JEE Main 2024 4th April Morning Shift Question Paper with Solutions

Acceleration (a) $=\frac{f-F}{m}$

When applied force became equal to $f_{\max }$, block will start moving.

As $F$ increases linearly, so acceleration of also moving block will increase linearly.

First, let's understand the relationship between the object distance ($u$), the image distance ($v$), and the focal length ($f$) of a convex lens, which is given by the lens formula:

Now, to find the error in the focal length ($\Delta f$) due to the errors in the measurements of $u$ and $v$ ($\Delta u$ and $\Delta v$, respectively), we have to differentiate the lens formula with respect to $u$ and $v$, keeping in mind the propagation of error.

By differentiating both sides of the lens formula with respect to $v$ and $u$, and also considering the negative reciprocal relation (given $f$ is a constant for a specific lens), we have:

Rearranging this equation to find $\Delta f$, we get:

Therefore, the correct option showing the error in the measurement of the focal length of the convex lens, taking into account the least counts ($\Delta u$ and $\Delta v$) of the measuring scales for the position of the object ($u$) and for the position of the image ($v$), is:

Option C: $f^2\left[\frac{\Delta \mathrm{u}}{\mathrm{u}^2}+\frac{\Delta \mathrm{v}}{\mathrm{v}^2}\right]$

The resistance of a platinum resistance thermometer varies linearly with temperature. The relation can be given by:

$R_t = R_0(1 + \alpha t)$

where:

• $R_t$ is the resistance at temperature $t$,


• $R_0$ is the resistance at 0°C (ice point),


• $\alpha$ is the temperature coefficient of resistance, and


• $t$ is the temperature in degrees Celsius.

In this question, we are given:

• Resistance at the ice point ($R_{0} = 8 \Omega$),


• Resistance at the steam point ($R_{100} = 10 \Omega$).

To find the temperature coefficient of resistance ($\alpha$), we use the resistance values at the ice and steam points:

$\alpha = \frac{R_{100} - R_0}{R_0 \times 100} = \frac{10\Omega - 8\Omega}{8\Omega \times 100} = \frac{2\Omega}{800\Omega} = \frac{1}{400} \text{ per } ^{\circ}C$

Now, to find the resistance $R_t$ at $400 ^{\circ}C$, we substitute $\alpha$, $R_0$, and $t = 400$ into the formula:

$R_t = R_0(1 + \alpha t) = 8\Omega(1 + \frac{1}{400} \times 400) = 8\Omega(1 + 1) = 8\Omega \times 2 = 16\Omega$

Hence, the resistance of the platinum wire at $400^{\circ}C$ is $16 \Omega$. The correct option is:

Option B: 16 $\Omega$

Given $\sigma_s=2 \lambda_e$

At point $P, E_S=\frac{\sigma_S}{2 \varepsilon_0}$

$$\begin{aligned} & E_I=\frac{\lambda_e}{2 \pi r \varepsilon_0} \\ & \frac{E_S}{E_I}=4 \pi: 1=\pi \sqrt{n}: 1 \end{aligned}$$

For value of $n=16$

To find the increase in temperature on the Fahrenheit scale, we use the relationship between the Celsius and Fahrenheit temperature scales. The formula to convert Celsius to Fahrenheit is:

$F = \frac{9}{5}C + 32$

However, since we are interested in the increase in temperature, we can ignore the "+ 32" part of the formula, because this constant does not affect the change in temperature, only the absolute temperatures. Thus, to find the increase in temperature on the Fahrenheit scale, we can use:

$\Delta F = \frac{9}{5}\Delta C$

Given that the increase in temperature is $40^{\circ}C$, we can substitute this value into the equation:

$\Delta F = \frac{9}{5} \times 40^{\circ}C$

$\Delta F = 9 \times 8$

$\Delta F = 72^{\circ}F$

Therefore, the increase in temperature on the Fahrenheit scale is $72^{\circ}F$. So, the correct answer is:

Option C $72^{\circ} \mathrm{F}$

Stopping potential is independent on intensity but photocurrent increases non-linearly on increasing intensity.

To solve this problem, we first note that for both the solid sphere and the hollow cylinder, the total mechanical energy is conserved as they roll up the inclined plane without slipping. The initial kinetic energy (comprised of both translational and rotational kinetic energy) is converted into potential energy at the maximum height.

