JEE Main 2024 5th April Morning Shift Question Paper with Solutions

The collision frequency ($Z$) of gas molecules is proportional to the square root of the absolute temperature ($T$) of the system. Mathematically, it can be represented as $Z \propto \sqrt{T}$. This implies that when the temperature changes, the collision frequency changes as well according to the formula:

$Z_1 = Z_0 \sqrt{\frac{T_1}{T_0}}$

where:

• $Z_1$ is the collision frequency at temperature $T_1$,
• $Z_0$ is the initial collision frequency at temperature $T_0$,
• $T_1$ is the final absolute temperature, and
• $T_0$ is the initial absolute temperature.

To solve the given problem, we first need to convert the provided temperatures from Celsius to Kelvin (since absolute temperature in Kelvin should be used):

• $T_0 = 27^{\circ}C = 27 + 273 = 300$ K
• $T_1 = 127^{\circ}C = 127 + 273 = 400$ K

Using the formula for collision frequency change and substituting the given values:

$Z_1 = Z_0 \sqrt{\frac{400}{300}} = Z_0 \sqrt{\frac{4}{3}}$

This can be simplified to:

$Z_1 = Z_0 \times \frac{2}{\sqrt{3}}$

Thus, the collision frequency of the system at $127^{\circ}C$ is $\frac{2}{\sqrt{3}}$ times the collision frequency at $27^{\circ}C$, making Option B the correct answer:

$\frac{2}{\sqrt{3}} \mathrm{Z}$

To find the resistance of the wire connecting the two plates of the capacitor, we need to apply the formula that relates the time constant $\tau$ (in seconds) of a capacitor-resistor (CR) circuit to the capacitance C (in Farads) and resistance R (in Ohms). The time constant $\tau$ is given by:

$\tau = R \times C$

The time constant also defines the time it takes for the voltage across the capacitor (and therefore the electric field between the plates, since they are directly related) to drop to approximately $\frac{1}{e}$ (where $e$ is the base of natural logarithms, approximately equal to 2.718) of its initial value. However, the question states that the electric field drops to one-third of its initial value. Using the natural logarithm properties, we can relate this decay process to the concept of the time constant.

The formula for the voltage (or electric field) across a discharging capacitor as a function of time $t$ is:

$V(t) = V_0 \times e^{-\frac{t}{RC}}$

where:

• $V(t)$ is the voltage across the capacitor at time $t$,
• $V_0$ is the initial voltage,
• $R$ is the resistance,
• $C$ is the capacitance, and
• $t$ is the time.

Given that the electric field drops to one-third of its initial value in $6.6 \mu \mathrm{s}$, we can set $V(t)$ to $\frac{1}{3}V_0$ and solve for $R$:

$\frac{1}{3}V_0 = V_0 \times e^{-\frac{6.6 \mu s}{R \times 1.5 \mu F}}$

By dividing both sides by $V_0$, we simplify to:

$\frac{1}{3} = e^{-\frac{6.6}{R \times 1.5}}$

Taking the natural logarithm of both sides to solve for $R$:

$\ln\left(\frac{1}{3}\right) = -\frac{6.6}{R \times 1.5}$

Given that $\ln\left(\frac{1}{3}\right) = \ln(3^{-1}) = -\ln(3) = -1.1$ (since $\log 3 = 1.1$ and using the natural log instead of common log), we have:

$-1.1 = -\frac{6.6}{R \times 1.5}$

Solving for $R$:

$R = \frac{6.6}{1.5 \times 1.1}$

Now, calculate the value of $R$:

$R = \frac{6.6}{1.65} = 4\, \Omega$

Therefore, the resistance of the wire connecting the two plates of the capacitor is $4 \, \Omega$.

The time period of a simple pendulum is given by the formula:

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

where $T$ is the time period, $l$ is the length of the pendulum, and $g$ is the acceleration due to gravity at the location of the pendulum.

The acceleration due to gravity changes with height above the Earth's surface. The acceleration due to gravity at a height $h$ above the Earth's surface can be expressed as:

$g' = g \left(\frac{R}{R + h}\right)^2$

where $g$ is the acceleration due to gravity at the surface of the Earth, $R$ is the radius of the Earth, and $h$ is the height above the Earth’s surface. Since the time period of the pendulum depends on the square root of the inverse of the acceleration due to gravity, any change in $g$ due to a change in height will affect the time period.

