MHT CET 2023 14th May Morning Shift (Physics & Chemistry) Question Paper with Solutions

For a light-emitting diode (LED) to emit light, it must be used in a forward-biased condition. Therefore, the correct answer is:

Option D : always used in forward biased condition

When an LED is forward biased, the electrons in the n-type material have enough energy to cross the p-n junction and recombine with holes in the p-type material. This recombination process results in the release of energy in the form of photons, which is the light that we see emitted from the LED. On the other hand, when an LED is in reverse bias, it does not emit light because the potential barrier at the p-n junction is increased, preventing electron-hole recombination.

To find the electric field at a distance of $0.2 \mathrm{~m}$ from the surface of the conducting sphere, we can use Gauss's Law. For a spherical charge distribution, the electric field at an external point is the same as if all the charge were concentrated at the center of the sphere.

First, we calculate the total charge $Q$ on the sphere. The charge density $\sigma$ is the charge per unit area, so multiplying it by the surface area $A$ gives us the total charge $Q$.

The surface area of a sphere is given by:

$A = 4\pi r^2$

Where $r$ is the radius of the sphere. Plugging in the given radius of $0.1 \mathrm{~m}$,

$A = 4\pi (0.1 \mathrm{~m})^2 = 0.04\pi \mathrm{~m}^2$

The charge density $\sigma$ is given as $1.8 \mu \mathrm{C} / \mathrm{m}^2$, which is $1.8 \times 10^{-6} \mathrm{C} / \mathrm{m}^2$. Now compute the total charge $Q$:

$Q = \sigma \cdot A = (1.8 \times 10^{-6} \mathrm{C} / \mathrm{m}^2)(0.04\pi \mathrm{~m}^2)$

$Q = (1.8 \times 10^{-6}) \cdot (0.04\pi) \mathrm{C} = 7.2 \times 10^{-8} \pi \mathrm{C}$

Gauss's Law states that the electric field $E$ multiplied by the surface area of an imaginary sphere $A'$ that encloses the charge is equal to the total charge $Q$ enclosed divided by the permittivity $\varepsilon_0$:

$E \cdot A' = \frac{Q}{\varepsilon_0}$

For a radial distance of $0.2 \mathrm{~m}$ from the surface, the actual radial distance from the center of the sphere will be the sum of the sphere's radius (0.1 m) and the distance from the surface (0.2 m), giving us:

$R = 0.1 \mathrm{~m} + 0.2 \mathrm{~m} = 0.3 \mathrm{~m}$

The area of the Gaussian surface $A'$ is now:

$A' = 4\pi (0.3 \mathrm{~m})^2 = 0.36\pi \mathrm{~m}^2$

Plugging the values back into the equation we have:

$E = \frac{7.2 \times 10^{-8} \pi \mathrm{~C}}{\varepsilon_0 \cdot 0.36\pi \mathrm{~m}^2}$

$E = \frac{7.2 \times 10^{-8}}{0.36\varepsilon_0}$

$E = \frac{7.2}{0.36} \times \frac{10^{-8}}{\varepsilon_0}$

$E = 20 \times \frac{10^{-8}}{\varepsilon_0} \mathrm{Vm}^{-1}$

$E = 2 \times \frac{10^{-7}}{\varepsilon_0} \mathrm{Vm}^{-1}$

Now, the value obtained is that the electric field $E$ at a radial distance of $0.2 \mathrm{~m}$ from the sphere's surface is equal to:

$E = \frac{2 \times 10^{-7}}{\varepsilon_0} \mathrm{Vm}^{-1}$

This matches Option C.

When a wave encounters an obstacle, such as a slit, that is comparable in size to its wavelength, diffraction occurs. This phenomenon can result in the formation of a pattern of bright and dark regions called fringes on the other side of the slit. In the case of a single-slit diffraction, both the width and the intensity of the fringes vary.

The central maximum is the brightest and the widest. As one moves away from the center towards the edges, the intensity of the fringes decreases and their width becomes narrower. Thus, the width and the intensity of the fringes are not constant. The intensity falls off more slowly than the width, but both are functions of the angle from the central maximum.

The diffraction pattern for a single-slit can be explained using the Huygens-Fresnel principle, where each point on a wavefront within the slit is considered as a source of secondary spherical wavelets, and the wavelets interfere with each other to produce the diffraction pattern.