Kinetic Energy for Each Body at the Start:

For the solid sphere, the moment of inertia $I$ is given by $I = \frac{2}{5}mr^2$, where $m$ is mass and $r$ is the radius of the sphere. The kinetic energy is the sum of translational kinetic energy $\left(\frac{1}{2}mv^2\right)$ and rotational kinetic energy $\left(\frac{1}{2}I\omega^2\right)$, where $\omega$ is the angular velocity. Since the sphere rolls without slipping, $v = r\omega$.

The total initial kinetic energy for the solid sphere is:

$KE_{\text{sphere}} = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{5}mr^2\right)\left(\frac{v}{r}\right)^2 = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$

For the hollow cylinder, the moment of inertia $I$ is $mr^2$. Thus, its total kinetic energy is:

$KE_{\text{cylinder}} = \frac{1}{2}mv^2 + \frac{1}{2}mr^2\left(\frac{v}{r}\right)^2 = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2$

Potential Energy at Maximum Height:

For both bodies, the potential energy at the maximum height is given by $PE = mgh$, where $h$ is the height reached.

Applying Conservation of Energy:

For the solid sphere, the energy conservation equation is:

$\frac{7}{10}mv^2 = mgh_1$

Solving for $h_1$ gives:

$h_1 = \frac{7v^2}{10g}$

For the hollow cylinder, the conservation of energy gives:

$mv^2 = mgh_2$

Thus, $h_2 = \frac{v^2}{g}$.

Finding the Ratio $h_1:h_2$:

The ratio of $h_1$ to $h_2$ is:

$\frac{h_1}{h_2} = \frac{\frac{7v^2}{10g}}{\frac{v^2}{g}} = \frac{7}{10}$

Therefore, the value of $n$, which represents the numerator in the ratio $\frac{n}{10}$, is $7$. Thus, the ratio of maximum heights $h_1:h_2$ reached by the solid sphere and the hollow cylinder, respectively, is $\frac{7}{10}$.

If speed $=0$

Then $P_1+\rho g h_1=P_2+\rho g h_2$

In given ventury tube,

$$\begin{aligned} & \qquad P_1+\rho g h+\frac{1}{2} \rho v_1^2=P_2+\frac{1}{2} \rho v_2^2 \\ & \Rightarrow \frac{1}{2} \rho\left(v_1^2-v_2^2\right)=\left(P_2-P_1\right)-\rho g h \\ & \Rightarrow v_1^2-v_2^2=\frac{2\left(P_2-P_1\right)}{\rho}-2 g h \end{aligned}$$

The given alternating current (AC) can be represented as $i=6+\sqrt{56} \sin (100 \pi t+\pi / 3)$ A, where $6$ is the DC component and $\sqrt{56} \sin (100 \pi t+\pi / 3)$ is the AC component of the current. The RMS (Root Mean Square) value of an alternating current is a measure of the equivalent direct current (DC) that will produce the same power in a resistor. The RMS value is mostly relevant for the AC component of the current, as the DC component's effective value is just its magnitude itself.

The RMS value of the total current is not straightforward because the presence of the DC component affects how we calculate the RMS value. However, when calculating RMS values for a signal consisting of a superposition of AC and DC components, one notable property is that the RMS value of the combined signal is the square root of the sum of the squares of the RMS values of the separate AC and DC components.

First, let's acknowledge the components separately:

• The DC component is: $6$ A


• The AC component is: $\sqrt{56} \sin (100 \pi t+\pi / 3)$ A

For the DC component, the RMS value is simply its magnitude:

$I_{RMS, DC} = 6$ A

For the AC component, the RMS value is calculated using the formula for the RMS value of a sinusoidal function, which is $I_{RMS} = \frac{I_{max}}{\sqrt{2}}$, where $I_{max}$ is the peak value of the current. In this case, $I_{max} = \sqrt{56}$.

Therefore, the RMS value of the AC component is:

$I_{RMS, AC} = \frac{\sqrt{56}}{\sqrt{2}} = \frac{\sqrt{56}}{\sqrt{2}} = \sqrt{\frac{56}{2}} = \sqrt{28}$ A.

Finally, to find the total RMS value of the current, combine the DC and AC components as follows:

$I_{RMS} = \sqrt{{(I_{RMS, DC})}^2 + {(I_{RMS, AC})}^2}$

Substituting the values:

$I_{RMS} = \sqrt{{(6)}^2 + {(\sqrt{28})}^2}$

$= \sqrt{36 + 28}$

$= \sqrt{64}$

$= 8$ A.

