The Question is from neet-physics , and it was created by the author VENKATESWARARAO

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The length of a simple pendulum is about 100 cm known to an accuracy  of 1 mm. Its period of oscillation is 2s determined by measuring the time for 100 oscillations using a clock of 0.1 s resolution. What is the accuracy in the determined value of g?

VENKATESWARARAO · Accepted Answer

## CORE INFORMATION\n✓ **Answer**: A\n## CONCEPT FRAMEWORK\n\t**Primary Concept**: Error Propagation in Physical Measurements\n\t**Related Topics**: Simple Pendulum, Period of Oscillation, Percentage Uncertainty\n\t**Prerequisites**: Formula for the period of a simple pendulum, understanding of uncertainty and error propagation\n## SOLUTION PATHWAY\n**Given Information**:\n\t* Length of pendulum (L) ≈ 100 cm = 1 m\n\t* Uncertainty in length (ΔL) = 1 mm = 0.001 m\n\t* Period of oscillation (T) = 2 s\n\t* Time for 100 oscillations (t) = 200 s\n\t* Uncertainty in time for 100 oscillations (Δt) = 0.1 s\n**Method & Steps**:\n\t1.\tFind the formula for g:\n\t\tReasoning: The period of a simple pendulum is given by $T = 2\pi\sqrt{\frac{L}{g}}$. We need to rearrange this to solve for g.\n\t\tFormula/Rule: $T = 2\pi\sqrt{\frac{L}{g}}$  => $T^2 = 4\pi^2\frac{L}{g}$ => $g = \frac{4\pi^2L}{T^2}$\n\t2.\tCalculate the uncertainty in the period (ΔT):\n\t\tWorking: The time for 100 oscillations is 200s with an uncertainty of 0.1 s. Therefore the time for 1 oscillation has uncertainty of 0.1/100 = 0.001s. Thus ΔT = 0.001s\n\t\tKey Point: The uncertainty in the period is the uncertainty in the time divided by the number of oscillations.\n\t3.\tCalculate the percentage uncertainty in g:\n\t\tReasoning: Using error propagation rules, the percentage uncertainty in g is the sum of the percentage uncertainty in L and twice the percentage uncertainty in T.\n\t\tWorking: \n\t\tPercentage uncertainty in L =  $ \frac{\Delta L}{L} \times 100 = \frac{0.001}{1} \times 100 = 0.1\%$\n\t\tPercentage uncertainty in T = $ \frac{\Delta T}{T} \times 100 = \frac{0.001}{2} \times 100 = 0.05\%$\n\t\tPercentage uncertainty in g = $ \frac{\Delta g}{g} \times 100 =  \frac{\Delta L}{L} \times 100 + 2 \times \frac{\Delta T}{T} \times 100 = 0.1\% + 2 \times 0.05\% = 0.1\% + 0.1\% = 0.2\%$\n\t4.\tFinal step:\n\t\tResult: The accuracy in the determined value of g is 0.2%.\n\t\tVerification: The result matches the correct answer provided.\n ## OPTION ANALYSIS\n[A] ✓: The calculated percentage uncertainty in g is 0.2%.\n[B] :x: This value is not consistent with the error propagation calculation.\n[C] :x: This value is not consistent with the error propagation calculation.\n[D] :x: This value is not consistent with the error propagation calculation.\n ## LEARNING AIDS\n\t**Common Mistakes**:\n\t\t1.[Error 1]: Forgetting to divide the uncertainty in time by the number of oscillations when calculating the uncertainty in the period.\n\t\t\t[How to avoid]:  Remember that the period is the time for one oscillation, not the total time for multiple oscillations.\n\t\t2.[Error 2]: Not multiplying the percentage uncertainty in the period by two because it is squared in the formula for g.\n\t\t\t[How to avoid]: Pay close attention to the exponents of the variables in the formula when applying error propagation rules.\n\t**Quick Tips**:\n\t\t1.[Helpful shortcut]: For quantities with the form $g=k\frac{A^m}{B^n}$, the percentage error in g is given by  $\% error(g) = m\% error(A) + n\% error(B)$\n\t\t2.[Memory aid]: Remember the error propagation rules, especially for exponents in the formula. The error is doubled for squared terms.

VENKATESWARARAO · Answer

The correct Option is "$0.2\,\% $"

VENKATESWARARAO · Answer

$0.5\,\% $

VENKATESWARARAO · Answer

$0.1\,\% $

VENKATESWARARAO · Answer

$2\% $

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