Calculate the length of the arc of a circle of radius $31.0 \text{ cm}$ which subtends an angle of $ \frac{ \pi}{6}$ at the centre.
We know that angle $ \theta = \frac{l}{r} \text{radian}$
Given, $ \theta = \frac{ \pi}{6} = \frac{l}{31 \text{ cm}}$
Hence, length $l = 31 \times \frac{ \pi}{6} \text{ cm} = \frac{31 \times 3.14}{6} \text{ cm} = 16.22 \text{ cm}$
Rounding off to three significant figures it would be $16.2 \text{ cm}$.
Calculate the solid angle subtended by the periphery of an area of $1\text{cm}^{2}$ at a point situated symmetrically at a distance of $5 \text{ cm}$ from the area.
We know that solid angle $ \Omega = \frac{\text{Area}}{(\text{Distance})^{2}}$
$= \frac{1 \text{cm}^{2}}{(5 \text{cm})^{2}}= \frac{1}{25} = 4 \times 10^{-2} \text{ steradian}$
(∵ Area = $1 \text{cm}^{2}$, distance = $5 \text{ cm}$)
Note We should not confuse, solid angle with plane angle $ \theta = \frac{l}{r} \text{ radian}$.
$y = A \sin(\omega t - kx)$, where $x$ is distance and $t$ is time. Write the dimensional formula of (i) $\omega$ and (ii) $k$.
Now, by the principle of homogeneity, i.e., dimensions of LHS and RHS should be equal, hence
$[LHS] = [RHS]$
$$ \Rightarrow $$$[L] = [A] = L$
As $\omega t - kx$ should be dimensionless, $[\omega t] = [kx] = 1$
$$ \Rightarrow $$$[\omega] T = [k] L = 1$
$$ \Rightarrow $$$[\omega] = T^{-1}$ and $[k] = L^{-1}$
Time for 20 oscillations of a pendulum is measured as $t_1 = 39.6$ s; $t_2 = 39.9$ s and $t_3 = 39.5$ s. What is the precision in the measurements? What is the accuracy of the measurement?
Given, $t_1 = 39.6$ s, $t_2 = 39.9$ s and $t_3 = 39.5$ s
Least count of measuring instrument = 0.1 s
(As measurements have only one decimal place)
Precision in the measurement = Least count of the measuring instrument = 0.1 s
Mean value of time for 20 oscillations is given by
$t = \frac{t_1 + t_2 + t_3}{3} = \frac{39.6 + 39.9 + 39.5}{3} = 39.7\text{ s}$
Absolute errors in the measurements
$\Delta t_1 = t - t_1 = 39.7 - 39.6 = 0.1\text{ s}$
$\Delta t_2 = t - t_2 = 39.7 - 39.9 = -0.2\text{ s}$
$\Delta t_3 = t - t_3 = 39.7 - 39.5 = 0.2\text{ s}$
Mean absolute error =
$\frac{|\Delta t_1| + |\Delta t_2| + |\Delta t_3|}{3} = \frac{0.1 + 0.2 + 0.2}{3} = \frac{0.5}{3} \approx 0.2\text{ s} \quad\text{(rounding off up to one decimal place)}$
$\therefore$ Accuracy of measurement = $\pm 0.2\text{ s}$
A new system of units is proposed in which unit of mass is α kg, unit of length β m and unit of time γ s. How much will 5J measure in this new system?
We know that dimension of energy $=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]$
Let $M_1, L_1, T_1$ and $M_2, L_2, T_2$ are units of mass, length and time in given two systems.
$$ \begin{array}{ll} & \therefore M_1=1 \mathrm{~kg}, L_1=1 \mathrm{~m}, T_1=1 \mathrm{~s} \\\\ & M_2=\alpha \mathrm{kg}, L_2=\beta \mathrm{m}, \mathrm{T}_2=\gamma \mathrm{s} \end{array} $$
The magnitude of a physical quantity remains the same, whatever be the system of units of its measurement i.e.,
$$ \begin{aligned} n_1 u_1 & =n_2 u_2 \\ n_2 & =n_1 \frac{u_1}{u_2}=n_1 \frac{\left[\mathrm{M}_1 L_1^2 \mathrm{~T}_1^{-2}\right]}{\left[\mathrm{M}_2 L_2^2 T_2^{-2}\right]}=5\left[\frac{\mathrm{M}_1}{\mathrm{M}_2}\right] \times\left[\frac{L_1}{\mathrm{~L}_2}\right]^2 \times\left[\frac{\mathrm{T}_1}{T_2}\right]^{-2} \\\\ & =5\left[\frac{1}{\alpha} \mathrm{kg}\right] \times\left[\frac{1}{\beta} \mathrm{m}\right]^2 \times\left[\frac{1}{\gamma} \mathrm{s}\right]^{-2} \\\\ & =5 \times \frac{1}{\alpha} \times \frac{1}{\beta^2} \times \frac{1}{\gamma^{-2}} \\ n_2 & =\frac{5 \gamma^2}{\alpha \beta^2} \end{aligned} $$
Thus, new unit of energy will be $\frac{\gamma^2}{\alpha \beta^2}$.