Given that the time period of the pendulum at a height $R$ above Earth's surface is $T_1$, and we're to find the time period $T_2$ at a height of $2R$, we can use the formula for acceleration due to gravity at different heights to express the relationship between $T_1$ and $T_2$.

For the initial case at height $R$:

$g_1 = g \left(\frac{R}{R + R}\right)^2 = g \left(\frac{R}{2R}\right)^2 = \frac{g}{4}$

For the new case at height $2R$:

$g_2 = g \left(\frac{R}{R + 2R}\right)^2 = g \left(\frac{R}{3R}\right)^2 = \frac{g}{9}$

The time period is proportional to the square root of the inverse of $g$, so:

$\frac{T_1}{T_2} = \sqrt{\frac{g_2}{g_1}} = \sqrt{\frac{\frac{g}{9}}{\frac{g}{4}}} = \sqrt{\frac{4}{9}} = \frac{2}{3}$

Therefore:

$T_1 = \frac{2}{3}T_2$

Rearranging this equation:

$3T_1 = 2T_2$

This corresponds to Option A.

For hollow sphere

$$\begin{aligned} & I_1=\frac{2}{3} m R^2=\left(\sqrt{\frac{2}{3}} R\right)^2 m \\ & \therefore \quad K_1=\sqrt{\frac{2}{3}} R \end{aligned}$$

For solid cylinder

$$\begin{aligned} & I_2=\frac{1}{4} m R^2+\frac{m}{3}(4 R)^2 \\ & =\left(\sqrt{\frac{67}{12}} R\right)^2 m \\ & \therefore \quad K_2=\sqrt{\frac{67}{12}} R \\ & \Rightarrow \frac{K_1}{K_2}=\sqrt{\frac{2}{3} \times \frac{12}{67}} \\ & =\sqrt{\frac{8}{67}} \end{aligned}$$

$\Rightarrow \quad x=67$

$$\begin{aligned} & \frac{\phi_{A / B}}{I_B}=M \\ & \Rightarrow \quad M=\frac{\mu_0 I_B \times \pi a^2}{2 b \times I_B} \\ & =\frac{\mu_0 \pi a^2}{2 b} \end{aligned}$$

Given that a body is moving on a frictionless plane and starts from rest, the motion can be assumed to be uniformly accelerated motion. The formula for the distance covered in uniformly accelerated motion from rest is given by $s = ut + \frac{1}{2}at^2$, where:

• $s$ is the distance covered,


• $u$ is the initial velocity (which is 0 since the body starts from rest),


• $a$ is the acceleration, and


• $t$ is the time.

Since the body starts from rest ($u=0$), the formula simplifies to $s = \frac{1}{2}at^2$.

The distance moved between $t = n - 1$ and $t = n$, denoted as $S_n$, can be found by calculating the distance covered by the end of time $n$ and subtracting the distance covered by the end of time $n-1$. Let's denote the total distance covered by time $n$ as $S(n)$, which according to the formula is $S(n) = \frac{1}{2}a n^2$. Thus, $S_n = S(n) - S(n-1)$.

Therefore, we have:

$S_n = \frac{1}{2}an^2 - \frac{1}{2}a(n-1)^2$

Simplifying this, we get:

$S_n = \frac{1}{2}a(n^2 - (n^2 - 2n + 1))$

This simplifies to:

$S_n = \frac{1}{2}a(2n - 1)$

Similarly,

$S_{n-1} = \frac{1}{2}a((n-1)^2 - (n-2)^2)$

Simplifying:

$S_{n-1} = \frac{1}{2}a((n-1)^2 - (n^2 - 4n + 4))$

Which further simplifies to:

$S_{n-1} = \frac{1}{2}a(2n - 3)$

Hence, the ratio $\frac{S_{n-1}}{S_n}$ can be calculated:

$\frac{S_{n-1}}{S_n} = \frac{\frac{1}{2}a(2n - 3)}{\frac{1}{2}a(2n - 1)} = \frac{2n - 3}{2n - 1}$

