Answer
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Hint: Oscillation is the motion which repeats again and again itself after equal intervals of time, of some measure about a central value or between two or more different states, or we can say that single swing or movement in one direction of an oscillation body.
Complete step by step answer:
Given values are: Least count of stopwatch is 1/5 sec = 0.2 sec.Time of 20 oscillation of a pendulum is 25sec. With the help of this value we can find out the percentage error. Calculate the time of one oscillation or the period (T) by dividing the total time by the number of oscillations as depicted below:
$T = \dfrac{t}{n}$
$t$ = total time
$n$ = number of oscillations
Number of oscillation (n) =20 (Given)
Now, we can calculate the error per oscillation using the following formula:
$
\vartriangle T = \left( {\dfrac{{least\, count \,error}}{{no.\,of \,oscillation}}} \right) \\
\Rightarrow \vartriangle T = \left( {\dfrac{{\dfrac{1}{5}}}{{20}}} \right) \\
$
Now, calculate the 1 oscillation time period i.e.:
$
T = \left( {\dfrac{t}{n}} \right) \\
\Rightarrow T = \left( {\dfrac{{25}}{{20}}} \right) \\
$
Finally calculate the % of error. We apply the following formula:
$
\left( {\dfrac{{\vartriangle T}}{T}\times 100\% } \right) \\
\Rightarrow \left( {\dfrac{{\dfrac{{0.2}}{{20}}}}{{\dfrac{{25}}{{20}}}}} \right)\times 100\% \\
\Rightarrow \left( {\dfrac{{0.2}}{{25}}\times 100} \right)\% \\
\Rightarrow \left( {\dfrac{2}{{250}}\times 100} \right)\% \\
\Rightarrow \left( {\dfrac{4}{5}} \right)\% \\
\therefore 0.8\%$
Correct answer is 0.8%.
Hence the answer is option B.
Note: It should be noted down that the least count of any measuring instrument refers to the smallest or accurate value that can be resolved on an instrument's scale. For example in the present case, the minimum value of time which can be measured on a stopwatch is 0.2 seconds. The time taken by a pendulum for completing 20 oscillations is 25 seconds, it’s not a single oscillation time.
Complete step by step answer:
Given values are: Least count of stopwatch is 1/5 sec = 0.2 sec.Time of 20 oscillation of a pendulum is 25sec. With the help of this value we can find out the percentage error. Calculate the time of one oscillation or the period (T) by dividing the total time by the number of oscillations as depicted below:
$T = \dfrac{t}{n}$
$t$ = total time
$n$ = number of oscillations
Number of oscillation (n) =20 (Given)
Now, we can calculate the error per oscillation using the following formula:
$
\vartriangle T = \left( {\dfrac{{least\, count \,error}}{{no.\,of \,oscillation}}} \right) \\
\Rightarrow \vartriangle T = \left( {\dfrac{{\dfrac{1}{5}}}{{20}}} \right) \\
$
Now, calculate the 1 oscillation time period i.e.:
$
T = \left( {\dfrac{t}{n}} \right) \\
\Rightarrow T = \left( {\dfrac{{25}}{{20}}} \right) \\
$
Finally calculate the % of error. We apply the following formula:
$
\left( {\dfrac{{\vartriangle T}}{T}\times 100\% } \right) \\
\Rightarrow \left( {\dfrac{{\dfrac{{0.2}}{{20}}}}{{\dfrac{{25}}{{20}}}}} \right)\times 100\% \\
\Rightarrow \left( {\dfrac{{0.2}}{{25}}\times 100} \right)\% \\
\Rightarrow \left( {\dfrac{2}{{250}}\times 100} \right)\% \\
\Rightarrow \left( {\dfrac{4}{5}} \right)\% \\
\therefore 0.8\%$
Correct answer is 0.8%.
Hence the answer is option B.
Note: It should be noted down that the least count of any measuring instrument refers to the smallest or accurate value that can be resolved on an instrument's scale. For example in the present case, the minimum value of time which can be measured on a stopwatch is 0.2 seconds. The time taken by a pendulum for completing 20 oscillations is 25 seconds, it’s not a single oscillation time.
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