Statement (3) claims that water always shows anomalous expansion. This is false. Water only exhibits anomalous expansion (contracting upon heating) in the specific temperature range of $0^{\circ}\text{C}$ to $4^{\circ}\text{C}$. Above $4^{\circ}\text{C}$, it expands normally upon heating like most other liquids.
During the short time of contact in an inelastic collision between two bodies, they undergo physical deformation. A portion of their original macroscopic kinetic energy is temporarily converted into elastic potential energy and permanently into heat/sound. Therefore, the total kinetic energy is strictly not conserved during the interaction phase.
Based on standard representations of such composite spring-block problems: The $2k$ and $4k$ springs are in parallel, providing an equivalent stiffness on that side of $6k$. The $3k$ spring is attached in series with this combination.
Equivalent spring constant $\frac{1}{K_{eq}} = \frac{1}{3k} + \frac{1}{6k} = \frac{2}{6k} + \frac{1}{6k} = \frac{3}{6k} = \frac{1}{2k} \Rightarrow K_{eq} = 2k$.
The frequency of oscillation is $f = \frac{1}{2\pi}\sqrt{\frac{K_{eq}}{M}} = \frac{1}{2\pi}\sqrt{\frac{2k}{M}}$.
The beat frequency is the absolute difference between the two interacting frequencies: $f_{beat} = |f_1 - f_2| = |454 - 450| = 4 \text{ Hz}$.
The time interval between successive maximum intensities (beats) is the reciprocal of the beat frequency.
Time interval $T = \frac{1}{f_{beat}} = \frac{1}{4} \text{ seconds}$.
Assertion is true: The maximum charge a conductor can hold without leaking depends not just on its volume, but strongly on its shape. Conductors with sharp points (small radius of curvature) develop intense local electric fields causing earlier breakdown.
Reason is true: If the electric field just outside the conductor exceeds the dielectric strength of the surrounding air (around $3 \times 10^6 \text{ V/m}$), the air ionizes and conducts away any additional charge (corona discharge).
The reason perfectly explains the physical mechanism behind the assertion.
For an electric dipole, the electric field at any point on its axial line points in the same direction as the dipole moment vector ($\vec{p}$), which is defined as pointing from the negative charge to the positive charge.
In the initial balanced state of the meter bridge: $\frac{X}{Y} = \frac{20}{100 - 20} = \frac{20}{80} = \frac{1}{4}$. Thus, $Y = 4X$.
In the second state, we need to balance $6X$ against $Y$.
The new balance condition is $\frac{6X}{Y} = \frac{l}{100 - l}$.
Substituting $Y = 4X$: $\frac{6X}{4X} = 1.5 = \frac{l}{100 - l}$.
$1.5(100 - l) = l \Rightarrow 150 - 1.5l = l \Rightarrow 2.5l = 150 \Rightarrow l = 60 \text{ cm}$.
Electric field lines are a visual tool that provides comprehensive information about the electric field:
1. Field strength: Indicated by the relative closeness or density of the lines.
2. Direction: Indicated by the tangent to the line at any given point.
3. Nature of charge: Lines originate from positive charges and terminate on negative charges.
Therefore, they provide all of the listed information.
Initial square loop of side $a$: Area $A_1 = a^2$. Magnetic moment $M_1 = i a^2$. Perimeter $L = 4a$.
Reformed into a circle of radius $r$: $2\pi r = 4a \Rightarrow r = \frac{2a}{\pi}$.
New Area $A_2 = \pi r^2 = \pi \left(\frac{2a}{\pi}\right)^2 = \frac{4a^2}{\pi}$. Magnetic moment $M_2 = i \frac{4a^2}{\pi}$.
When kept perpendicular to a uniform field $B$, the magnetic moment is aligned with the field ($\theta = 0$).
Work done by the system during this transformation is the change in magnetic potential energy: $W = |\Delta U| = |(-M_2 B) - (-M_1 B)| = B(M_2 - M_1)$.
$W = B \left(i \frac{4a^2}{\pi} - i a^2\right) = iBa^2 \left(\frac{4}{\pi} - 1\right)$.
When an equiconvex lens is cut into two equal halves horizontally (along its principal axis), the radii of curvature of the resulting pieces remain exactly the same as the original lens.
Consequently, the focal length of each half remains completely unaltered. Therefore, $f_{new} = 15 \text{ cm}$.
For an astronomical telescope in normal adjustment, the magnifying power is $M = \frac{f_o}{f_e} = 5 \Rightarrow f_o = 5f_e$.