Therefore, the RMS value of the current is $8$ A.

To determine which of the options is NOT correct, we need to analyze the dimensional consistency of each term in the given equation of the stationary wave:

$y = 2a \sin \left(\frac{2\pi nt}{\lambda}\right) \cos \left(\frac{2\pi x}{\lambda}\right).$

Let's break down the dimensions for each relevant term:

1. Analyzing $\mathrm{nt}$:

The argument of the sine function $\frac{2\pi nt}{\lambda}$ must be dimensionless. Therefore, the dimensions of $\mathrm{nt}$ should be the same as the dimensions of $\lambda$ (wavelength), which is [L].

Thus, the dimensions of $nt$ should be [L].
Hence, Option A is correct.

2. Analyzing $n$:

From the above analysis, since $\mathrm{nt}$ has the dimension [L] and $t$ (time) has the dimension [T], it follows that:

$n = \frac{\text{Dimension of } nt}{\text{Dimension of } t} = \frac{[L]}{[T]} = [\mathrm{LT}^{-1}].$

Hence, Option B is also correct.

3. Analyzing $x$:

In the argument of the cosine function $\frac{2\pi x}{\lambda}$, since it must be dimensionless, the dimensions of $x$ should be the same as the dimensions of $\lambda$ (wavelength), which is [L].

Hence, the dimensions of $x$ should be [L].
Therefore, Option C is correct.

4. Analyzing $\frac{n}{\lambda}$:

From the dimensions we have determined for $n$ and $\lambda$:

$\frac{n}{\lambda} = \frac{[\mathrm{LT}^{-1}]}{[L]} = [\mathrm{T}^{-1}].$

Thus, the dimensions of $\frac{n}{\lambda}$ should be [T^-1], not [T].
This indicates that Option D is NOT correct.

Conclusion:

Option D is the correct answer since it is NOT dimensionally correct.

Here $D_1$ is forward wise while $D_2$ is reversed wise

So net resistance between end, $R=\frac{15 \times 10}{25}=6 \Omega$

In order to identify the correct nuclear fragments resulting from the fission of uranium-235 by a neutron, $\left({ }_0^1 \mathrm{n}\right) + \left({ }_{92}^{235} \mathrm{U}\right)$, we need to apply the conservation of mass number and atomic number. These conservation laws tell us that the sum of mass numbers (top numbers, representing the total count of protons and neutrons) and the sum of atomic numbers (bottom numbers, representing the total count of protons) before and after the fission must be equal. Let's apply these rules to each option:

For the uranium-235 fission reaction, before the reaction, the total mass number is $235 + 1 = 236$ and the total atomic number is $92$ (since a neutron doesn't contribute to the atomic number).

Option A: ${ }_{56}^{140} \mathrm{Xe}+{ }_{38}^{94} \mathrm{Sr}+3{ }_0^1 \mathrm{n}$

• Total mass number after = $140 + 94 + 3(1) = 234 + 3 = 237$
• Total atomic number after = $56 + 38 = 94$

Option B: ${ }_{51}^{153} \mathrm{Sb}+{ }_{41}^{99} \mathrm{Nb}+3{ }_0^1 \mathrm{n}$

• Total mass number after = $153 + 99 + 3(1) = 252 + 3 = 255$
• Total atomic number after = $51 + 41 = 92$

Option C: ${ }_{56}^{144} \mathrm{Ba}+{ }_{36}^{89} \mathrm{Kr}+4{ }_0^1 \mathrm{n}$

• Total mass number after = $144 + 89 + 4(1) = 233 + 4 = 237$
• Total atomic number after = $56 + 36 = 92$

Option D: ${ }_{56}^{144} \mathrm{Ba}+{ }_{36}^{89} \mathrm{Kr}+3{ }_0^1 \mathrm{n}$

• Total mass number after = $144 + 89 + 3(1) = 233 + 3 = 236$
• Total atomic number after = $56 + 36 = 92$

By examining the conservation of mass numbers and atomic numbers, Option D is identified as the correct option because both the total mass number and atomic number after the reaction match exactly with the total mass number and atomic number before the reaction. Therefore, the nuclear fission fragments and the neutron count for the reaction are accurately represented by ${ }_{56}^{144} \mathrm{Ba}+{ }_{36}^{89} \mathrm{Kr}+3{ }_0^1 \mathrm{n}$.

