JEE Main 2023 15th April Morning Shift Question Paper with Solutions

The gravitational force between two particles of mass $m$ separated by a distance $r$ is given by:

$F = \frac{Gm^2}{r^2}$

where $G$ is the gravitational constant. In this problem, the two particles are moving in a circular orbit of radius $a$ under the influence of their mutual gravitational attraction. Therefore, the gravitational force between the two particles provides the necessary centripetal force to keep them in circular motion.

The centripetal force required for a particle of mass $m$ moving in a circle of radius $a$ with angular speed $\omega$ is given by:

$F_{\text{centripetal}} = m\omega^2a$

Setting the gravitational force equal to the centripetal force, we get:

$\frac{Gm^2}{r^2} = m\omega^2a$

Substituting $r = 2a$ (since the two particles are separated by a distance equal to twice the radius of the circle), we get:

$\frac{Gm^2}{(2a)^2} = m\omega^2a$

Simplifying, we get:

$\omega^2 = \frac{Gm}{4a^3}$

Taking the square root of both sides, we get:

$\omega = \sqrt{\frac{Gm}{4a^3}}$

Therefore, the angular speed of each particle is $\sqrt{\frac{Gm}{4a^3}}$.

The position of a particle as a function of time is given by $x=\left(5 t^{2}-4 t+5\right) \mathrm{m}$.

To find the magnitude of the velocity of the particle at $t=2\,\mathrm{s}$, we first need to find the velocity of the particle as a function of time.

The velocity $v$ is the time derivative of the position $x$:

$v = \frac{dx}{dt}$

Taking the derivative of $x$ with respect to $t$, we get:

$v = \frac{dx}{dt} = 10t - 4\,\mathrm{m/s}$

Now we can find the velocity of the particle at $t=2\,\mathrm{s}$ by plugging in $t=2$:

$v(2\,\mathrm{s}) = 10(2) - 4\,\mathrm{m/s} = 16\,\mathrm{m/s}$

Therefore, the magnitude of the velocity of the particle at $t=2\,\mathrm{s}$ is:

$\boxed{|v(2\,\mathrm{s})| = 16\,\mathrm{m/s}}$

To find the direction of the net force experienced by the particle, we need to find the acceleration vector of the particle and then use Newton's second law, which states that the net force on an object is equal to its mass times its acceleration vector.

The position vector of the particle is given by:

$\vec{r} = (10t\hat{i} + 15t^2\hat{j} + 7\hat{k})\,\text{m}$

Differentiating $\vec{r}$ twice with respect to time $t$, we get the acceleration vector:

$\vec{a} = \frac{d^2\vec{r}}{dt^2} = \frac{d}{dt}(10\hat{i} + 30t\hat{j}) = 30\hat{j}\,\text{m/s}^2$

Therefore, the acceleration vector is $\vec{a} = 30\hat{j}\,\text{m/s}^2$.

Using Newton's second law, the net force on the particle is given by:

$\vec{F}_{net} = m\vec{a}$

where $m$ is the mass of the particle.

Since we are only interested in the direction of the net force, we can ignore the magnitude of the acceleration and focus on its direction, which is along the positive $y$-axis.

To find the angle of minimum deviation for the liquid in the equilateral hollow prism, we can use the formula for the angle of minimum deviation based on the refractive index $n$ and the prism angle $A$:

$n = \frac{\sin \frac{A + \delta_m}{2}}{\sin \frac{A}{2}}$

In this case, the refractive index $n = \sqrt{2}$, and since the prism is equilateral, the prism angle $A = 60^\circ$. Now, we can substitute these values into the formula and solve for the angle of minimum deviation $\delta_m$:

$\sqrt{2} = \frac{\sin \frac{60^\circ + \delta_m}{2}}{\sin \frac{60^\circ}{2}}$

First, let's find the sine of half the prism angle:

$\sin \frac{60^\circ}{2} = \sin 30^\circ = \frac{1}{2}$

Now, we can substitute this value into the formula:

$\sqrt{2} = \frac{\sin \frac{60^\circ + \delta_m}{2}}{\frac{1}{2}}$

To isolate the sine term, we multiply both sides by $\frac{1}{2}$:

$\frac{\sqrt{2}}{2} = \sin \frac{60^\circ + \delta_m}{2}$

Now, we can find the angle inside the sine function:

$\frac{60^\circ + \delta_m}{2} = \sin^{-1} \frac{\sqrt{2}}{2} = 45^\circ$

Finally, we can solve for the angle of minimum deviation $\delta_m$:

$60^\circ + \delta_m = 2 \cdot 45^\circ$

$\delta_m = 90^\circ - 60^\circ = 30^\circ$

The angle of minimum deviation for the liquid in the equilateral hollow prism is $30^\circ$.

$$\begin{aligned} & \mathrm{A}_{\mathrm{y}}=\mathrm{A} \cos 30^{\circ}=2 \sqrt{3} \\ & \Rightarrow \mathrm{A} \frac{\sqrt{3}}{2}=2 \sqrt{3} \\ & \Rightarrow \mathrm{A}=4 \end{aligned}$$

Now $A_x=A \sin 30^{\circ}=4 \times \frac{1}{2}=2$

In the circuit $I=\frac{9}{3}=3 \mathrm{~A}$

$\mathrm{V}_{\mathrm{C}}-\mathrm{V}_{\mathrm{A}}=2 \times 1.5=3~~......(I)$

$\mathrm{V}_{\mathrm{C}}-\mathrm{V}_{\mathrm{B}}=4 \times 1.5=6~~......(II)$

$\mathrm{Eq}^{\mathrm{n}}(\mathrm{II})-\mathrm{Eq}^{\mathrm{n}}(\mathrm{I})$

$\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}=6-3=3 \text { Volt }$

$\lambda = \frac{h}{\sqrt{2mE}}$

where $h$ is Planck's constant, $m$ is the mass of the particle, and $E$ is its kinetic energy.

We are given that the de Broglie wavelength of an electron with kinetic energy $E$ is $\lambda$, and we want to find the de Broglie wavelength of the same electron when its kinetic energy becomes $\frac{E}{4}$.

To do this, we can use the formula for the de Broglie wavelength again, but with the new kinetic energy $\frac{E}{4}$:

$\lambda' = \frac{h}{\sqrt{2m\left(\frac{E}{4}\right)}} = \frac{2h}{\sqrt{2mE}} = 2\lambda$

where we have used the fact that $\sqrt{\frac{1}{4}} = \frac{1}{\sqrt{4}} = \frac{1}{2}$ to simplify the expression.

Therefore, the de Broglie wavelength of the electron when its kinetic energy becomes $\frac{E}{4}$ is twice its original value, or $\boxed{2\lambda}$.

By conservation of mechanical energy

$$\begin{aligned} & \mathrm{U}_{\mathrm{i}}+\mathrm{K}_{\mathrm{i}}=\mathrm{U}_{\mathrm{f}}+\mathrm{K}_{\mathrm{i}} \\ & -\frac{\mathrm{GMm}}{2 \mathrm{R}}+0=-\frac{\mathrm{GMm}}{\mathrm{R}}+\frac{1}{2} \mathrm{mv}^2 \\ & \frac{\mathrm{GMm}}{2 \mathrm{R}}=\frac{1}{2} \mathrm{mv}^2 \\ & \mathrm{v}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}=\sqrt{\mathrm{gR}} \end{aligned}$$

To calculate the work done by the force $F = 5x$ in moving the block from $x = 2m$ to $x = 4m$, we can use the formula for work done by a variable force:

$W = \int_{x_1}^{x_2} F(x) dx$

In this case, $F(x) = 5x$, $x_1 = 2m$, and $x_2 = 4m$. Now, we can substitute these values into the formula and evaluate the integral:

$W = \int_{2}^{4} 5x dx$

To evaluate the integral, we find the antiderivative of $5x$:

$\int 5x dx = \frac{5}{2}x^2 + C$

Now, we can find the work done by evaluating the antiderivative at the limits of integration:

$W = \left[\frac{5}{2}x^2\right]_{2}^{4} = \frac{5}{2}(4^2) - \frac{5}{2}(2^2)$

$W = \frac{5}{2}(16) - \frac{5}{2}(4) = 40 - 10 = 30 \mathrm{J}$

The work done by the force in moving the block from $x = 2m$ to $x = 4m$ is 30 J.