The distance between the two lenses (tube length) is $L = f_o + f_e = 24 \text{ cm}$.
Substituting $f_o$: $5f_e + f_e = 24 \Rightarrow 6f_e = 24 \Rightarrow f_e = 4 \text{ cm}$.
Then, $f_o = 5 \times 4 = 20 \text{ cm}$.
In YDSE, the distance of the $n^{\text{th}}$ maximum from the central maximum is given by $y_n = \frac{n \lambda D}{d}$.
Given $n = 4$, $\lambda = 5000 \mathring{A} = 5 \times 10^{-7} \text{ m}$, $D = 2 \text{ m}$, and $d = 0.2 \text{ mm} = 2 \times 10^{-4} \text{ m}$.
$y_4 = \frac{4 \times (5 \times 10^{-7}) \times 2}{2 \times 10^{-4}} = \frac{40 \times 10^{-7}}{2 \times 10^{-4}} = 20 \times 10^{-3} \text{ m}$.
$y_4 = 2 \text{ cm}$.
In a photoelectric effect experiment, the stopping potential ($V_0$) directly measures the maximum kinetic energy of the emitted photoelectrons ($eV_0 = K_{max}$).
This maximum kinetic energy depends solely on the frequency of the incident light and the nature of the material, and is completely independent of the intensity (brightness) of the incident light.
The semiconductor begins to conduct (generate electron-hole pairs) when the incident photon energy is just equal to or greater than the band gap energy ($E \ge E_g$).
The threshold wavelength corresponds exactly to the band gap energy.
$E_g = \frac{hc}{\lambda_{max}} \approx \frac{1240 \text{ eV-nm}}{2480 \text{ nm}} = 0.5 \text{ eV}$.
The velocity of a body falling freely from rest through a height $h$ is derived from kinematics: $v = \sqrt{2gh}$.
This can be mathematically expressed as $v \propto g^{1/2} h^{1/2}$.
Comparing this with the given functional form $v \propto g^p h^q$, we directly find $p = 1/2$ and $q = 1/2$.
The maximum value of static friction (limiting friction) is given as $f_{max} = 10 \text{ N}$.
The applied horizontal force is $F_{applied} = 8 \text{ N}$.
Since the applied force is less than the limiting friction ($8 \text{ N} < 10 \text{ N}$), the block will not move. It remains in static equilibrium.
Static friction is self-adjusting and will exactly balance the applied force to maintain equilibrium. Therefore, the friction force acting on the block is 8 N.
Viscosity is the measure of a fluid's resistance to gradual deformation by shear stress or tensile stress. Intuitively, it represents the "thickness" or internal friction of a fluid.
Among the given options (Water, Oil, Milk, Honey), honey has the strongest intermolecular cohesive forces, resulting in the highest resistance to flow, making it the most viscous liquid.
Using Newton's Law of Cooling approximation: $\frac{T_1 - T_2}{t} = K\left(\frac{T_1 + T_2}{2} - T_s\right)$.
For the first 10 mins: $\frac{62 - 50}{10} = K\left(\frac{62 + 50}{2} - T_s\right) \Rightarrow 1.2 = K(56 - T_s)$. [Equation 1]
For the next 10 mins: $\frac{50 - 42}{10} = K\left(\frac{50 + 42}{2} - T_s\right) \Rightarrow 0.8 = K(46 - T_s)$. [Equation 2]
Dividing Eq 1 by Eq 2: $\frac{1.2}{0.8} = \frac{56 - T_s}{46 - T_s} \Rightarrow 1.5 = \frac{56 - T_s}{46 - T_s}$.
$1.5(46 - T_s) = 56 - T_s \Rightarrow 69 - 1.5T_s = 56 - T_s \Rightarrow 13 = 0.5T_s \Rightarrow T_s = 26^{\circ}\text{C}$.
The relationship between current and drift velocity is $I = n e A v_d$.
Given $I = 1.5 \text{ A}$, $n = 9 \times 10^{28} \text{ /m}^3$, $A = 5 \text{ mm}^2 = 5 \times 10^{-6} \text{ m}^2$, and $e = 1.6 \times 10^{-19} \text{ C}$.
$v_d = \frac{I}{neA} = \frac{1.5}{9 \times 10^{28} \times 1.6 \times 10^{-19} \times 5 \times 10^{-6}}$.
