Question

Length of athe metal rod of temperature 27°C is 4.256m.find the temperature at which the length of the same rod increases to 4.268m (alpha for iron =1.2x10^-5 k^-1)

Answer 1

Given:

Length of metal rod,

= 4.256m

Temperature,

= 27°C

Increment in the length,

= 4.268m

Alpha,

= $1.2\times 10^{-5} k^{-1}$

To find:

Temperature at which the length of a rod increase = ?

Solution:

As we know,

⇒ $\alpha = \frac{\Delta l}{l \Delta \theta}$

On substituting the given values in the above formula, we get

⇒ $1.2\times 10^{-5}=\frac{4.268-4.256}{4.256\times (T-27)}$

⇒       $T-27=\frac{1.2\times 10^{-7}}{1.2\times 10^5\times 4.256}$

⇒               $T=27+.25\times 10^{-7}$  

⇒               $T=27$

Thus the temperature is "27".