We are told that the kinetic energy of a simple harmonic oscillator oscillates with an angular frequency of 176 rad/s and we need to find the frequency of the SHM oscillator itself.

For a particle executing SHM with displacement $x = A\sin(\omega_0 t)$, where $\omega_0$ is the angular frequency of the SHM and $A$ is the amplitude.

Differentiating with respect to time gives the velocity: $v = A\omega_0\cos(\omega_0 t)$

This leads to the kinetic energy: $KE = \tfrac{1}{2}mv^2 = \tfrac{1}{2}mA^2\omega_0^2\cos^2(\omega_0 t)$

Using the identity $\cos^2(\theta) = \tfrac{1 + \cos(2\theta)}{2}$, we obtain:

$KE = \tfrac{1}{2}mA^2\omega_0^2 \times \tfrac{1 + \cos(2\omega_0 t)}{2} = \tfrac{mA^2\omega_0^2}{4}\left(1 + \cos(2\omega_0 t)\right)$

It follows that the kinetic energy has a constant part and an oscillating part $\cos(2\omega_0 t)$, so its angular frequency is $2\omega_0$, which is twice the angular frequency of the SHM.

Since the KE oscillates at $\omega_{KE} = 176$ rad/s, we set:

$2\omega_0 = 176$

$\omega_0 = \frac{176}{2} = 88 \text{ rad/s}$

Using $\omega_0 = 2\pi f$, we have:

$f = \frac{\omega_0}{2\pi}$

Substituting $\pi = \frac{22}{7}$ yields:

$f = \frac{88}{2 \times \frac{22}{7}} = \frac{88}{\frac{44}{7}} = \frac{88 \times 7}{44} = \frac{616}{44} = 14 \text{ Hz}$

The frequency of the simple harmonic oscillator is 14 Hz.

The correct answer is Option (4): 14.

Two identical metallic spheres each of radius $R$, charge $Q$, and mass $m$ are initially separated by a center-to-center distance of $4R$ and are given an initial speed $u$ towards each other; we seek the minimum value of $u$ for them to just touch.

We denote the initial separation as $d_i = 4R$ and the final separation upon just touching as $d_f = 2R$.

The interaction between the spheres involves electrostatic repulsion with potential energy $U_e = \frac{kQ^2}{d}$ and gravitational attraction with potential energy $U_g = -\frac{Gm^2}{d}$, so that the total potential energy at separation $d$ is $U(d) = \frac{kQ^2}{d} - \frac{Gm^2}{d}.$

Since the spheres have equal mass and equal and opposite velocities, they momentarily come to rest at the point of touching, making the final kinetic energy zero. By conservation of energy, $\text{Initial KE} + \text{Initial PE} = \text{Final KE} + \text{Final PE}.$

The initial kinetic energy is $2\times \frac12 mu^2 = mu^2$ and the initial potential energy at $d_i=4R$ is $\frac{kQ^2}{4R} - \frac{Gm^2}{4R}$. The final potential energy at $d_f=2R$ is $\frac{kQ^2}{2R} - \frac{Gm^2}{2R}$, and the final kinetic energy is zero. Hence, $mu^2 + \frac{kQ^2}{4R} - \frac{Gm^2}{4R} = \frac{kQ^2}{2R} - \frac{Gm^2}{2R}.$

Rearranging to isolate $mu^2$ gives $mu^2 = \frac{kQ^2}{2R} - \frac{Gm^2}{2R} - \frac{kQ^2}{4R} + \frac{Gm^2}{4R} = kQ^2\Bigl(\frac{1}{2R}-\frac{1}{4R}\Bigr) - Gm^2\Bigl(\frac{1}{2R}-\frac{1}{4R}\Bigr) = \frac{1}{4R}\bigl(kQ^2 - Gm^2\bigr).$ Dividing by $m$ yields $u^2 = \frac{kQ^2 - Gm^2}{4mR} = \frac{kQ^2}{4mR}\Bigl(1 - \frac{Gm^2}{kQ^2}\Bigr),$ and thus $u = \sqrt{\frac{kQ^2}{4mR}\Bigl(1 - \frac{Gm^2}{kQ^2}\Bigr)}.$

The correct answer is Option (4): $\sqrt{\frac{kQ^{2}}{4mR}\left(1-\frac{Gm^{2}}{kQ^{2}}\right)}$.