The width of the diffraction fringes can be described mathematically by the formula that gives the position of the minima: $d \sin \theta = m \lambda$ where: $d$ is the width of the slit, $\theta$ is the diffraction angle, $m$ is the order of the minimum (with $m = \pm1, \pm2, \pm3, ...$), and $\lambda$ is the wavelength of the light.

As $m$ increases, the value of $\sin \theta$ for the minima also increases, but not linearly, and thus the spacing between fringes changes. The intensity distribution of the single-slit diffraction pattern is governed by the intensity function, which shows that the intensity of the fringes falls off as a function of the angle $\theta$.

The correct answer is : Option D - unequal width and unequal intensity.

In case of conical pendulum, the tension in the string provides the necessary centripetal force.

$\therefore \quad \mathrm{T}=\mathrm{mr} \omega^2=\mathrm{mL} \omega^2\quad$ .... (Here, $\mathrm{r}=\mathrm{L}$)

For $\mathrm{n}$ identical capacitors connected in series, the equivalent capacitance is, $\mathrm{C}_{\mathrm{s}}=\frac{\mathrm{C}}{\mathrm{n}}$

Similarly, for $\mathrm{m}$ identical capacitors connected parallel to each other, the equivalent capacitance is, $\mathrm{C}_{\mathrm{p}}=\mathrm{mC}$

Assuming the two combinations are connected in series, the net capacitance,

$$\begin{aligned} & \frac{1}{\mathrm{C}_{\text {net }}}=\frac{1}{\mathrm{mC}}+\frac{\mathrm{n}}{\mathrm{C}}=\frac{11}{10} \mu \mathrm{F} \quad \ldots .\left(\because \mathrm{C}_{\text {net }}=\frac{10}{11} \mu \mathrm{F}\right) \\ & \therefore \text { for } C=2 \mu \mathrm{F} \text {, } \\ & \frac{11 \times 2}{10}=\frac{1}{m}+n \\ & \therefore \quad \frac{1}{m}+n=\frac{11}{5} \quad \text{..... (i)}\\ \end{aligned}$$

Substituting the values for $m$ and $n$ in equation (i) from each option, the correct answer can be found to be (A).

$$\begin{array}{ll} & h \propto \frac{1}{r} \\ \therefore \quad & \text { Mass of water in a capillary, } m=\pi r^2 h \rho \\ \therefore \quad & m \propto r^2 h \\ \therefore \quad & m \propto r^2/r \\ \therefore \quad & \frac{m_2}{m}=\frac{\frac{r}{3}}{r} \\ \therefore \quad & m_2=\frac{m}{3} \end{array}$$

Let $R_1$ be the resistance of $3 \mathrm{~m}$ long wire connected in the left gap.

For meter-bridge, $\frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{l_1}{l_2}$

$$\begin{aligned} & \therefore \quad \frac{\mathrm{R}_1}{8}=\frac{3}{2} \\ & \therefore \quad \mathrm{R}_1=\frac{3}{2} \times 8=12 \Omega \end{aligned}$$

Length of the wire corresponding to the resistance of $1 \Omega$ is $l=\frac{3}{12}=0.25 \mathrm{~m}$

$\rho=\frac{M}{V}$;

Given: three objects are of same materials $\rho$ is same. Also, $\mathrm{M}=$ same,

$\therefore \quad$ volume of objects is same.

For constant volume, amongst the given objects surface area of the plate is maximum.

Hence according to Stefan's law $\frac{\mathrm{dQ}}{\mathrm{dt}} \propto \mathrm{AT}^4$

As $A_{\text {plate }}$ is maximum, the plate will cool fastest and reaches room temperature fast.

$$\begin{aligned} & \mathrm{e}=-\frac{\mathrm{d} \phi}{\mathrm{dt}}=-\frac{\left(\phi_2-\phi_1\right)}{\mathrm{t}}=-\frac{(0-\mathrm{NBA})}{\mathrm{t}} \\ \therefore \quad 0.1 & =\frac{50 \times 2 \times 10^{-2} \times 10^{-2}}{\mathrm{t}} \end{aligned}$$

