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1.Units, Dimensions and Measurement
normal
The period of oscillation of a simple pendulum is $T=2 \pi \sqrt{L / g}$ Measured value of $L$ is $20.0 \;cm$ known to $1\; mm$ accuracy and time for $100$ oscillations of the pendulum is found to be $90 \;s$ using a wrist watch of $1\; s$ resolution. What is the accuracy in the determination of $g in \% ?$
A$5$
B$4$
C$2$
D$3$
Solution
Answer $g=4 \pi^{2} L / T^{2}$
Here, $T=\frac{t}{n}$ and $\Delta T=\frac{\Delta t}{n} .$ Therefore, $\frac{\Delta T}{T}=\frac{\Delta t}{t}$
The errors in both $L$ and $t$ are the least count errors. Therefore, $(\Delta g / g)=(\Delta L / L)+2(\Delta T / T)$
$=\frac{0.1}{20.0}+2\left(\frac{1}{90}\right)=0.027$
Thus, the percentage error in $g$ is
$100(\Delta g / g)=100(\Delta L / L)+2 \times 100(\Delta T / T)$
$=3 \%$
Here, $T=\frac{t}{n}$ and $\Delta T=\frac{\Delta t}{n} .$ Therefore, $\frac{\Delta T}{T}=\frac{\Delta t}{t}$
The errors in both $L$ and $t$ are the least count errors. Therefore, $(\Delta g / g)=(\Delta L / L)+2(\Delta T / T)$
$=\frac{0.1}{20.0}+2\left(\frac{1}{90}\right)=0.027$
Thus, the percentage error in $g$ is
$100(\Delta g / g)=100(\Delta L / L)+2 \times 100(\Delta T / T)$
$=3 \%$
Standard 11
Physics