Let the mass of both the solid sphere and the solid cylinder be $M$, and let their common radius be $R$. The moment of inertia $I$ of a solid sphere and a solid cylinder are given by:

$I_{sph} = \frac{2}{5}MR^2$

$I_{cyl} = \frac{1}{2}MR^2$

The radius of gyration $k$ is related to the moment of inertia $I$ by the formula $I = Mk^2$. Therefore, we can find the radius of gyration for both the solid sphere and the solid cylinder using their respective moments of inertia:

$k_{sph}^2 = \frac{I_{sph}}{M} = \frac{2}{5}R^2$

$k_{cyl}^2 = \frac{I_{cyl}}{M} = \frac{1}{2}R^2$

Now, let's find the ratio of their radius of gyrations:

$\frac{k_{sph}}{k_{cyl}} = \frac{2}{\sqrt{x}}$

Squaring both sides:

$\frac{k_{sph}^2}{k_{cyl}^2} = \frac{4}{x}$

Substituting the expressions for $k_{sph}^2$ and $k_{cyl}^2$:

$\frac{\frac{2}{5}R^2}{\frac{1}{2}R^2} = \frac{4}{x}$

Simplifying and solving for $x$:

$\frac{2}{5} \cdot \frac{2}{1} = \frac{4}{x}$

$\frac{4}{5} = \frac{4}{x}$

Thus, $x = 5$.

We can use the Young-Laplace equation to find the difference in pressure inside and outside the air bubble due to surface tension:

$\Delta P = 2 \frac{T}{R}$

where $\Delta P$ is the pressure difference, $T$ is the surface tension, and $R$ is the radius of the bubble.

Plugging in the given values:

$\Delta P = 2 \frac{0.075 \mathrm{~Nm}^{-1}}{1.0 \mathrm{~mm}} = 2 \frac{0.075 \mathrm{~Nm}^{-1}}{10^{-3} \mathrm{~m}} = 150 \mathrm{~Pa}$

Now, we need to account for the hydrostatic pressure due to the depth of the bubble below the free surface:

$P_{hydrostatic} = \rho g h$

where $\rho$ is the density of the liquid, $g$ is the acceleration due to gravity, and $h$ is the depth below the free surface.

Plugging in the given values:

$P_{hydrostatic} = 1000 \mathrm{~kg} \mathrm{~m}^{-3} \cdot 10 \mathrm{~ms}^{-2} \cdot 0.1 \mathrm{~m} = 1000 \mathrm{~Pa}$

So, the total pressure difference inside the bubble compared to atmospheric pressure is the sum of the pressure difference due to surface tension and hydrostatic pressure:

$\Delta P_{total} = \Delta P + P_{hydrostatic} = 150 \mathrm{~Pa} + 1000 \mathrm{~Pa} = 1150 \mathrm{~Pa}$

The average kinetic energy per molecule of a gas is given by the expression $\frac{3}{2}kT$, where $k$ is the Boltzmann constant and $T$ is the temperature of the gas in kelvins. Since the temperature of the mixture is given in Celsius, we need to convert it to kelvins by adding 273.15 to get $303.15$ K.

Let us assume that the total mass of the mixture is $3x$ (where $x$ is a constant), then the mass of hydrogen and argon in the mixture will be $2x$ and $x$ respectively, according to the given ratio.

The number of moles of hydrogen and argon can be calculated using their respective masses and molar masses, which are 1 g/mol and 39.9 g/mol respectively. Therefore:

Number of moles of hydrogen = $\frac{2x}{1~\mathrm{g/mol}} = 2x$ mol

Number of moles of argon = $\frac{x}{39.9~\mathrm{g/mol}} = \frac{x}{39.9}$ mol

The total number of moles of gas in the mixture is the sum of the number of moles of hydrogen and argon:

Total number of moles of gas = $2x + \frac{x}{39.9} = \frac{79.9x}{39.9}$ mol

The average kinetic energy per molecule of hydrogen is:

$\frac{3}{2}kT_{\mathrm{H_2}} = \frac{3}{2} \times 1.38 \times 10^{-23} \times 303.15~\mathrm{K} = 6.12 \times 10^{-21}~\mathrm{J}$

The average kinetic energy per molecule of argon is:

$\frac{3}{2}kT_{\mathrm{Ar}} = \frac{3}{2} \times 1.38 \times 10^{-23} \times 303.15~\mathrm{K} = 6.12 \times 10^{-21}~\mathrm{J}$