For $n = 10$,

$\frac{S_{n-1}}{S_n} = \frac{2(10) - 3}{2(10) - 1} = \frac{20 - 3}{20 - 1} = \frac{17}{19}$

Given that the ratio is represented as $\left(1 - \frac{2}{x}\right)$, we have:

$\frac{17}{19} = 1 - \frac{2}{x}$

Solving this equation for $x$ gives:

$1 - \frac{17}{19} = \frac{2}{x}$

$\frac{2}{19} = \frac{2}{x}$

Thus:

$x = 19$

$\text { K.E }=\frac{G M m}{2 a} \quad \text{(II)}$

$U_G=\frac{-G M m}{a} \quad \text{(I)}$

$M . E=\frac{-G M m}{2 a} \quad \text{(IV)}$

$\text { and Escape Energy }=\frac{G m}{r} \quad \text{(III)}$

For bulb to glow, there should be low-high combination across the bulb.

If $A=1, B=0, C=0$ & $D=0$

$\overline{\overline{1}+1}=0 \quad \& \quad \overline{0}=1$

So, bulb will glow.

To find the angle between the vector $\vec{Q}$ and the resultant of $(2 \vec{Q}+2 \vec{P})$ and $(2 \vec{Q}-2 \vec{P})$, we first find the resultant vector of $(2 \vec{Q}+2 \vec{P})$ and $(2 \vec{Q}-2 \vec{P})$.

The resultant vector of $(2 \vec{Q}+2 \vec{P})$ and $(2 \vec{Q}-2 \vec{P})$ can be simply found by adding these two vectors:

Now, we need to find the angle between the vector $\vec{Q}$ and this resultant vector $4\vec{Q}$. Since the resultant vector is just a scaled version of $\vec{Q}$, they are in the same direction. The angle between any vector and another vector that is a scaled version of the first vector is always $0^\circ$, because they are parallel to each other.

Therefore, the angle between $\vec{Q}$ and the resultant of $(2 \vec{Q}+2 \vec{P})$ and $(2 \vec{Q}-2 \vec{P})$ is $0^\circ$.

So, the correct option is:

Option B: $0^\circ$

$E=\frac{1}{2}(25+30+45)(100)^2 \quad \text{.... (i)}$

$\text { Also, } \frac{9}{x} E=\frac{1}{2} \frac{1}{\left(\frac{1}{25}+\frac{1}{30}+\frac{1}{45}\right)}(100)^2 \quad \text{.... (ii)}$

From (i) and (ii)

$x=86$

Work done in both cases is equal to $-m g \Delta h$

$\therefore \quad \text { Ratio }=1: 1$

$$\begin{aligned} x & =\frac{3 \lambda D}{d} \\ & =\frac{3 \times 5000 \times 10^{-10} \times 200 \times 10^{-2}}{0.3 \times 10^{-3}} \\ & =10 \mathrm{~mm} \end{aligned}$$

To find the ratio of the Coulombic force to the gravitational force between an electron and a proton in a hydrogen-like system, we use the formulae for both forces and then divide them.

The Coulombic (electrostatic) force, $F_C$, between two charges is given by Coulomb's law:

$F_C = k \frac{|q_1 q_2|}{r^2}$

where $k$ is Coulomb's constant ($8.987 \times 10^9 \, \text{Nm}^2\text{C}^{-2}$), $q_1$ and $q_2$ are the magnitudes of the charges, and $r$ is the distance between the charges. For a proton and an electron, $q_1 = q_2 = e$, where $e$ is the elementary charge ($1.602 \times 10^{-19} \, \text{C}$).

The gravitational force, $F_G$, between two masses is given by Newton's law of universal gravitation:

$F_G = G \frac{m_1 m_2}{r^2}$

where $G$ is the gravitational constant ($6.674 \times 10^{-11} \, \text{Nm}^2\text{kg}^{-2}$), and $m_1$ and $m_2$ are the masses of the two objects. For a proton and an electron, $m_p \approx 1.673 \times 10^{-27} \, \text{kg}$ and $m_e \approx 9.109 \times 10^{-31} \, \text{kg}$, respectively.

