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  • To find the value of ‘g’ using simple pendulum. T = 2.00 sec; l = 50.0 cm was measured. The maximum permissible error in ‘g’ is :  <div class='q

To find the value of ‘g’ using simple pendulum. T = 2.00 sec; l = 50.0 cm was measured. The maximum permissible error in ‘g’ is :

 

Option: 1

1.4%


Option: 2

1.1%


Option: 3

1.5%


Option: 4

1.2%


Answers (1)

best_answer

 

Time period of a simple pendulam -

T=Km^{a}l^{b}g^{0}

Equating exponents of similar quantities

a=0     b=1/2    c=-1/2

\therefore T= 2\pi \sqrt{l/g}

- wherein

T= time \: period

l= length

g=\: acceleration \: due\: to\: gravity

 

 

Error in diffrence (x=a-b) -

\Delta x= \pm \left ( \Delta a+\Delta b \right ) (minimum absolute error in x )

- wherein

\Delta a= absolute error in measurement of a

\Delta b= absolute error in measurement of b

\Delta x= absolute error in measurement of x

 

 

 

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

\left ( \frac{\Delta g}{g} \right )_{max}= \frac{\Delta l}{l}+2\frac{\Delta T}{T}

                    == \left ( \frac{0.1}{50.0}+2\frac{0.01}{2.00} \right )\times 100% = 1.2%

Posted by

Rishabh

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