Physics Test - 12

Given,

According to the mass action law, the product of hole concentration \(n_{h}\) and free electron concentration \({n}_{{e}}\) remains constant at constant temperature.

Let \({n}_{{h}}\) be the concentration of the holes before doping and \({n}_{{e}}\) be the concentration of the free electrons before doping.

Let \(n_{h}\) be the concentration of the holes after doping and \(n_{e}\) be the concentration of the free electrons after doping.

\(n_{i}=1.5 \times 10^{16} m^{-3}\) and \(n_{h}=4.5 \times 10^{22} m^{-3}\)

Now, we know that in an extrinsic semiconductor,

\(n_{e} n_{h}=\left(n_{i}\right)^{2}\)

\(n_{e} \times 4.5 \times 10^{22}=\left(1.5 \times 10^{16}\right)^{2}\)

\(n_{e}=\frac{2.25 \times 10^{32}}{4.5 \times 10^{22}}\)

\(n_{e}=5 \times 10^{9}\)m\(^{-3}\)

Given,

Mass of the book \(( m )=10 kg\)

Coefficient of friction \(\left(\mu_{ s }\right)=0.5\)

External force applied \(( F )=\) \(50 N\)

Normal reaction force is given by,

\( N = mg \)

\(\therefore N =10 \times 9.8=98 N\)

Where \(g\) is the acceleration due to gravity.

Limiting force of friction is given by,

\( f _{ s }=\mu_{ s } N \)

\(\therefore( f_s )_{\max }=0.5 \times 98=49 N\)

As \(F

In an isobaric process,

\(P=\) Constant...(1)

The ideal gas equation is given as,

\(P V=n R T \ldots \text { (2) }\)

Where \(P=\) pressure, \(V=\) volume, \(T=\) temperature, \(n=\) number of moles, and \(R=\) gas constant

\(V \propto T_{\ldots}(3)\)

By equation 3 we can say for an isobaric process the volume of the gas is directly proportional to the temperature.

In an expansion process the volume of the gas increases, so the temperature also increases.

Given:

\({L}=50 {mH}=50 \times 10^{-3} {H}\) and \({I}=4 {~A}\)

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

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

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

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

\(U=0.4 {~J}\)

Fundamental quantities are those quantities which cannot be expressed or measured in terms of any other physical quantities. The seven fundamental quantities along with their dimensions are as follows:

The angle of diffraction for the \(n^{\text {th }}\) minima is given as:

\(\sin \theta \approx \theta \approx \frac{n \lambda}{b}\)

\(\theta \propto \lambda\)....(1)

The angle of diffraction for the \(n^{\text {th }}\) maxima is given as,

\(\sin \theta \approx \theta \approx \frac{(2 n+1) \lambda}{2 b}\)

\(\theta \propto \lambda\)...(2)

We know that the wavelength of the red light is more than the blue light, so by equation 1 and equation 2 it is clear that when the red light is replaced by blue light the angle of diffraction will decrease, so diffraction bands become narrower and get crowded together.

Number of oscillations = 45

Total time taken = 5 seconds

Time period \(=\frac{\text {Total time taken}}{\text { Number of oscillations }}=\frac{5}{45}=\frac{1}{9} \mathrm{~s}\)

Frequency \(=\frac{1}{\text { Time period }}=\frac{1}{\frac{1}{9}} =9 \mathrm{~Hz}\)

Given that:

In loop \((1)\)

\(12-500 \mathrm{i}_{1}-\mathrm{Ri}_{1}=0\)

\(\Rightarrow 12=\mathrm{i}_{1}(500+\mathrm{R}) \ldots(\mathrm{i})\)

In loop (2)

\(12-500 \mathrm{i}_{1}-2=0\)

\(\Rightarrow 10=500 \mathrm{i}_{1}\)

Or \(\mathrm{i}_{1}=\frac{1}{50} \mathrm{~A} \ldots\) (ii)

From equations (i) and (ii)

\(12 \times \frac{1}{\mathrm{i}_{1}}=(500+\mathrm{R})\)

\(\Rightarrow 12 \times 50=500+\mathrm{R}\)

\(\Rightarrow \mathrm{R}=100 \Omega\)

We know that,

De Broglie wavelength is given by:

\(\lambda=\frac{{h}}{{p}}\)

Writing momentum as a a function of kinetic energy and mass

\(\lambda=\frac{{h}}{\sqrt{2 {Em}}}\)

Kinetic energy = potential through which it is accelerated times the charge on it

\(\lambda=\frac{{h}}{\sqrt{2 {~V} {qm}}}\)

\(\frac{\lambda_{\text {proton }}}{\lambda_{\text {alpha }}}=\sqrt{\frac{{q}_{{alpha}} {m}_{{alpha}}}{{qproton}_{\text {proton }}}}\)

\({q}_{{alpha}}=4 {e}\)

\({q}_{{proton}}={e}\)

\({m}_{{alpha}}=4 {~m}_{\text {proton }}\)

So, \(\frac{\lambda_{\text {proton }}}{\lambda_{\text {alpha }}}=\sqrt{\frac{8}{1}}=2 \sqrt{2}\)