Therefore, the ratio of the average kinetic energy per molecule of argon to hydrogen is:

$\frac{K_{\mathrm{Ar}}}{K_{\mathrm{H_2}}} = \frac{\frac{3}{2}kT_{\mathrm{Ar}}}{\frac{3}{2}kT_{\mathrm{H_2}}} = \frac{6.12 \times 10^{-21}~\mathrm{J}}{6.12 \times 10^{-21}~\mathrm{J}} = 1$

Hence, the ratio of average kinetic energy per molecule of argon to hydrogen is $1$.

$$\begin{aligned} & \text { As } \frac{1}{\lambda}=\mathrm{RZ}^2\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right] \\ & \frac{1}{\lambda_1}=\mathrm{R}(1)^2\left[\frac{1}{(2)^2}-\frac{1}{(3)^2}\right]=\mathrm{R}\left(\frac{5}{36}\right)~~......(i) \\ & \& \frac{1}{\lambda_2}=\mathrm{R}(1)^2\left[\frac{1}{(3)^2}-\frac{1}{(4)^2}\right]=\mathrm{R}\left(\frac{7}{144}\right) ~~......(ii) \end{aligned}$$

(ii) $\div$ (i) gives

$$\begin{aligned} & \frac{\lambda_1}{\lambda_2}=\frac{7 / 144}{5 / 36}=\frac{7}{20}=\frac{7}{4 \times 5} \\ & \therefore \mathrm{n}=5 \end{aligned}$$

Statement I is incorrect, as the equivalent resistance of resistors in a series combination is always greater than the largest individual resistance used in the combination. This is because the total voltage drop across the resistors in series is equal to the sum of the voltage drops across each resistor, and the current through each resistor is the same. Therefore, the resistance of the entire series combination is the sum of the individual resistances.

Statement II is also incorrect, as the resistivity of most materials changes with temperature. This is because temperature affects the mobility of charge carriers in a material, which in turn affects its resistivity. For example, the resistivity of metals generally increases with temperature, while the resistivity of semiconductors generally decreases with temperature.

Therefore, the correct answer is Both Statement I and Statement II are false

The formula for the magnetic field due to a moving charge is given by:

$\mathbf{B}=\frac{\mu_0}{4 \pi} \frac{q v \sin \theta}{r^2}$

where $\mu_0$ is the permeability of free space, $q$ is the charge of the moving particle, $v$ is the speed of the particle, $\theta$ is the angle between the velocity vector and the position vector from the particle to the point where we want to calculate the magnetic field, and $r$ is the distance between the particle and the point where we want to calculate the magnetic field.

In this case, we're interested in the magnetic field produced by the electron moving in a circular orbit around the nucleus of a hydrogen atom. Since the orbit is circular, the angle between the velocity vector and the position vector is 90 degrees, so $\sin \theta = 1$. We can substitute the known values into the formula to find the magnetic field:

$\mathbf{B}=\frac{\mu_0}{4 \pi} \frac{q v \sin \theta}{r^2} = \frac{\mu_0}{4 \pi} \frac{e v}{r^2}$

where $e$ is the charge of an electron. We know that the radius of the orbit is $0.52 \mathrm{~A}^{\circ}$, which is equivalent to $0.52 \times 10^{-10} \mathrm{m}$.

Substituting the values, we get:

$\mathbf{B}=\frac{\mu_0}{4 \pi} \frac{e v}{r^2} =\frac{10^{-7} \times 1.6 \times 10^{-19} \times 6.76 \times 10^6}{0.52 \times 0.52 \times 10^{-20}} = 40 ~\mathrm{T}$

This means that the magnetic field produced by the electron moving in a circular orbit around the nucleus of a hydrogen atom is 40 tesla, which is an incredibly strong magnetic field.

At a large distance $r$ from the center of a short electric dipole, the electric field on the equatorial plane can be approximated as:

$E = \frac{1}{4\pi\epsilon_0}\frac{2p}{r^3}$

where $p$ is the dipole moment of the electric dipole, and $\epsilon_0$ is the permittivity of free space.

This formula is derived using the concept of electric dipole moment, which is defined as:

$\vec{p} = q\vec{d}$

where $q$ is the magnitude of the electric charge, and $\vec{d}$ is the separation vector between the positive and negative charges of the dipole. The electric field at a point on the equatorial plane of the dipole is due to the electric field of the positive and negative charges at that point.