The circuit can be simplified as

$$\begin{aligned} & R_{a c}=\frac{6 \times 2}{8}=\frac{3}{2} \Omega \\ & \begin{aligned} & i=1 \mathrm{Amp} . \\ & V_{e f}=\left(\frac{i}{2}\right)^2 \\ & \quad=1 \mathrm{~V} \end{aligned} \end{aligned}$$

The distance covered by a body in the $n^{\text{th}}$ second can be found using the equation:

$S_{n} = u + \dfrac{1}{2}a(2n-1)$

where,

• $S_{n}$ is the distance covered in the $n^{\text{th}}$ second,
• $u$ is the initial velocity,
• $a$ is the acceleration, and
• $n$ is the nth second.

The distance covered in the $n^{\text{th}}$ second is given as $102.5 \, \text{m}$, so we have:

$102.5 = u + \dfrac{1}{2}a(2n-1)$ ---- (1)

For the $(n + 2)^{\text{th}}$ second, the distance covered is:

$S_{n+2} = u + \dfrac{1}{2}a(2(n+2)-1)$

Substituting $n + 2$ in place of $n$, we get:

$115.0 = u + \dfrac{1}{2}a(2n+3)$ ---- (2)

Subtracting equation (1) from equation (2), we get:

$115.0 - 102.5 = \dfrac{1}{2}a(2n + 3 - 2n + 1)$

$12.5 = \dfrac{1}{2}a(4)$

So, solving for $a$ gives:

$a = \dfrac{12.5 \times 2}{4} = \dfrac{25}{4} = 6.25 \, \text{m/s}^2$

Therefore, the acceleration of the body is:

$6.25 \, \text{m/s}^2$

Which corresponds to Option A: $6.25 \, \text{m/s}^2$.

Field at centre due to arc, $I=\frac{2 G M \pi}{L^2}$

$\therefore \quad$ Net force on mass, $F=\frac{2 G M m \pi}{L^2}$

To measure the internal resistance of a battery using a potentiometer, we need to understand the principle behind it. The potentiometer is used to measure the voltage across the battery (emf) under different conditions. The balance length corresponds to the emf of the battery when no current is drawn, whereas under load, it corresponds to the terminal voltage.

Let's denote the emf of the battery by $E$ and the internal resistance by $r$. According to Ohm's law, when a resistance $R$ is connected across the battery, the terminal voltage $V$ is given by:

$V = E - Ir$

where $I$ is the current through the circuit.

From the problem, the balance length $l$ is proportional to the voltage across the potentiometer wire. Thus, we can write:

$\frac{V_1}{V_2} = \frac{l_1}{l_2}$

In the first condition, when $R = 10 \Omega$ and the balance length $l_1 = 500 \, \text{cm}$:

$V_1 = E - I_1 r$

For the second condition, when $R = 1 \Omega$ and the balance length $l_2 = 400 \, \text{cm}$:

$V_2 = E - I_2 r$

Given that:

$\frac{V_1}{V_2} = \frac{500}{400} = \frac{5}{4}$

Now let's denote the internal emf of the battery as $E$. We can use Ohm’s Law in the calculation of $I_1$ and $I_2$:

$I_1 = \frac{E}{R_1 + r} = \frac{E}{10 + r}$

$I_2 = \frac{E}{R_2 + r} = \frac{E}{1 + r}$

Now, rewriting the values of $V_1$ and $V_2$ we have:

$V_1 = E - I_1 r = E \left(1 - \frac{r}{10 + r}\right) = E \frac{10}{10 + r}$

$V_2 = E - I_2 r = E \left(1 - \frac{r}{1 + r}\right) = E \frac{1}{1 + r}$

Using the ratio:

$\frac{V_1}{V_2} = \frac{\frac{10E}{10 + r}}{\frac{E}{1 + r}} = \frac{10 \cdot (1 + r)}{1 \cdot (10 + r)} = \frac{10 + 10r}{10 + r}$

Given:

$\frac{V_1}{V_2} = \frac{5}{4}$

So, we can set up the equation:

$\frac{10 + 10r}{10 + r} = \frac{5}{4}$

Cross multiplying gives:

$4(10 + 10r) = 5(10 + r)$

Expanding both sides:

$40 + 40r = 50 + 5r$

Rearranging terms to solve for $r$:

$40r - 5r = 50 - 40$

$35r = 10$

$r = \frac{10}{35} = \frac{2}{7} \approx 0.285 \, \Omega$

Since this value is closest to $0.3 \Omega$, the correct option is:

Option B: $0.3 \Omega$

When we have a combination of identical lenses in contact, the effective power ($P_{\text{eff}}$) of the combination can be calculated as the sum of the powers of all the individual lenses. This is because the lenses are in direct contact, and their powers effectively add up.

