Question

An iron rod of lenght 30cm is heated through 50 kelvin calculate its increase in lenght (linear expansivity of iron=1.2 ×10-^5k-^1
solve

Answer 1

Given,

• Length of iron rod, $\sf{L=0.3\ m.}$

• Temperature change during heating, $\sf{\Delta T=50\ K.}$

• Linear expansivity of iron, $\sf{\alpha=1.2\times10^{-5}\ K^{-1}.}$

We have to find increase in length of iron rod, $\sf{\Delta L.}$

The increase in length of a rod due to thermal expansion is given by,

$\longrightarrow\sf{\Delta L=L\,\alpha\,\Delta T}$

Then,

$\longrightarrow\sf{\Delta L=0.3\times1.2\times10^{-5}\times50}$

$\longrightarrow\sf{\underline{\underline{\Delta L=1.8\times10^{-4}\ m}}}$

$\longrightarrow\sf{\underline{\underline{\Delta L=0.018\ cm}}}$

Answer 2

$\huge\sf{Answer:}$

☆Given:

⇏ An iron rod of length 30cm is heated through 50 kelvin.

☆ Find:

⇏ Calculate its increase in length [linear expansivity of iron = 1.2 ×10-^5 k-^1].

☆ According to the question:

⇏ Length = (L)

⇏Temperature = (∆)

⇏ Linear expansivity = (ā)

☆ Note:

⇏ L, ∆, ā are variables, these variables helps us to use them at the time when we need to use the following properties such as Length, Temperatures, Linear expansivity, We use these variables at the place of these properties in our equation to make the calculations easier, Also these symbols means the same values as per the properties.

☆ Example: In the following equation below we are using several symbols such as (∆L), (∆), (L), (ā) these are used as variables which helps us to make the equation easier.

☆ Calculations:

⇏ Increase length of rod = Thermal expansion

⇏ ∆L = Lā∆T

⇏ ∆$\sf L = 0.3 \times 1.2 \times 10-^5 \times 50$

⇏ ∆$\sf T = 1.8 \times 10-^4 m$

⇏ ${\sf{\underline{\boxed{\green{\sf{\triangle T = 0.018\: cm}}}}}}$

Therefore, 0.018 is it's length.