We analyze both statements about Young's double slit experiment.

Statement I: The angular separation of fringes will increase as the screen is moved away from the plane of the slits.

The angular fringe separation is given by $\Delta\theta = \frac{\lambda}{d}$, where $\lambda$ is the wavelength and $d$ is the slit separation.

This expression is independent of the distance $D$ between the slits and the screen. Therefore, moving the screen away does not change the angular separation.

Statement I is false.

Statement II: The angular separation of fringes will increase when the monochromatic source is replaced by another monochromatic source of higher wavelength.

Since $\Delta\theta = \frac{\lambda}{d}$, increasing $\lambda$ increases $\Delta\theta$.

Statement II is true.

Therefore, Statement I is false but Statement II is true.

The correct answer is Option B.

We begin with a battery with EMF $E$ and internal resistance $r$ connected across an external resistance $R$, and we need to find the value of $R$ for which the power consumed in $R$ is maximum.

Applying Ohm's law to the complete circuit, the current is:

$I = \frac{E}{R + r}$

The power consumed in the external resistance $R$ is:

$P = I^2 R = \left(\frac{E}{R + r}\right)^2 R = \frac{E^2 R}{(R + r)^2}$

To find its maximum, we differentiate with respect to $R$ using the quotient rule, with numerator $E^2 R$ and denominator $(R + r)^2$:

$\frac{dP}{dR} = E^2 \cdot \frac{(R + r)^2 \cdot 1 - R \cdot 2(R + r)}{(R + r)^4}$

This simplifies to:

$= E^2 \cdot \frac{(R + r) - 2R}{(R + r)^3} = E^2 \cdot \frac{r - R}{(R + r)^3}$

Setting $\frac{dP}{dR} = 0$ gives:

$r - R = 0$

$R = r$

For $R < r$, $\frac{dP}{dR} > 0$ (power is increasing)

For $R > r$, $\frac{dP}{dR} < 0$ (power is decreasing)

Since the derivative changes from positive to negative at $R = r$, this confirms it is a maximum.

This result is known as the Maximum Power Transfer Theorem: maximum power is delivered to the external load when the load resistance equals the internal resistance of the source.

The correct answer is Option (1): $R = r$.

In second branch capacitor acts as an open circuit when in steady state.

The upper diode (in the 3 Ω branch) points left. Its cathode is at a higher
potential than its anode, meaning it is reverse-biased and acts as an
open circuit

The lower diode (in the 4 Ω branch) points right. Its anode is at a higher
potential, meaning it is forward-biased and acts as a short circuit (assuming an ideal diode)

$R_{\text{eq}} = 1\,\Omega + 4\,\Omega = 5\,\Omega$

$I = \frac{V}{R_{\text{eq}}} = \frac{2.5}{5} = 0.5\text{ A}$

The voltage across the capacitor, $V_{C}$ , is exactly equal to the potential difference between the left and
right common nodes.

$V_C = I \times 4\,\Omega$
$V_C = 0.5 \times 4 = 2\text{ V}$

$Q = C \times V_C$

$Q = 5\,\mu\text{F} \times 2\text{ V}$
$Q = 10\,\mu\text{C}$

Energy of electron in orbit: $E_n = \frac{E_0}{n^2}$ (where $E_0$ is ground state energy, negative).

Given: $E_n = -0.04 E_0$, where $E_0$ is the magnitude of ground state energy.

So $\frac{E_0}{n^2} = 0.04 E_0$ (in terms of magnitudes).

$n^2 = \frac{1}{0.04} = 25$

$n = 5$

Angular momentum: $L = \frac{nh}{2\pi}$

$\frac{2\pi L}{h} = n = 5$

The answer is Option 3: 5.

We are given that the river width is 200 m, the river speed is 18 km/h (west to east), and the boat speed in still water is 36 km/h. We need to find the minimum round-trip time and the displacement along the river bank.

First, we convert the speeds to meters per second: the river speed is $v_r = 18 \text{ km/h} = 18 \times \frac{5}{18} = 5$ m/s, and the boat speed in still water is $v_b = 36 \text{ km/h} = 36 \times \frac{5}{18} = 10$ m/s.

For the minimum crossing time, the boat should head straight across (perpendicular to the banks), so the component of its velocity perpendicular to the banks is the full 10 m/s.

