Physics Test - 7

1 unit of electric energy: When one-kilowatt load works for 1 hour then the energy consumed is called 1 unit of electricity.

\(1\) Unit of electricity \(=1 KWh =1000\) Watt-hour \(=3.6 \times 10^{6} J\)

1 Kilo-watt \(=1000\) Watt

1 Watt: The energy consumption rate of 1 joule per second is called 1 watt.

Since \(1 hr =3600\) sec

1 Kilo-watt \(=1000\) Watt

\(1 KWh =1000\) Watt-hour \(=1000 \times 3600\) W.s \(=3.6 \times 10^{6}\) J

Given:

Fill factor of the solar cell = 0.65

Maximum power obtained when sun light is illuminated = 65 mW

Power delivered by the solar cells is given by:

\(\mathrm{P}_{\max }=(\mathrm{F} \cdot \mathrm{F}) \times \mathrm{V}_{\mathrm{OC}} \cdot {\mathrm{I_{SC}}}\)

\(\mathrm{V}_{\mathrm{OC}}=\) Open Circuit Voltage

\(\mathrm{I}_{\mathrm{SC}}=\) Short Circuit Current and \( \mathrm{f}\)

\(\mathrm{ F.F. }=\) Fill Factor

The fill factor of solar cells is:

\(\mathrm{F} . \mathrm{F}=\frac{\text { maximum power obtained }}{\mathrm{V}_{\mathrm{oc}} \times \mathrm{I}_{\mathrm{sc}}}\)

Put the given values in above formula:

\(0.65=\frac{65 \times 10^{-3}}{V_{o c} \times I_{s c}}\)

\(\Rightarrow V_{o c} \times I_{s c}=\frac{65 \times 10^{-3}}{0.65}=100 \mathrm{~mW}\)

\(\therefore\) (1) and (3) satisfy the result i.e.,

\(\mathrm{V}_{\mathrm{oc}} \times I_{\mathrm{Sc}}=40 \mathrm{~mA} \times 2.5 \mathrm{~V}=100 \mathrm{mV}\)

\(\mathrm{V}_{\mathrm{oc}} \times \mathrm{I}_{\mathrm{sc}}=50 \mathrm{~mA} \times 2 \mathrm{~V}=100 \mathrm{mV}\)

Given,

Magnetic moment of a magnetic needle, \(m=6.7 \times 10^{-2} \mathrm{Am}^{2}\)

Moment of inertia \(I=7.5 \times 10^{-6} \mathrm{~kg} \mathrm{~m}{ }^{2}\)

Magnetic needle performs \(10\) complete oscillations in \(6.70 \mathrm{~s}\).

The time period of oscillation is,

\(T=\frac{6.70}{10}=0.67 \mathrm{~s}\)

As we know, magnetic field is,

\(B=\frac{4 \pi^{2} I}{m T^{2}}\)

\(=\frac{4 \times(3.14)^{2} \times 7.5 \times 10^{-6}}{6.7 \times 10^{-2} \times(0.67)^{2}}\)

\(=0.01 \) T

Let the charge distribution be as shown in figure:

\(\therefore V_{A}-V_{B}=\frac{q}{C_{1}}-E+\frac{q}{C_{2}}\)

or, \( \left(V_{A}-V_{B}\right)+E=q\left(\frac{1}{C_{1}}+\frac{1}{C_{2}}\right)\)

\( \left(V_{A}-V_{B}\right)+E=\frac{{q}\left({C}_{2}+C_{1} \right)}{{C}_{1} {C}_{2}}\)

\(\therefore  {q}=\frac{\left[\left({V}_{{A}}-{V}_{{B}}\right)+{E}\right] {C}_{1} {C}_{2}}{{C}_{1}+{C}_{2}}\)

Voltage across \(C_{1}\) is \(V_{1}=\frac{q}{C_{1}}\)

\(=\frac{ \left [\left(V_{A}-V_{B}\right)+E\right] C_{2}}{C_{1}+C_{2}}\) 

\(=\frac{(5+10) 2.0}{1.0+2.0}\)

\(=10\) Volt

Voltage across \(C_{2}\) is \(V_{2}=\frac{q}{C_{2}}\)

\(=\frac{ \left [\left(V_{A}-V_{B}\right)+E\right] C_{1}}{C_{1}+C_{2}}\) 

\(=\frac{(5+10) 1.0}{1.0 \times 2.0}\)

\(=5\) Volt

The idea of displacement current was introduced to the current for making ampere circuital law consistent.