Given that the effective power of a combination of 5 identical convex lenses is $25 \mathrm{D}$, we can use the formula for the effective power of the combination:

$P_{\text{eff}} = nP$

where:

• $P_{\text{eff}}$ is the effective power of the combination,
• $n$ is the number of lenses, and
• $P$ is the power of each individual lens.

Given $n = 5$ and $P_{\text{eff}} = 25 \mathrm{D}$, we can solve for $P$, the power of each lens:

$25 \mathrm{D} = 5P$

Dividing both sides by 5:

$P = \frac{25 \mathrm{D}}{5} = 5 \mathrm{D}$

The power of a lens ($P$) is related to its focal length ($f$) by the equation:

$P = \frac{1}{f}$

where $P$ is in diopters (D) and $f$ is in meters. Thus:

$5 \mathrm{D} = \frac{1}{f}$

Solving for $f$ gives:

$f = \frac{1}{5} \text{ meters} = \frac{1}{5} \times 100 \text{ cm} = 20 \text{ cm}$

Therefore, the focal length of each of the convex lenses is 20 cm. So, the correct answer is Option B: 20 cm.

To understand which answer is correct, we must consider the relationship between voltage and current in various types of circuits, namely circuits with pure inductors, pure capacitors, pure resistors, and a combination of inductors and capacitors (LC circuits).

In a purely resistive circuit, the voltage and current are in phase, meaning when the voltage is maximum, the current is also maximum. Therefore, option C (pure resistor) cannot result in the instantaneous current being zero when the instantaneous voltage is maximum.

In a purely inductive circuit, the current lags the voltage by $90^\circ$, or in other words, when the voltage is at its maximum, the current is zero. This is because the inductor opposes changes in current, leading to this phase difference.

In a purely capacitive circuit, the current leads the voltage by $90^\circ$. This means when the voltage is at its maximum value, the current through the capacitor is zero because the current reaches its maximum or minimum before the voltage does.

In an LC circuit (inductor-capacitor), under certain conditions like at its resonant frequency, the circuit behaves as if it is purely resistive in nature where voltage and current are in phase. However, it's more nuanced because the question likely implies a condition not at resonance but rather at a general characteristic of LC circuits. In LC circuits, aside from the resonance condition, the presence of both inductive and capacitive components can lead to scenarios where the inductive and capacitive reactances cancel each other, particularly in the case where oscillations are involved, and indeed, there can be moments when the voltage is maximum and the current is minimum (zero in the ideal case) due to the energy being alternately stored in the magnetic field of the inductor and the electric field of the capacitor.

Thus, reflecting on the scenarios:

• Pure inductor - Fits the description: instantaneous current is zero when the instantaneous voltage is maximum.
• Pure capacitor - Also fits the description for the reason described, taking into account ideal conditions.
• Pure resistor - Does not fit as voltage and current are in phase.
• Combination of an inductor and capacitor (LC circuit) - Can fit the description under certain non-resonance conditions where the behavior of energy exchange between L and C at a specific instant can result in the instant current being zero when the voltage is maximum, but this is more complex and dependent on specific conditions unlike the straightforward lag or lead in purely inductive or capacitive circuits.

Therefore, the correct answer is Option C: A, B, and D only.

To determine the tension for the elastic spring's length of $(3a - 2b)$, we can use Hooke's law which states that the force exerted by a spring is proportional to the extension or compression of the spring from its natural length. Mathematically, Hooke's law is given by:

$F = k \cdot \Delta x$

where:

• $F$ is the force exerted by the spring.
• $k$ is the spring constant (a measure of the stiffness of the spring).
• $\Delta x$ is the displacement from the natural length.