The time for one crossing is then $t_1 = \frac{\text{width}}{v_{\perp}} = \frac{200}{10} = 20$ s, and the round-trip time is $T = 2 \times 20 = 40$ s.

Next, during each crossing the river current carries the boat downstream by an amount $\text{drift} = v_r \times t_1 = 5 \times 20 = 100$ m. On the return trip the situation is the same, so the total drift is $\text{total drift} = 100 + 100 = 200$ m in the direction of the river flow.

Thus, the minimum time is 40 s and the displacement along the river bank is 200 m, corresponding to Option (3): 40 s and 200 m.

Adding 52.01 m + 153.2 m + 0.123 m with significant figures.

In addition, the result should have the same number of decimal places as the quantity with the least decimal places.

52.01 has 2 decimal places, 153.2 has 1 decimal place, 0.123 has 3 decimal places.

Least decimal places = 1 (from 153.2).

Sum = 52.01 + 153.2 + 0.123 = 205.333

Rounded to 1 decimal place: 205.3 m.

The answer is Option 4: 205.3 m.

Split the question into two parts: $\vec{B}$ at point O by infinitely long wire and by circular loop.

 for straight wire :

$\vec{B}_1 = \frac{\mu_0 I}{2\pi r} \hat{i}$

By the Right-Hand Rule,the magnetic field at O points outward, in the +x direction ($\hat{i}$)

for circuar loop :$\vec{B}_2 = \frac{\mu_0 I}{2r} (-\hat{i}) = -\frac{\mu_0 I}{2r} \hat{i}$

According to the Right-Hand Rule for circular loops, the magnetic field at the centre points
inward, into the page, which is the −x direction ($-\hat{i}$)

The total magnetic field at the centre $O$ is the vector sum of $\vec{B}_1$ and $\vec{B}_2$:
$\vec{B}_{\text{net}} = \vec{B}_1 + \vec{B}_2$
$\vec{B}_{\text{net}} = \frac{\mu_0 I}{2\pi r} \hat{i} - \frac{\mu_0 I}{2r} \hat{i}$

Taking $\frac{\mu_0 I}{2r}$ as a common factor:
$\vec{B}_{\text{net}} = \frac{\mu_0 I}{2r} \left( \frac{1}{\pi} - 1 \right) \hat{i}$

We need to find the temperature at which the r.m.s. speed of hydrogen molecules equals the r.m.s. speed of oxygen molecules at $47°C$.

The root mean square (r.m.s.) speed of gas molecules is given by:

$v_{rms} = \sqrt{\frac{3RT}{M}}$

where $R$ is the universal gas constant, $T$ is the absolute temperature in Kelvin, and $M$ is the molar mass of the gas.

Converting the temperature of oxygen to Kelvin gives

$T_{O_2} = 47 + 273 = 320 \text{ K}$.

Setting the r.m.s. speeds equal, we have

$\sqrt{\frac{3R \times 320}{M_{O_2}}} = \sqrt{\frac{3R \times T_{H_2}}{M_{H_2}}}$. Squaring both sides (the $3R$ cancels):

$\frac{320}{M_{O_2}} = \frac{T_{H_2}}{M_{H_2}}$.

Given that $\frac{M_{O_2}}{M_{H_2}} = \frac{32}{2} = 16$, so $M_{O_2} = 32$ and $M_{H_2} = 2$:

$\frac{320}{32} = \frac{T_{H_2}}{2}$. Hence

$10 = \frac{T_{H_2}}{2} \quad\Longrightarrow\quad T_{H_2} = 20 \text{ K}$.

Converting back to Celsius,

$T_{H_2} = 20 - 273 = -253°C$.

The r.m.s. speed of oxygen at $47°C$ equals the r.m.s. speed of hydrogen at $-253°C$.

The correct answer is Option (3): $-253$.

$x(t) = t^2 + t + 1$ m, mass = 2 kg. Find work done from $t = 2$ to $t = 3$.

$v(t) = \frac{dx}{dt} = 2t + 1$ m/s

$v(2) = 5$ m/s, $v(3) = 7$ m/s

Work = Change in KE = $\frac{1}{2}m(v_3^2 - v_2^2) = \frac{1}{2}(2)(49 - 25) = 24$ J

The answer is Option 2: 24 J.

Let the refracting angle (apex angle) of the prism be $A$.

