Question

A clock is calibrated at a temperature of \[{20^ \circ }C\]. Assume that the pendulum is a thin brass rod of negligible mass with a heavy box attached to the end\[\left( {{\alpha _{brass}} = 19 \times {{10}^{ - 6}}/K} \right)\].
A. On a hot day at \[{30^ \circ }C\] the clock gains 8.2s
B. On a hot day at \[{30^ \circ }C\] the clock loses 8.2s
C. On a cool day at \[{10^ \circ }C\] the clock gains 8.2s
D. On a cool day at \[{10^ \circ }C\] the clock loses 8.2s

Answer 1

Hint: Use the property of a simple pendulum to find the gain or loss of the time of a calibrated clock with heat transfer from the surroundings.

Complete step by step solution:
Given the clock is calibrated at a temperature \[{20^ \circ }C\] and \[{\alpha _{brass}} = 19 \times {10^{ - 6}}/K\]
The time period of the pendulum is given as \[T = 2\pi \sqrt {\dfrac{l}{g}} \]
Where $l$ is the length of bob from hinge and $g$ is the gravitational force.
In a pendulum, as the temperature increases, the length of the pendulum increases, and also time period increases, which means the clock loses its calibration. Also, when the temperature is reduced, the pendulum length decreases, and also time period decreases, and the clock gains time.
So from the above observation of pendulum properties, we can say Option (B) and Option (C) can be correct.
Case 1. First check for the Option (B)
\[
  \dfrac{{\vartriangle t}}{t} = \dfrac{1}{2}\alpha \left( {\vartriangle T} \right) \\
   = \dfrac{1}{2}\alpha \left( {30 - 20} \right) \\
   = \dfrac{1}{2}\alpha \times 10 \\
   = 5\alpha \\
   = 5\left( {19 \times {{10}^{ - 6}}} \right) \\
   = 95 \times {10^{ - 6}} \\
 \]
Total numbers of second in the whole day a clock loses\[ = 24 \times 60 \times 60 = 8400s\]
Hence total time clock loses is given as:
 \[
  \dfrac{{\vartriangle t}}{t} = 95 \times {10^{ - 6}} \\
  \vartriangle t = 95 \times {10^{ - 6}} \times t \\
   = 95 \times {10^{ - 6}} \times 8400 \\
   = 8.2\sec \\
 \]
So option B is correct.
Case 2. First check for the Option (C)
\[
  \dfrac{{\vartriangle t}}{t} = \dfrac{1}{2}\alpha \left( {\vartriangle T} \right) \\
   = \dfrac{1}{2}\alpha \left( {20 - 10} \right) \\
   = \dfrac{1}{2}\alpha \times 10 \\
   = 5\alpha \\
   = 5\left( {19 \times {{10}^{ - 6}}} \right) \\
   = 95 \times {10^{ - 6}} \\
 \]
Total numbers of second in the whole day a clock loses\[ = 24 \times 60 \times 60 = 8400s\]
Hence total time clock loses is given as:
 \[
  \dfrac{{\vartriangle t}}{t} = 95 \times {10^{ - 6}} \\
  \vartriangle t = 95 \times {10^{ - 6}} \times t \\
   = 95 \times {10^{ - 6}} \times 8400 \\
   = 8.2\sec \\
 \]
So option C is correct.

Hence, options B and C are correct.

Note: It is to be noted here that the difference in the temperature from the calibrated temperature should always be positive as there will be no motion of the pendulum when the temperature goes negative.