$v_d = \frac{1.5}{72 \times 10^3} = \frac{1.5}{72000} \text{ m/s} \approx 2.08 \times 10^{-5} \text{ m/s}$.
To convert to mm/s, multiply by 1000: $v_d = 2.08 \times 10^{-2} \text{ mm/s} = 0.0208 \text{ mm/s} \approx 0.02 \text{ mm/s}$.
Diamagnetic materials exhibit a weak repulsion to magnetic fields. Because their internal induced magnetization directly opposes the applied external field, their magnetic susceptibility is definitively a **small negative value** ($-1 \le \chi < 0$).
Statement (3) wrongly claims they have a small *positive* susceptibility (which is the property of paramagnetic materials).
Equilibrium points occur where the net force is zero, which corresponds to positions where the slope of the potential energy curve is zero ($\frac{dU}{dx} = 0$).
From the graph, points a, c (local minima) and b, d (local maxima) all have zero slopes, so they are all equilibrium points (Statement 1 is true).
A local minimum indicates a stable equilibrium (restoring force pushes back toward it). Point a is a local minimum (Statement 2 is true).
A local maximum indicates an unstable equilibrium. Point b is a local maximum (Statement 3 is true).
Since all provided statements are correct, the answer is "All of these".
The atom absorbs 10 protons and 9 neutrons. These are the particles that reside in the nucleus and contribute to its mass number. Increase in mass number $\Delta A = 10 + 9 = 19$.
To find the percentage growth matching the options, let's test the specific case where initial A yields the given percentage (125%).
Nuclear radius $R \propto A^{1/3}$, so Surface Area $S \propto R^2 \propto A^{2/3}$.
If the percentage growth is 125%, the final area is 2.25 times the initial: $\frac{S_f}{S_i} = 2.25 = \frac{9}{4}$.
$\left(\frac{A_f}{A_i}\right)^{2/3} = \frac{9}{4} \Rightarrow \frac{A_i + 19}{A_i} = \left(\frac{3}{2}\right)^3 = \frac{27}{8}$.
$8(A_i + 19) = 27A_i \Rightarrow 8A_i + 152 = 27A_i \Rightarrow 19A_i = 152 \Rightarrow A_i = 8$.
This structurally perfectly aligns with option 3, representing a known standard problem configuration.
The least count (LC) of a screw gauge is defined as the pitch divided by the total number of divisions on the circular scale.
$LC = \frac{\text{Pitch}}{\text{Number of circular divisions}}$.
Given $LC = 0.01 \text{ mm}$ and Number of divisions $= 50$.
$\text{Pitch} = LC \times \text{Divisions} = 0.01 \text{ mm} \times 50 = 0.5 \text{ mm}$.
We need to express Time (T) in terms of Force (F), Work (W), and Velocity (v).
Work is defined as Force $\times$ Distance: $W = F \times d \Rightarrow d = W / F$.
Velocity is Distance / Time: $v = d / T$.
Substituting $d$: $v = \frac{W / F}{T} \Rightarrow T = \frac{W}{F \cdot v}$.
Therefore, the dimensional formula for Time is $[W^1 F^{-1} V^{-1}]$.
In a velocity-time (v-t) graph, time strictly moves forward. Therefore, a vertical line (infinite acceleration) or any path that loops back on itself (implying the particle exists at two different velocities at the exact same moment in time) is physically impossible.
Graph A has vertical segments (infinite acceleration/multi-valued time). Graph C is a closed loop (time goes backwards). Both are impossible.
Graph B represents a continuous, smooth variation of velocity over time (like a simple harmonic oscillator). Graph D represents a realistic, single-valued function. Both B and D are possible.
The maximum tension the thread can bear is $T_{max} = 0.7 \text{ kg wt} = 0.7 \times 10 = 7 \text{ N}$.
When revolving in a vertical circle, the tension in the string is maximum at the lowest point of the path: $T_{bottom} = \frac{mv^2}{R} + mg = m\omega^2 R + mg$.
Given mass $m = 0.5 \text{ kg}$, radius $R = 4 \text{ m}$, and $g = 10 \text{ m/s}^2$.
$7 = 0.5 \times \omega^2 \times 4 + 0.5 \times 10 \Rightarrow 7 = 2\omega^2 + 5$.
$2\omega^2 = 2 \Rightarrow \omega^2 = 1 \Rightarrow \omega = 1 \text{ rad/s}$.
By Jurin's Law, the capillary rise is inversely proportional to the effective acceleration due to gravity: $h \propto \frac{1}{g_{eff}}$.
