Physics Test - 16

Dimensions of energy \(E=\left[M L^{2} T^{-2}\right]\)

Dimensions of volume \(v=\left[L^{3}\right]\)

Dimensions of force \(F=\left[M L T^{-2}\right]\)

Dimensions of area \(A=\left[L^{2}\right]\)

Dimensions of voltage \(V=\left[M L^{2} T^{-3} A^{-1}\right]\)

Dimensions of charge \(q=[A T]\)

Dimensions of angular momentum \(L=\left[M L^{2} T^{-1}\right]\)

\(\therefore\) Dimensions of \(\frac{E}{v}=\frac{\left[M L^{2} T^{-2}\right]}{\left.L^{3}\right]}=\left[M L^{-1} T^{-2}\right]\)

Dimensions of \(\frac{F}{A}=\frac{\left[M L T^{-2}\right]}{\left[L^{2}\right]}=\left[M L^{-1} T^{-2}\right]\)

Dimensions of \(\frac{V_{q}}{v}=\frac{\left[M L^{2} T^{-3} A^{-1}\right][A T]}{\left[L^{3}\right]}=\left[M L^{-1} T^{-2}\right]\)

Dimensions of angular momentum is \(\left[M L^{2} T^{-1}\right]\)

while other three has dimensions \(\left[M L^{-1} T^{-2}\right]\)

The relative permittivity of the water is highest among the other materials given.

Permittivity describes the amount of charge needed to generate one unit of electric flux in a particular medium. Accordingly, a charge will yield more electric flux in a medium with low permittivity than in a medium with high permittivity. Thus, permittivity is the measure of a material's ability to resist an electric field.

Relative permittivity is the ratio of its absolute permittivity ' \(\epsilon\) ' to free space (empty of matter) permittivity ' \(\epsilon_{0}\) ' i.e.,

\(\epsilon_{r}=\frac{\epsilon}{\epsilon_{0}}\)

We have the total area over which the transmission is viewed as

\(\mathrm{A}=2 \pi \mathrm{hR}\)

\(=2 \times 3.14 \times 0.1 \times 6.37 \times 10^{3} \mathrm{~km}^{2}\)\(=4000 \mathrm{~km}^{2}\)

And the population density is given as \(1000 / \mathrm{km}^{2}\).

Thus, we get the population covered as:

\(=4000 \times 1000=40 \times 10^{5}\)

The angle of incidence at which a beam of unpolarized light falling on a transparent surface is reflected as a beam of completely plane polarised light is called polarising or Brewster angle. It is denoted by \(\mathrm{i}_{P}\).

Brewster's law: It states that when a ray is passed through some transparent medium having refractive index \(\mu\) at any particular angle of incidence, the reflected ray is completely polarized; and the angle between reflected and refracted ray is \(90^{\circ}\).

\(\mu=\tan \theta_{\mathrm{B}}\)

Where \(\mu=\) refractive index and \(\theta_{\mathrm{B}}\) is Brewster's angle or polarizing angle \(\left(\mathrm{i}_{\mathrm{p}}\right)\).

Given that:

The light is incident on the glass.

The refractive index of glass \((\mu)=\frac{3 }{2}\)

The angle of incidence \(\left(\mathrm{i}_{\mathrm{p}}\right)=?\)

\(\frac{3}{2}=\tan \theta \)

\(\Rightarrow \theta=\tan ^{-1}\left(\frac{3}{2}\right)\) \(\Rightarrow \theta=56.30^{\circ} \approx 57^{\circ}\)

Mass of water in the first tube is

m =${\mathrm{\pi r}}^2\mathrm{h\rho }$

Now, surface tension$\sigma =\frac{h\rho gr}{2}=\frac{h\text{'}\rho gr\text{'}}{2}$

where h' is the height to which water rises in the second tube and r' its radius. since r' = 2r, h' =$\frac{h}{2}$ Therefore, the mass of water in the second capillary tube is

m' =${\mathrm{\pi r}}^2\mathrm{h}\text{'}\mathrm{\rho }=\mathrm{\pi }{\left(2\mathrm{r}\right)}^2\frac{\mathrm{h}}{2}\mathrm{\rho }$

=$2\pi r^{2}h\mathrm{\rho }$= 2m = 2 × 5 = 10g

The area of the hysteresis loop is proportional to the net energy absorbed per unit volume by the material, as it is taken over a complete cycle of magnetization.

