In a simple pendulum experiment for determination of acceleration due to gravity (g), time taken for 20 oscillation 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:

  • Option 1)

    0.7%

  • Option 2)

    0.2%

  • Option 3)

    3.5%

  • Option 4)

    6.8%

Answers (1)

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

\frac{\Delta g}{g}=\frac{\Delta \l}{\l}+2\frac{\Delta T}{T}

     =\frac{0.1}{55}+2 \times \frac{1}{30}=0.0685

\frac{\Delta g}{g}\times 100= 0.068 \times 100=6.85 \%

=7 %


Option 1)

0.7%

Option 2)

0.2%

Option 3)

3.5%

Option 4)

6.8%

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