$\therefore \quad$ The value of '$t$' is,

$\qquad\mathrm{t}=\frac{10^{-2}}{0.1}=0.1 \mathrm{~s}$

Tension in the string is given as,

$$\begin{aligned} & \mathrm{T}^{\prime}=\mathrm{mg} \cos \theta+\frac{\mathrm{mv}^2}{l} \\ & \mathrm{~T}_{\max }^{\prime}=\mathrm{mg}+\frac{\mathrm{mv}^2}{l} \\ \end{aligned}$$

Now, time period $\mathrm{T}=2 \pi \sqrt{\frac{l}{g}}$

$\omega=\frac{2 \pi}{\mathrm{T}}=\sqrt{\frac{\mathrm{g}}{l}}$

Maximum Velocity is given as $\mathrm{v}_{\max }=A \omega$

Substituting the values

$$\begin{aligned} \mathrm{T}_{\max }^{\prime} & =\mathrm{mg}+\frac{\mathrm{m}(\mathrm{A} \omega)^2}{l} \\ & =\mathrm{mg}+\frac{\frac{\mathrm{mA}^2 \mathrm{~g}}{l}}{l} \\ & =\mathrm{mg}+\frac{\mathrm{mA}^2 \mathrm{~g}}{l^2} \\ & =\mathrm{mg}\left(1+\left(\frac{\mathrm{A}}{l}\right)^2\right) \end{aligned}$$

In semiconductors at room temperature, the behavior of electrons across the valence and conduction bands is crucial for understanding their electrical properties. Let's clarify the options given and identify the correct answer:

• Option A: the valence band is completely filled

At absolute zero temperature, the valence band is indeed completely filled with electrons, and the conduction band is empty. However, at room temperature, thermal energy excites some electrons from the valence band to the conduction band, creating holes (absence of electrons) in the valence band. Thus, while the valence band may still be considered "full" in a broader sense, it's more accurate to say it contains holes due to the excitation of electrons.

• Option B: the conduction band is completely filled

This is incorrect for semiconductors at room temperature. The conduction band becomes partially filled due to electron excitation from the valence band, but it is not completely filled under normal conditions.

• Option C: the condition band is partially filled and valence band is partially empty

This statement introduces a misunderstanding. While the concept of the conduction band being partially filled is correct, saying the valence band is "partially empty" might be misleading. The valence band is less than fully occupied due to the presence of holes, but it's a result of electron movement, not an inherent emptiness.

• Option D: the valence band is completely filled and the conduction band is partially filled

This option might seem correct at a glance, but it misrepresents the state of the valence band at room temperature. The valence band has holes (unoccupied states) due to electrons being excited to the conduction band, which contradicts the notion of being "completely filled."

At room temperature, semiconductors have their valence band almost filled with some electrons excited to the conduction band, creating holes in the valence band. The conduction band is partially filled with these excited electrons. This scenario is crucial for the semiconductor's ability to conduct electricity, as both the excited electrons in the conduction band and the holes left behind in the valence band contribute to electrical conduction.

Given the explanations, the precise wording should reflect that the valence band has electrons but also contains holes due to thermal excitation, and the conduction band is partially filled with these excited electrons. The options provided do not perfectly capture this nuanced state. However, understanding semiconductors involves recognizing the excitation of electrons from the valence to the conduction band, leaving the valence band less than fully occupied (due to holes) and the conduction band partially occupied by electrons at room temperature.

At end points of the half revolution magnitude of the velocity is same but it directs in opposite direction.

$$\begin{array}{ll} \therefore & \Delta \mathrm{V}=\mathrm{V}-(-\mathrm{V}) \\ \therefore & \Delta \mathrm{V}=2 \mathrm{~V} \end{array}$$

Time taken to complete the half revolution is

$\mathrm{t}=\frac{\pi \mathrm{r}}{\mathrm{V}}$

Average acceleration is, $a=\frac{\Delta V}{t}=\frac{2 V}{\frac{\pi r}{V}}$

$\therefore \quad a=\frac{2 V^2}{\pi r}$

In order to stop the tiger, the momentum of the bullets fired must be equal to the momentum of the tiger.

$$\begin{array}{ll} \therefore & \mathrm{MV}=\mathrm{nmv} \\ \therefore & \mathrm{n}=\frac{\mathrm{MV}}{\mathrm{mv}}=\frac{60 \times 12}{50 \times 10^{-3} \times 240} \\ \therefore & \mathrm{n}=60 \end{array}$$

