Physics Test - 15

\(X_{c}=X\)

When both capacitance frequency are doubled the reactance will become

\(X_{C}^{\prime}=\frac{1}{2 \pi \times(2 f) \times(2 C)}\)

\(=\frac{1}{4} \times \frac{1}{2 \pi f C}\)

\(=\frac{X}{4}\)

Given:

\(\mathrm{I}_{\mathrm{c}}=120 \mathrm{~mA}, \mathrm{I}_{\mathrm{b}}=2 \mathrm{~mA}\)

As we know,

\( \mathrm{I}_{\mathrm{e}}=\mathrm{I}_{\mathrm{b}}+\mathrm{I}_{\mathrm{c}}\) (emitter current)

\(\Rightarrow \mathrm{I}_{\mathrm{e}}=120+2=122 \mathrm{~mA}\)

Resistance gain \(=3\)

d.c power gain \(=(\text { current gain })^{2} \times \text {resistance gain}\)

Put the given values in above formual,

\(=\left(\frac{\mathrm{I}_{\mathrm{c}}}{\mathrm{I}_{\mathrm{b}}}\right)^{2} \times\) resistance gain

\(=\left(\frac{120}{2}\right)^{2} \times 3\)

\(=(60)^{2} \times 3\)

\(=3600 \times 3\)

\(=10800\)

Given that,

Charge, \((Q)=200 {nC}\)

Separation between the plates, \(({d})=5 {~mm}\)

Area of the plate, \((A)=10\) sq.cm.

Relative permeability of air \(\left(\epsilon_{r}\right)=1\)

We know that:

Capacitance,\(C=\frac{\epsilon_{0} \epsilon_{r} A}{d}\)

\(C=\frac{8.85 \times 10^{-12} \times 1 \times 10 \times 10^{-4}}{5 \times 10^{-3}}\)

\(C=1.77 {pF}\)

Also, we know that:

\(Q=C V\)

\(V=\frac{Q}{C}\)

\(=\frac{2 \times 10^{-9}}{1.77 \times 10^{-12}}\)

\(V=1130 {~V}\)

Given,

Light of frequency \(1.5\) times the threshold frequency.

\(v=\)Light of frequency

\(v_0=\) Threshold frequency

\(v=\frac{3}{2} v_0\)

\(v^{\prime}=\frac{v}{2}=\frac{3}{2} v_0\)

\(\because v^{\prime}< v _0\)

When the frequency of incident light is halved of its original value i.e. \(1.5 v_{0}\) then it becomes less than the threshold value. In that case no photoelectric effect takes place. No photoelectrons would be emitted. The photoelectric current becomes zero.

Given:

The angle of polarization for glass, \(\mathrm{i}_{\mathrm{g}}=58^{\circ}\)

The angle of polarization for water, \(\mathrm{i}_{w}=53^{\circ}\)

According to Brewster's law, the angle of polarization is related to the refractive index of the transparent medium:

\(\tan \left(i_{p}\right)=\mu_{g} \ldots .(i)\)

and when the medium is water:

\(\tan \left(i_{w}\right)=\mu w \ldots .(ii)\)

Now, when the glass is in water, the relative refractive index of the system is

\(\frac{\mu_{g}}{\mu_{w}}=\frac{\tan \left(i_{g}\right)}{\tan \left(i_{w}\right)} \)

\(\Rightarrow \frac{\mu_{g}}{\mu_{w}}=\mu_{g w}=\frac{\tan \left(58^{\circ}\right)}{\tan \left(53^{\circ}\right)}\)

Let the angle of polarization be ip when a glass is in water. Then we get

\(\tan \left(i_{p}\right)=\mu_{g w}=\frac{\tan \left(58^{\circ}\right)}{\tan \left(53^{\circ}\right)}\)

or, \(\tan ^{-1}\left[\frac{\tan \left(58^{\circ}\right)}{\tan \left(53^{\circ}\right)}\right]\)

