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physics ncert solutions class 11th Physics Class 11th Thermal Properties of Matter

11.9 A brass wire 1.8 m long at 27°C is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of –39 °C, what is the tension developed in the wire, if its diameter is 2.0 mm ? Co-efficient of linear expansion of brass = 2.0 × 10^–5 K–1; Young’s modulus of brass = 0.91 × 10¹¹ Pa.

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Payal Gupta | Contributor-Level 10

5 months ago

11.9 Initial temperature, $T_{1}$ = 27°C, Length of the wire $l_{1}$ at $T_{1}$ = 1.8 m

Final temperature, $T_{2}$ = -39°C

Diameter of the wire, d = 2.0 mm = 2 $\times 10^{- 3}$ m

Coefficient of linear expansion of brass, $α = 2 \times 10^{- 5}$ /K

Youngs’ modulus of brass, Y = 0.91 $\times 10^{11}$ Pa

Let the tension developed be F

We know Youngs’ modulus = $\frac{S t r e s s}{S t r a i n}$ = $\frac{\frac{F}{A}}{\frac{? L}{L}} = Y$

Y $\times \frac{? L}{L}$ = $\frac{F}{A}$ or F = $\frac{A Y ? L}{L}$

Here, A = cross-sectional area of the wire = $\frac{π}{4} d^{2}$ = $\frac{π}{4} {(2 \times 10^{- 3})}^{2}$ $m^{2}$ = 3.1416 $\times 10^{- 6} m^{2}$

Now $? L$ can be written as $? L$ = $α L (T_{2} - T_{1})$ = ( $2 \times 10^{- 5}) \times L \times (- 39 - 27)$ = -1.32 $\times 10^{- 3}$ L

Substituting all values, we get

F = (3.1416 $\times 10^{- 6} \times 0.91 \times 10^{11} \times (- 1.32 \times 10^{- 3})$ &n

...more

11.9 Initial temperature, $T_{1}$ = 27°C, Length of the wire $l_{1}$ at $T_{1}$ = 1.8 m

Final temperature, $T_{2}$ = -39°C

Diameter of the wire, d = 2.0 mm = 2 $\times 10^{- 3}$ m

Coefficient of linear expansion of brass, $α = 2 \times 10^{- 5}$ /K

Youngs’ modulus of brass, Y = 0.91 $\times 10^{11}$ Pa

Let the tension developed be F

We know Youngs’ modulus = $\frac{S t r e s s}{S t r a i n}$ = $\frac{\frac{F}{A}}{\frac{? L}{L}} = Y$

Y $\times \frac{? L}{L}$ = $\frac{F}{A}$ or F = $\frac{A Y ? L}{L}$

Here, A = cross-sectional area of the wire = $\frac{π}{4} d^{2}$ = $\frac{π}{4} {(2 \times 10^{- 3})}^{2}$ $m^{2}$ = 3.1416 $\times 10^{- 6} m^{2}$

Now $? L$ can be written as $? L$ = $α L (T_{2} - T_{1})$ = ( $2 \times 10^{- 5}) \times L \times (- 39 - 27)$ = -1.32 $\times 10^{- 3}$ L

Substituting all values, we get

F = (3.1416 $\times 10^{- 6} \times 0.91 \times 10^{11} \times (- 1.32 \times 10^{- 3})$ = -377.37 N

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