Physics Test - 69

The heat given to the substance during the phase change is called latent heat. It can be either the latent heat of fusion or the latent heat of vaporization depending upon the phase change.

Electromagnetic waves: It is a transverse wave defined as composed of oscillating electrical and magnetic fields perpendicular to each other.

The speed of electromagnetic waves is the same as of speed of light, i.e., \(c=3 \times 10^{8}\) \({m} / {s}\)

Mathematically,

Speed of \(E M W=\frac{1}{\sqrt{\epsilon_{0} \mu_{0}}}=3 \times 10^{8} {~m} / {s}\)

Or, the relation between the speed of electromagnetic wave and dielectric constant and vacuum permeability is:

Speed of \(E M W=\frac{1}{\left(\epsilon_{0} \mu_{0}\right)^{\frac{1}{2}}}\)

where,

\(\epsilon_{0}=8.854 \times 10^{-12} {~F} / {m}\) (vacuum permittivity or dielectric constant)

\(\mu_{0}=4 \pi \times 10^{-7} {H} / {m}\) \((\frac{\text{vacuum permeability}}{{\text{magnetic constant}}})\)

Given,

\(KE _{2}=9 KE _{1}\)

We know that the kinetic energy of a body is given as,

\( K E=\frac{1}{2} m v^{2}\)...(1)

The linear momentum of a body is given as,

\( P = mv\)...(2)

By equation (1) and equation (2) the relation between the kinetic energy and the momentum is given as,

\( K E=\frac{P^{2}}{2 m}\)...(3)

By equation (3) for the initial position,

\( K E_{1}=\frac{P_{1}^{2}}{2 m}\)...(4)

By equation (3) for the final position,

\( K E_{2}=\frac{P_{2}^{2}}{2 m} \)

\(\Rightarrow 9 \times K E_{1}=\frac{P_{2}^{2}}{2 m}\)...(5)

By equation (4) and equation (5),

\(\frac{P_{2}^{2}}{2 m} \times \frac{2 m}{P_{1}^{2}}=\frac{9 \times K E_{1}}{K E_{1}}\)

\(\Rightarrow \frac{P_{2}^{2}}{P_{1}^{2}}=9 \)

\(\Rightarrow \frac{P_{2}}{P_{1}}=3 \)

\(\Rightarrow P_{2}=3 P_{1}\)

So, if the kinetic energy of a body is increased nine times then the momentum of the body will be increased by three times.

The spin quantum number is used to determine the orientation of the spin of the electron i.e. either a spin up or spin down.This orientation is referred to with \(+1 / 2\) for a spin up and \(-1 / 2\) for spin down.So, an electron spin quantum number can have only two values.

The drift velocity of charge carriers in the material of constant cross-sectional area \(A\) is given as:

\(v=\frac{I}{nAq}\) 

(Here \(I\) is the current flowing through the material, \(n\) is the charge-carrier density, and \(q\) is the charge on the charge-carrier). 

We have the ratio of \(I\) as \(\frac{7}{4}\) and ratio of \(n\) as \(\frac{7}{5}\). 

\(I=n A e v_{d}\) or \(I \propto n v_{d}\)

\(\therefore \frac{{I}_{{e}}}{{I}_{{h}}}=\frac{{n}_{{e}} V_{{e}}}{{n}_{{h}} {V}_{{h}}}\) or \(\frac{{n}_{{e}}}{{n}_{{h}}}\)

\(=\frac{I_{e}}{I_{h}} \times \frac{V_{{h}}}{V_{e}}\)

\(=\frac{7}{4} \times \frac{5}{7}=\frac{5}{4}\)

Time period of simple pendulum in air

\(T=2 \pi \sqrt{\frac{l}{g}}\)

When it is suspended between vertical plates of a charged parallel plate capacitor, then acceleration due to electric field,

\(a=\frac{q E}{m}\)

This acceleration is acting horizontally and acceleration due to gravity is acting vertically. So, effective acceleration

\(g^{\prime}=\sqrt{g^{2}+a^{2}}\)

\(=\sqrt{g^{2}+\left(\frac{q E}{m}\right)^{2}}\)

Thus, \(T^{\prime}=2 \pi \sqrt{\frac{l}{\sqrt{g^{2}+\left(\frac{q E}{m}\right)^{2}}}}\)

Given,

The horizontal component of the earth's magnetic field, \(H_{E}=0.26 \mathrm{G}\)

The dip angle, \(\theta=60^{\circ}\)

As we know,

\(B_{E}=\frac{H_{E}}{\cos \theta}\)

\(B_{E}=\frac{H_{E}}{\cos 60^{\circ}}\)

\(=\frac{0.26}{(\frac 1 2)}=0.52 \mathrm{G}\)

Given:

Length of the rectangular wire, l = 8 cm = 0.08 m

Width of the rectangular wire, b = 2 cm = 0.02 m

Hence, area of the rectangular loop,

A = lb

= 0.08 × 0.02

= 16 × 10−4 m²

Magnetic field strength, B = 0.3 T

Velocity of the loop, v = 1 cms^-1 = 0.01 ms^-1

Emf developed in the loop is given as:

e = Blv

= 0.3 × 0.08 × 0.01 = 2.4 × 10⁻⁴V

= 0.02 V

Angle of deviation \(=\delta=i+e-A\)

\(i=0,\)

So, \(r_{1}=0\)

Therefore, \(r_{2}=60\) \(deg\) (angle of incidence of second face)

\(r_{2}>\) critical angle for glass-air \(=\sin ^{-1} \frac{1}{1.5}=41.81\) \(deg\).

It will be totally internally reflected and fall on the base of prism at \(90\) deg to it. It will emerge with angle of emergence \(\mathrm{e}=0\) \(deg\).

\(\delta=0+0-60 ~deg=60 ~deg .\)

Initial momentum of system, \(\mathrm{p}_{\mathrm{i}}=0\)

Let \(p_1\) and \(p_2\) be the momentum of broken nuclei of masses \(m_1\) and \(5 \mathrm{~m}_1\) respectively.

\(\mathrm{p}_{\mathrm{f}}=\mathrm{p}_1+\mathrm{p}_2\)

From the conservation of momentum

\(\mathrm{p}_{\mathrm{i}}=\mathrm{p}_{\mathrm{f}} \)

\(0=\mathrm{p}_1+\mathrm{p}_2 \)

\(\mathrm{p}_1=-\mathrm{p}_2\)

From de Broglie relation, wavelength

\(\lambda_1=\frac{\mathrm{h}}{\mathrm{p}_1} \text { and } \lambda_2=\frac{\mathrm{h}}{\mathrm{p}_2} \)

\(\left|\lambda_1\right|=\left|\lambda_2\right| \)

\(\lambda_1=\lambda_2=\lambda .\)