The magnetic force (F) experienced by a moving charged particle (q) in a uniform magnetic field (B) is given by, \(F=q v B \sin \theta\)
\(\Rightarrow B=\frac{F}{q v \sin \theta}\) where, \(\theta\) is angle between the magnetic field and the velocity of the particle.
The dimensions of charge, force and velocity are
\([q]=[Q]\), \([F]=\left[M L T^{-2}\right]\), \([\mathrm{v}]=\left[\mathrm{LT}^{-1}\right]\)
\(\sin \theta\) is a dimensionless quantity. Now, the dimension of magnetic field is
\([B]=\frac{[F]}{|q| v \mid \sin \theta]}=\frac{\left[M L T^{-2}\right]}{[Q]\left[L T^{-1}\right]}=\left[M T^{-1} Q^{-1}\right]\)
The energy stored in a capacitor is given by \(E=\frac{q^{2}}{2 C}\) where, \(C\) is capacitance.
The dimension of energy is \([E]=\left[M L^{2} T^{-2}\right]\)
Thus, the dimension of capacitance is a\([C]=\frac{\left[q^{2}\right]}{[E]}\) \(\ =\frac{\left[Q^{2}\right]}{\left[M L^{2} T^{-2}\right]}\) \(\ =\left[M^{-1} L^{-2} T^{2} Q^{2}\right.]\)
Given: \(\mathrm{X}=3 \mathrm{YZ}^{2}\) where, \(\mathrm{X}\) and \(\mathrm{Z}\) have the dimensions of capacitance and magnetic field respectively. Thus, \([X]=[C]=\left[M^{-1} L^{-2} T^{2} Q^{2}]\right.\)
\([z]=[B]=\left[M T^{-1} Q^{-1}\right]\)
Therefore, the dimension of \(Y\) is \([Y]=\frac{[X]}{[Z]^{2}}\)
\(\Rightarrow[Y]=\frac{\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{2} Q^{2}\right]}{\left[\mathrm{MT}^{-1} \mathrm{Q}^{-1}\right]^{2}}\) \(\ =\frac{\left[M^{-1} L^{-2} T^{2} Q^{2}\right]}{\left|M^{2} T^{-2} Q^{-2}\right|}\) \(\ =\left[M^{-3} L^{-2} T^{4} Q^{4}\right]\)

Germanium and Silicon are the most preferable material whose electrical properties lie in between semiconductors and insulators. 

The energy band diagram of the semiconductor is shown where the conduction band is empty and the valence band is completely filled but the forbidden gap between the two bands is very small which is about 1eV.

Volume of cuboid = l×b×h

= 1.8×1.5×1.2 cm³

= 2.70×1.2

= 3.240 cm³

Using concept of significant figures, product is reported in number of significant figures present in measurement which has least number of significant figures, here all measurement have 2 significant figures.

So, volume = 3.2 cm³

Given,

\(K = as ^{2}\)

Where \(K =\) kinetic energy, \(s =\) distance covered and \(a=\) acceleration.

Since the kinetic energy of the particle is changing, so we can say that the speed of the particle is also changing.

In a non-uniform circular motion, two forces will act on the particle \(F_{c}\) and \(F_{t}\).

Where \(F_{c}=\) centripetal force and \(F_{t}=\) tangential force.

The centripetal force acts toward the center and the tangential force acts tangentially.

So, the net force is given as,

\( F=\sqrt{F_{c}^{2}+F_{t}^{2}}\)...(1)

Centripetal force is given as,

\( F_{c}=\frac{m v^{2}}{R}\)...(2)

Where \(m=\) mass, \(R=\) radius and \(v=\) velocity

\(K = as ^{2} \)...(3) (Given)

We know that,

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

By equation (3) and equation (4), we get

\( mv ^{2}=2 as ^{2} \)...(5)

\(\Rightarrow v=s \sqrt{\frac{2 a}{m}}\)

By equation (2) and equation (5), we get

\( F_{c}=\frac{2 a s^{2}}{R}\)...(6)

Tangential acceleration is given as,

\( a_{t}=\frac{d v}{d t} \)

\(\Rightarrow a_{t}=\frac{d v}{d s} \cdot \frac{d s}{d t}\)

\(\Rightarrow v=\frac{d s}{d t}\)

By equation (5),

\( a_{t}=v \sqrt{\frac{2 a}{m}} \)

\(\Rightarrow a_{t}=\frac{2 a s}{m}\)...(7)

The tangential force is given as,

\( F _{ t }= ma _{ t } \)

\(\Rightarrow F _{ t }=2 as\)...(8)

By equation (1), (6) and (8), we get

\( F=\sqrt{\left(\frac{2 a{s}^{2}}{R}\right)^{2}+(2 a s)^{2}}\)

\(\therefore F=2 a s\left[\frac{s^{2}}{R^{2}}+1\right]^{\frac{1}{2}}\) or \( F=2 a s\left[1+\frac{s^{2}}{R^{2}}\right]^{\frac{1}{2}}\)

Given that:

The radius of the outer shell = b

The radius of the inner shell = a

The inner surface of the outer shell has charge +Q.

The outer surface of the inner shell has induced charge −Q.

