Question

answer to the question​

Answer 1

Answer:

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Explanation:

Here's the Answer and explanation:

given

T₁= 0 ⁰C

T₂ = 20 ⁰C

α = 5 x 10⁻⁵ / ⁰C

We know that

ΔL = L' - L

Also longitudinal strain is equal to the following:

=>   ΔL/L = α x ΔT

now, ΔT= T₂ -T₁

=> 20 -0 = 20 ⁰C

As strain  α x ΔT, we get the anser as: 5 x 20 x 10⁻⁵ = 100 x 10⁻⁵ = 10⁻³

This is a dimensionless quantity...

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Answer 2

$\bigstar$ Solution:

Initial length $=2m$

Change in length after change in transfer $=\alpha l \triangle T$

$\implies 5 \times 10^{-5} \times 2 \times 20$

$\implies 2 \times 10^{-3} m$

Longitudinal strain = $\frac{\triangle \: l}{l} = \frac{2 \times 10^{-3} }{2}$

$\implies 1 \times 10^{-3}$

$\implies \boxed{\underline{\underline{10^{-3} }}}$

Hence, the longitudinal strain developed is $10^{-3}$