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Units and Measurements — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsUnits and Measurements1 Mark
Question
The period of oscillation of a simple pendulum is $\text{T}=2\pi\sqrt{\frac{\text{L}}{\text{g}}}.$ Measured value of L is 20cm known to 1mm accuracy and time for 100 oscillations of the pendulum is found to be 90s using a wrist watch of 1s resolution. What is the accuracy in the determination of g?OR
The length, breadth and thickness of a rectangular sheet of metal are 4.234m, 1.005m and 2.01cm respectively. Find the area and volume of the sheet to correct significant figures.
✓

Answer

$\text{T}=2\pi\sqrt{\frac{\text{L}}{\text{g}}},$ Squaring both sides and rearranging for G, We have $\text{g}=\frac{4\pi^2\text{L}}{\text{T}^2}$ $\therefore\frac{\Delta\text{g}}{\text{g}}=\frac{\Delta\text{L}}{\text{L}}+\frac{2\Delta\text{T}}{\text{T}}$ $\Delta\text{L}=1\text{mm}=0.1\text{cm},$ To calculate $\frac{\Delta\text{T}}{\text{T}}$ $\text{t}=\text{nt}$ $\therefore\frac{\Delta\text{T}}{\text{T}}=\frac{\Delta\text{t}}{\text{t}}$ Putting, this value $\therefore\frac{\Delta\text{g}}{\text{g}}=\frac{\Delta\text{L}}{\text{L}}+\frac{2\Delta\text{t}}{\text{t}}$ $=\frac{0.1}{20}+2\frac{1}{90}=0.027=0.03$ Hence accurcy is 3%.OR
$\mathrm{L}=4.234 \mathrm{~m}, \mathrm{~B}=1.005 \mathrm{~m}, \mathrm{~d}=2.01 \mathrm{~cm}=2.01 \times 10^{-2} \mathrm{~m}$ Area of metal sheet $=\mathrm{L} \times \mathrm{B}=4.234 \times 1.005=4.25517 \mathrm{~m}^2$. Since both length and breadth have four significant figure area of metal sheet is given by $4.255 \mathrm{~m}^2$. Volume $=$ area $<$ thickness $=0.0855 \mathrm{~m}^3$

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