Let's denote the natural (unstretched) length of the spring as $l_0$. Given that the spring's length is $a$ under a tension of $3 \mathrm{~N}$, and its length is $b$ under a tension of $2 \mathrm{~N}$, we can write the equations as:

$3 \mathrm{~N} = k \cdot (a - l_0)$

$2 \mathrm{~N} = k \cdot (b - l_0)$

Next, we need to find the length displacement $(3a - 2b)$ in terms of the natural length $l_0$. We express the displacements as:

$x = (3a - 2b) - l_0$

To solve for this, let's express $a - l_0$ and $b - l_0$ from the given conditions:

$a - l_0 = \frac{3 \mathrm{~N}}{k}$

$b - l_0 = \frac{2 \mathrm{~N}}{k}$

Substitute these into the displacement equation:

$x = (3a - 2b) - l_0 = 3 \left( \frac{3 \mathrm{~N}}{k} + l_0 \right) - 2 \left( \frac{2 \mathrm{~N}}{k} + l_0 \right) - l_0$

On simplifying, we get:

$x = 3 \frac{3 \mathrm{~N}}{k} + 3 l_0 - 2 \frac{2 \mathrm{~N}}{k} - 2 l_0 - l_0$

$x = \frac{9 \mathrm{~N}}{k} + 3 l_0 - \frac{4 \mathrm{~N}}{k} - 3 l_0$

$x = \frac{5 \mathrm{~N}}{k}$

Finally, by Hooke's Law, the force corresponding to this displacement is:

$F = k \cdot x = k \cdot \frac{5 \mathrm{~N}}{k} = 5 \mathrm{~N}$

Therefore, the tension required for the elastic spring's length $(3a - 2b)$ is 5 N.

Let's denote the magnitude of the smaller force as $F$, hence the magnitude of the larger force is $3F$. The resultant force $\vec{R}$ is equal in magnitude to the larger force, which means $|\vec{R}| = 3F$. When two forces $\vec{F}_1$ and $\vec{F}_2$ act on a body, the magnitude of their resultant $\vec{R}$ can be found using the law of vector addition:

$|\vec{R}| = \sqrt{|\vec{F}_1|^2 + |\vec{F}_2|^2 + 2|\vec{F}_1||\vec{F}_2|\cos\theta}$,

where $\theta$ is the angle between $\vec{F}_1$ and $\vec{F}_2$. Given that in our case $|\vec{R}| = 3F$, $|\vec{F}_1| = F$ and $|\vec{F}_2| = 3F$, by substituting these values into the equation, we get:

$3F = \sqrt{F^2 + (3F)^2 + 2(F)(3F)\cos\theta}$

$9F^2 = F^2 + 9F^2 + 6F^2\cos\theta$

Simplifying this equation by subtracting $10F^2$ from both sides gives:

$-F^2 = 6F^2\cos\theta$

Dividing both sides by $-F^2$ gives:

$-1 = -6\cos\theta$

Therefore, $\cos\theta = \frac{1}{6}$.

It is given that the angle between $\vec{F}_1$ and $\vec{F}_2$ is $\cos^{-1}\left(\frac{1}{n}\right)$, hence comparing this with the above result, we find that $n = 6$. Therefore, $|n| = 6$.

To determine the magnetic field induction of the given electromagnetic wave, we need to use the relationship between the electric field $\overrightarrow{\mathrm{E}}$ and the magnetic field $\overrightarrow{\mathrm{B}}$ in an electromagnetic wave. For an electromagnetic wave propagating in vacuum, the following relation holds:

$\overrightarrow{\mathrm{B}} = \frac{\overrightarrow{\mathrm{E}} \times \hat{\mathrm{k}}}{\mathrm{c}}$

where:

• $\overrightarrow{\mathrm{E}}$ is the electric field.
• $\hat{\mathrm{k}}$ is the unit vector in the direction of propagation of the wave.
• $\mathrm{c}$ is the speed of light in a vacuum.

Given the electric field:

$\overrightarrow{\mathrm{E}}=\hat{i} 40 \cos \omega(\mathrm{t}-z / \mathrm{c}) \mathrm{NC}^{-1}$

The wave is propagating in the $z$-direction, so $\hat{\mathrm{k}} = \hat{z}$. The unit vector $\hat{\mathrm{i}}$ represents the $x$-direction.