Since the charges are equal in magnitude and opposite in sign, their electric fields at a point on the equatorial plane cancel out along the axis of the dipole, leaving only the component perpendicular to the axis.

This perpendicular component of the electric field is proportional to the dipole moment $p$ and inversely proportional to the cube of the distance $r$ from the center of the dipole.

Therefore, the electric field due to a short electric dipole at a large distance $r$ from the center of the dipole on the equatorial plane varies with distance as:

$\boxed{E \propto \frac{1}{r^3}}$

where the proportionality constant is $\frac{1}{4\pi\epsilon_0}$.

The correct options are:

(A) Restoring force is directly proportional to the displacement. - True (this is a defining characteristic of SHM)

(B) The acceleration and displacement are opposite in direction. - True (the acceleration is proportional to the displacement but in the opposite direction)

(C) The velocity is maximum at mean position. - True (the velocity is zero at the extreme positions and reaches a maximum at the mean position)

(D) The acceleration is minimum at extreme points. - False (the acceleration is maximum at the extreme points and zero at the mean position)

Given that the rod is rotating at 210 rpm, we first convert this to radians per second:

$\omega = 210 \cdot \frac{2\pi \mathrm{rad}}{60 \mathrm{s}} = 22 \mathrm{rad/s}$

Now, we can find the linear velocity $v$ of the tip of the rod:

$v = \omega r$

where $r$ is the length of the rod (0.2 m).

$v = 22 \mathrm{rad/s} \cdot 0.2 \mathrm{m} = 4.4 \mathrm{m/s}$

Now, we can find the emf developed between the center and the ring using the formula:

$\epsilon = \frac{1}{2} B\ell v$

where $B$ is the magnetic field (0.2 T), $\ell$ is the length of the rod (0.2 m), and $v$ is the linear velocity (4.4 m/s).

$\epsilon = \frac{1}{2} \cdot 0.2 \mathrm{T} \cdot 0.2 \mathrm{m} \cdot 4.4 \mathrm{m/s} = 0.088 \mathrm{V}$

To express this value in mV, we can simply multiply it by 1000:

$\epsilon = 0.088 \mathrm{V} \cdot 1000 = 88 \mathrm{mV}$

So the emf developed between the center and the ring is 88 mV.

When light passes through a narrow slit, it diffracts and produces a diffraction pattern on a screen. The pattern consists of a central bright maximum flanked by a series of alternating bright and dark fringes.

The condition for the first minimum in the diffraction pattern is given by :

$a\sin\theta = \lambda$

where $a$ is the width of the slit, $\theta$ is the angle of diffraction, and $\lambda$ is the wavelength of the incident light.

In this problem, we are given that the wavelength of the incident light is $\lambda = 600\,\mathrm{nm}$, and the angle of diffraction for the first minimum is $\theta = 30^\circ$. We want to find the width of the slit, $a$, for which this occurs.

Substituting the given values into the condition for the first minimum, we get :

$a\sin 30^\circ = 600\,\mathrm{nm}$

Simplifying, we get :

$a\cdot \frac{1}{2} = 600\,\mathrm{nm}$

Multiplying both sides by 2, we get :

$a = 1200\,\mathrm{nm} = \boxed{1.2\,\mu\mathrm{m}}$

Therefore, the width of the slit for which the first minimum appears at $\theta = 30^\circ$ is $\boxed{1.2\,\mu\mathrm{m}}$.

When the switch is closed, the circuit is a simple DC circuit and the current in the circuit is given by Ohm's law:

$I = \frac{V}{R} = \frac{12\text{V}}{6\Omega} = 2\text{A}$.

When the switch is opened, the current in the circuit drops to zero instantaneously.

However, the magnetic field generated by the current in the coil does not disappear immediately, and it continues to produce a back EMF that opposes the change in current.

This back EMF induces a voltage across the coil that can be calculated using Faraday's law of induction: $\mathcal{E} = -L\frac{\Delta I}{\Delta t}$, where $\mathcal{E}$ is the induced voltage, $L$ is the inductance of the coil, and $\Delta I/\Delta t$ is the rate of change of current in the coil.