The kinetic energy of the rod while passing through the mean position will be,

$$\begin{aligned} \text { K.E. } & =\frac{1}{2} \mathrm{I} \omega^2 \\ & =\frac{1}{2} \frac{\mathrm{m} l^2}{3} \times \omega^2=\frac{\mathrm{m} l^2 \omega^2}{6} \end{aligned}$$

The angle of contact, also known as the contact angle, is the angle at which a liquid/vapor interface meets the solid surface. It quantifies the wettability of a solid surface by a liquid via the Young equation:

where:

• $\theta$ is the contact angle
• $\sigma_{SG}$ is the interfacial tension between the solid and gas
• $\sigma_{SL}$ is the interfacial tension between the solid and liquid
• $\sigma_{LG}$ is the interfacial tension between the liquid and gas

When a soluble impurity is added to a liquid, it affects the surface tension of the liquid-gas interface ( $\sigma_{LG}$ ). If the solute has a surface-active property, meaning it tends to accumulate at the surface, it typically reduces the surface tension of the liquid.

A lower $\sigma_{LG}$ (surface tension of the liquid) influences the balance of forces and consequently could decrease the contact angle if the impurity does not significantly affect the solid-liquid and solid-gas interfacial tensions. A decreased contact angle means the liquid spreads out more on the surface, leading to better wetting.

So, when considering the effects of a soluble impurity in a liquid, and assuming the impurity is surface-active and does not significantly affect the solid-liquid and solid-gas interfacial tensions, the correct answer is:

This is because the addition of the impurity decreases the surface tension of the liquid, typically resulting in a smaller contact angle and increased wetting.

As there is no resonant frequency between $315 \mathrm{~Hz}$ and $420 \mathrm{~Hz}$, let $315 \mathrm{~Hz}$ be $\mathrm{n}^{\text {th }}$ overtone and $420 \mathrm{~Hz}$ be $(\mathrm{n}+1)^{\text {th }}$ overtone.

Now, $v=\frac{\mathrm{nv}}{2 l}$ ..... (i)

$\therefore \quad 315=\frac{\mathrm{nv}}{2 l} \text { and } 420=\frac{(\mathrm{n}+1) \mathrm{v}}{2 l}$

Taking the ratio,

$$\begin{aligned} & \frac{315}{420}=\frac{n}{n+1} \\ \therefore \quad & 315 n+315=420 n \\ \therefore \quad & n=3 \end{aligned}$$

The resonant frequency is $\nu_0=\frac{v}{2 l}$

Therefore, from equation (i) we get,

$\nu_0=\frac{\nu}{\mathrm{n}}=\frac{315}{3}=105 \mathrm{~Hz}$

For an open organ pipe, the length of the pipe with end correction is given as:

$$\begin{aligned} & \mathrm{L}=l+2 \mathrm{e}=l+2 \times 0.6 \mathrm{r} \\ & \mathrm{L}=l+1.2 \mathrm{r} \end{aligned}$$

$\therefore$ The fundamental frequency of open pipe is:

$$\begin{aligned} & \mathrm{f}=\frac{\mathrm{v}}{2 \mathrm{~L}} \\ & \mathrm{f}=\frac{\mathrm{v}}{2(l+1.2 \mathrm{r})} \end{aligned}$$

When in contact, the equivalent power of two thin lenses is,

$\mathrm{P}_1+\mathrm{P}_2=10$ ..... (i)

When the lenses are separated by $0.25 \mathrm{~m}$,

$P_1+P_2-0.25 P_1 P_2=6$

$$\begin{array}{ll} \therefore & 0.25 \mathrm{P}_1 \mathrm{P}_2=4 \\ \therefore & \mathrm{P}_1 \mathrm{P}_2=16 \\ & \mathrm{P}_1-\mathrm{P}_2=\sqrt{\left(\mathrm{P}_1+\mathrm{P}_2\right)^2+4 \mathrm{P}_1 \mathrm{P}_2} \\ \therefore & \mathrm{P}_1-\mathrm{P}_2=\sqrt{36}=6 \quad \text{..... (ii)} \end{array}$$

Adding equations (i) and (ii), we get,

$$\begin{aligned} P_1 & =8 \mathrm{D} \\ \therefore \quad P_2 & =10-8=2 \mathrm{D} \end{aligned}$$