The ratio of the Coulombic to the gravitational force is therefore:

$\frac{F_C}{F_G} = \frac{k \frac{|e^2|}{r^2}}{G \frac{m_p m_e}{r^2}} = \frac{k e^2}{G m_p m_e}$

Plugging in the values:

$\frac{F_C}{F_G} = \frac{(8.987 \times 10^9) (1.602 \times 10^{-19})^2}{(6.674 \times 10^{-11}) (1.673 \times 10^{-27}) (9.109 \times 10^{-31})}$

$\frac{F_C}{F_G} = \frac{(8.987 \times 10^9) \times (2.568 \times 10^{-38})}{(6.674 \times 10^{-11}) \times (1.523 \times 10^{-57})}$

$\frac{F_C}{F_G} \approx 2.3 \times 10^{39}$

Therefore, the ratio of the Coulombic force to the gravitational force between an electron and a proton in a hydrogen-like system is on the order of $10^{39}$, making Option B the correct answer.

To find the energy released when three helium nuclei combine to form a carbon nucleus, we first need to understand that this process is essentially nuclear fusion, forming a heavier nucleus from lighter ones. The mass defect in this fusion process is the key to calculating the energy released, according to Einstein's equation $E = \Delta mc^2$, where $E$ is the energy released, $\Delta m$ is the mass defect, and $c$ is the speed of light.

The atomic mass of a helium nucleus (also known as an alpha particle) is $4.002603 \, \text{u}$.

1. Calculate the total initial mass of three helium nuclei:

$\text{Total initial mass} = 3 \times 4.002603 \, \text{u} = 12.007809 \, \text{u}$

2. The atomic mass of a carbon nucleus formed by the fusion of three helium nuclei is not directly given, but we can infer it's approximately $12 \, \text{u}$, based on knowledge of isotopes and considering that the question appears to simplify the carbon nucleus to a mass number of 12 (common carbon-12 isotope).

3. Calculate the mass defect ($\Delta m$):

$\Delta m = \text{Total initial mass} - \text{Final mass}$

$\Delta m = 12.007809 \, \text{u} - 12 \, \text{u} = 0.007809 \, \text{u}$

4. Convert the mass defect to energy. Given $1 \, \text{u} = 931 \, \text{MeV/c}^2$, the energy released is calculated using the formula $E = \Delta mc^2$:

$E = 0.007809 \, \text{u} \times 931 \, \text{MeV/u} = 7.271839 \, \text{MeV}$

Since the question asks for the answer in the format of $\times 10^{-2} \, \text{MeV}$, we convert the energy released:

$7.271839 \, \text{MeV} = 727.1839 \times 10^{-2} \, \text{MeV}$

Therefore, the energy released in this reaction is approximately $727.1839 \times 10^{-2} \, \text{MeV}$. The exact value might differ slightly depending on how the atomic mass of the carbon nucleus is considered or rounded in specific scenarios, but based on the information provided, this is a suitable approximation.

$$\begin{aligned} & \left(V_{r m s}\right)_c=\frac{V_{r m s}}{Z} X_c \\ & =\frac{50}{\sqrt{(500-100)^2+300^2}} \times 500 \\ & =50 \mathrm{~V} \end{aligned}$$

$$\begin{aligned} & i L B=m g \text { and } L=\frac{A R}{\rho} \\ & \begin{aligned} \therefore B & =\frac{m g \rho}{i A R} \\ & =\frac{0.5 \times 10 \times 2 \times 10^{-6}}{2 \times 10 \times 10^{-6} \times 1} \\ & =0.5 \mathrm{~T} \end{aligned} \end{aligned}$$

$$\begin{aligned} \Delta Q & =\Delta U+w \\ & =\pi(140)^2 \times 10^3 \times 10^{-6} \\ & =61.6 \mathrm{~J} \end{aligned}$$

In the context of an electron orbiting around a nucleus with a positive charge of $\mathrm{Ze}$, we are dealing with classical physics approximations and the electrostatic force between the electron and the nucleus. In such a setup, the electron's potential energy (U) is due to electrostatic interaction, and it is given by Coulomb's law:

$U = -\frac{kZe^2}{r}$

Where:

• $U$ is the potential energy of the electron,
• $k$ is Coulomb's constant,
• $Z$ is the atomic number (number of protons in the nucleus),
• $e$ is the charge of an electron, and
• $r$ is the radius of the orbit of the electron around the nucleus.

