Physics Test - 5

Given:

\(C=3 \times 10^{8}  \mathrm{~ms}^{-1}\)

\(\Delta \lambda=0.02 \%\)

We know that wavelength is expressed as \(\lambda\). So now the change in the wavelength is given as follows:

\(\frac{\Delta \lambda}{\lambda}=\frac{v}{C}\)

Here \(v\) is the velocity and \(C\) represents the speed of light. Hence we can express \(\mathrm{v}\) as:

\(v=\frac{\Delta \lambda}{\lambda} C\)

Now we have to put the values in the above expression, to get:

\(v=\frac{0.02}{100} \times 3 \times 10^{8} \mathrm{~ms}^{-1} \)

\(v=60 \mathrm{~kms}^{-1}\)

Given,

\(\mathrm{n_e} =5 \times 10^{12} \) cm\(^{-3}\)

\(\mathrm{n_p}=8 \times 10^{13} \) cm\(^{-3}\)

The conductivity of a pure semiconductor sample,

\(\sigma=\mathrm{e}\left[\mathrm{n}_{\mathrm{e}} \mu_{\mathrm{e}}+\mathrm{n}_{\mathrm{h}} \mu_{\mathrm{h}}\right]\)

\(\Rightarrow \sigma=1.6 \times 10^{-19}\left[5 \times 10^{18} \times 2.3+8\right.\left.\times 10^{19} \times 0.01\right]\)

\(\Rightarrow \sigma=1.968 \Omega^{-1} \mathrm{~m}^{-1}\)

Displacement current \(=\frac{{dQ}}{{dt}}=\frac{{d}({CV})}{{dt}}\)

\(={C} \frac{{dV}}{{dt}}\)

\(={C}\left({d} \cdot \frac{{dE}}{{dt}}\right)\)

\(=\frac{{A} \epsilon_{0}}{{~d}} \cdot {d} \frac{{dE}}{{dt}}\)

\(={A} \epsilon_{0} \frac{{dE}}{{dt}}\)

\(=\pi {r}^{2} {e}_{0} \frac{{dE}}{{dt}}\)

\(=14 {~A}\)

Given,

\(a=6 \mathrm{~mm}=6 \times 10^{-3} \mathrm{~m}\), and \(\lambda=600 \mathrm{~nm}=600 \times 10^{-9} \mathrm{~m}\)

\( Z_{F}=\frac{a^{2}}{\lambda}\)

\( Z_{F}=\frac{\left(6 \times 10^{-3}\right)^{2}}{600 \times 10^{-9}}\)

\( Z_{F}=\frac{36 \times 10^{-9}}{600 \times 10^{-9}}\)

\( Z_{F}=\frac{36 \times 10^{-9}}{600 \times 10^{-9}}\)

\( Z_{F}=60 \mathrm{~m}\)

The focal length of the lens kept in the transparent medium is given as:

\(\frac{1}{f}=\left(\frac{\mu_{2}}{\mu_{1}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)

\(\mu_{1}\) is the focal length of the transparent medium, \(\mu_{2}\) is the focal length of the lens, \(R_{1}\) and \(R_{2}\) are the radii of two curved surfaces of lens.

In an expression, if the denominator is increased, the value of the number will decrease.

From the expression of the focal length of the lens, \(\frac{1}{f}\) will decrease, if the value of \(\mu_{1}\) (refractive index of medium ) increases. 

If \(\frac{1}{f}\) will decrease, the \(\mathrm{f}\) will increase. So, \(\mathrm{f}\) will be maximum for the maximum value of the refractive index. The maximum refractive index among options is \(1.6\). So, \(1.6\) is the answer.

The efficiency of a Carnot engine depends on temperatures of source and sink.

Carnot theorem states that heat engines working between two heat reservoirs are less efficient than the Carnot heat engine that is operating between the same reservoirs.

Carnot engine is a reversible heat engine working between two reservoirs and has the maximum possible efficiency. It is based on the theoretical thermodynamic cycle proposed by Leonard Carnot.

The efficiency of the Carnot Heat engine is given by:

\(\eta=1-\frac{T_{L}}{T_{H}} \)

\(\eta=\frac{T_{H}-T_{L}}{T_{H}}\)

Where \({T}_{{H}}\) is the hot reservoir temperature and \({T}_{{L}}\) is the cold reservoir temperature.

The efficiency of the Carnot cycle is the function of source and sink temperature. 

Carnot cycle efficiency depends upon the absolute temperature range of operation.

Given,

Threshold wavelength is \(2480 \mathrm{~nm}\).

So energy associated with this wavelength is threshold energy which will be same as energy of bandgap for the electron to jump in that bandgap.

\(E_{\text {bandgap }}=\frac{h c}{\lambda}=\frac{\left(6.626 \times 10^{-34}\right) \times\left(3 \times 10^{8}\right)}{\left(2480 \times 10^{-9}\right)}\)

\(=\frac{\left(6.626 \times 10^{-34}\right) \times\left(3 \times 10^{8}\right)}{\left(2480 \times 10^{-9}\right) \times\left(1.6 \times 10^{-19}\right)} \mathrm{eV}\)

\(=0.5 \mathrm{eV}\)

\(v=\frac{1}{2 l} \sqrt{\frac{M g}{m}}\)

Where M= mass of stone.

If ρ is the density of the stone and V its volume, then m = ρV.

When the stone is wholly immersed in water of density ρ' the effective weight of the stone.

M'g = (M-Vρ') = (Vρ-Vρ')g = \(V \rho'\left(\frac{\rho}{\rho'}-1\right) g\)

Now, \(v=\frac{1}{2 l} \sqrt{\frac{\rho V g}{m}}\)

And, \(v'=\frac{1}{2 l} \sqrt{\frac{V \rho' ([\frac{p}{p'}]-1) g}{m}}\)

Given, l = 40 cm and l' = 30 cm. Also v = v', which gives

\(\frac{\rho}{\rho'}=\frac{16}{7}\)

In a steady flow reversible adiabatic process, work done is equal to change in enthalpy.

The first law of thermodynamics:

This law states that the heat and mechanical work are mutually convertible. According to this law, a definite amount of mechanical work is needed to produce a definite amount of heat and vice versa.

From first Law of Thermodynamics,

\(q=\Delta u+p d v\) for a closed system and,

\(q=\Delta h-v d p\) for an open system

And for a reversible adiabatic process \(q=0\), i.e. \(\Delta \mathrm{h}=\) vdp

From Steady Flow Energy Equation:

\(\left(h_{1}+\frac{v_{1}^{2}}{2}+g z_{1}\right)+q=\left(h_{2}+\frac{v_{2}^{2}}{2}+g z_{2}\right)+w\)

When potential and kinetic energy changes are zero or negligible,

\(h_{1}+q=h_{2}+w\)

Given reversible adiabatic process so \(q=0\),

\(w=h_{1}-h_{2}=\Delta h\)

Photoelectric effect

\({K} {E}_{\max }={h v}-{W}_{{o}}\)

Here, \(W_{o}\) is the work function of the metal

For \({KE} \geq 0\)

\({hv} \geq {W}_{{o}}\)

The minimum energy required \(={W}_{{o}}\)

The work function of the metal, a characteristic property of the metal.