Equation for displacement for a particle performing S.H.M. is,

$y=A \cos \omega t=L \cos \left(\frac{2 \pi}{T} \times t\right)$

But time period of a simple pendulum,

$$\begin{aligned} T & =2 \pi \sqrt{\frac{L}{g}} \\ \therefore \quad y & =L \cos \left(\frac{2 \pi}{2 \pi \sqrt{\frac{L}{g}}} \times t\right)=L \cos \left(\sqrt{\frac{g}{L}} \times t\right) \end{aligned}$$

$\therefore \quad$ The linear displacement is,

$s=y \theta=L \theta \cos \left[\sqrt{\frac{g}{L}} \times t\right]$

Magnetic moment is given as $M=I A$

$\therefore \quad \mathrm{M}=\mathrm{I}\left(\pi \mathrm{R}^2\right)$

Now, length of the wire is, $L=2 \pi R$

$$\begin{aligned} \therefore \quad R & =\frac{L}{2 \pi} \\ \Rightarrow \quad M & =I \pi\left(\frac{L}{2 \pi}\right)^2 \\ \therefore \quad L & =\sqrt{\frac{4 M \pi}{I}} \end{aligned}$$

According to Bohr's second postulate,

$\frac{\mathrm{nh}}{2 \pi}=\mathrm{mvr}_{\mathrm{n}}$

$\therefore$ de-Broglie wavelength, $\lambda_n=\frac{h}{m v}=\frac{2 \pi r_n}{n}$

Also, $\mathrm{r}_{\mathrm{n}} \propto \mathrm{n}^2$

$\therefore \quad$ The de-Broglie wavelength of the electron in the $4^{\text {th }}$ orbit is:

$$\begin{aligned} &\lambda_4=\frac{2 \pi r_4}{4}=\frac{2 \pi \times(16 r)}{4} \\ &\therefore \quad \lambda_4 =8 \pi r \end{aligned}$$

The equation for torque is given by:

$\tau=\mathrm{NIAB} \sin \theta$

As axis of coil and solenoid are perpendicular to each other, $\theta=90^{\circ}$ and $\sin \theta=1$

$$\begin{aligned} \therefore \quad \tau & =\mu_0 \mathrm{nI}_1 \times \text { NIA } \\ & =4 \pi \times 10^{-7} \times \frac{500}{0.4} \times 3 \times 10 \times 0.4 \times \pi \times 10^{-2} \\ \tau & =6 \times 10^{-4} \mathrm{Nm} \end{aligned}$$

The secondary coil experiences an emf as a result of the changing magnetic field. Therefore, a changing magnetic field is primarily responsible for the transformer voltage that is induced in a transformer's secondary coil.

$$\begin{aligned} & \text { Given, } y=12 \sin (5 t-4 x) \\ & \therefore \quad y=12 \sin 2 \pi\left(\frac{5 t}{2 \pi}-\frac{4 x}{2 \pi}\right) \end{aligned}$$

Comparing above eq. with, $y=A \sin 2 \pi\left(\frac{x}{\lambda}-\frac{t}{T}\right)$

We get, $\lambda=\frac{2 \pi}{4}$

Relation between phase difference and path difference is

$$\begin{aligned} \Delta \phi & =\frac{2 \pi}{\lambda} \Delta \mathrm{x} \\ \therefore \quad \frac{\pi}{2} & =\frac{2 \pi}{\left(\frac{2 \pi}{4}\right)} \Delta \mathrm{x} \\ \therefore \quad \Delta \mathrm{x} & =\frac{\pi}{8} \end{aligned}$$

For an isothermal process, $\Delta \mathrm{U}=0$. According to first law of thermodynamics,

$\Delta \mathrm{Q}=\Delta \mathrm{U}+\mathrm{dW}$

For an adiabatic process,

$$\begin{aligned} \Delta Q & =0 \\ \therefore \quad \Delta U & =-d W \end{aligned}$$

$\therefore \quad$ It is an adiabatic process.

