Physics Test - 10

Given,

The work function of a substance,

\(=4.0{eV}\)

The minimum energy of incident radiations, required to eject the electrons from the metallic surface is defined as the work function of that surface.

\(W_{0}=h v_{0}=\frac{h c}{\lambda_{0}}\)

where \(\lambda_{0}=\) threshold wavelength.

And\(W_{0}(e V)\) =Work function in electron volt,

\(\lambda_{0}=\frac{\mathrm{hc}}{\mathrm{W}}\)\(=\frac{\left(6.62 \times 10^{-34}\right)\left(3 \times 10^{8}\right)}{4 \times 1.6 \times 10^{-19}}\)

\(=3.10 \times 10^{-7} \mathrm{~m}=310 \times 10^{-9} \mathrm{~m}=310 \mathrm{~nm}\)

The \(P-V\) diagram of adiabatic process, isothermal process and isobaric process. 

Work done in process= area enclosed by \(P-V\) diagram with volume axis.Since area under the curve is maximum for adiabatic process, so work done on the gas will be maximum for adiabatic process.

In a transverse waves particle of the medium vibrate up and down in the vertical direction whereas it is propagating along the horizontal direction. So, in a transverse wave, a crest is a part where particle rises from its mean position and has the maximum positive displacement whereas trough is a part where particle dips below mean position and has the maximum negative displacement.

The correct sentence is that work has only magnitude but no direction.

• Work is donewhen the body moves with the application of external force or the moving body stop after theexternal forceapplied.
• Work done is thedot productof force and displacement.
• W = F ×d
  1. Where F = force applied
  2. d = displacement
• The dot product of vector quantities is always scalarwhich means work is having only magnitude but no direction.

Given,

\(\mathrm{L}=50 \mathrm{mH}\)

\(=50 \times 10^{-3} \mathrm{H}\)

\(\mathrm{I}=4 \mathrm{~A}\)

Now,

We know that,

Then,

\(U=\frac{1}{2}LI^{2}\)

The magnetic potential energy of inductor \((U)\) is:

\(U=\frac{1}{2} \times 50 \times 10^{-3} \times(4)^{2}\)

\(\Rightarrow U=\frac{1}{2} \times 50 \times 10^{-3} \times 16\)

\(\Rightarrow U=400 \times 10^{-3} \mathrm{H}\)

\(\Rightarrow U=0.4 \mathrm{~J}\)

Since the proton and electron, both are moving upward and the direction of the magnetic field is perpendicular to the plane of paper and inward, so we can say that both the particles are moving perpendicular to the magnetic field.

When the velocity of a charged particle is perpendicular to a magnetic field, it describes a circle.

Therefore we can say that both the particles will move in a circular path.

The direction of the magnetic force on the moving charged particle in the magnetic field depends on the direction of velocity and the direction of the magnetic field but it is independent of the nature of the charge. 

By the right-hand thumb rule, we can find out that the direction of the magnetic force. So the direction of the magnetic force on the proton and the electron will be such that it tends to rotate both the particles in the anticlockwise direction when viewed in the direction of the magnetic field.

We know the relation between frequency \((f)\) wavelength \((\lambda)\) and speed \((v)\):

\(v=f \lambda\)

When light passes from rarer medium to denser medium frequency remain the same but speed decrease \(v=\) constant

\(\therefore \lambda\) decreases

• Option (B) is correct because in refraction frequency is the same and the wavelength decreases as light travel from rarer to denser.
• Option (A) is incorrect because frequency doesn't change in refraction.
• Option (C) is incorrect because wavelength decreases as light travel from rarer to denser.
• Option (D) is incorrect because frequency doesn't change in refraction.

Free body diagram is given as,

If \(F_{1}=0\)

So contact force will only have Normal which is equal to \(\mathrm{Mg}\).

So \(F\) will be \(\mathrm{Mg}\).

Now as \(\mathrm{F}_{1}\) increases \(f_r\) keeps increasing till it reaches limiting value i.e., \(\mu \mathrm{N}=\mu \mathrm{mg}\).

So, in that case, contact force \(F\) will be,

\(\sqrt{N^{2}+f_{r}^{2}}=\sqrt{(M g)^{2}+(\mu M g)^{2}}\)

\(F=M g \sqrt{1+\mu^{2}}\)

This means contact force \(F\) will lie between \({Mg} \leq  {F} \leq {Mg} \sqrt{1+\mu^{2}}\).

Recombination rate \((\mathrm{R})=\frac{P_{n}}{\tau_{P}}\)

\(\begin{aligned}&P_{n}=P_{n o}+d P_{n} \\&=\frac{10^{20}}{10^{15}}+\left(4 \times 10^{5}\right) \mathrm{cm}^{-3}\end{aligned}\)

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

\(\mathrm{R}=\frac{P_{n}}{z_{P}}=\frac{5 \times 10^{5}}{10^{-6}}=5 \times 10^{11} \mathrm{~cm}^{-3} \mathrm{~s}^{-1}\)