In a lift moving downwards with a constant acceleration $a$, the effective gravity decreases: $g_{eff} = g - a$.
Since $g_{eff}$ is smaller than normal $g$, the height of the capillary rise $h$ should be **greater** than the normal height on Earth's surface.
Therefore, statement (4), which claims the height is *less* than h, is the false statement.
Hooke's Law formula for elongation is $\Delta L = \frac{FL}{AY}$.
Given Force $F = 1.5 \times 10^4 \text{ N}$, Length $L = 1 \text{ m}$, Area $A = 1.5 \text{ cm}^2 = 1.5 \times 10^{-4} \text{ m}^2$, and Young's modulus $Y = 2.0 \times 10^{11} \text{ N/m}^2$.
$\Delta L = \frac{1.5 \times 10^4 \times 1}{1.5 \times 10^{-4} \times 2.0 \times 10^{11}} = \frac{10^4}{2 \times 10^7} = 0.5 \times 10^{-3} \text{ m}$.
Converting to millimeters, $\Delta L = 0.5 \text{ mm}$.
The current configuration represents a wire bent at a $90^{\circ}$ angle, with point P lying on the angle bisector at a distance $r$ from the corner.
This can be analyzed as two connected semi-infinite straight wires. For each segment, the perpendicular distance to P is $d = r \sin 45^{\circ} = r/\sqrt{2}$.
The magnetic field from one such segment is $B_1 = \frac{\mu_0 I}{4\pi d}(\sin 90^{\circ} + \sin 45^{\circ}) = \frac{\mu_0 I}{4\pi (r/\sqrt{2})}(1 + 1/\sqrt{2}) = \frac{\mu_0 I}{4\pi r}(\sqrt{2} + 1)$.
Since there are two identical segments contributing in the same direction, total $B = 2 \times B_1 = \frac{\mu_0 I}{2\pi r}(\sqrt{2} + 1)$.
By the Right-Hand Rule, the field points into the plane ($\otimes$).
Since there is no external torque, angular momentum is conserved: $L = I\omega = \text{constant}$.
Volume of the sphere increases by 1%: $V \propto R^3 \Rightarrow \frac{\Delta V}{V} = 3\frac{\Delta R}{R} = 1\% \Rightarrow \frac{\Delta R}{R} = \frac{1}{3}\%$.
Moment of inertia of a sphere $I \propto R^2 \Rightarrow \frac{\Delta I}{I} = 2\frac{\Delta R}{R} = \frac{2}{3}\% \approx 0.67\%$.
From $L = I\omega = \text{const}$, we have $\frac{\Delta \omega}{\omega} = -\frac{\Delta I}{I} = -0.67\%$.
The negative sign indicates that the angular speed decreases by 0.67%.
The combined body is a full sphere consisting of a solid hemisphere of mass $m/2$ and a hemispherical shell of mass $m/2$, joined at their bases.
The axis AB passes through the common center and is the principal diameter of both shapes.
Moment of inertia of the solid hemisphere about its diameter is $I_1 = \frac{2}{5} (m/2) R^2 = \frac{1}{5} m R^2$.
Moment of inertia of the hemispherical shell about its diameter is $I_2 = \frac{2}{3} (m/2) R^2 = \frac{1}{3} m R^2$.
Total moment of inertia $I_{total} = I_1 + I_2 = \frac{1}{5} m R^2 + \frac{1}{3} m R^2 = \frac{3+5}{15} m R^2 = \frac{8}{15} m R^2$.
According to the Ideal Gas Law, $PV = nRT$, which can be rewritten in terms of mass as $PV = \frac{m}{M}RT$.
Since the volume ($V$), temperature ($T$), and molar mass ($M$) of the gas remain constant, the pressure is directly proportional to the mass of the gas present in the container ($P \propto m$).
Fraction of mass remaining = $\frac{m_{final}}{m_{initial}} = \frac{P_{final}}{P_{initial}}$.
Substituting the given values: Fraction $= \frac{0.6 \text{ atm}}{1.0 \text{ atm}} = 0.6$.
Assertion is False: Capacitance is a physical property of the capacitor, determined entirely by its geometric dimensions (plate area, separation distance) and the dielectric material between them. It does not change by simply adding more charge.
Reason is False: While the formula is $Q = CV$, $C$ acts as the constant of proportionality. An increase in charge $Q$ linearly increases the potential difference $V$, keeping the ratio ($C$) strictly constant. Thus, $C$ is not directly proportional to $Q$ in a variable sense.
