A steel wire of length 20 cm and uniform cross-section 1 mm 2 is tied rigidly at both the ends. The temperature of the wire is altered from 40° C to 20° C. Coefficient of linear expansion of steel is α=1.1 × 10-5 ° C -1 and Y for steel is 2.0 × 1011 Nm 2 the tension in the wire is

Question

A steel wire of length 20 cm and uniform cross-section 1 mm 2 is tied rigidly at both the ends. The temperature of the wire is altered from 40° C to 20° C. Coefficient of linear expansion of steel is α=1.1 × 10-5 ° C -1 and Y for steel is 2.0 × 1011 Nm 2 the tension in the wire is (A) 2.2 × 106 N (B) 16 N (C) 8 N (D) 44 N

Tardigrade Admin · Accepted Answer

Increase in length due to rise in temperature Δ L=α L Δ T As, Y=(F L/A Δ L) So, F=(Y A Δ L/L)=(Y A × α L Δ T/L)=Y A α Δ T ∴ F=2 × 1011 × 10-6 × 1.1 × 10-5 × 20=44 N

$2.2 \times 1 0^{6} N$

16 N

8 N

44 N

Increase in length due to rise in temperature $Δ L = αL Δ T$
As, $Y = \frac{F L}{A Δ L}$
So, $F = \frac{Y A Δ L}{L} = \frac{Y A \times αL Δ T}{L} = Y A α Δ T$
$∴ F = 2 \times 1 0^{11} \times 1 0^{- 6} \times 1.1 \times 1 0^{- 5} \times 20 = 44 N$

Q. A steel wire of length 20cm and uniform cross-section 1mm2 is tied rigidly at both the ends. The temperature of the wire is altered from 40∘C to 20∘C. Coefficient of linear expansion of steel is α=1.1×10−5∘C−1 and Y for steel is 2.0×1011Nm2 the tension in the wire is

2.2×106N

16 N

8 N

44 N

Increase in length due to rise in temperature ΔL=αLΔT As, Y=AΔLFL​ So, F=LYAΔL​=LYA×αLΔT​=YAαΔT ∴F=2×1011×10−6×1.1×10−5×20=44N

$2.2 \times 1 0^{6} N$

Increase in length due to rise in temperature $Δ L = αL Δ T$
As, $Y = \frac{F L}{A Δ L}$
So, $F = \frac{Y A Δ L}{L} = \frac{Y A \times αL Δ T}{L} = Y A α Δ T$
$∴ F = 2 \times 1 0^{11} \times 1 0^{- 6} \times 1.1 \times 1 0^{- 5} \times 20 = 44 N$