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Question: An iron scale is calibrated at \( 0^\circ C \) . The length of a zinc rod is measured to be 100 cm b...

An iron scale is calibrated at 0∘C0^\circ C . The length of a zinc rod is measured to be 100 cm by the scale when the rod and the scale both are at 0∘C0^\circ C . What will be the length of the rod as measured by the scale when both are at 100∘C100^\circ C . Given: αiron=1.2×10−5  ∘C−1{\alpha _{iron}} = 1.2 \times {10^{ - 5}}\;^\circ {C^{ - 1}} and αzinc=2.6×10−4∘C−1{\alpha _{zinc}} = 2.6 \times {10^{ - 4}}^\circ {C^{ - 1}} .

Explanation

Solution

In this solution, we will use the concept of expansion of the length of an object due to the heating of the object. The length of the rod will be due to the difference in the change in the lengths of the rods.

Formula used In this solution, we will use the following formula:
- L′=L(1+αΔT)L' = L(1 + \alpha \Delta T) where ΔL\Delta L is the change in length of the rod of length LL , coefficient of linear expansion α\alpha due to temperature change ΔT\Delta T .

Complete step by step answer:
We’ve been given that an iron scale is calibrated at 0∘C0^\circ C and that is used to measure the length of a rod as 100 cm. This is the case when both the rods are at 0∘C0^\circ C .
Now, we have been told that both the rods are heated to 100∘C100^\circ C . We know that whenever a rod is heated, it will expand its length. Hence when both the rods are heated to 100∘C100^\circ C , both their lengths will increase. So, the distance between the 0 cm and the 100 cm marks on the iron rod that we use to measure the length of the zinc rod will expand. Also, at the same time, the length of the zinc rod will increase. Then the new length of the zinc rod will be the difference of change in lengths of the rod. So, we can find the new length of the rod as
Lz′=(Lz0(1+αzΔT))−(LI0(1+αIΔT)){L_z}' = \left( {{L_{z0}}(1 + {\alpha _z}\Delta T)} \right) - \left( {{L_{I0}}(1 + {\alpha _I}\Delta T)} \right)
We can consider only the length of the iron rod corresponding to the length of the zinc rod so, Lz0=LI0{L_{z0}} = {L_{I0}} . So the above equation can be written as
Lz′=Lz0(1+(αz−αI)ΔT){L_z}' = {L_{z0}}\left( {1 + ({\alpha _z} - {\alpha _I})\Delta T} \right)
Substituting the value of αiron=1.2×10−5  ∘C−1{\alpha _{iron}} = 1.2 \times {10^{ - 5}}\;^\circ {C^{ - 1}} and αzinc=2.6×10−4∘C−1{\alpha _{zinc}} = 2.6 \times {10^{ - 4}}^\circ {C^{ - 1}} ,we get
Lz′=100(1+(2.6×10−5−1.2×10−5)×(100−0)){L_z}' = 100\left( {1 + (2.6 \times {{10}^{ - 5}} - 1.2 \times {{10}^{ - 5}}) \times (100 - 0)} \right)
Which is simplified to
Lz′=100.14{L_z}' = 100.14 cm
Hence the new length of the zinc rod as measured by the iron rod will be Lz′=100.14{L_z}' = 100.14 cm.

Note:
We do not need to know the length of the iron rod in this case. As we have done, we can assume the length of the iron rod to be equal to the length of the zinc rod when measuring the change in the length due to temperature change.