Question

A rod, a sheet and a cube of steel are taken whose temperature is increased from 30 degree Celsius to 80 degree Celsius. If length of rod is 10cm, length of each side of sheet is 10 cm and length of each side of cube is also 10cm. Calculate increment in length, area and volume of rod, sheet and cube respectively (co efficient of linear expansion of steel is 1.2*10^-5 degree Celsius

Answer 1

$\huge\red{\underline{{\boxed{\textbf{QUESTION}}}}}$

A rod, a sheet and a cube of steel are taken whose temperature is increased from 30 degree Celsius to 80 degree Celsius. If length of rod is 10cm, length of each side of sheet is 10 cm and length of each side of cube is also 10cm. Calculate increment in length, area and volume of rod, sheet and cube respectively (co efficient of linear expansion of steel is 1.2*10^-5 degree Celsius

$\huge\red{\underline{{\boxed{\textbf{ANSWER}}}}}$

Given that the coefficient of linear expansion is

$\alpha = 1.2×10^{5} per \: degree \: celius$

Change in length is

∆L= L α (∆θ)

$10 \times 1.2 \times 10 {}^{5} \times (80 - 30) \: \: cm$

$1.2 \times 5 \times 10 {}^{3} \: \: cm$

$change \: in \: L = 6 \times 10 {}^{3} \: \: cm$

Change in area of square is,

∆A=A 2α (∆θ)

$10 {2}^{} \times 2 \times 1.2 \times (10 - 5) \times (80 - 30)$

$2.4 \times 5 \times 10 {}^{2}$

$change \: in \: area \: = 0.12 \: cm {}^{2}$

Change in volume of the cube is,

∆V=V3α(∆θ)

$103 \times 3 \times 1.2 \times 10 {}^{ - 5} \times (80 - 30)$

$3.6 \times 5\times 10 {}^{ -1}$

$change \: in \: volume = 1.8 \: cm \: {}^{2}$

@HarshPratapSingh

Answer 2

Answer:

A rod, a sheet and a cube of steel are taken whose temperature is increased from 30 degree Celsius to 80 degree Celsius. If length of rod is 10cm, length of each side of sheet is 10 cm and length of each side of cube is also 10cm. Calculate increment in length, area and volume of rod, sheet and cube respectively (co efficient of linear expansion of steel is 1.2*10^-5 degree Celsius

Given that the coefficient of linear expansion is

\alpha  = 1.2×10^{5} per \: degree \: celius

Change in length is

∆L= L α (∆θ)

10 \times 1.2 \times 10 {}^{5}  \times (80 - 30) \:  \: cm

1.2 \times 5 \times 10 {}^{3}  \:  \: cm

change \: in \: L = 6 \times 10 {}^{3}  \:  \: cm

Change in area of square is,

∆A=A 2α (∆θ)

10 {2}^{}  \times 2 \times 1.2 \times (10 - 5) \times (80 - 30)

2.4 \times 5 \times 10 {}^{2}  

change \: in \: area \:  = 0.12 \: cm {}^{2}  

Change in volume of the cube is,

∆V=V3α(∆θ)

103 \times 3 \times 1.2 \times 10 {}^{ - 5}  \times (80 - 30)

3.6  \times 5\times 10 {}^{ -1}  

change \: in \: volume = 1.8 \: cm \:  {}^{2}  

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Explanation: