Length of the

\begin{tabular}{|c|c|c|}
\hline
S.No. & \begin{tabular}{l}
Time taken for 20 \\ oscillations (s)
\end{tabular} & \begin{tabular}{l}
Time period \\ (s)
\end{tabular} \\
\hline
1 & 42 & 2.1 \\
\hline
2 & & \\
\hline
3 & & \\
\hline
\end{tabular}

Fill in the missing data in the table.



Answer :

Let's analyze the information provided and fill in the missing details for the table.

We know the following:
- For the first entry (S.No 1), the time taken for 20 cycles is 42 seconds, and the time period (time per cycle) is 2.1 seconds.

Assuming that the time period for the second entry (S.No 2) is the same as the time period for the first entry:

1. Time Period Consistency:
- For S.No 1, the time period is 2.1 seconds. We will assume this is also true for S.No 2.
- Therefore, the time period for S.No 2 is likewise 2.1 seconds.

2. Calculating the Time Taken for 20 Cycles for S.No 2:
- Since the time period is the same, and the second entry must complete 20 cycles as well, we need to determine the total time taken for these 20 cycles.
- Using the time period of 2.1 seconds per cycle, the total time taken for 20 cycles is [tex]\(2.1 \, \text{seconds/cycle} \times 20 \, \text{cycles} = 42 \, \text{seconds}\)[/tex].

Now, we have the complete table filled out:

[tex]\[ \begin{tabular}{|c|c|c|} \hline S.No. & \begin{tabular}{l} Time taken for 20 \\ (s) \end{tabular} & \begin{tabular}{l} Time period \\ (s) \end{tabular} \\ \hline 1 & 42 & 2.1 \\ \hline 2 & 42 & 2.1 \\ \hline \end{tabular} \][/tex]

Summarizing:

- For S.No 1:
- Time taken for 20 cycles: 42 seconds
- Time period: 2.1 seconds

- For S.No 2:
- Time taken for 20 cycles: 42 seconds
- Time period: 2.1 seconds