Question

The compensated pendulum of a clock consists of an isosceles triangular frame of base length ${l_1}$ and expansivity ${\alpha _1}$ and slides of length ${l_2}$ and expansivity ${\alpha _2}$. The pendulum is supported, as shown in Figure. Find the ratio $\dfrac{{{l_1}}}{{{l_2}}}$ so that the length of the pendulum may remain unchanged at all temperatures.

A. $\dfrac{{{l_1}}}{{{l_2}}} = 2\sqrt {\dfrac{{{a_2}}}{{{a_1}}}}$
B. $\dfrac{{{l_1}}}{{{l_2}}} = \dfrac{{{a_2}}}{{{a_1}}}$
C. $\dfrac{{{l_1}}}{{{l_2}}} = \dfrac{{{a_2}}}{{2{a_1}}}$
D. $\dfrac{{{l_1}}}{{{l_2}}} = \dfrac{{2{a_2}}}{{{a_1}}}$

Answer 1

Three basic thermal expansions are linear expansion, superficial expansion and cubical expansion. Here, in the question we need to determine the ratio $\dfrac{{{l_1}}}{{{l_2}}}$ so that the length of the pendulum may remain unchanged at all temperatures. For this the amount a material expands or contracts per unit length due to a one-degree change in temperature.

Complete step by step answer:
Let C be the base of the triangle,
Now draw a perpendicular bisector on the side AB from the vertex C

Here $\cos \theta = \dfrac{{BM}}{{BC}} = \dfrac{{\dfrac{1}{2}BM}}{{BC}} = \dfrac{{{l_1}}}{{2{l_2}}}$
Now when the triangle expands,
Let A’B’C be the new triangle and draw a perpendicular AN from A to A’C

Here AA’=x increase in the length of AB$= \dfrac{1}{2}{l_1}{\alpha _1}t$, where t is the increase in temperature and
A’N= increase in the length of AB$= {l_2}{\alpha _2}t$
Since the change in angle after the expansion is very small, hence we can write
$\angle BAC = \angle B'A'C = \theta$
Hence
$\cos \theta = \dfrac{{A'N}}{{AA'}} - - (i)$
Where $\cos \theta = \dfrac{{{l_1}}}{{2{l_2}}}$
Hence substituting the value in equation (i), we get
$\dfrac{{{l_1}}}{{2{l_2}}} = \dfrac{{{l_2}{\alpha _2}t}}{{\dfrac{1}{2}{l_1}{\alpha _1}t}}$
By solving

$$\dfrac{{{l_1}}}{{2{l_2}}} = \dfrac{{2{l_2}{\alpha _2}}}{{{l_1}{\alpha _1}}} \\\ {\left( {\dfrac{{{l_1}}}{{{l_2}}}} \right)^2} = 4\dfrac{{{\alpha _2}}}{{{\alpha _1}}} \\\ \dfrac{{{l_1}}}{{{l_2}}} = \sqrt {4\dfrac{{{\alpha _2}}}{{{\alpha _1}}}} \\\ = 2\sqrt {\dfrac{{{\alpha _2}}}{{{\alpha _1}}}} \\\$$

Hence the ratio $\dfrac{{{l_1}}}{{{l_2}}}$so that the length of the pendulum may remain unchanged at all temperatures $= 2\sqrt {\dfrac{{{\alpha _2}}}{{{\alpha _1}}}}$

So, the correct answer is “Option A”.

Note:
Expansion corresponds to change in length, area and volume of the substance where linear expansivity is the increase in the length of a substance for per unit length of the substance for one degree Celsius rise in temperature.