We know that:

The potential difference between the two shells is given by,

\(V=\frac{Q}{4 \pi \epsilon_{0} a}-\frac{Q}{4 \pi \epsilon_{0} b}\)

Where \(\epsilon_{0}\) is the permittivity of free space,

\(V=\frac{Q}{4 \pi \epsilon_{0}}\left[\frac{1}{a}-\frac{1}{b}\right]\)

\(\frac{Q}{V}=\frac{4 \pi \epsilon_{0}}{\left[\frac{1}{a}-\frac{1}{b}\right]}\)

\(C=4 \pi \epsilon_{0} \left(\frac{a b}{b-a}\right)\)

Basically, a coaxial cable consists of a hollow (outer) cylindrical conductor surrounding a single (inner) conductor along is the axis. The two conductors are well insulated from each other. The electric field (E) and magnetic field (H) at the cross-sections are shown by solid lines and dotted lines, respectively. The outer conductor acts as the shield and minimizes interference.

Different kinds of dielectric materials, such as Teflon and polythene are covered over copper wire, it acts as a spacer. In the transmission of power through coaxial cable, the dielectric medium separating the inner conductor from outer one plays a vital role. These dielectric materials are good insulators only at low frequencies. As the frequency increase, the energy loss becomes significant That is why a coaxial cable can be used effectively for transmission up to a frequency of 20 MHz.

A steady signal flowing in a wire uniformly distributes itself throughout the cross-section of the wire. A high-frequency signal, on the other hand, distributes itself uniformly, there being a concentration of current on the outer surface of the conductor. If the frequency of the current is very high, the current is almost wholly confined to the surface layers. This is called 'Skin effect".

Given:

\(\mathrm{V}_{\mathrm{d}}=7.5 \times 10^{-4}\)

\(\mathrm{E}=3 \times 10^{-10}\)

Mobility of charged particle

\(\mu=\frac{\left|\mathrm{v}_{\mathrm{d}}\right|}{\mathrm{F}}=\frac{7.5 \times 10^{-4}}{3 \times 10^{-10}}=2.5 \times 10^6 \mathrm{~m}^2 \mathrm{~V}^{-1} \mathrm{~s}^{-1}\)

Let's calculate the field due to one half by way of integration. (This is just to demonstrate ; otherwise the result is obviously half the field due to a solenoid extending to large length on both sides.)
Consider a rig element of angular width \(d \theta\) as shown in the figure.\(\mathrm x =a \tan \theta \)
\(\therefore \mathrm { d x =a \sec ^2 \theta d \theta}\)
Number of turns in width \(\mathrm {d x}\) is \(\mathrm {n d x}\)
Field due to ring at \(\mathrm P\) is along the axis given by\(\mathrm {d B =\frac{\mu_0 \mu_1 n d x I a^2}{2\left(a^2+x^2\right)^{3 / 2}}=\frac{\mu_0 \mu_1 n I a^2}{2} \frac{\left(a \sec { }^2 \theta d \theta\right)}{(a \sec \theta)^3}\left[\because \quad \sqrt{a^2+x^2}=a \sec \theta\right]} \)
\(\therefore \mathrm { d B =\frac{\mu_0 \mu_1 n I}{2} \cos \theta d \theta} \)
\(\therefore \mathrm { B_1 =\frac{\mu_0 \mu_1 n I}{2} \int_0^{\pi / 2} \cos \theta d \theta=\frac{1}{2} \mu_0 \mu_1 n I}\)
Similarly, field due to second half will be\(\mathrm {B_2=\frac{1}{2} \mu_0 \mu_2 n I}\)
Resultant field at \(\mathrm P\) is:\(\mathrm {B=B_1+B_2=\frac{1}{2}\left(\mu_1+\mu_2\right) \mu_0 n I}\)

Given:

Depth of the fish from the surface of the water \(=4 \mathrm{~m}\)

Height of the flame \(\mathrm{BC}=2 \mathrm{~m}\)

Refractive index of the water, \(\mu=\frac{4}{3}\)

We have to find the the apparent height of the flame from the eyes of the fish. The ray diagram is shown below:

When viewed from the rarer medium, the apparent height is

Apparent height \(=\frac{\text { Real height }}{\text { Refractive index of the denser medium }}\)

But, when viewed from the denser medium, the apparent height is

Apparent height \(=\) Real height \(\times\) Refractive index of the densermedium

\(A C=B C \times \mu \)

Put the given valuesin above formula:

\(A C=2 \times \frac{4}{3}=\frac{8}{3} \)

\(\Rightarrow A C=\frac{8}{3}\)

Thus, the total apparent height of the flame from the eyes of the fish is

\(A D=A C+C D \)

\(\Rightarrow A D=\frac{8}{3}+4 \)

\(=\frac{20}{3} m\)