-->

Mechanical Properties Of Fluids

Question
CBSEENPH11020032

A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0 °C. What is the change in the diameter of the hole when the sheet is heated to 227 °C? Coefficient of linear expansion of copper = 1.70 × 10–5 K–1.

Solution
Given,
Initial temperature, T1 = 27.0°C
Diameter of the hole at T1d1 = 4.24 cm
Final temperature, T2 = 227°C 
Diameter of the hole at Td

Co-efficient of linear expansion of copper, αCu= 1.70 × 10–5K–1
Relation between coefficient of superficial expansion and change in temperature is given by, 
space space space space space space space fraction numerator Change space in space Area space left parenthesis increment straight A right parenthesis over denominator Original space Area space left parenthesis straight A right parenthesis end fraction space equals space beta space increment T

rightwards double arrow space fraction numerator open parentheses begin display style fraction numerator pi d subscript 2 squared over denominator 4 end fraction end style space minus begin display style fraction numerator pi d subscript 1 squared over denominator 4 end fraction end style space close parentheses over denominator begin display style fraction numerator pi d subscript 1 squared over denominator 4 end fraction end style end fraction space equals space fraction numerator increment straight A over denominator A end fraction space

therefore space fraction numerator increment straight A over denominator A end fraction space equals fraction numerator d subscript 2 squared space minus space d subscript 1 squared over denominator d subscript 1 squared end fraction

But comma space given space that space beta space equals space 2 alpha space

Therefore comma space

space space space space space space space space fraction numerator straight d subscript 2 squared space minus space straight d subscript 1 squared over denominator straight d subscript 1 squared end fraction space equals space space 2 straight alpha space increment straight T
rightwards double arrow space space space space space fraction numerator straight d subscript 2 squared space over denominator straight d subscript 1 squared end fraction minus space 1 space equals space 2 straight alpha space left parenthesis straight T subscript 2 space minus space straight T subscript 1 right parenthesis space

rightwards double arrow space fraction numerator straight d subscript 2 squared space over denominator 4.24 space squared end fraction space equals space 2 cross times 1.7 cross times 10 to the power of negative 5 end exponent space left parenthesis 227 minus 27 right parenthesis space plus space 1

rightwards double arrow space space space space space straight d subscript 2 squared space equals space 17.98 space cross times space 1.0068 space equals space 18.1 space

therefore space space space space space space straight d subscript 2 space equals space 4.2544 space cm space

So comma space change space in space diameter space equals space straight d subscript 2 space minus space straight d subscript 1 space equals space 4.2544 space minus space 4.24 space

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 0.0144 space cm

Therefore comma space

Increase space in space diameter space equals space 1.44 space cross times space 10 to the power of negative 2 end exponent space cm