Question
A bar of iron is $$10\,cm$$ at $${20^ \circ }C.$$ At $${19^ \circ }C$$ it will be ($$\alpha $$ of iron $$ = 11 \times {10^{ - 6}}{/^ \circ }C$$ )
A.
$$11 \times {10^{ - 6}}\,cm\,{\text{longer}}$$
B.
$$11 \times {10^{ - 6}}\,cm\,{\text{shorter}}$$
C.
$$11 \times {10^{ - 5}}\,cm\,{\text{shorter}}$$
D.
$$11 \times {10^{ - 5}}\,cm\,{\text{longer}}$$
Answer :
$$11 \times {10^{ - 5}}\,cm\,{\text{shorter}}$$
Solution :
$$\eqalign{
& L = {L_0}\left( {1 + \alpha \Delta \theta } \right) \Rightarrow \frac{{{L_1}}}{{{L_2}}} = \frac{{1 + \alpha {{\left( {\Delta \theta } \right)}_1}}}{{1 + \alpha {{\left( {\Delta \theta } \right)}_2}}} \cr
& \Rightarrow \frac{{10}}{{{L_2}}} = \frac{{1 + 11 \times {{10}^{ - 6}} \times 20}}{{1 + 11 \times {{10}^{ - 6}} \times 19}} \Rightarrow {L_2} = 9.99989 \cr} $$
⇒ Length is shorten by $$10 - 9.99989 = 0.00011 = 11 \times {10^{ - 5}}\,cm.$$