Question

A copper rod of $88cm$ and an aluminium rod of unknown length have their increase in length independent of increase in temperature. The length of aluminium rod is:
 (\[{\alpha _{Cu}} = 1.7 \times {10^{ - 5}}{K^{ - 1}}\;\]And\[{\alpha _{Al}} = 2.2 \times {10^{ - 5}}{K^{ - 1}}\])
a. $6.8cm$
b. $113.9cm$
c. $88cm$
d. $68cm$

Answer 1

Hint: We can solve this question using concept, Change in length of any rod due to thermal expansion is given by $\Delta l = l \times \alpha \Delta T$, where $\Delta l$ represents change in length and $\Delta T$ represents change in temperature.

Complete step by step answer:
Every object has the property of expansion. It can be expanded by length area or volume. When the length of an object is expanded due to thermal changes, it is considered as linear expansion. Similarly, if expansion is in the area or volume, it would be defined as area expansion and volume expansion respectively.
If the initial temperature of surrounding in ${T_1}$ and final temperature given is ${T_2}$, then the change in temperature will be \[\left( {{T_2} - {T_1}} \right)\] and it is given as $\Delta T$
If the initial length of rod is $l$ and final length is $l'$, then the change in length is given by $\Delta l$.
If the substance is in the form of a long rod, then for small change in temperature,$\Delta T$ the fractional change in length, \[\Delta l/l\] is directly proportional to $\Delta T$. And is given as:
$\Delta l/l = \alpha \Delta T$--$(1)$
$\Delta l = l' - l$--$(2)$
Expanding equation $(1)$ with help of equation $(2)$:
$
  \Delta l = l\alpha \Delta T \\
  l' = l + l\alpha \Delta T \\
  l' = l(1 + \alpha \Delta T) \\
 $
Where,$\alpha $ varies from substance to substance
Now as per question:
Change in length of copper rod is given as:
${l_{cu}}' = {l_{cu}}(1 + {\alpha _{cu}}\Delta T)$--$(3)$
Change in length of aluminium rod is given as:
\[{l_{al}}' = {l_{al}}(1 + {\alpha _{al}}\Delta T)\]--$(4)$
Subtract $(4) - (3)$ , we get:
${l_{al}}' - {l_{cu}}' = {l_{al}} - {l_{cu}} + ({l_{al}}{\alpha _{al}} - {l_{cu}}{\alpha _{cu}})\Delta T - - - (5)$
According to question, increase in length is independent of increase in temperature, therefore equating $\Delta T$ part to $0$, we get:
\[{l_{al}}{\alpha _{al}} = {l_{cu}}{\alpha _{cu}}\]
Put the values from the question:
${l_{al}} \times 2.2 \times {10^{ - 5}}{k^{ - 1}} = 1.7 \times {10^{ - 5}}{k^{ - 1}} \times 88cm$
Solving this, we get:
\[{l_{al}} = 68cm\]

Hence, the correct answer is option (D).

Note: In this type of question please remember that the $\Delta T$ part can be put zero, if and only if, increase in length, area or volume is temperature independent. And $\Delta $ is used to represent very small change.