Net force $F_{net} = F - f_k = ma \Rightarrow 7 - \mu_k(2)(10) = 2a \Rightarrow a = \frac{7 - 20\mu_k}{2}$
Distance $s = \frac{1}{2}at^2 = \frac{1}{2}\left(\frac{7 - 20\mu_k}{2}\right)(100) = 25(7 - 20\mu_k)$
Work done against friction $W_f = f_k \times s = 250$
$20\mu_k \times 25(7 - 20\mu_k) = 250 \Rightarrow 500\mu_k(7 - 20\mu_k) = 250 \Rightarrow 2\mu_k(7 - 20\mu_k) = 1$
$40\mu_k^2 - 14\mu_k + 1 = 0 \Rightarrow \mu_k = \frac{14 \pm \sqrt{196 - 160}}{80} = \frac{14 \pm 6}{80}$
$\mu_k = \frac{20}{80} = 0.25$ or $\mu_k = \frac{8}{80} = 0.1$. Since 0.1 is in the options, the coefficient of kinetic friction is 0.1.
For a uniform chain of mass $M$ and length $L$, let the hanging part be $1/n$ of its length. (Interpreted from standard problem context despite typographical errors in the source).
Mass of hanging part $m_1 = \frac{M}{n}$. Mass of part on table $m_2 = M - \frac{M}{n} = M\left(\frac{n-1}{n}\right)$.
For equilibrium, the weight of the hanging part must be balanced by the maximum static friction on the table.
$m_1 g = \mu m_2 g \Rightarrow \frac{M}{n}g = \mu M\left(\frac{n-1}{n}\right)g \Rightarrow 1 = \mu(n-1) \Rightarrow \mu = \frac{1}{n-1}$.
By conservation of mechanical energy:
$\frac{1}{2}mv^2 - \frac{GMm}{R} = \frac{1}{2}mv_{\infty}^2$
Given projection speed $v = nv_e$, and we know escape speed $v_e = \sqrt{\frac{2GM}{R}}$, so $\frac{GMm}{R} = \frac{1}{2}mv_e^2$.
$\frac{1}{2}m(nv_e)^2 - \frac{1}{2}mv_e^2 = \frac{1}{2}mv_{\infty}^2 \Rightarrow v_{\infty}^2 = v_e^2(n^2 - 1)$
$v_{\infty} = v_e\sqrt{n^2 - 1} = 11.2\sqrt{n^2 - 1} \text{ km/s}$.
The moment of inertia of a rigid body is not just dependent on its total mass, but heavily depends on how that mass is distributed relative to the specific axis of rotation. Therefore, it changes if the axis of rotation changes.
Conservative force $F = -\frac{dU}{dx}$.
Given $U(x) = -6x^2 + 2x$
$\frac{dU}{dx} = -12x + 2$
$F = -(-12x + 2) = 12x - 2$
At $x = 0$, $F = 12(0) - 2 = -2 \text{ N}$.
Assertion is true: In uniform circular motion, the speed is constant, and the radius is constant, so the angular velocity $\vec{\omega}$ remains constant in magnitude and direction (perpendicular to the plane).
Reason is false: Linear momentum is a vector ($\vec{p} = m\vec{v}$). While its magnitude is constant, its direction continuously changes, so the linear momentum vector is not constant.
The expression for the energy of an electron in the $n^{\text{th}}$ orbit of a hydrogen atom is $E_n = -\frac{me^4}{8\epsilon_0^2 h^2 n^2}$.
Therefore, the quantity $\frac{me^4}{\epsilon_0^2 h^2}$ has the dimensions of Energy.
Dimensional formula for Energy is $[ML^2T^{-2}]$.
Initial velocity $\vec{V} = a\hat{i} + b\hat{j}$. So horizontal velocity $u_x = a$ and vertical velocity $u_y = b$.
Range $R = \frac{2u_x u_y}{g} = \frac{2ab}{g}$.
Maximum height $H = \frac{u_y^2}{2g} = \frac{b^2}{2g}$.
Given $R = 4H \Rightarrow \frac{2ab}{g} = 4\left(\frac{b^2}{2g}\right) \Rightarrow \frac{2ab}{g} = \frac{2b^2}{g} \Rightarrow a = b$.
The rate at which the speed is increasing is the tangential acceleration: $a_t = \frac{dv}{dt} = 5 \text{ m/s}^2$.
Angular acceleration $\alpha = \frac{a_t}{r}$.
$\alpha = \frac{5}{500} = 0.01 \text{ rad/s}^2$.
Moment of inertia of a solid sphere about its diameter is $I_{cm} = \frac{2}{5}MR^2$.