The negative sign indicates that the potential energy is negative because the electron and nucleus attract each other.

The total energy (E) of the electron in orbit is the sum of its kinetic energy (K) and its potential energy (U). Since the electron is in a stable orbit, its kinetic energy can be shown to be exactly half the magnitude of its potential energy but positive:

$K = -\frac{1}{2}U$

Therefore,

$E = K + U = -\frac{1}{2}U + U = \frac{1}{2}U$

To find a relation between total energy (E) and potential energy (U), we rearrange the equation as follows:

$2E = U$

This is to say, the total energy (E) is half the magnitude of potential energy (U) but negative, and the correct relationship between them, when looking for a positive proportionality, yields to $2E = U$. Hence, the correct option is:

Option C: $2 \mathrm{E} = \mathrm{U}$

Let's calculate the maximum displacement current between the plates of the capacitor when an alternating voltage is applied directly across it.

The formula for the capacitive reactance $X_C$ of a capacitor is given by:

$X_C = \frac{1}{2\pi fC}$

where

• $f$ is the frequency of the alternating voltage, and
• $C$ is the capacitance of the capacitor.

Given:

• The amplitude of the voltage ($V_{max}$) = $40$ V
• Frequency ($f$) = $4000$ Hz (or $4 \mathrm{~kHz}$)
• Capacitance ($C$) = $12 \mu \mathrm{F} = 12 \times 10^{-6} \mathrm{~F}$

First, we calculate $X_C$:

$X_C = \frac{1}{2\pi(4000)(12 \times 10^{-6})} \approx \frac{1}{2\pi \cdot 4 \cdot 12 \cdot 10^{-3}} \approx \frac{1}{96\pi \cdot 10^{-3}}$

$X_C \approx \frac{1}{3.14 \cdot 96 \cdot 10^{-3}} \approx \frac{1}{0.30144} \approx 3.316 \Omega$

Now, the maximum current ($I_{max}$) in the circuit can be calculated using Ohm's law, considering the maximum voltage across the capacitor and its capacitive reactance:

$I_{max} = \frac{V_{max}}{X_C}$

$I_{max} = \frac{40}{3.316} \approx 12.06 \mathrm{~A}$

Hence, the maximum displacement current between the plates of the capacitor is nearly:

Option D: 12 A

Both statements given above have implications relating to the phenomena of capillarity, which involves the interaction between a liquid and a solid (in this case, a capillary tube). Let's break down each statement for clarity.

Statement I: When a capillary tube is dipped into a liquid and the liquid neither rises nor falls in the capillary, it suggests a scenario where the adhesive forces (between the liquid and the solid) and the cohesive forces (among the liquid molecules) are in perfect balance. The contact angle, which is the angle formed by the tangent to the liquid surface at the point of contact with the wall of the tube, plays a crucial role here. A contact angle of $0^{\circ}$ implies complete wetting, meaning the liquid spreads out to maximize contact with the solid. However, the statement that the liquid neither rises nor falls specifically with a contact angle of $0^{\circ}$ seems inaccurate. In reality, when the contact angle is $0^{\circ}$, it denotes perfect wetting, and the liquid tends to rise in the capillary tube. Thus, the precision of Statement I could be disputed based on the typical behavior of liquids in the context of capillarity and contact angles.

Statement II: The contact angle really is a property of both the solid and the liquid. It depends on the nature of the solid surface (whether it's hydrophilic or hydrophobic) and the type of liquid. This is because the contact angle reflects the degree of interaction between the liquid and solid surfaces, which is influenced by characteristics such as surface tension of the liquid and the surface energies of both the liquid and solid. Thus, Statement II accurately describes the nature of the contact angle as being dependent on both the material of the solid and the liquid.

Considering the analysis above:

• Statement I suggests a specific scenario that does not align accurately with the principles of capillarity and wettability, particularly at a contact angle of $0^{\circ}$, which typically would result in the liquid rising in the capillary, not staying neutral.
• Statement II accurately reflects the factors influencing the contact angle between a liquid and a solid surface.