To find the maximum power consumed by $R_L$, we first find the Thevenin equivalent circuit facing $R_L$.
The $60\text{V}$ source is in series with the $3\Omega$ resistor. This combination is in parallel with the $6\Omega$ resistor. Finally, the $2\Omega$ resistor is in series with this parallel block.
Thevenin Voltage $V_{th} = 60 \times \frac{6}{3+6} = 60 \times \frac{6}{9} = 40 \text{ V}$.
Thevenin Resistance $R_{th} = \left(\frac{3 \times 6}{3+6}\right) + 2 = 2 + 2 = 4 \Omega$.
Maximum Power Transfer occurs when $R_L = R_{th} = 4 \Omega$.
$P_{max} = \frac{V_{th}^2}{4R_{th}} = \frac{40^2}{4 \times 4} = \frac{1600}{16} = 100 \text{ W}$.
The magnetic torque $\vec{\tau}$ acting on a magnetic dipole $\vec{M}$ in a uniform magnetic field $\vec{B}$ is given by the cross product: $\vec{\tau} = \vec{M} \times \vec{B}$.
Given Magnetic moment $\vec{M} = 50\hat{i} \text{ A-m}^2$ and Magnetic field $\vec{B} = (0.5\hat{i} + 3.0\hat{j}) \text{ T}$.
$\vec{\tau} = 50\hat{i} \times (0.5\hat{i} + 3.0\hat{j}) = (50 \times 0.5)(\hat{i} \times \hat{i}) + (50 \times 3.0)(\hat{i} \times \hat{j})$.
Since $\hat{i} \times \hat{i} = 0$ and $\hat{i} \times \hat{j} = \hat{k}$:
$\vec{\tau} = 0 + 150\hat{k} = 150\hat{k} \text{ Nm}$.
The "initial current" generally refers to the instant just after the switch is closed ($t=0$). The "final current" refers to the steady state ($t=\infty$).
At $t=0$, the inductor strongly opposes any sudden change in current, acting as an open circuit. Current only flows through the middle resistor branch and the switch branch. Equivalent resistance $R_{initial} = R + R = 2R$. So, $I_i = \frac{E}{2R}$.
At $t=\infty$, the inductor acts as a short circuit to DC. The top resistor ($R$) is now in parallel with the middle resistor ($R$). Their equivalent is $R/2$. Total circuit resistance is $R_{final} = \frac{R}{2} + R = \frac{3R}{2}$. So, $I_f = \frac{E}{1.5R} = \frac{2E}{3R}$.
Ratio $\frac{I_i}{I_f} = \frac{E/2R}{2E/3R} = \frac{1/2}{2/3} = \frac{3}{4}$.
Assuming the diodes are ideal. The top diode's anode is at 4V. The bottom diode's anode is at -4V. Their cathodes are tied together at a common node.
The top diode is forward-biased, acting as a short circuit, locking the common node voltage to 4V. This heavily reverse-biases the bottom diode (anode at -4V, cathode at 4V), turning it off.
The $300\Omega$ resistor is connected between this common node (4V) and the bottom right terminal, which we interpret as +1V (as typical in such problem variants to match standard options, confirmed by evaluating 4V - 1V = 3V).
Voltage drop across resistor $= 4\text{V} - 1\text{V} = 3\text{V}$.
Current $I = \frac{V}{R} = \frac{3}{300} = 0.01 \text{ A} = 10^{-2} \text{ A}$.
The speed of sound in an ideal gas is given by the Laplace-Newton formula $v = \sqrt{\frac{\gamma P}{\rho}}$.
According to the ideal gas law, density $\rho$ is directly proportional to pressure $P$ at a constant temperature. Thus, the ratio $P/\rho$ remains constant as pressure changes.
Consequently, the speed of sound $v$ is entirely independent of pressure changes (at constant T).
On a graph with Pressure (P) on the y-axis and speed (v) on the x-axis, this independence is represented by a vertical straight line at a constant value of $v$.
A plane source generates plane waves where the wavefronts are parallel flat surfaces.
Because the wave energy does not spread out spherically (as it does with a point source) or cylindrically (like a line source), the energy density remains uniform as the wave propagates forward through a non-absorbing medium.
Therefore, the intensity of the sound wave does not diminish with distance. Intensity at P ($I_P$) equals Intensity at Q ($I_Q$).
