JEE MAIN - Physics (2019 - 8th April Evening Slot - No. 7)

In a simple pendulum experiment for determination of acceleration due to gravity (g), time taken for 20 oscillations is measured by using a watch of 1 second least count. The mean value of time taken comes out to be 30 s. The length of pendulum is measured by using a meter scale of least count 1 mm and the value obtained is 55.0 cm. The percentage error in the determination of g is close to :-
0.2%
3.5%
0.7%
6.8%

Explanation

Time period of a pendulum (T) = $$2\pi \sqrt {{l \over g}} $$

$$ \Rightarrow $$ T2 = $$4{\pi ^2}{l \over g}$$

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

Fractional change

$$\left( {{{dg} \over g}} \right) \times 100 = \left( {{{dl} \over l}} \right) \times 100 - \left( {2{{dT} \over T}} \right) \times 100$$

$$ \therefore $$ Maximum possible percentage error,

$$\left( {{{dg} \over g}} \right) \times 100 = \left( {{{dl} \over l}} \right) \times 100 + \left( {2{{dT} \over T}} \right) \times 100$$

Error in time period(dT) = least count of time = 1 second

and T = 30 second

Error in length(dl) = least count of length = 1 mm

and $$l$$ = 55.0 cm

$$ \therefore $$ $$\left( {{{dg} \over g}} \right) \times 100 =$$ $$\left( {{{0.1} \over {55}}} \right) \times 100 + 2\left( {{1 \over {30}}} \right) \times 100$$ = 6.8%

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