The pressure that has to be applied to the ends of a steel wire of length 10 cm to keep its length constant when its temperature is raised by 100° C is: (For steel Young's modulus is 2 × 1011 Nm-2 and coefficient of thermal expansion is 1.1 × 10-5 K-1)

Question

The pressure that has to be applied to the ends of a steel wire of length 10 cm to keep its length constant when its temperature is raised by 100° C is: (For steel Young's modulus is 2 × 1011 Nm-2 and coefficient of thermal expansion is 1.1 × 10-5 K-1) (A) 2.2 × 108 Pa (B) 2.2 × 109 Pa (C) 2.2 × 107 Pa (D) 2.2 × 106 Pa

Tardigrade Admin · Accepted Answer

As length is constant, Strain =(Δ L/L)=α Δ Q Now pressure=stress = Upsilon × strain =2 × 1011 × 1.1 × 10-5 × 100 =2.2 × 108 Pa

$2.2 \times 1 0^{8} P a$

$2.2 \times 1 0^{9} P a$

$2.2 \times 1 0^{7} P a$

$2.2 \times 1 0^{6} P a$

As length is constant,
Strain $= \frac{Δ L}{L} = α Δ Q$
Now pressure=stress $= Υ \times$ strain
$= 2 \times 1 0^{11} \times 1.1 \times 1 0^{- 5} \times 100$
$= 2.2 \times 1 0^{8} P a$

Q. The pressure that has to be applied to the ends of a steel wire of length 10cm to keep its length constant when its temperature is raised by 100∘C is : (For steel Young's modulus is 2×1011Nm−2 and coefficient of thermal expansion is 1.1×10−5K−1)

2.2×108Pa

2.2×109Pa

2.2×107Pa

2.2×106Pa

As length is constant, Strain =LΔL​=αΔQ Now pressure=stress =Υ× strain =2×1011×1.1×10−5×100 =2.2×108Pa

$2.2 \times 1 0^{8} P a$

$2.2 \times 1 0^{9} P a$

$2.2 \times 1 0^{7} P a$

$2.2 \times 1 0^{6} P a$

As length is constant,
Strain $= \frac{Δ L}{L} = α Δ Q$
Now pressure=stress $= Υ \times$ strain
$= 2 \times 1 0^{11} \times 1.1 \times 1 0^{- 5} \times 100$
$= 2.2 \times 1 0^{8} P a$