Embibe Experts Solutions for Chapter: Physical World and Measurement, Exercise 3: Exercise - 3

The energy of a system as a function of time $t$ is given as $E\left(t\right)=A^2\exp \left(-\alpha t\right)$, where $\alpha =0.2{ \mathrm{s}}^{-1}$. The measurement of $A$ has an error of $1.25\%$. If the error in the measurement of time is $1.50\%,$ the percentage error in the value of $E\left(t\right)$ at $t=5 \mathrm{s}$ is

In an experiment to determine the acceleration due to gravity $g$, the formula used for the time period of a period of a periodic motion is, $T=2\pi \sqrt{\frac{7\left(R-r\right)}{5g}}$. The values of $R$ and $r$ are measured to be $\left(60\pm 1\right) \mathrm{mm}$ and $\left(10\pm 1\right) \mathrm{mm}$, respectively. In five successive measurements, the time period is found to be $0.52 \mathrm{s}\text{,} 0.56 \mathrm{s}\text{,} 0.57 \mathrm{s}\text{,} 0.54 \mathrm{s}$ and $0.59 \mathrm{s}$. The least count of the watch used for the measurement of time period is $0.01 \mathrm{s}$. Which of the following statement(s) is (are) true?

A person measures the depth of a well by measuring the time interval between dropping a stone and receiving the sound of impact with the bottom of the well. The error in his measurement of time is $\mathrm{\delta }T=0.01$ second and he measures the depth of the well to be $L=20 \mathrm{meters}$. Take the acceleration due to gravity $g=10{ \mathrm{m} \mathrm{s}}^{-2}$ and the velocity of sound is $300{ \mathrm{m} \mathrm{s}}^{-1}$. Then the fractional error in the measurement $\frac{\mathrm{\delta }L}{L}$ is closest to

The respective number of significant figures for the numbers $23.023, 0.0003$ and $2.1 \times {10}^{–3}$ are

The current-voltage relation of the diode is given by $I=\left(e^{1000\mathrm{V}/\mathrm{T}}-1\right) \mathrm{mA}$, where the applied voltage $V$ is in $\mathrm{volt}$ and the temperature $T$ is in $\mathrm{kelvin}$. If a student makes an error measuring $\pm 0.01 \mathrm{V}$ while measuring the current of $5 \mathrm{mA}$ at $300 \mathrm{K}$, what will be the error in the value of current in $\mathrm{mA}$?

The period of oscillation of a simple pendulum is $T=2\mathrm{\pi }\sqrt{\frac{\mathrm{l}}{\mathrm{g}}}$. The measured value of $l$ is $20.0\mathrm{cm}$ known to $1\mathrm{mm}$ accuracy and time for $100$ oscillation for the pendulum is founded to be $90\mathrm{s}$ using a wrist watch of 1st resolution. The accuracy in the determination of $g$ is

A student measures the time period of $100$ oscillations of a simple pendulum four times. That data set is $90 \mathrm{s}, 91 \mathrm{s}, 95 \mathrm{s}$ and $92 \mathrm{s}$. If the minimum division in the measuring clock is $1 \mathrm{s}$, then the reported mean time should be:

The density of a material in the shape of a cube is determined by measuring three sides of the cube and its mass. If the relative errors in measuring the mass and length are, respectively, $1.5 \%$ and $1\%$, the maximum error in determining the density is,