The magnetic field induction is given by:

$\overrightarrow{\mathrm{B}} = \frac{(\hat{\mathrm{i}} 40 \cos \omega(\mathrm{t}-z / \mathrm{c})) \times \hat{\mathrm{z}}}{\mathrm{c}}$

The cross product $\hat{\mathrm{i}} \times \hat{\mathrm{z}}$ yields $\hat{\mathrm{j}}$ (the unit vector in the $y$-direction):

$\overrightarrow{\mathrm{B}} = \hat{\mathrm{j}} \frac{40 \cos \omega(\mathrm{t}-z / \mathrm{c})}{\mathrm{c}}$

Therefore, the magnetic field induction is:

$\overrightarrow{\mathrm{B}}=\hat{\mathrm{j}} \frac{40}{\mathrm{c}} \cos \omega(\mathrm{t}-z/\mathrm{c})$

The correct answer is:

Option A:

$\overrightarrow{\mathrm{B}}=\hat{j} \frac{40}{\mathrm{c}} \cos \omega(\mathrm{t}-z / \mathrm{c})$

In Young's double slit experiment, the condition for constructive interference (bright fringes) is given by:

$d \sin \theta = n \lambda$

where:

• $d$ is the distance between the slits
• $\theta$ is the angle of the fringe relative to the central maximum
• $n$ is the order of the fringe (an integer)
• $\lambda$ is the wavelength of the light

We are given two different wavelengths:

$\lambda_1 = 450 \, \text{nm}$

$\lambda_2 = 650 \, \text{nm}$

For the fringes produced by these two wavelengths to overlap, the path difference must be an integer multiple of both wavelengths. This means:

$d \sin \theta = m \lambda_1 = n \lambda_2$

where $m$ and $n$ are the orders of the fringes for $\lambda_1$ and $\lambda_2$, respectively.

To find the minimum order of fringe $n$ for $\lambda_2$ that coincides with a fringe for $\lambda_1$, we need to find the least common multiple (LCM) of these wavelengths in terms of their smallest integers. This can be formulated as:

$m \lambda_1 = n \lambda_2$

Dividing both sides by $\lambda_1$ and $\lambda_2$, we get:

$\frac{m}{\lambda_2} = \frac{n}{\lambda_1}$

Cross-multiplying, we get:

$m \lambda_1 = n \lambda_2$

Using the given wavelengths:

$m \times 450 = n \times 650$

Simplifying this equation, we get:

$\frac{m}{n} = \frac{650}{450}$

$\frac{m}{n} = \frac{13}{9}$

For the fringes to overlap, $m$ and $n$ must be integers. The smallest integers that satisfy this ratio are:

$m = 13$

$n = 9$

Therefore, the minimum order of fringe produced by $\lambda_2$ which overlaps with the fringe produced by $\lambda_1$ is:

$n = 9$

To determine the nature of the motion of the particle given by its coordinates in the $x$-$y$ plane, we analyze the given equations for $x$ and $y$ in terms of time $t$:

• $x = 2 + 4t$
• $y = 3t + 8t^2$

Firstly, the equation for $x$ is of the form $x = x_0 + vt$, where $x_0 = 2$ is the initial position and $v = 4$ is the constant velocity along the $x$-axis. This suggests a uniform motion along the $x$-axis because the velocity remains constant with time.

Secondly, the equation for $y$ is a second-degree polynomial in $t$, which indicates a parabolic path. The presence of the $t^2$ term ($8t^2$) signifies acceleration since the position along the $y$-axis is changing at a rate that itself changes over time.

The equation for $y$ can show two types of motion depending on the terms:

1. If it was of the form $y = y_0 + vt$, it would indicate uniform motion.
2. If it was of the form $y = y_0 + vt + \frac{1}{2}at^2$, where $a$ would represent acceleration, it would indicate uniformly accelerated motion. The presence of the $8t^2$ term here plays a similar role, indicating that the motion is uniformly accelerated in the $y$-direction due to the constant acceleration implied by this term.

Since the motion in the $y$ direction is determined by a quadratic equation, and the path of the particle depends on both the $x$ and $y$ coordinates, the motion of the particle is not along a straight line but rather follows a parabolic path due to the quadratic (second-degree) dependence on time in the $y$-coordinate.

Additionally, the acceleration is not changing with time, as deduced from the constant coefficient of the $t^2$ term in the $y$ equation, indicating uniform acceleration. Therefore, the motion is uniformly accelerated and follows a parabolic path.

Hence, the correct answer is:

Option D: uniformly accelerated having motion along a parabolic path.