Question

A meter scale made of steel reads accurately at \[{25^ \circ }C\]. Suppose in an experiment an accuracy of\[0.06mm\]in\[1m\] is required, the range of temperature in which the experiment can be performed with this meter scale is (coefficient of linear expansion of steel is \[11 \times {10^{ - 6}}{/^ \circ }C\]).
(A) \[{19^ \circ }C\]to\[{31^ \circ }C\]
(B) \[{25^ \circ }C\] to \[{32^ \circ }C\]
(C) \[{18^ \circ }C\]to \[{25^ \circ }C\]
(D) \[{18^ \circ }C\]to\[{32^ \circ }C\]

Answer 1

Hint : Whenever an object is heated then it experiences some changes in it. The process of heating any metal or alloy is called thermal expansion, where thermal means providing heat. The changes in an object can include change in its length which depends on its original length and change in temperature.

Complete Step by Step Solution:
Step I:
Here linear expansion of steel occurs as per which if there is change in any one dimension that is length, opposed to change in its expanded volume. The coefficient of linear expansion can vary with temperature.
Step II:
The formula for linear expansion is written as
\[\Delta l = l\alpha \Delta T\]---(i)
Step III:
Given length, \[l = 1m\]
Change in temperature, \[\Delta T = ?\]
Change in length, \[\Delta l = 0.06mm = 0.6 \times {10^{ - 3}}m\]
Coefficient of linear expansion\[\alpha = 11 \times {10^{ - 6}}{/^ \circ }C\]
Given temperature, \[T = {25^ \circ }C\]
Step IV:
In order to find the change in temperature, substitute the values in equation (i) and solve
\[\Delta T = \dfrac{{\Delta l}}{{l \propto }}\]
\[\Delta T = \dfrac{{0.6 \times {{10}^{ - 3}}}}{{1 \times 11 \times {{10}^{ - 6}}}}\]
\[\Delta T = 0.0545 \times {10^{ - 3 + 6}}\]
\[\Delta T = 0.545 \times {10^{ - 3}}\]
\[\Delta T = 5.45\]
Step V:
Range of temperature can be calculated by using \[(T + \Delta T)\] and \[(T - \Delta T)\]
\[T + \Delta T = 25 + 5.45\]
\[\Delta T = {31^ \circ }C\]
\[T - \Delta T = 25 - 5.45 = {19^ \circ }C\]
Step VI:
The range of temperature in which the experiment was performed on the meter scale is between \[{19^ \circ }C\]to\[{31^ \circ }C\].

Option A is the right answer.

Note:
An object expands or its size increases if there is an increase in temperature and its size decreases and the object contracts if the temperature decreases. Therefore, in order to compare the abilities of different objects or solids, it is important to know the coefficient of linear expansion and the range in which the object was heated. When exposed to heat, there is a change in volume, surface area or length of the object.