An experiment is performed to obtain the value of acceleration due to gravity g by using a simple pendulum of length L. In this experiment time for 100 oscillations is measured by using a watch of 1 second least count and the value is 90.0 seconds. The length L is measured by using a meter scale of least count 1 mm and the value is 20.0 cm. The error in the determination of g would be :

  • Option 1)

    1.7%

  • Option 2)

    2.7%

  • Option 3)

    4.4%

  • Option 4)

    2.27%

 

Answers (1)

As discussed in @7226

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

\frac{\Delta g}{g}\times 100= \left ( \frac{\Delta l}{l}\times 100 \right )+2\left ( \frac{\Delta T}{T} \times 100\right )

                      = \left ( \frac{0.1}{20}\times 100 \right )+2\left ( \frac{0.01}{9} \times 100\right )

                      = 0.5+2\times \frac{10}{9}= 0.5+2.2=2.7\% 


Option 1)

1.7%

This option is incorrect

Option 2)

2.7%

This option is correct

Option 3)

4.4%

This option is incorrect

Option 4)

2.27%

This option is incorrect

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