In this case, we know that the induced voltage is $20\text{V}$ and the rate of change of current is

$\Delta I/\Delta t = -2\text{A}/(1\text{ms}) = -2\times 10^3\text{A/s}$.

Substituting these values into the equation above, we get: $20\text{V} = -L\times(-2\times 10^3\text{A/s})$.

Solving for $L$, we get: $L = \frac{20\text{V}}{2\times 10^3\text{A/s}} = 0.01\text{H}$.

Therefore, the inductance of the coil is $0.01\text{H}$, or 10 mH.

On $\mathrm{P}-\mathrm{V}$ scale area of loop $=$ work done

$$\begin{aligned} & \Rightarrow \mathrm{W}=+\frac{1}{2}(2) \times 300 \\ & \mathrm{~W}=300 \mathrm{~J} \end{aligned}$$

The correct matching of the electromagnetic waves with their corresponding wavelength ranges is:

(A) Microwave --> (IV) 0.1 m to 1 mm

(B) Ultraviolet --> (I) 400 nm to 1 nm

(C) X-Ray --> (II) 1 nm to $10^{-3}$ nm

(D) Infra-red --> (III) 1 mm to 700 nm

Therefore, the correct option is (A-IV, B-I, C-II, D-III).

Let $A$ be the cross-sectional area of the first wire, and let $Y$ be its Young's modulus.

The strain in the wire is given by $\epsilon = \frac{l}{L}$, where $l$ is the increase in length. The stress in the wire is given by $\sigma = \frac{f}{A}$.

According to Hooke's law, the stress is proportional to the strain, so we have $\sigma = Y \epsilon$. Solving for $f$, we get $f = \frac{YA}{l}$.

The second wire has twice the length and four times the cross-sectional area of the first wire, so its cross-sectional area is $4A$ and its Young's modulus is still $Y$.

When a force of $2f$ is applied to this wire, the stress in the wire is $\sigma = \frac{2f}{4A} = \frac{f}{2A}$.

Using Hooke's law again, we have $\sigma = Y \epsilon$. Solving for $\epsilon$, we get $\epsilon = \frac{\sigma}{Y} = \frac{f}{2AY}$.

The increase in length of the second wire is given by $\Delta l = \epsilon \cdot 2L = \frac{f \cdot 2L}{2AY}$.

Substituting the expression for $f$ that we derived earlier, we get $\Delta l = \frac{YL \cdot 2A \cdot l}{2AY \cdot L} = \boxed{l}$.

Therefore, the increase in length of the second wire is the same as the increase in length of the first wire, which is $l$.

$d_{\max} = \sqrt{2Rh_t} + \sqrt{2Rh_r}$

where $d_{\max}$ is the maximum distance between the antennas, $R$ is the radius of the earth, $h_t$ is the height of the transmitting antenna above the earth's surface, and $h_r$ is the height of the receiving antenna above the earth's surface.

Substituting the given values, we get:

$d_{\max} = \sqrt{2\times 6400\,\mathrm{km}\times 180\,\mathrm{m}} + \sqrt{2\times 6400\,\mathrm{km}\times 245\,\mathrm{m}} = 104\,\mathrm{km}$

Therefore, the maximum distance between the transmitting and receiving antennas for line of sight communication is $\boxed{104\,\mathrm{km}}$.

The decay of a radioactive nucleus is an exponential process, and the fraction of the original sample that remains after time $t$ is given by:

$N(t) = N_0 e^{-\lambda t}$

where $N_0$ is the initial number of nuclei, $N(t)$ is the number of nuclei remaining after time $t$, and $\lambda$ is the decay constant, which is related to the half-life $T_{1/2}$ by the equation:

$\lambda = \frac{\ln(2)}{T_{1/2}}$

In this problem, the half-life of the nucleus is given as 5 years, so we have:

$\lambda = \frac{\ln(2)}{5~\mathrm{yrs}} \approx 0.1386~\mathrm{yr^{-1}}$

We are asked to find the fraction of the original sample that would decay in 15 years. At $t=15$ years, the fraction of nuclei remaining is:

$N(15) = N_0 e^{-\lambda (15~\mathrm{yrs})}$

To find the fraction that has decayed, we subtract this expression from 1, since the fraction that remains plus the fraction that has decayed must add up to 1:

Fraction decayed = $1 - N(15) = 1 - N_0 e^{-\lambda (15~\mathrm{yrs})}$

We know that the half-life of the nucleus is 5 years, which means that the fraction of nuclei remaining after one half-life is 1/2. Therefore, after 3 half-lives (which is equivalent to 15 years), the fraction of nuclei remaining is:

$N(15) = N_0 \left(\frac{1}{2}\right)^3 = \frac{N_0}{8}$

Plugging this into the equation for the fraction of nuclei that have decayed, we get:

Fraction decayed = $1 - N_0 e^{-\lambda (15~\mathrm{yrs})} = 1 - \frac{N_0}{8} = \frac{7N_0}{8}$

Therefore, the fraction of the original sample that would decay in 15 years is $\frac{7}{8}$

To find the speed of the transverse waves produced in the string, we can use the formula for the fundamental frequency of a vibrating string:

$f = \frac{1}{2L} \cdot v$

where $f$ is the fundamental frequency, $L$ is the length of the string, and $v$ is the speed of the transverse waves.

First, we are given the mass of the string ($m = 18g$) and the linear mass density ($\mu = 20g/m$). We can find the length of the string by dividing the mass by the linear mass density:

$L = \frac{m}{\mu} = \frac{18g}{20g/m} = 0.9m$

Now we can plug in the values for the fundamental frequency ($f = 50Hz$) and the length of the string ($L = 0.9m$) into the formula:

$50Hz = \frac{1}{2(0.9m)} \cdot v$

To isolate $v$, we multiply both sides by $2(0.9m)$:

$v = 50Hz \cdot 2(0.9m) = 90 \mathrm{ms}^{-1}$

The speed of the transverse waves produced in the string is $90 ~\mathrm{ms}^{-1}$.

Alternate Method:

To find the speed of the transverse waves produced in the string, we can use the formula for the fundamental frequency of a vibrating string:

$f_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$

where $f_1$ is the fundamental frequency, $L$ is the length of the string, $T$ is the tension in the string, and $\mu$ is the linear mass density of the string.

We're given that the fundamental frequency $f_1 = 50 ,\text{Hz}$, the mass of the string $m = 18 ,\text{g}$, and the linear mass density $\mu = 20 ,\text{g/m}$. To find the speed of the transverse waves, we need to find the tension $T$ and the length $L$ of the string.

First, let's find the length $L$ of the string using the mass and linear mass density:

$L = \frac{m}{\mu} = \frac{18 ,\text{g}}{20 ,\text{g/m}} = 0.9 ~\text{m}$

Now, we can rearrange the formula for the fundamental frequency to solve for the tension $T$:

$T = \mu \left(\frac{2Lf_1}{1}\right)^2$

Substitute the known values:

$T = 20 ,\text{g/m} \cdot \left(\frac{2 \cdot 0.9 ~\text{m} \cdot 50 ~\text{Hz}}{1}\right)^2$

$T = 20 ,\text{g/m} \cdot (90 ~\text{m/s})^2$

$T = 20 ,\text{g/m} \cdot 8100 ~\text{m}^2/\text{s}^2$

$T = 162000 ,\text{g m}/\text{s}^2$

Now, we can find the speed of the transverse waves $v$ using the formula:

$v = \sqrt{\frac{T}{\mu}}$

Substitute the known values:

$v = \sqrt{\frac{162000}{20}}$

$v = \sqrt{8100} = 90~ \text{m/s}$

The speed of the transverse waves produced in the string is $90 ~\text{m/s}$.

For voltmeter

$$\begin{aligned} & \mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}_{\mathrm{g}}}-\mathrm{G} \\ & =\frac{50}{10^{-3}}-54 \approx 50 \mathrm{k} \Omega~(\mathrm{A}) \end{aligned}$$

For ammeter

$\mathrm{S}=\frac{\mathrm{I}_{\mathrm{g}} \mathrm{G}}{\mathrm{I}-\mathrm{I}_{\mathrm{g}}}=\frac{10^{-3} \times 54}{(10-1) \times 10^{-3}}=6 \Omega~(\mathrm{C})$

In the circuit $\mathrm{D}_1$ and $\mathrm{D}_3$ are forward biased and $\mathrm{D}_2$ is reverse biased.


$\therefore \mathrm{I}=\frac{10}{20 / 3}=\frac{3}{2} \mathrm{~A}=1.5 \mathrm{~A}$