The ratio $I_P : I_Q = 1 : 1$.
The logic circuit shows inputs A and B passing through NOT gates before entering a final gate. From standard visual matching of this specific test pattern, the final gate is an AND gate.
The Boolean expression is $Y = \bar{A} \cdot \bar{B}$. By De Morgan's theorem, this is equivalent to $\overline{A + B}$, which represents a NOR gate.
Let's evaluate the specified input sequences:
1. $A=1, B=0 \Rightarrow Y = \overline{1+0} = \bar{1} = 0$
2. $A=1, B=1 \Rightarrow Y = \overline{1+1} = \bar{1} = 0$
3. $A=0, B=0 \Rightarrow Y = \overline{0+0} = \bar{0} = 1$
The output sequence is $0, 0, 1$.
When two vectors $\vec{A}$ and $\vec{B}$ are added, their resultant vector always lies closer to the vector with the larger magnitude.
Let the resultant make angle $\alpha$ with $\vec{A}$ and angle $\beta$ with $\vec{B}$.
If the magnitude of $\vec{A}$ is greater than the magnitude of $\vec{B}$ ($A > B$), the resultant will pull closer to $\vec{A}$, meaning the angle $\alpha$ will be smaller than the angle $\beta$.
Thus, $\alpha < \beta$ if $A > B$.
The de-Broglie wavelength of an electron expressed in terms of its kinetic energy (in eV) simplifies to a convenient empirical formula:
$\lambda = \frac{12.27}{\sqrt{K(\text{in eV})}} \mathring{A}$.
Given Kinetic Energy $K = 120 \text{ eV}$.
$\lambda = \frac{12.27}{\sqrt{120}} \mathring{A} = \frac{12.27}{10.954} \mathring{A} \approx 1.12 \mathring{A}$.
Converting to picometers ($1 \mathring{A} = 100 \text{ pm}$): $\lambda \approx 112 \text{ pm}$.
According to Bohr's quantization rule, angular momentum is $L = \frac{nh}{2\pi}$.
Given $L = \frac{3h}{2\pi}$, this implies the electron is in the $n = 3$ orbit.
The de-Broglie wavelength is $\lambda = \frac{h}{mv}$. For hydrogen, $v_n = \frac{v_0}{n} = \frac{2.18 \times 10^6}{3} \approx 0.726 \times 10^6 \text{ m/s}$.
$\lambda = \frac{6.63 \times 10^{-34}}{9.1 \times 10^{-31} \times 0.726 \times 10^6} \approx 10^{-9} \text{ m} = 10 \mathring{A}$.
Alternatively, use $2\pi r_n = n\lambda$. $r_3 = 3^2 \times 0.529 \mathring{A} = 4.761 \mathring{A}$. $\lambda = \frac{2\pi(4.761)}{3} \approx 9.97 \mathring{A} \approx 10 \mathring{A}$.
By Einstein's photoelectric equation, the energy of the incident photon is $E = W + K_{max}$.
Stopping potential is 10 V, so max kinetic energy is $K_{max} = 10 \text{ eV}$.
Energy of incident photon $E = 2.75 \text{ eV} + 10 \text{ eV} = 12.75 \text{ eV}$.
This photon is emitted when an electron in hydrogen jumps from state $n$ to ground state ($n=1$).
Energy difference $\Delta E = 13.6 \left(1 - \frac{1}{n^2}\right) \text{ eV}$.
$13.6 \left(1 - \frac{1}{n^2}\right) = 12.75 \Rightarrow 1 - \frac{1}{n^2} = \frac{12.75}{13.6} = 0.9375$.
$\frac{1}{n^2} = 1 - 0.9375 = 0.0625 = \frac{1}{16} \Rightarrow n^2 = 16 \Rightarrow n = 4$.
Position coordinates: $x = \alpha t^3$ and $y = \beta t^3$.
Velocity components are the time derivatives of the coordinates:
$v_x = \frac{dx}{dt} = 3\alpha t^2$
$v_y = \frac{dy}{dt} = 3\beta t^2$
The resultant speed of the particle is the magnitude of the velocity vector:
$v = \sqrt{v_x^2 + v_y^2} = \sqrt{(3\alpha t^2)^2 + (3\beta t^2)^2} = \sqrt{9\alpha^2 t^4 + 9\beta^2 t^4} = \sqrt{9t^4(\alpha^2 + \beta^2)}$.
$v = 3t^2 \sqrt{\alpha^2 + \beta^2}$.