From the geometry of the triangle, the sum of the angles is $180^\circ$:

$A + 45^\circ + \theta = 180^\circ$

$A = 135^\circ - \theta \quad \text{--- (Equation 1)}$

1.Refraction at the first surface

The incident ray is parallel to the base. Because the left base angle is $45^\circ$, the angle between the incident ray and the prism face is $45^\circ$. Therefore, the angle of incidence $i_1$ with respect to the normal is:
$i_1 = 90^\circ - 45^\circ = 45^\circ$
Applying Snell's Law at the first interface (air to glass):
$n_1 \sin i_1 = n_2 \sin r_1$
$1 \cdot \sin(45^\circ) = \sqrt{2} \cdot \sin(r_1)$
$r_1 = 30^\circ$

2: Refraction at the second surface

The emergent ray grazes the second surface, meaning the angle of emergence $e$ is $90^\circ$.
Applying Snell's Law at the second interface (glass to air):
$n_2 \sin r_2 = n_1 \sin e$
$\sqrt{2} \cdot \sin(r_2) = 1 \cdot \sin(90^\circ)$
$\sin(r_2) = \frac{1}{\sqrt{2}}$
$r_2 = 45^\circ$

3: Calculating the angle $\theta$

For a prism, the apex angle $A$ is equal to the sum of the internal angles of refraction:
$A = r_1 + r_2$
$A = 30^\circ + 45^\circ = 75^\circ$

Substitute the value of $A$ back into Equation 1:
$75^\circ = 135^\circ - \theta$
$\theta = 135^\circ - 75^\circ$
$\theta = 60^\circ$

To keep the mass stuck to the wall, three primary forces are at play:

Normal Force ($N$): Acts horizontally toward the axis, providing the centripetal force \longrightarrow\ $N = M\omega^2R$

Weight ($Mg$): Acts vertically downward.

Friction ($f$): Acts vertically upward to oppose the weight \longrightarrow\ $f = Mg$

The friction force is limited by the coefficient of friction ($\mu$):

$f \leq \mu N$

$Mg \leq \mu(M\omega^2R)$

$g \leq \mu\omega^2R$

$\omega^2 \geq \frac{g}{\mu R}$

$\omega_{min} = \sqrt{\frac{g}{\mu R}}$

Two liquids with the same densities and surface tensions $T_1$ and $T_2$ rise to heights $h_1$ and $h_2$, respectively, in capillary tubes of radii $r_1 > r_2$.

We first recall the capillary rise formula:

$h = \frac{2T\cos\theta}{\rho g r}$

Assuming that both liquids have the same contact angle (so that $\cos\theta = 1$) and the same density, we can rearrange to find

$T = \frac{h\rho g r}{2}$, which shows that $T \propto hr$.

Because the surface tensions are equal ($T_1 = T_2$), it follows that

$h_1 r_1 = h_2 r_2$, and since $r_1 > r_2$, we have $h_1 < h_2$.

This conclusion matches Option C, which states $h_1 < h_2$ and $T_1 = T_2$.

The correct answer is Option C.

The pulley consists of a thin rim and two rods (length $L = 2R$).

Rim: $I_{rim} = MR^2$

Two Rods (about center): $I_{rods} = 2 \times \left( \frac{1}{12} M L^2 \right) = 2 \times \left( \frac{1}{12} M (2R)^2 \right) = \frac{2}{3} MR^2$

Total $I$: $I = MR^2 + \frac{2}{3} MR^2 = \frac{5}{3} MR^2$

Equations:

Force on $M$: $Mg - T_2 = Ma$

Force on $m$: $T_1 - mg = ma$

Torque on Pulley: $(T_2 - T_1)R = I\alpha \implies T_2 - T_1 = \frac{I a}{R^2}$

$(Mg - Ma) - (mg + ma) = \frac{\frac{5}{3} MR^2 \cdot a}{R^2}$

$(M - m)g - (M + m)a = \frac{5}{3} Ma$

$(M - m)g = \left( M + m + \frac{5}{3} M \right) a$

$(M - m)g = \left( \frac{8}{3} M + m \right) a$

$a = \frac{(M - m)g}{\frac{8}{3} M + m}$

Car A: mass $10^3$ kg, speed 72 km/h = 20 m/s. Car B: mass $10^3$ kg, speed 36 km/h = 10 m/s. Separation = 10 m.

Angular momentum of A with respect to B:

In B's frame, A has velocity $20 - 10 = 10$ m/s (same direction) and perpendicular distance = 10 m.