Therefore, the correct option based on the given statements and their analysis would be:

Option B: Statement I is false but Statement II is true.

$$\begin{aligned} & T_1-(4+6+10) g=(4+6+10)(2) \\ & \Rightarrow T_1=20(10+2) \\ & \quad=240 \mathrm{~N} \end{aligned}$$

Let's dig into how the magnetic field behaves in a coaxial cable in relation to the given options. The principle to consider here is Ampère's law, which states that the magnetic field in a loop surrounding a current is proportional to the amount of current enclosed. When we apply this to a coaxial cable, we must look at different regions within the cable.

Option A: Inside the outer conductor

The magnetic field within the outer conductor is not zero because the current in the outer conductor itself contributes to the magnetic field in that region. However, considering the symmetric distribution of current and the geometry of the coaxial cable, there may be a varying magnetic field within the conductor depending on the distance from the center axis.

Option B: Outside the cable

Outside the coaxial cable, the net current enclosed by a path enclosing both conductors is zero because the current in the inner conductor flows in the opposite direction to the equally magnitude current in the outer conductor. These currents being equal and opposite in direction cancel each other out, leading to a net enclosed current of zero. According to Ampère's law, if the net enclosed current is zero, the magnetic field in that space is also zero. Therefore, the magnetic field is zero outside the cable.

Option C: In between the two conductors

In the space between the two conductors, the magnetic field is not zero. This region only encloses the current from the inner conductor. The magnetic field in this region is due to the current in the inner conductor and follows the right-hand rule, which would result in concentric circles of magnetic field around the inner conductor. Since only the inner conductor's current contributes to the magnetic field in this space, Ampère's law suggests that there is a non-zero magnetic field in this region.

Option D: Inside the inner conductor

Within the inner conductor, the magnetic field is not necessarily zero. Like within the outer conductor, the magnetic field inside the inner conductor will depend on the distribution of the current within that conductor. Utilizing the formula derived from Ampère's law for a cylindrical conductor with a uniform current distribution, the magnetic field inside the conductor increases linearly with the distance from the center axis up to the conductor's surface.

In conclusion, the correct option, based on Ampère's law and the principle that the net current enclosed determines the magnetic field outside the current's path, is:

Option B: Outside the cable

This is because the equal and opposite currents in the inner and outer conductors cancel each other, leading to a net magnetic field of zero outside the coaxial cable.

$$\begin{aligned} R_{\mathrm{eq}}= & 12 \Omega \\ \text { and, } I & =\frac{E}{R_{\mathrm{eq}}} \\ & =\frac{12}{12} \\ & =1 \mathrm{~A} \end{aligned}$$

To find the arithmetic mean of the time periods recorded, we need to sum up the values and then divide by the number of readings. Let's calculate the sum first:

$4.62 \mathrm{~s} + 4.632 \mathrm{~s} + 4.6 \mathrm{~s} + 4.64 \mathrm{~s}$

Adding these values together:

$4.62 + 4.632 + 4.6 + 4.64 = 18.492 \mathrm{~s}$

Now, we divide this sum by the number of readings, which is 4:

$\frac{18.492 \mathrm{~s}}{4} = 4.623 \mathrm{~s}$

So, the arithmetic mean of these readings is $4.623 \mathrm{~s}$. However, we need to consider the significant figures. The least number of significant figures among the readings is 2 (from 4.6 s). Hence, the mean should also be represented with 2 significant figures.

In this case, the correct answer with proper significant figures is:

Option B

4.6 s

$\frac{1}{3} \mathrm{~A} \longrightarrow \frac{1}{2} \mathrm{div}$

$\frac{1}{2} \mathrm{~A} \longrightarrow \frac{2}{3} \mathrm{div}$

$$\text { Figure of merit }=\frac{\Delta i}{\Delta \theta} \quad \begin{aligned} & \frac{\frac{1}{2}--}{\frac{2}{3}-\frac{1}{2}} \\ &=0.5 \\ &=5 \times 10^{-1} \mathrm{~A} / \mathrm{div} \end{aligned}$$

$$\begin{aligned} & V_0=\frac{h}{e} f-\frac{h}{e} f_0 \\ & \Rightarrow \text { Slope }=\frac{h}{e} \text { (S-I is correct) } \\ & \because\left(f_0\right)_1<\left(f_0\right)_2 \\ & \therefore \quad(\mathrm{S}-I I \text { is incorrect) } \\ & \therefore \quad \text { Option (3) is correct. } \end{aligned}$$

To determine the dimension of the quantity $\sqrt{uG}$, we first need to understand the dimensions of both the gravitational constant (G) and the energy density (u).

