Question

A thin copper wire of length l increase in length by 1% when heated from 0°c to 100°c. if a thin copper plate of area 2l x l is heated from 0°c to 100°c, the percentage increase in its area would be

Answer 1

 Given:

 ΔT=100-0=100 °C
ΔL/L=length =L=1%=1/100=0.01

ΔL= αLΔT
ΔL/L= αx100
0.01=αx100
α=0.01/100=1× 10^-4

Given Area of copper plate = 2LxL=2L^2
By thermal expansion theory
ΔA=βAΔT
ΔA/A=βΔT
but
β=2 α
ΔA/A=2αΔT
=2×1×10^-4×100
=2×10^-2
ΔA/A×100=2×10^-2×100=2%

Therefore the percentage increase in its area is 2%

Answer 2

Answer:

ΔT=100-0=100 °C

ΔL/L=length =L=1%=1/100=0.01

ΔL= αLΔT

ΔL/L= αx100

0.01=αx100

α=0.01/100=1× 10^-4

Given Area of copper plate = 2LxL=2L^2

By thermal expansion theory

ΔA=βAΔT

ΔA/A=βΔT

but

β=2 α

ΔA/A=2αΔT

=2×1×10^-4×100

=2×10^-2

ΔA/A×100=2×10^-2×100=2%

Therefore the percentage increase in its area is 2%

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