$L = m \times v_{rel} \times d = 10^3 \times 10 \times 10 = 10^5$ J.s

The answer is Option 1: $10^5$ J.s.

A capacitor $C$ charged to voltage $V_0$ is disconnected from the battery and connected across an inductor $L$. We need to find the time $t$ at which 25% of the initial energy in the capacitor is transferred to the inductor.

We recall the LC oscillation equations. In an LC circuit, the charge on the capacitor oscillates as:

$q(t) = Q_0\cos(\omega t)$

where $Q_0 = CV_0$ is the initial charge, and the angular frequency of oscillation is:

$\omega = \frac{1}{\sqrt{LC}}$

Next, we write the expressions for the energies. The initial energy stored in the capacitor is:

$U_0 = \frac{1}{2}CV_0^2 = \frac{Q_0^2}{2C}$

At time $t$, the energy in the capacitor is:

$U_C = \frac{q^2}{2C} = \frac{Q_0^2\cos^2(\omega t)}{2C} = U_0\cos^2(\omega t)$

By conservation of energy (no resistance, so no energy loss), the energy in the inductor is:

$U_L = U_0 - U_C = U_0 - U_0\cos^2(\omega t) = U_0\sin^2(\omega t)$

Applying the condition that 25% of the initial energy is in the inductor gives:

$U_L = 0.25 \times U_0$

$U_0\sin^2(\omega t) = 0.25 \times U_0$

$\sin^2(\omega t) = 0.25$

$\sin(\omega t) = 0.5$

(Taking the first positive solution)

Finally, solving for $t$ yields:

$\omega t = \frac{\pi}{6}$

since $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2} = 0.5$.

$t = \frac{\pi}{6\omega} = \frac{\pi}{6 \times \frac{1}{\sqrt{LC}}} = \frac{\pi\sqrt{LC}}{6}$

Therefore, the correct answer is Option (3): $\frac{\pi\sqrt{LC}}{6}$.

From the diagram, there are two parallel paths between A and B:

• bottom branch: just R
• top branch: 2R in series with X → total = 2R + X

So equivalent resistance:

$R_{eq}=\frac{\left(R(2R+X)\right)}{(R+2R+X)}$

Given: Req = X

So,

$R(2R+X)/(3R+X)=X$

Cross-multiply:

$R(2R+X)=X(3R+X)$

Expand:

$2R^2+RX=3RX+X^2$

Bring everything to one side:

$2R^2+RX−3RX−X^2=0$
$2R^2−2RX−X^2=0$

Rearrange:

$X^2+2RX−2R^2=0$

Now solve quadratic in X:

$X=[−2R\pm\sqrt{(4R^2+8R^2)}]/2$
$X=[−2R\pm\sqrt{\ 12R^2}]/2$
$X=[−2R\pm2\sqrt{3}R]/2$

$X=−R\pm\sqrt{3}R$

Physical resistance must be positive, so:

$X=R(\sqrt{3}−1)$

potentiometer works on the principle of wheatstone bridge.

 Length of segment AP = $x$
 Length of segment PB = $50 - x$

resistance of potentiometer wire directly prorportional to length($R = \rho \frac{l}{A}$)

The balancing condition for the Wheatstone bridge is:

$\frac{R_1}{R_2} = \frac{\text{Length of AP}}{\text{Length of PB}}$
$\frac{R_1}{R_2} = \frac{x}{50 - x}$

$\frac{6}{4} = \frac{x}{50 - x}$

$x$=30cm

Terminal velocity of a sphere: $v_0 = \frac{2r^2(\sigma - \rho)g}{9\eta}$, so $\eta = \frac{2r^2(\sigma - \rho)g}{9v_0}$.

Ignoring errors in $\sigma, \rho, g$:

$\frac{\Delta\eta}{\eta} = 2\frac{\Delta r}{r} + \frac{\Delta v_0}{v_0}$

(Since $\eta \propto r^2/v_0$, the maximum relative error is $2\frac{\Delta r}{r} + \frac{\Delta v_0}{v_0}$.)

The answer is Option 1: $\frac{2\Delta r}{r} + \frac{\Delta v_0}{v_0}$.

The stopping potential gives the maximum kinetic energy of the photoelectrons: K = eV max 0 From the graph, V = 5V 0 Hence, K = 5eV max Using Einstein's photoelectric equation, hν = ϕ+K max Given, ϕ = 2eV Therefore, hν = 2+5 = 7eV Now, the saturation current is due to the photoelectrons emitted per second. If n is the number of incident photons per second, then due to the eficiency, n = 10−5n e Therefore, I = en = e(10−5n) s e Given, I = 10μA = 10−5A s Thus, 10−5 = e(10−5)n n = 1 e The power of incident light is P = nhν Since, hν = 7eV = 7eJ we get P = 1 ×7e e P = 7W Therefore, P = 7W Hence, the correct option is B. 

Consider a small circular ring of radius r and thickness dr. Its charge is dq = σ(2πrdr) Since the disc rotates with angular velocity ω, the time taken for one revolution is T = 2π ω Hence, the current due to this small ring is dI = dq T dI = σ(2πrdr)ω 2π dI = σωrdr The magnetic field at the centre due to a circular ring is dB = μ 0 dI 2r Therefore, dB = μ 0 σωrdr 2r dB = μ 0 σω dr 2 The annular disc extends from r = R to r = 2R. Hence, B = ∫ 2R dB R B = μ 0 σω ∫ 2R dr 2 R B = μ 0 σω (2R−R) 2 B = μ 0 σωR 2 Given, B = μ 0 σωR n Comparing, 1 = 1 n 2 Therefore, n = 2 Hence, the correct answer is 2. 

The dimensional formula of surface tension is [surface tension] = [force] [length] Therefore, [surface tension] = MLT−2L−1 = MT−2 Given, X = cahbGc The dimensions of the quantities are [c] = LT−1 [h] = ML2T−1 [G] = M−1L3T−2 Hence, [X] = (LT−1)a(ML2T−1)b(M−1L3T−2)c Comparing the powers of M,L,T with MT−2, b−c = 1 a+2b+3c = 0 a+b+2c = 2 The quantity asked is a+2b+3c From the dimensional equation itself, a+2b+3c = 0 Hence, the correct answer is 0. 

The object is in water while the mirror is in air. Therefore, first convert the velocities into the same medium using the relative refractive index. For water with respect to air, μ = μo = 4/3 = 4 rel μ obser ver 1 3 Hence, the apparent depth is d′ = d μ rel Therefore, the apparent velocity of the object as seen from air is v′ = 8 = 6ms−1 o 4/3 Equivalently, when we consider the mirror from the water side, its apparent velocity is obtained using the inverse relative refractive index: μ = 1 = 3 rel 4/3 4 Thus, v′ = 6 = 8ms−1 m 3/4 Now both the object and mirror are considered in the same medium, and both have velocity v = v′ = o m 8ms−1 For a plane mirror, v = 2v −v i m o Therefore, v = 2(8)−8 i v = 8ms−1 i Hence, the image and the object have the same velocity. Therefore, the velocity of image with respect to object is v = v −v i/o i o v = 8−8 = 0 i/o Thus, 0ms−1 Hence, the correct answer is 0. 

Let the inductance of each inductor be L and the total current entering the circuit be I. Since L and L are identical and connected in parallel, the current divides equally between them. 2 3 I = I = I 2 3 2 The magnetic energy stored in an inductor carrying current I is U = 1LI2 2 For L , U = 1LI2 1 1 2 For L 2 , U L 2 = 2 1 L( I 2 ) 2 U = 1LI2 L 2 8 Similarly, for L , U = 1LI2 3 L 3 8 Therefore, the total magnetic energy stored in the circuit is U = U +U +U t 1 L 2 L 3 U = 1LI2 + 1LI2 + 1LI2 t 2 8 8 U = 4+1+1LI2 t 8 U = 3LI2 t 4 Hence, U U L t 2 = 8 1 4 3 L L I I 2 2 Ut = 6 U L2 Hence, the correct answer is 6. 

This reaction proceeds via the Sₙ1 mechanism.
The rate depends on the stability of the benzyl carbocation formed after departure of Br.

Key Idea:
Electron-donating groups increase carbocation stability and hence increase rate.
Electron-withdrawing groups decrease carbocation stability and hence decrease rate.
Resonance effects (para position) are stronger than inductive effects (meta position).

Analysis:

(b) p-NH₂
Strong +M effect, donates electron density by resonance, gives maximum stabilization → highest reactivity

(a) p-OH
Also shows +M effect but weaker than NH₂ → slightly lower reactivity

(d) m-NO₂
Shows -I effect only (no resonance at meta position) → moderate destabilization

(c) p-NO₂
Strong -M and -I effect, withdraws electron density strongly → least stable carbocation

Final Order:
b > a > d > c

Answer: p-NH₂ > p-OH > m-NO₂ > p-NO₂

We need to identify which compound exhibits optical isomerism and find its carbon percentage.

(a) n-propyl chloride: $CH_3CH_2CH_2Cl$ — no chiral center.

(b) iso-propyl chloride: $(CH_3)_2CHCl$ — no chiral center (the carbon bonded to Cl has two identical methyl groups).

(c) sec-butyl chloride: $CH_3CH(Cl)CH_2CH_3$ — has a chiral center at C-2 (bonded to H, Cl, CH₃, C₂H₅ — all different). Exhibits optical isomerism.

(d) neo-pentyl chloride: $(CH_3)_3CCH_2Cl$ — no chiral center.

sec-butyl chloride is $C_4H_9Cl$. Molecular formula: $C_4H_9Cl$.

Molar mass = 4(12) + 9(1) + 35.5 = 48 + 9 + 35.5 = 92.5 g/mol

% Carbon = $\frac{48}{92.5} \times 100 = 51.89\% \approx 52\%$

The answer is Option 2: 52.

1.00 g of compound X gives 1.79 g of $Mg_2P_2O_7$ (magnesium pyrophosphate).

Molar mass of $Mg_2P_2O_7 = 2(24) + 2(31) + 7(16) = 48 + 62 + 112 = 222$ g/mol.

Each mole of $Mg_2P_2O_7$ contains 2 moles of P.

Moles of $Mg_2P_2O_7 = \frac{1.79}{222}$

Moles of P = $\frac{2 \times 1.79}{222} = \frac{3.58}{222}$

Mass of P = $\frac{3.58}{222} \times 31 = \frac{110.98}{222} = 0.4999 \approx 0.50$ g

% P = $\frac{0.50}{1.00} \times 100 = 50\%$

The answer is Option 4: 50.

We need to evaluate two statements about Crystal Field Stabilization Energy (CFSE) and magnetic moments of coordination compounds.

First we consider Statement I, which asserts that CFSE of $[Cr(H_2O)_6]^{2+}$ is greater than that of $[Mn(H_2O)_6]^{2+}$. Chromium(II) has atomic number 24 and electronic configuration $[Ar]3d^4$, making $Cr^{2+}$ a $d^4$ ion, while manganese(II) has atomic number 25 and electronic configuration $[Ar]3d^5$, so $Mn^{2+}$ is a $d^5$ ion.

Since $H_2O$ is a weak field ligand, both complexes are high‐spin octahedral. The $d^4$ configuration of $Cr^{2+}$ gives $t_{2g}^3 e_g^1$, and the $d^5$ configuration of $Mn^{2+}$ gives $t_{2g}^3 e_g^2$.

Using CFSE = $(-0.4n_{t_{2g}} + 0.6n_{e_g})\Delta_o$, we find for $Cr^{2+}$: CFSE = $(-0.4\times3 + 0.6\times1)\Delta_o = (-1.2 + 0.6)\Delta_o = -0.6\Delta_o$, whereas for $Mn^{2+}$: CFSE = $(-0.4\times3 + 0.6\times2)\Delta_o = (-1.2 + 1.2)\Delta_o = 0$. Because the magnitude of CFSE for $Cr^{2+}$ is $0.6\Delta_o$ and for $Mn^{2+}$ it is zero, CFSE of $[Cr(H_2O)_6]^{2+}$ is indeed greater, so Statement I is true.

Next we analyze Statement II, which compares the spin‐only magnetic moments of potassium ferricyanide and sodium ferrocyanide. Potassium ferricyanide is $K_3[Fe(CN)_6]$, containing $Fe^{3+}$ ($d^5$), while sodium ferrocyanide is $Na_4[Fe(CN)_6]$, containing $Fe^{2+}$ ($d^6$). Because $CN^-$ is a strong field ligand, both complexes are low‐spin octahedral. Thus $Fe^{3+}$ ($d^5$) adopts $t_{2g}^5 e_g^0$ with one unpaired electron, and $Fe^{2+}$ ($d^6$) adopts $t_{2g}^6 e_g^0$ with no unpaired electrons.