Let the time period of a seconds pendulum is \[2.5\,s\] . Tell by how much time will the clock be behind in $10\,hr$ .
A. $2.5\,hr$
B. $2\,hr$
C. $1.5\,hr$
D. None
Answer
531.3k+ views
Hint: A pendulum is a swinging weight suspended from a pivot. Second's pendulum is a unique pendulum with a continuous period of time. Second's pendulum is a unique pendulum with a fixed period of time. The tail of the pendulum of this pendulum reaches its extreme point twice in one cycle. As a result, the temporal span of a second's pendulum is two seconds. It has a frequency of \[0.5{\text{ }}hertz\].
Complete step by step answer:
We already know that when we talk about a second’s pendulum, the time period is going to be\[2{\text{ }}seconds\] . That is why it is referred to as a second's pendulum.
The time period for second’s pendulum $ = 2s$
And the time period given to us in the question $ = 2.5s$
Now, for seconds pendulum, $s$ ,
$s = \dfrac{{10 \times 60 \times 60}}{2} = 18000\sec $
For $2.5s$ pendulum, $s'$
$s' = \dfrac{{10 \times 60 \times 60}}{{2.5}} = 14400\sec $
Now, we observe that,
$s = 14400 + 3600 = 18000\sec \\
\Rightarrow s = s' + 1\,hr \\
\therefore s' = s - 1\,hr $
Thus, the clock will be behind by $1\,hr$ in $10\,hr$.
Hence, the correct option is D.
Note: It should be noted that the seconds pendulum (also known as the Royal pendulum), which is \[0.994{\text{ }}m{\text{ }}\left( {39.1{\text{ }}in} \right)\] long and swings once every second, became frequently utilised in high-quality clocks. The grandfather clocks, which were initially produced by William Clement around 1680, were long narrow clocks fashioned around these pendulums.
Complete step by step answer:
We already know that when we talk about a second’s pendulum, the time period is going to be\[2{\text{ }}seconds\] . That is why it is referred to as a second's pendulum.
The time period for second’s pendulum $ = 2s$
And the time period given to us in the question $ = 2.5s$
Now, for seconds pendulum, $s$ ,
$s = \dfrac{{10 \times 60 \times 60}}{2} = 18000\sec $
For $2.5s$ pendulum, $s'$
$s' = \dfrac{{10 \times 60 \times 60}}{{2.5}} = 14400\sec $
Now, we observe that,
$s = 14400 + 3600 = 18000\sec \\
\Rightarrow s = s' + 1\,hr \\
\therefore s' = s - 1\,hr $
Thus, the clock will be behind by $1\,hr$ in $10\,hr$.
Hence, the correct option is D.
Note: It should be noted that the seconds pendulum (also known as the Royal pendulum), which is \[0.994{\text{ }}m{\text{ }}\left( {39.1{\text{ }}in} \right)\] long and swings once every second, became frequently utilised in high-quality clocks. The grandfather clocks, which were initially produced by William Clement around 1680, were long narrow clocks fashioned around these pendulums.
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