For electromagnets and transformers, energy loss should be low.

i.e., thin hysteresis curves.

Also, \(|\mathrm{B}| \rightarrow \mathrm{0}\) when \(\mathrm{H}=0\) and \(|\mathrm{H}|\) should be small when \(\mathrm{B} \rightarrow 0\)

For a Hydrogen like species, the energy of each orbit is expressed by the Bohr's model as\(\mathrm{E}=-\frac{\mathrm{E}_{0}}{\mathrm{n}^{2}}\)

Here, \(-E_{0}\) is the energy of the first orbit.

Given that \(\mathrm{E}(3)-\mathrm{E}(2)=6.8 \mathrm{eV}\)

Thus, \(\mathrm{E}_{0}\left(-\frac{1}{9}+\frac{1}{4}\right)=6.8\)

\(\Rightarrow \mathrm{E}_{0}=48.96 \mathrm{eV}\)

Hence, in order to shift electron from first orbit to infinity, energy required is\(\mathrm{E}=48.96 \mathrm{eV}\).

Wavelength is given by \(\lambda=\frac{h c}{E}\)

\(=\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{48.96 \times 1.602 \times 10^{-18}} \approx 0.252 \times 10^{-8} \mathrm{~m}=25.27 \mathring{A}\)

Case 1: Stopping potential for \(\lambda_1=V_1\)

\(eV_1=h\nu\Rightarrow\frac{hc}{\lambda_1}\)

\(eV_1=\frac{hc}{\lambda_1}\quad\ldots\)(1)

Case 2: Stoping potential for \(\lambda_2=V_2\)

\(eV_2=h\nu\)

\(eV_2=\frac{hc}{\lambda_2}\)

According to the question,

\(V_2>V_1\)

\(\therefore\frac{hc}{\lambda_2}>\frac{hc}{\lambda_1}\)

\(\Rightarrow \lambda_1>\lambda_2\)

Minimum current required by the silicon diode \(= 1mA = 10^{-3} A\)

At knee point voltage across the diode is \(0.7 V\).

Hence voltage across resistance \(R\) is \(5-0.7=4.3 V\)

Using \(V= iR\)

\( \Rightarrow 4.3=1 \times 10^{-3} \times R\)

\( \Rightarrow 4.3 k \Omega\)

Acceleration toward +vex-direction in first case is:

\(\mathrm{v}-\mathrm{u}=\mathrm{at}\)

\(a=\frac{6}{1}=6 \mathrm{~m} / \mathrm{s}^{2}\)

When field reserved acceleration changed to

\(a=-6 m / s^{2}\)

\(\mathrm{x}=\frac{1}{2} a \mathrm{t}_{1}^{2}\)

\(=\frac{1}{2}\times 6 \times 1=3\)

\(\mathrm{y}=\mathrm{vt}+\frac{1}{2} \mathrm{at} \mathrm{2}_{2}^{2}\)

\(=6\times 2+\frac{1}{2}(-6)\times 2^{2}\)

\(y=12-12=0\)

\(\mathrm{x}_{2}=\) distance traveled is \(1 \) sec

\(\mathrm{x}_{2}=6 \times1-\frac{1}{2}(6) \times1\)

\(=3 \mathrm{m}\)

Total distance cover \(=\mathrm{x}_{1}+2 \mathrm{x}_{2}=9 \mathrm{m}\)

Total dispalcement \(=3 \mathrm{m}\)

Average velocity \(=\frac{\text { displacement cover }}{\text { time taken }}\)

\(=\frac{3}{3}=1\)m/s

Average speed \(=\frac{\text { distance cover }}{\text { time taken }}\)

\(=\frac{9}{3}=3\)m/s