\(\oint \vec{B} . d \vec{l}=\mu_{0}\left(i_{c}+i_{d}\right)\)\(\quad\) (Modified Ampere circuital law)

Where, \(\mu_{0} \) is the permittivity of free space, \({i}_{d}\) is the displacement current, and \({i}_{c}\) is the conduction current. \(\oint \vec{B} \cdot d \vec{l}\) is line integral of the magnetic field over the closed-loop.

The expression for displacement current is given by,

\( i_{d}=\epsilon_{0} \frac{d \phi_{E}}{d t}\)

Where \(\phi_{{E}}\) is the flux of the electric field through the area bounded by the closed curve, \({i}_{{d}}\) is the displacement current, and \(\epsilon_{0} \) is the permittivity of free space.

In the region (I)

• Fermi level is above the mid-gap so material is n-type.

In the region (II)

• The Fermi level is very near to conduction band.
• Thus, this region is more doped than 1.
• So n⁺ region.

In region (III)

• The fermi level is inside the valence band. So this is degenerate so p⁺⁺.

In general capacitance of parallel plate, the capacitor is given by:

\(\mathrm{C}=\frac{\mathrm{k} \epsilon_{0} \mathrm{A}}{\mathrm{~d}}\)

Where \(\mathrm{C}\) is capacitance, \(\mathrm{k}\) is the relative permittivity of dielectric material, \(\epsilon_{0}\) is the permittivity of free space constant, \(A\) is an area of plates and \(d\) is the distance between them.

Therefore, the capacitance of parallel plates is increased by the insertion of a dielectric material. Further, the capacitance is inversely proportional to the electric field between the plates, and hence the presence of the dielectric decreases the effective electric field.

Magnetic field due to a circular coil of radius a is given by:

\(\mathrm{B}=\frac{\mu_{0} \mathrm{Ia}^{2}}{2\left(\mathrm{x}^{2}+\mathrm{a}^{2}\right)^{\frac{3}{2}}}\)

For \({x} \gg \mathrm{a}\),

\(\mathrm{B \approx \frac{\mu_{0} I a^{2}}{2 x^{3}}} \)

\(\mathrm{\Rightarrow B \propto a^{2}}\)

As radius is doubled, the magnetic field becomes four times.

Lyman seriesof hydrogen spectrum corresponds to ultraviolet region.

The Lyman series: It includes the lines emitted by transitions of the electron from an outer orbit of quantum number n₂ > 1 to the 1st orbit of quantum number n₁ = 1.All the energy wavelengths in the Lyman series lie in the ultraviolet band.

Given: A solenoid of inductance \(L\) and resistance \(R\) is connected to a battery. The time is taken for the magnetic energy to reach \(\frac 14\) of its maximum value is

Initial magnetic energy \(=\frac{1}{2} L i ^{2}\)

Final magnetic energy \(=\frac{1}{4}\) of initial magnetic energy

\(=\frac{1}{4}\left(\frac{1}{2} L i ^{2}\right)\)

For final energy to become \(\frac{1}{4}\) of initial magnetic energy final current have to become \(\frac 12\) of initial current.

We have a relation \(i = i _{0}\left(1- e ^{\frac{- Rt }{ L }}\right)\)

\(t =\) time, \(L=\) inductance, \(R =\) resistance

Substituting the value of current here we get

\(\frac{i_{o}}{2}=i_{0}\left(1-e^{\frac{-R t}{L}}\right)\)

\(\Rightarrow\frac{1}{2}=\left(1-e^{\frac{-R t}{L}}\right)\)

\(\Rightarrow\frac{-1}{2}=\left(- e ^{\frac{- Rt }{ L }}\right)\) (taking \(\ln\) both sides)

\(\Rightarrow\ln \left(\frac{1}{2}\right)=\frac{- Rt }{ L }\)

\(\Rightarrow t =\frac{ L }{ R } \ln ( 2 )\)