The gravitational constant $G$ has dimensions given by:

$[G] = M^{-1}L^{3}T^{-2}$

where $M$ stands for mass, $L$ for length, and $T$ for time.

Energy density $u$ is defined as the energy per unit volume. Since energy has dimensions of $ML^{2}T^{-2}$ (from the dimension of work or energy, which is force times distance, and force itself has dimension $MLT^{-2}$), and volume has dimensions of $L^{3}$, the dimensions of energy density would be:

$[u] = \frac{ML^{2}T^{-2}}{L^{3}} = M L^{-1} T^{-2}$

Now, we find the dimensions of $\sqrt{uG}$ by multiplying the dimensions of $u$ and $G$, and then taking the square root:

$[\sqrt{uG}] = \sqrt{[u][G]} = \sqrt{(M L^{-1} T^{-2})(M^{-1}L^{3}T^{-2})} = \sqrt{L^{2}T^{-4}} = LT^{-2}$

So, the dimension of $\sqrt{uG}$ is $LT^{-2}$, which corresponds to acceleration (length per square time).

Now, let's match this with the provided options:

• Option A (Gravitational potential) has dimensions of $[L^{2}T^{-2}]$, not matching our target of $LT^{-2}$.
• Option B (Pressure gradient per unit mass) would have dimensions of $[M^{-1}L^{-2}T^{-2}][L^{-1}]$ ($Pressure\ Gradient = \frac{Pressure}{Length} = \frac{ML^{-1}T^{-2}}{L}$, and then divided by mass, $M$), which simplifies to $L^{-3}T^{-2}M^{-1}$, not matching.
• Option C (Energy per unit mass) has dimensions $ML^{2}T^{-2}M^{-1}$ which simplifies to $L^{2}T^{-2}$, also not a match for the target dimension.
• Option D (Force per unit mass) has dimensions $MLT^{-2}M^{-1}$ which simplifies directly to $LT^{-2}$, an exact match for our target dimension.

Thus, the correct answer is Option D (Force per unit mass), which has the same dimensions as that of $\sqrt{\mathrm{uG}}$.

The breaking stress of a wire, denoted as $\sigma$, is the maximum tension per unit area it can withstand before it breaks. It is given as:

$\sigma = \frac{F}{A}$

where $F$ is the breaking force and $A$ is the cross-sectional area of the wire. The weight of the wire, when it's on the verge of breaking, equals the maximum force $F$ it can sustain. The weight of an object is given by:

$W = mg$

where $m$ is the mass of the object and $g$ is the acceleration due to gravity. Since the wire's mass, $m$, can also be expressed in terms of its density ($\rho$), length ($L$), and area ($A$) as:

$m = \rho V = \rho A L$

We can substitute this expression in the equation for weight:

$W = \rho A L g$

On a planet where the acceleration due to gravity is $\frac{1}{3}$rd of that on Earth, we substitute $g_{\text{planet}} = \frac{1}{3}g_{\text{Earth}} = \frac{1}{3} \times 10 \, \text{m/s}^2 = \frac{10}{3} \, \text{m/s}^2$. The breaking force $F$ which is equivalent to the weight at the breaking point, is therefore given by:

$F = \rho A L g_{\text{planet}}$

Since the breaking stress $\sigma$ is also $F/A$, we can set the two expressions equal to find the maximum length $L$ of the wire before breaking:

$\sigma = \frac{\rho A L g_{\text{planet}}}{A}$

$\sigma = \rho L g_{\text{planet}}$

Solving for $L$:

$L = \frac{\sigma}{\rho g_{\text{planet}}}$

Given that $\sigma = 1.2 \times 10^8 \, \text{N/m}^2$, $\rho = 6 \times 10^4 \, \text{kg/m}^3$, and $g_{\text{planet}} = \frac{10}{3} \, \text{m/s}^2$, we can substitute these values into the formula: