31 Jan 2024 Shift 1

We need to find the percentage error in measuring the resistance of a wire, given percentage errors in length and diameter measurements. The resistance of a wire is given by $R = \frac{\rho L}{A}$, where $\rho$ is the resistivity, $L$ is the length, and $A$ is the cross-sectional area. For a circular wire with diameter $d$, $A = \frac{\pi d^2}{4}$. Therefore, $R = \frac{\rho L}{\frac{\pi d^2}{4}} = \frac{4\rho L}{\pi d^2}$. To determine the propagation of error for $R = \frac{4\rho L}{\pi d^2}$ (with $\rho$ treated as a constant), we take the natural logarithm: $\ln R = \ln(4\rho/\pi) + \ln L - 2\ln d$. Differentiating yields $\frac{\Delta R}{R} = \frac{\Delta L}{L} + 2\frac{\Delta d}{d}$, where absolute values are taken to represent maximum error. Given that $\frac{\Delta L}{L}\times100 = 0.1\%$ and $\frac{\Delta d}{d}\times100 = 0.1\%$, it follows that $\frac{\Delta R}{R}\times100 = 0.1\% + 2\times0.1\% = 0.3\%$. The percentage error in measuring the resistance is 0.3%. Hence the correct answer is Option (2): 0.3%.

$F = ax^2 + bt^{1/2}$. $[F] = MLT^{-2}$. $[a] = [F]/\$[x^2]\$ = MLT^{-2}/L^2 = ML^{-1}T^{-2}$. $[b] = [F]/\$[t^{1/2}]\$ = MLT^{-2}/T^{1/2} = MLT^{-5/2}$. $[b^2/a] = M^2L^2T^{-5}/(ML^{-1}T^{-2}) = ML^3T^{-3}$. The answer is Option (1): $[ML^3T^{-3}]$.

The time-distance relation is given as $t = \alpha x^{2} + \beta x$, where $\alpha$ and $\beta$ are constants. First differentiate with respect to $x$: $\frac{dt}{dx} = 2\alpha x + \beta$ $-(1)$ Velocity is $v = \frac{dx}{dt} = \frac{1}{\dfrac{dt}{dx}} = \frac{1}{2\alpha x + \beta}$ $-(2)$ Acceleration is $a = \frac{dv}{dt}$. Using the chain rule, $a = \frac{dv}{dx}\,\frac{dx}{dt} = v \,\frac{dv}{dx}$ $-(3)$ Differentiate $v$ from $(2)$ with respect to $x$: $\frac{dv}{dx} = \frac{d}{dx}\left[(2\alpha x + \beta)^{-1}\right]\$ = -1 \cdot (2\alpha)\,(2\alpha x + \beta)^{-2} = -\frac{2\alpha}{(2\alpha x + \beta)^{2}}$ $-(4)$ Substitute $(2)$ and $(4)$ into $(3)$: $a = v\left(-\frac{2\alpha}{(2\alpha x + \beta)^{2}}\right) = -\frac{2\alpha\,v}{(2\alpha x + \beta)^{2}}$ $-(5)$ From $(2)$, $v = \dfrac{1}{2\alpha x + \beta} \;\Rightarrow\; 2\alpha x + \beta = \frac{1}{v}$ $-(6)$ Insert $(6)$ into $(5)$ to eliminate $x$: $a = -\frac{2\alpha\,v}{\left(\dfrac{1}{v}\right)^{2}} = -2\alpha\,v \, v^{2} = -2\alpha\,v^{3}$ Thus the required relation is $a = -2\alpha v^{3}$. Option A is correct.

According to Newton's second law, the net force on the ball is: $ma = mg - F_b - 6\pi\eta rv$ Initially, when the ball is dropped ($v=0$), the acceleration is at its maximum. As the velocity increases, the upward viscous drag also increases, which reduces the net force and causes the acceleration ($a = \frac{dv}{dt}$) to decrease over time. Eventually, the upward forces balance the weight, and the acceleration becomes zero. At this point, the ball moves with a constant terminal velocity. Among the given options, graph (B) accurately represents this behavior.

Now lets write the equations for both the block So for the 10 Kg block $Mg\sin\theta\ _1-\mu\ _sMg\cos\theta\ _1-T=Ma$ $Mg\sin\ 53^{\circ\ }-0.25Mg\cos53^{\circ\ }-T=Ma$ $100\times \ \frac{4}{5}-0.25\times \ 100\times \ \frac{3}{5}-T=20$ $80-T=15$ $T=45$ Now for the block of mass m $T-mg\sin\theta\ _2-\mu\ _smg\cos\ \theta\ _2=ma$ $T-mg\sin37^{\circ\ }-0.25mg\cos37^{\circ\ }=m\left(2\right)$ $45-mg\times \ \frac{3}{5}-0.25mg\times \ \frac{4}{5}\ =m\times \ 2$ $mg\times \ \frac{3}{5}+0.25mg\times \ \frac{4}{5}\ +m\times \ 2=45$ $6m+2m+2m=45$ $10m=45$ $m=4.5\ Kg$

A coin is placed on a rotating disc at a distance $r$ from the center. The coefficient of friction between the coin and the disc is $\mu$. We need to find the maximum angular velocity so the coin does not slip. For the coin to move \in a circle on the disc, it needs a centripetal force directed toward the center. The only horizontal force acting on the coin is the static friction between the coin and the disc. The centripetal force required for circular motion is $F_{centripetal} = m\omega^2 r$, where $m$ is the mass of the coin and $\omega$ is the angular velocity. The maximum static friction force available is $f_{max} = \mu N = \mu mg$, where $N = mg$ is the normal force (since the disc is horizontal). For the coin not to slip, the required centripetal force must not exceed the maximum friction, which gives $m\omega^2 r \leq \mu mg$. At the maximum angular velocity, the two forces are equal: $m\omega_{max}^2 r = \mu mg$. The mass $m$ cancels from both sides, yielding $\omega_{max}^2 = \dfrac{\mu g}{r}$ and hence $\omega_{max} = \sqrt{\dfrac{\mu g}{r}}$. The correct answer is Option (3): $\sqrt{\dfrac{\mu g}{r}}$ .

An artillery piece of mass $M_1$ fires a shell of mass $M_2$ horizontally, and we seek the ratio of the kinetic energy of the artillery to that of the shell immediately after firing. Before firing, both the artillery and the shell are at rest, so the total momentum is zero. By conservation of momentum, the total momentum after firing must also be zero: $M_1 v_1 + M_2 v_2 = 0$ where $v_1$ is the recoil velocity of the artillery and $v_2$ is the velocity of the shell. It follows that $M_1 v_1 = -M_2 v_2$ and taking magnitudes gives $|v_1| = \frac{M_2 |v_2|}{M_1}$. The kinetic energy of the artillery is $KE_1 = \frac{1}{2}M_1 v_1^2 \quad (\text{artillery})$ and that of the shell is $KE_2 = \frac{1}{2}M_2 v_2^2 \quad (\text{shell})$. Their ratio is $\frac{KE_1}{KE_2} = \frac{\frac{1}{2}M_1 v_1^2}{\frac{1}{2}M_2 v_2^2} = \frac{M_1}{M_2} \cdot \frac{v_1^2}{v_2^2}$. Substituting $v_1 = \frac{M_2 v_2}{M_1}$ gives $\frac{KE_1}{KE_2} = \frac{M_1}{M_2} \cdot \frac{M_2^2 v_2^2}{M_1^2 v_2^2} = \frac{M_1}{M_2} \cdot \frac{M_2^2}{M_1^2} = \frac{M_2}{M_1}$. An alternative derivation notes that both objects have the same magnitude of momentum ($p = M_1|v_1| = M_2|v_2|$), and since $KE = \frac{p^2}{2M}$ one obtains $\frac{KE_1}{KE_2} = \frac{p^2/(2M_1)}{p^2/(2M_2)} = \frac{M_2}{M_1}$. The correct answer is Option (2): $\frac{M_2}{M_1}$ .

Four identical particles of mass $m$ are at the corners of a square of side $a$. The net gravitational force on one particle due to the other three is given as $\frac{(2\sqrt{2}+1)}{32}\frac{Gm^2}{L^2}$. We need to find $a$ \in terms of $L$. Consider the particle at one corner. The other three particles are two adjacent particles at distance $a$ (along the sides) and one diagonal particle at distance $a\sqrt{2}$. Each adjacent particle exerts a force: $F_{adj} = \frac{Gm^2}{a^2}$. The two adjacent forces are perpendicular to each other along the two sides of the square, so by the Pythagorean theorem the resultant is $F_{resultant,adj} = \sqrt{F_{adj}^2 + F_{adj}^2} = F_{adj}\sqrt{2} = \frac{\sqrt{2}\cdot Gm^2}{a^2}$. This resultant points along the diagonal toward the diagonally opposite corner. The diagonal particle exerts a force $F_{diag} = \frac{Gm^2}{(a\sqrt{2})^2} = \frac{Gm^2}{2a^2}$, which also points along the same diagonal direction. Since all forces are along the diagonal, the total net force is $F_{net} = \frac{\sqrt{2}\cdot Gm^2}{a^2} + \frac{Gm^2}{2a^2} = \frac{Gm^2}{a^2}\left(\sqrt{2} + \tfrac{1}{2}\right) = \frac{(2\sqrt{2}+1)}{2}\cdot\frac{Gm^2}{a^2}$. Equating this to the given expression $\frac{(2\sqrt{2}+1)}{2}\cdot\frac{Gm^2}{a^2} = \frac{(2\sqrt{2}+1)}{32}\cdot\frac{Gm^2}{L^2}$, the common factor $(2\sqrt{2}+1)\,Gm^2$ cancels, giving $\frac{1}{2a^2} = \frac{1}{32L^2}$, and hence $2a^2 = 32L^2$, $a^2 = 16L^2$, $a = 4L$. The correct answer is Option (2): $4L$ .

Two conductors have the same resistance $R_0$ at $0°C$ but different temperature coefficients $\alpha_1$ and $\alpha_2$. The resistance at temperature $T$ can be written as $R = R_0(1 + \alpha T)$, so for these conductors: $R_1 = R_0(1 + \alpha_1 T), \quad R_2 = R_0(1 + \alpha_2 T)$. For the series combination the total resistance is $R_s = R_1 + R_2 = R_0(1 + \alpha_1 T) + R_0(1 + \alpha_2 T) = 2R_0\left(1 + \frac{\alpha_1 + \alpha_2}{2}T\right)$, so the equivalent resistance at $0°C$ is $R_{s,0} = 2R_0$ and the effective temperature coefficient becomes $\alpha_s = \frac{\alpha_1 + \alpha_2}{2}$. In the parallel combination one has $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{R_0(1+\alpha_1 T)} + \frac{1}{R_0(1+\alpha_2 T)}$. Using the linear approximation $(1 + \alpha T)^{-1} \approx (1 - \alpha T)$ for small $\alpha T$ gives $\frac{1}{R_p} \approx \frac{1}{R_0}(1 - \alpha_1 T) + \frac{1}{R_0}(1 - \alpha_2 T) = \frac{2}{R_0}\left(1 - \frac{\alpha_1 + \alpha_2}{2}T\right)$ and hence $R_p \approx \frac{R_0}{2} \cdot \frac{1}{1 - \frac{\alpha_1+\alpha_2}{2}T} \approx \frac{R_0}{2}\left(1 + \frac{\alpha_1 + \alpha_2}{2}T\right)$, so the equivalent resistance at $0°C$ is $R_{p,0} = \frac{R_0}{2}$ and the effective temperature coefficient is $\alpha_p = \frac{\alpha_1 + \alpha_2}{2}$. Key insight: When both conductors have the same initial resistance $R_0$, the effective temperature coefficient is $\frac{\alpha_1 + \alpha_2}{2}$ for both series and parallel combinations. The correct answer is Option (2): $\frac{\alpha_1+\alpha_2}{2}, \frac{\alpha_1+\alpha_2}{2}$ .

For an ideal gas, $PV=nRT$ For an isobaric process (P= constant), $V=\frac{nR}{P}T$ This is a straight line on a V-T graph, with slope $slope=\frac{nR}{P}$ Since same mass of gas is used, n is constant, so slope is inversely proportional to pressure: $\text{slope}\propto\frac{1}{P}$ Line $P_2$​ is steeper than line $P_1$, so $\frac{nR}{P_2}>\frac{nR}{P_1}$ which gives $P_1>P_2$

We need to identify which parameter remains the same for molecules of all gases at a given temperature. According to the kinetic theory of gases, the average kinetic energy of a gas molecule depends only on the absolute temperature: $\overline{KE} = \frac{3}{2}k_B T$. This expression is independent of the mass or nature of the gas. The average speed, given by $\bar{v} = \sqrt{\frac{8k_BT}{\pi m}}$, depends on the molecular mass $m$, so different gases have different speeds at the same temperature. The average momentum, calculated as $m\bar{v} = m\sqrt{\frac{8k_BT}{\pi m}} = \sqrt{\frac{8mk_BT}{\pi}}$, also depends on $m$ and therefore varies between gases. Mass itself obviously differs for different gas molecules. In contrast, the average kinetic energy, given by $\frac{3}{2}k_BT$, depends only on temperature and is thus the same for all gases at a given temperature, regardless of their mass or chemical identity. The correct answer is Option (1): kinetic energy.

Fundamental of closed pipe: $f_c = \frac{v}{4L_c}$. First overtone of open pipe: $f_o = \frac{2v}{2L_o} = \frac{v}{L_o}$. $\frac{v}{4L_c} = \frac{v}{L_o} \Rightarrow L_c = \frac{L_o}{4} = \frac{60}{4} = 15$ cm. The answer is Option (4): 15 cm.

Two charges $q$ and $3q$ are separated by a distance $r$. We need to find the point between them where the resultant electric field is zero. Let the charge $q$ be at position O and the charge $3q$ at position $r$; let the point where the electric field vanishes be at a distance $x$ from $q$ (and hence at distance $(r - x)$ from $3q$). Since both charges are positive, the zero-field point lies between them where their fields are oppositely directed. At distance $x$ from $q$, its field points away from $q$ (toward $3q$), while the field due to $3q$ at distance $(r-x)$ points away from $3q$ (toward $q$). Setting their magnitudes equal gives: $\frac{kq}{x^2} = \frac{k(3q)}{(r-x)^2}$ Canceling $kq$ yields $\frac{1}{x^2} = \frac{3}{(r-x)^2}$. Cross-multiplying gives $(r-x)^2 = 3x^2$, and taking the positive root (since $x>0$ and $r-x>0$) leads to $r - x = \sqrt{3}\,x$, hence $r = x(1 + \sqrt{3})$ and $x = \frac{r}{1 + \sqrt{3}}$ Numerically, with $\sqrt{3}\approx1.732$, this yields $x\approx\frac{r}{2.732}\approx0.366r$, confirming the point is between the charges and closer to the smaller charge $q$, as expected. The correct answer is Option (3): $\frac{r}{1 + \sqrt{3}}$

The total magnetic force on the wire is the vector sum of the forces on its segments. The forces on the two straight sections within the field are equal in magnitude and opposite in direction, so they cancel each other. The net force is solely the force on the semicircular portion. This force is equivalent to the force on a straight conductor of length $2R$ connecting the endpoints of the semicircle. This effective length vector is $\vec{L}_{eff} = 2R\hat{i}$.

From the diagram, the magnetic field is perpendicular to the plane of the wire and pointing outwards, which corresponds to the +z direction. Thus, $\vec{B} = B\hat{k}$. Note that the textual description $\vec{B}=B_0 \hat{j}$ contradicts the diagram and the options; we proceed based on the diagram.

The magnetic force $\vec{F}$ is given by the formula $\vec{F} = i(\vec{L}_{eff} \times \vec{B})$.

Using the vector cross product rule $\hat{i} \times \hat{k} = -\hat{j}$, we get:

$|emf| = N\frac{d\Phi}{dt} = N \cdot A \cdot \frac{\Delta B}{\Delta t}$. $A = \pi(0.01)^2 = \pi \times 10^{-4}$. $\Delta B = 5000-3000 = 2000$ T. $\Delta t = 2$ s. $22 = N \times \pi \times 10^{-4} \times 1000 = N \times 0.1\pi$. $N = \frac{22}{0.1\pi} = \frac{220}{\pi} \approx 70$. The answer is Option (2): 70.

The total average energy density, $<u>$, of an electromagnetic wave is the \sum of the average energy densities of the electric field ($<u_E>$) and the magnetic field ($<u_B>$). The energy is shared equally, so $<u> = <u_E> + <u_B> = 2<u_E>$. The average electric energy density is $<u_E> = \frac{1}{4}\varepsilon_0 E_0^2$, where $E_0$ is the amplitude of the electric field.

Therefore, the total average energy density is given by the formula:
$<u> = \frac{1}{2} \varepsilon_0 E_0^2$
We are given the electric field amplitude $E_0 = 50 \text{ V m}^{-1}$ and the permittivity of free space $\varepsilon_0 = 8.85 \times 10^{-12} \text{ C}^2 \text{N}^{-1}\text{m}^{-2}$. Note that the frequency is not needed for this calculation.

Substituting the given values into the formula:
$<u> = \frac{1}{2} \times (8.85 \times 10^{-12}) \times (50)^2$
$<u> = \frac{1}{2} \times 8.85 \times 10^{-12} \times 2500 = 1.10625 \times 10^{-8} \text{ J m}^{-3}$
This value is approximately $1.106 \times 10^{-8} \text{ J m}^{-3}$.

A prism has apex angle $A$ and refractive index $\mu = \cot\frac{A}{2}$. We need to find the angle of minimum deviation $\delta_m$. We begin by recalling the prism formula for minimum deviation. For a prism, the refractive index is related to the apex angle and minimum deviation by: $\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$ Next, we substitute $\mu = \cot\frac{A}{2}$ into this formula: $\cot\frac{A}{2} = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$ Since $\cot\frac{A}{2} = \frac{\cos\frac{A}{2}}{\sin\frac{A}{2}}$, we can write: $\frac{\cos\frac{A}{2}}{\sin\frac{A}{2}} = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$ Multiplying both sides by $\sin\frac{A}{2}$ yields: $\cos\frac{A}{2} = \sin\left(\frac{A + \delta_m}{2}\right)$ Then, using the identity $\cos\theta = \sin\left(90° - \theta\right)$, we have: $\sin\left(90° - \frac{A}{2}\right) = \sin\left(\frac{A + \delta_m}{2}\right)$ Equating the arguments leads to: $90° - \frac{A}{2} = \frac{A + \delta_m}{2}$ Multiplying both sides by 2 gives: $180° - A = A + \delta_m$ Solving for $\delta_m$, we find: $\delta_m = 180° - 2A$ The correct answer is Option (4): $\delta_m = 180° - 2A$ .

Photoelectric effect: $eV_s = \frac{hc}{\lambda} - \phi$. $8e = \frac{hc}{\lambda} - \phi$ ... (1). $2e = \frac{hc}{3\lambda} - \phi$ ... (2). From (1)-(2): $6e = \frac{hc}{\lambda}(1-1/3) = \frac{2hc}{3\lambda} \Rightarrow \frac{hc}{\lambda} = 9e$. From (1): $\phi = 9e - 8e = e$. Threshold: $\frac{hc}{\lambda_0} = \phi = e$. $\lambda_0 = \frac{hc}{e} = 9\lambda \cdot \frac{e}{e} = 9\lambda$. The answer is Option (3): $9\lambda$.

Lyman series: $\frac{1}{\lambda} = R(1 - 1/n^2)$. First member (n=2): $\frac{1}{\lambda} = R(1-1/4) = \frac{3R}{4}$. Second member (n=3): $\frac{1}{\lambda_2} = R(1-1/9) = \frac{8R}{9}$. $\frac{\lambda_2}{\lambda} = \frac{3R/4}{8R/9} = \frac{27}{32}$. $\lambda_2 = \frac{27}{32}\lambda$. The answer is Option (1): $\frac{27}{32}\lambda$.

The inputs A and B are first processed by two individual NOT gates, yielding the intermediate outputs $\bar{A}$ and $\bar{B}$, respectively. These signals then serve as the inputs for a two-input NAND gate. The Boolean expression for the final output Y is the NAND of these intermediate signals:
$Y = \overline{\bar{A} \cdot \bar{B}}$
According to De Morgan's theorem, $\overline{X \cdot Y} = \bar{X} + \bar{Y}$. Applying this theorem to the expression gives:
$Y = \overline{(\bar{A})} + \overline{(\bar{B})}$
Using the law of double complementation, which states that $\overline{\bar{X}} = X$, the expression simplifies to:
$Y = A + B$
This final expression, $Y = A + B$, represents the logical OR operation.

A body falls freely from height $H$, hits an inclined plane at height $h$, and after a perfectly elastic impact, its velocity becomes horizontal. We need to find $\frac{H}{h}$ for which the total time to reach the ground is maximized. We start by finding the velocity at the inclined plane. The body falls from height $H$ to height $h$, so it falls through a distance $(H - h)$. Using energy conservation (or kinematics): $v = \sqrt{2g(H - h)}$ Next, we analyze the motion after the perfectly elastic impact. After the perfectly elastic impact, the direction of velocity changes to horizontal (magnitude remains the same). From height $h$ above the ground, the body undergoes projectile motion with: - Initial horizontal velocity: $v = \sqrt{2g(H-h)}$ - Initial vertical velocity: 0 (horizontal direction) Then, we calculate the total time. Time for free fall from $H$ to $h$ (falling distance $H - h$): $t_1 = \sqrt{\frac{2(H-h)}{g}}$ Time for projectile fall from height $h$ to ground (vertical free fall from rest): $t_2 = \sqrt{\frac{2h}{g}}$ The total time is: $T = t_1 + t_2 = \sqrt{\frac{2(H-h)}{g}} + \sqrt{\frac{2h}{g}} = \sqrt{\frac{2}{g}}\left(\sqrt{H-h} + \sqrt{h}\right)$ To maximize the total time with respect to $h$, we consider: We need to maximize $f(h) = \sqrt{H - h} + \sqrt{h}$ for $0 \leq h \leq H$. Taking the derivative and setting it to zero: $f'(h) = \frac{-1}{2\sqrt{H-h}} + \frac{1}{2\sqrt{h}} = 0$ $\frac{1}{2\sqrt{h}} = \frac{1}{2\sqrt{H-h}}$ $\sqrt{h} = \sqrt{H - h}$ $h = H - h$ $2h = H$ $h = \frac{H}{2}$ Finally, we find the ratio: $\frac{H}{h} = \frac{H}{H/2} = 2$ The answer is $\boxed{2}$.

