Physics Test - 14

The frequency of electromagnetic waves

\(v=9 \sqrt{\text { maximum electron density }}\)

or \(v=9 \sqrt{d_{\max }}\)

\(\Rightarrow 9 \sqrt{2} \times 10^{6}=9 \sqrt{d_{\max }}\)

\(\Rightarrow \sqrt{2} \times 10^{6}=\sqrt{d_{\max }}\)

\(d_{\max }=2 \times 10^{12} m^{-3}\)

Given a mass of copper slab \(=2 \) g contains \(2 \times 10^{22}\) atoms.

We have to find the fraction of electrons that must be removed from sphere to give it \(+2 \mu \)C charge.

Charge on nucleus of each atom \(=29 \) e

\(\therefore\) Net charge on \(2 \) gm sphere \(=(29 ) e \times\left(2 \times 10^{22}\right)=5.8 \times 10^{23} \) ec

\(\therefore\) No of electrons on sphere \(=5.8 \times 10^{23}\)

\(\therefore\) Number of electrons removed to give \(2 \mu c\) charge \(=\frac{ q }{ e }\)

\(=\frac{2 \times 10^{-6}}{1.6 \times 10^{-19}} \)

\(=1.25 \times 10^{13}\)

Fraction of electrons removed \(=\frac{1.25 \times 10^{13}}{\text { Total number of electrons in sphere }}\)

\(=\frac{1.25 \times 10^{13}}{29 \times 2 \times 10^{22}} \)

\(=2.16 \times 10^{-11}\)

Given:

Mass of spaceship (m) = 1000 kg

\(g = 10 ~m/s^2\)

R (radius of earth) = 6400 km

The energy needed for the spaceship to out into free space must be equal to gravitational potential energy.

\(E=\frac{G M m}{R} \)

\(=\left(\frac{G M}{R^{2}}\right) m R \)

\(=m g R \) \(~~[\because g=\left(\frac{G M}{R^{2}}\right)\)]

\(=1000 \times 10 \times 6400 \times 10^{3} \)

\(=6.4 \times 10^{10} \mathrm{~J}\)

Infrared radiation was discovered in 1800 by William Herschel.

He did so with a simple experiment in which he dispersed sunlight through a prism and placed a thermometer at the location of each colour.He noticed that the thermometer temperature increased when he did this, which was not really unexpected since sunlight carries warmth.However when he placed the thermometer past the red end of the spectrum – where there was no visible sunlight – the thermometer’s temperature still increased.

Herschel had discovered infrared radiation – radiation beyond the red end of the visible spectrum.Wavelength: 700 nanometers to 1 millimeter.

Given: Height of the needle, \(h _{1}=4.5 ~cm\)

Object distance, \(u =-12~ cm\)

The focal length of the convex mirror, \(f =15~ cm \)

Image distance, \(v\) 

The value of \(v\) can be obtained using the mirror formula. \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

\(\Rightarrow \frac{1}{v}+\frac{1}{-12}=\frac{1}{15}\)

\(\Rightarrow \frac{1}{v}=\frac{1}{12}+\frac{1}{15}\)

\(\Rightarrow \frac{1}{v}=\frac{9}{60}\)

\(\therefore v \approx 6.7 ~cm\)

Hence, the image of the needle is \(6.7 ~cm\) away from the mirror. Also, it is on the other side of the mirror. The image size is given by the magnification formula. 

\(m =\frac{ h ^{\prime}}{ h }=-\frac{ v }{ u }\)

\(h ^{\prime}=\frac{6.7 \times 4.5}{12}\)

\(\Rightarrow h ^{\prime}=+2.5 ~cm\)

So, \(m =\frac{2.5}{4.5}\)

\(\Rightarrow m =0.56\)

The height of the image is \(2.5~ cm\). The positive sign indicates that the image is erect, virtual, and diminished. If the needle is moved farther from the mirror, the size of the image will reduce gradually.

\(4\left(x_{1} t\right)=0.5 \sin \left(\frac{5 \pi}{4} x\right) \cos (2 \omega \pi t)\)

\(y_{1}=A \sin (\kappa x+\omega t)\)

\(y_{2}=A \sin (-\omega t+\kappa x)\)

\(y_{1}+y_{2}=A[\sin \kappa x \cos \omega t+\cos \kappa x \sin \omega t \sin \kappa x \cos \omega t-\cos \kappa x \sin \omega t]\)

\(y_{1}+y_{2}=2 A \sin \kappa x \cos \omega t\)

\(v =\frac{\omega}{\kappa}\)

\(=\frac{2 \omega \pi}{5 \pi / 4}\)

\(=160 m / s\)

Kirchhoff's Law describes the enthalpy of a reaction's variation with temperature changes. In general, enthalpy of any substance increases with temperature, which means both the products and the reactants' enthalpies increase. The overall enthalpy of the reaction will change if the increase in the enthalpy of products and reactants is different.

At constant pressure, the heat capacity is equal to change in enthalpy divided by the change in temperature.

\(c_{p}=\frac{\Delta H}{\Delta T}\)

Therefore, if the heat capacities do not vary with temperature then the change in enthalpy is a function of the difference in temperature and heat capacities. The amount that the enthalpy changes by is proportional to the product of temperature change and change in heat capacities of products and reactants.

In the given question, we have to calculate the quantum number which is given as 'n' at the excited state.

Here, the spectral lines are considered as the dark or light coloured lines which are formed when the hydrogen atom absorbs energy and excites to the excited level from the ground level.

As a result of excitation the electrons emit the light of different colours.

Now, in the question, it is given that the wavelength of the photon is \(\lambda\), whereas the value of \(f\) the Rydberg constant is fixed that is \(R\).

Now, by using the formula of Rydberg equation we will calculate the value of \(\mathrm{n}_{2}\) whose value is given as \(n\).

The Rydberg equation is:

\(\frac{1}{\lambda}={R}\left(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}\right) \)

\(\frac{1}{\lambda {R}}=\frac{{n}^{2}-1}{{n}^{2}} \ldots . \text { (i) }\)

The equation (i) can also be written as:

\(\lambda {R}\left({n}^{2}-1\right)={n}^{2}\)

\(\lambda {Rn}^{2}-{n}^{2}=\lambda {R}\)

\({n}^{2}(\lambda {R}-1)=\lambda {R}\)

\({n}^{2}=\frac{\lambda {R}}{(\lambda {R}-1)} \)

\({n}=\sqrt{\frac{\lambda {R}}{\lambda {R}-1}}\)

So, we can see that the quantum number, \(n\) is equal to the under the root of the ratio of the \(\lambda {R}\) and \(\lambda {R}-1\)

Mutual induction between the two coils of area \({A}\), number of turns \({N}_{1}\) and \({N}_{2}\) with the length of secondary or primary \(l\) is given by,

\( M=-\frac{e_{2}}{\frac{d I_{1}}{d t}}=-\frac{e_{1}}{\frac{d I_{2}}{d t}}\)

Emf induced in coil 1 is given by \( e_{1}=-L_{1} \frac{d I_{1}}{d t}\)

Emf induced in coil 2 is given by \( e_{2}=-L_{2} \frac{d I_{2}}{d t}\)

If all the flux of coil 2 links coil 1 and vice versa, then

\( M^{2}=\frac{e_{1} e_{2}}{\left(\frac{d_{i}}{d t}\right)\left(\frac{d_{i_{2}}}{d t}\right)}\)

\(M^{2}=L_{1} L_{2}\)

\(M=\sqrt{L_{1} L_{2}}\)