Question

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Answer 1

Question:

A student performs an experiment to determine using formula $g= \frac{4 \pi^2l}{T^2}$ ; here l(length of the wire) approximately equal to 1m, $\Delta l$ represents error in measurement of length $l$,$\Delta T$represents error in measurement of time, n is number of oscillation. Then for which of the data of the measurement of g will be most accurate ?

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \begin{array}{ c c c c}\rm \bf \Delta l \: & \rm \bf \Delta \: t & \rm \bf n & \rm \bf Amplitude~ of ~oscillation\\ \\ \rm A) 5mm& \rm 0.2sec& \rm 10& \rm 5mm\\\\\rm B)5mm& \rm 0.2sec & \rm 20 & \rm 5mm\\\\ \rm C)5mm& \rm 0.1sec & \rm 20& \rm 1mm \\\\ \rm D) 1mm & \rm 1sec & \rm 50 & \rm 1mm \\\\\end{array}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}$

Answer:

We are given:

$g= \frac{4 \pi^2l}{T^2}$

We have:

$\frac{\Delta g} {g} = \frac{ \Delta l }{l} + \frac{2 \Delta T} {T}$ This was experimentally formulated.

For accurate result, error in measurement should be less.

So, $\bf \Delta l$ and $\bf \Delta T$ should be minimum.

Now, coming to the option, we have $\bf \Delta l = 1mm$ in option D which is the lowest among all,

Also, $\bf \Delta T = 1mm$ for option D.

And, it's obvious that we will get more accurate if we do a certain experiment repeatedly,

In option D, we have n = 50.

This shows that for option D, the measurement of g will be most accurate.

Answer 2

Answer is in the attachment...

hope it helps...✨✨

$\huge \sf {\orange {\underline {\red{\underline {❥Advi }}}}}$