A solid circular disc of mass $m = 50$ kg rolls along a horizontal floor with center-of-mass speed $v = 0.4$ m/s. We need to find the work done to stop it. We start by recalling that for a body rolling without slipping, the total kinetic energy is the \sum of translational and rotational kinetic energies: $KE_{total} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$ Here, $I$ is the moment of inertia about the center of mass and $\omega$ is the angular velocity. Next, the rolling condition gives $v = R\omega$, so $\omega = \frac{v}{R}$, and for a solid disc, $I = \frac{1}{2}mR^2$. Substituting into the rotational kinetic energy term yields: $\frac{1}{2}I\omega^2 = \frac{1}{2} \cdot \frac{1}{2}mR^2 \cdot \frac{v^2}{R^2} = \frac{1}{4}mv^2$ Therefore, the total kinetic energy becomes: $KE_{total} = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$ Substituting the numerical values gives: $KE_{total} = \frac{3}{4} \times 50 \times (0.4)^2 = \frac{3}{4} \times 50 \times 0.16 = \frac{3}{4} \times 8 = 6 \text{ J}$ By the work-energy theorem, the work done to stop the disc equals the total kinetic energy (\in magnitude). The answer is $\boxed{6}$ J.

$\Delta P = \rho g h$, $\frac{\Delta V}{V} = \frac{\Delta P}{B}$. $0.02\% = 0.0002 = \frac{\rho g h}{B} = \frac{10^3 \times 10 \times h}{9 \times 10^8}$. $h = \frac{0.0002 \times 9 \times 10^8}{10^4} = \frac{1.8 \times 10^5}{10^4} = 18$ m. The answer is $\boxed{18}$.

A particle in SHM with amplitude $A$ has its speed tripled at the instant when its displacement is $\frac{2A}{3}$. We need to find the new amplitude $A' = \frac{nA}{3}$. We start by recalling the velocity-displacement relation \in SHM. For a particle \in SHM with angular frequency $\omega$ and amplitude $A$: $v^2 = \omega^2(A^2 - x^2)$ At the instant $x = \frac{2A}{3}$, the original speed is found as follows: $v^2 = \omega^2\left(A^2 - \frac{4A^2}{9}\right) = \omega^2 \cdot \frac{9A^2 - 4A^2}{9} = \frac{5\omega^2 A^2}{9}$ Next, when the speed is tripled to $3v$ while the displacement remains $x = \frac{2A}{3}$ and the angular frequency $\omega$ is unchanged, the new amplitude $A'$ satisfies: $(3v)^2 = \omega^2(A'^2 - x^2)$ $9v^2 = \omega^2\left(A'^2 - \frac{4A^2}{9}\right)$ Substituting the value of $v^2$ gives: $9 \times \frac{5\omega^2 A^2}{9} = \omega^2\left(A'^2 - \frac{4A^2}{9}\right)$ $5\omega^2 A^2 = \omega^2\left(A'^2 - \frac{4A^2}{9}\right)$ Dividing both sides by $\omega^2$ leads to: $5A^2 = A'^2 - \frac{4A^2}{9}$ $A'^2 = 5A^2 + \frac{4A^2}{9} = \frac{45A^2 + 4A^2}{9} = \frac{49A^2}{9}$ Finally, solving for $A'$ and determining $n$ gives: $A' = \sqrt{\frac{49A^2}{9}} = \frac{7A}{3}$ Since $A' = \frac{nA}{3}$, it follows that $n = 7$. The answer is $\boxed{7}$.