The potential energy of a molecule on the surface of a liquid is higher compared to the molecules inside the liquid. This is because molecules on the surface are not surrounded by as many neighboring molecules as those inside, meaning they experience fewer attractive forces. Hence, the correct answer is :

To calculate the velocity of a particle in simple harmonic motion (S.H.M.), we use the derivative of the displacement function with respect to time. The displacement function given is:

The velocity $v$ is the first derivative of displacement $x$ with time $t$:

Let's compute the derivative:

We need to calculate this velocity at time $t = \frac{\phi}{\omega}$:

Therefore, the correct answer is Option B:

The total effective resistance of an LCR (inductor-capacitor-resistor) AC circuit is known as impedance, which is a complex quantity comprising the actual resistance and the reactances due to the inductance and capacitance. However, the reciprocal of impedance is called admittance. Admittance is a measure of how easily a circuit will allow electrical current to flow.

In mathematical terms, if the impedance $Z$ is a complex number represented as :

where $R$ is the resistance, $X$ is the reactance, and $j$ is the imaginary unit.

The admittance $Y$ is the reciprocal of the impedance :

So the correct answer to the question is :

Option B : admittance

The given wave equation is:

At t = 0 , the displacement Y is given as 5 cm. So, let's find the phase $\alpha$ using this information :

$5 = 10 \sin(\alpha)$

$\frac{5}{10} = \frac{1}{2} = \sin(\alpha)$

Since $\sin(30^\circ) = \frac{1}{2}$, we can say that $\alpha = 30^\circ$ or $\alpha = \frac{\pi}{6}$ radians, considering that sine is positive in the first and second quadrants, and seeing that $\alpha$ represents a phase shift, it usually takes the smallest positive angle that satisfies the equation.

Now, let's calculate the total phase at $t = 7.5$ s :

Plug in the values:

$\mathrm{Total \ phase} = \frac{2 \pi \cdot 7.5}{30} + \frac{\pi}{6}$

$\mathrm{Total \ phase} = \frac{2 \pi \cdot 1}{4} + \frac{\pi}{6} = \frac{\pi}{2} + \frac{\pi}{6}$

To add these fractions, we first find a common denominator :

Now we simplify the fraction :

$\mathrm{Total \ phase} = \frac{2\pi}{3}$

Therefore, the total phase at $t = 7.5$ s is $\frac{2\pi}{3}$ radians, which corresponds to Option D.

The slit width has no effect on the intensity of the principal maximum in the single-slit diffraction pattern.

$\therefore$ The intensity will be same, $\mathrm{I}_0$.

Opaque body does not transmit any radiation directly,

$$\begin{array}{ll} \therefore & t=0 \\ \therefore & a+r=1 \end{array}$$

$$\begin{aligned} & \frac{g_d}{g_h}=\frac{g\left[1-\frac{d}{R}\right]}{g\left[1-\frac{2 h}{R}\right]} \\ & \frac{g_d}{g_h}=\frac{R-d}{R-2 h} \end{aligned}$$

$$\begin{aligned} & \mathrm{d} \propto \mathrm{n} \lambda \\ \therefore \quad & \frac{\mathrm{d}_2}{\mathrm{~d}_1}=\frac{\mathrm{n}_2 \lambda_2}{\mathrm{n}_1 \lambda_1}=\frac{6 \lambda_2}{8 \lambda_1} \\ \therefore \quad & \frac{\mathrm{d}_2}{\mathrm{~d}_1}=\frac{3 \lambda_2}{4 \lambda_1} \end{aligned}$$

According to Kirchhoff's first law, at point $\mathrm{P}$,

$$\begin{aligned} & 5 \mathrm{~A}+4 \mathrm{~A}+\mathrm{I}-5 \mathrm{~A}-3 \mathrm{~A}=0 \\ & \mathrm{I}=-1 \mathrm{~A} \end{aligned}$$

$\therefore \quad I=-1 A$

1 A current flows from $\mathrm{P}$ to $\mathrm{Q}$.

Wavelength of Balmer series is given as, $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right)$ where $\mathrm{n}_1=2$

For series limit, $\mathrm{n}_2=\infty$

$\therefore \quad \frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{4}-\frac{1}{\infty}\right)=\mathrm{R}\left(\frac{1}{4}\right)$

Now, $v=\frac{c}{\lambda}$

$\therefore \quad v=\frac{R c}{4}$

The radius of the $\mathrm{n}^{\text {th }}$ Bohr orbit is, $r_n \propto n^2$.