\circ} C, Y_{2} = 10^{9} N / m^{2}, α_{2} = 6 \times 10^{- 4} /^{\circ} C$ ,

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and young's modulus of elasticity

Y_{1}

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Two rods of different materials, having coefficients of linear expansion $α_{1}$ and $α_{2}$ and Young's moduli $Y_{1}$ and $Y_{2}$ respectively, are fixed between two rigid walls. The rods are heated to the same temperature. There is no bending of the rods. If $α_{1} : α_{2} = 2 : 3$ , the thermal stress developed in the two rods will be equal provided $Y_{1} : Y_{2}$ is equal to

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The correct option is B 120 N If F is the horizontal force on the wire, then (El0AY1+Fl0FY2)=α1ΔT+α2ΔT Putting values gives F = 120 N.

The correct option is B
120 N

If F is the horizontal force on the wire, then

$(\frac{E l_{0}}{A Y_{1}} + \frac{F l_{0}}{F Y_{2}}) = α_{1} Δ T + α_{2} Δ T$

Putting values gives F = 120 N.