Two clocks are being tested against a standard clock located in a national laboratory. At $12: 00: 00$ noon by the standard clock, the readings of the two clocks are
$\begin{array}{ccc} & \text {Clock} 1 & \text {Clock} 2 \\ \text { Monday } & 12: 00: 05 & 10: 15: 06 \\ \text { Tuesday } & 12: 01: 15 & 10: 14: 59 \\ \text { Wednesday } & 11: 59: 08 & 10: 15: 18 \\ \text { Thursday } & 12: 01: 50 & 10: 15: 07 \\ \text { Friday } & 11: 59: 15 & 10: 14: 53 \\ \text { Saturday } & 12: 01: 30 & 10: 15: 24 \\ \text { Sunday } & 12: 01: 19 & 10: 15: 11\end{array}$
If you are doing an experiment that requires precision time interval measurements, which of the two clocks will you prefer?
The range of variation over the seven days of observations is $162 \;s$ for clock $1$ , and $31 \,s$ for clock $2 .$ The average reading of clock $1$ is much closer to the standard time than the average reading of clock $2 .$ The important point is that a clock's zero error is not as significant for precision work as its vartation, because a zero-error can always be easily corrected. Hence clock $2$ is to be preferred to clock $1$
The relative error in resistivity of a material where
resistance $= 1.05 \pm 0.01\, \Omega$
diameter $= 0.60 \pm 0.01\, mm$
length $= 75.3 \pm 0.1 \,cm$ is
In order to determine the Young's Modulus of a wire of radius $0.2\, cm$ (measured using a scale of least count $=0.001\, cm )$ and length $1 \,m$ (measured using a scale of least count $=1\, mm$ ), a weight of mass $1\, kg$ (measured using a scale of least count $=1 \,g$ ) was hanged to get the elongation of $0.5\, cm$ (measured using a scale of least count $0.001\, cm$ ). What will be the fractional error in the value of Young's Modulus determined by this experiment? (in $\%$)
A physical quantity $A$ is related to four observable $a,b,c$ and $d$ as follows, $A = \frac{{{a^2}{b^3}}}{{c\sqrt d }}$, the percentage errors of measurement in $a,b,c$ and $d$ are $1\%,3\%,2\% $ and $2\% $ respectively. What is the percentage error in the quantity $A$ ......... $\%$
The period of oscillation of a simple pendulum is $T=2\pi \sqrt {\frac{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. The accuracy in the determination of $g$ is ........ $\%$
The random error in the arithmetic mean of $100$ observations is $x$; then random error in the arithmetic mean of $400$ observations would be