The period of oscillation of a simple pendulum is <math> <mrow> <mi>T</mi> <mo>=</mo> <mn>2</mn> <mi>π</mi> <mroot> <mrow> <mfrac> <mrow> <mi>L</mi> </mrow> <mrow> <mi>g</mi> </mrow> </mfrac> </mrow> <mrow></mrow> </mroot> <mo>.</mo> </mrow> </math> Measured value of ‘L’ is 1.0m from meter scale having a minimum division of 1mm and time of one-complete oscillation is 1.95s measured from stopwatch of 0.01s resolution. The percentage error in the determination of ‘g’ will be:

T = 2 π L g o r T 2 = 4 π 2 ( L g ) ⇒ g = 4 π 2 ( L T 2 ) ⇒ Δ g g % = ( Δ L L + 2 Δ T T ) % = [ 1 m m 1 m + 2 × 0 . 0 1 1 . 9 5 ] × 1 0 0 % = ( 0 . 0 0 1 + 0 . 0 1 0 2 ) × 1 0 0 = 1 . 1 3 %

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Physics Class 11th Units and Measurement

The period of oscillation of a simple pendulum is $T = 2 π \sqrt[]{\frac{L}{g}} .$ Measured value of ‘L’ is 1.0m from meter scale having a minimum division of 1mm and time of one-complete oscillation is 1.95s measured from stopwatch of 0.01s resolution. The percentage error in the determination of ‘g’ will be:

Option 1 -

1.33%

Option 2 -

1.13%

Option 3 -

1.03%

Option 4 -

1.30%

3 Views | Posted 2 weeks ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

2 weeks ago

Correct Option - 2

Detailed Solution:

$T = 2 π \sqrt[]{\frac{L}{g}} o r T^{2} = 4 π^{2} (\frac{L}{g})$

$\Rightarrow g = 4 π^{2} (\frac{L}{T^{2}})$

$\Rightarrow \frac{Δ g}{g} % = (\frac{Δ L}{L} + \frac{2 Δ T}{T}) % = [\frac{1 m m}{1 m} + \frac{2 \times 0.01}{1.95}] \times 100 % = (0.001 + 0.0102) \times 100 = 1.13 %$

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