Question

If weight of 1.5kg is suspended on a
were then increast of length is 1%
then linear Strain is -​

Answer 1

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Linear strain is defined as the ratio of change in length to the original length.

Let, the original length of the wire is l

Now upon loading the length of the wire increased by 1%, therefore the length of the wire becomes.

$\longrightarrow \sf l' = l + \bigg ( \frac{1}{100} \bigg)l \\ \\\longrightarrow \sf l' = 1.01l$

Change in length of the wire

$\longrightarrow \sf Δl=1.01l−l \\ \\ \longrightarrow \sf ∆l =0.01l$

Therefore the linear strain in the wire is

$\longrightarrow \boxed{\sf Strain = \frac{∆l}{l}} \\ \\\longrightarrow \sf Strain = \frac{0.01l}{l} \\ \\ \longrightarrow \sf Strain =0.01$

Answer 2

Explanation:

Linear strain is defined as the ratio of change in length to the original length.

Let, the original length of the wire is l

Now upon loading the length of the wire increased by 1%, therefore the length of the wire becomes.

\begin{gathered} \longrightarrow \sf l' = l + \bigg ( \frac{1}{100} \bigg)l \\ \\\longrightarrow \sf l' = 1.01l\end{gathered}

⟶l

′

=l+(

100

1

)l

⟶l

′

=1.01l

Change in length of the wire

\begin{gathered} \longrightarrow \sf Δl=1.01l−l \\ \\ \longrightarrow \sf ∆l =0.01l\end{gathered}

⟶Δl=1.01l−l

⟶∆l=0.01l

Therefore the linear strain in the wire is

\begin{gathered}\longrightarrow \boxed{\sf Strain = \frac{∆l}{l}} \\ \\\longrightarrow \sf Strain = \frac{0.01l}{l} \\ \\ \longrightarrow \sf Strain =0.01\end{gathered}

⟶

Strain=

l

∆l

⟶Strain=

l

0.01l

⟶Strain=0.01