Using parallel axis theorem, I about a tangent is $I_T = I_{cm} + MR^2 = \frac{2}{5}MR^2 + MR^2 = \frac{7}{5}MR^2$.
Equating to $Mk^2$, we get $k^2 = \frac{7}{5}R^2 \Rightarrow k = \sqrt{\frac{7}{5}}R$.
By conservation of energy, the kinetic energy projected equals the change in potential energy to reach height $h$.
$\Delta U = \frac{mgh}{1 + h/R}$. Given $h = R/2$.
$KE = \Delta U = \frac{mg(R/2)}{1 + (R/2)/R} = \frac{mgR/2}{3/2} = \frac{mgR}{3}$.
Since $g = \frac{GM}{R^2}$, $KE = \frac{m(\frac{GM}{R^2})R}{3} = \frac{GmM}{3R}$.
Resistivity $\rho = \frac{R \cdot A}{l} = \frac{R \cdot \pi r^2}{l}$.
Percentage error $\frac{\Delta\rho}{\rho}\% = \frac{\Delta R}{R}\% + 2\frac{\Delta r}{r}\% + \frac{\Delta l}{l}\%$.
$\frac{\Delta\rho}{\rho}\% = \left(\frac{10}{100} \times 100\right) + 2\left(\frac{0.05}{0.35} \times 100\right) + \left(\frac{0.2}{15} \times 100\right)$
$= 10\% + 2(14.28\%) + 1.33\% = 10 + 28.56 + 1.33 = 39.89\% \approx 39.9\%$.
For an isobaric process, pressure $P$ is constant. From Ideal Gas Law ($PV = nRT$), if volume $V$ is doubled, absolute temperature $T$ must also be doubled.
Internal energy $U = nC_vT$. Since $T$ doubles, the internal energy also doubles.
Excess pressure $P = \frac{2T}{R}$, which means $P \propto \frac{1}{R}$.
Given $P_1 = 4P_2 \Rightarrow \frac{1}{R_1} = \frac{4}{R_2} \Rightarrow R_2 = 4R_1$, or $\frac{R_1}{R_2} = \frac{1}{4}$.
Mass of a drop $M = \text{Volume} \times \text{Density} \propto R^3$.
Ratio $\frac{M_1}{M_2} = \left(\frac{R_1}{R_2}\right)^3 = \left(\frac{1}{4}\right)^3 = \frac{1}{64}$.
Assertion is True: Two isothermal curves can never intersect because if they did, the gas would simultaneously exist at two different temperatures at the exact same pressure and volume state, which is physically impossible.
Reason is True: Adiabatic changes may occur slowly if the system is perfectly insulated. However, this reason does not explain why isothermal curves do not intersect.
For pipe $P_1$ (closed), first overtone frequency $f_1 = \frac{3v}{4L_1}$.
For pipe $P_2$ (open), second overtone frequency $f_2 = \frac{3v}{2L_2}$.
Since they resonate with the same tuning fork, $f_1 = f_2 \Rightarrow \frac{3v}{4L_1} = \frac{3v}{2L_2}$.
$\frac{1}{2L_1} = \frac{1}{L_2} \Rightarrow \frac{L_1}{L_2} = \frac{1}{2}$.
The reciprocal of the bulk modulus of a material represents its fractional change in volume per unit increase in pressure, which is defined as its compressibility.
The given differential equation is $\frac{4d^2y}{dt^2} + 9y = 4$.
Rewriting it in the standard SHM form $\frac{d^2y}{dt^2} = -\omega^2 y + C$:
$\frac{d^2y}{dt^2} + \frac{9}{4}y = 1$. The coefficient of $y$ is $\omega^2$.
$\omega^2 = \frac{9}{4} \Rightarrow \omega = \frac{3}{2} \text{ rad/s}$.
Using the principle of thermometry: $\frac{\text{Reading} - \text{LFP}}{\text{UFP} - \text{LFP}} = \text{Constant}$.
$\frac{C - 0}{100 - 0} = \frac{T - (-5)}{95 - (-5)}$
Given $C = 65$: $\frac{65}{100} = \frac{T + 5}{100} \Rightarrow 65 = T + 5 \Rightarrow T = 60$.
Velocity of sound in a gas is $v = \sqrt{\frac{\gamma RT}{M}}$.
For Helium (He): $\gamma_1 = 5/3$, $M_1 = 4$.
For Hydrogen ($H_2$): $\gamma_2 = 7/5$, $M_2 = 2$.
$\frac{v_{He}}{v_{H_2}} = \sqrt{\frac{\gamma_1 / M_1}{\gamma_2 / M_2}} = \sqrt{\frac{(5/3)/4}{(7/5)/2}} = \sqrt{\frac{5/12}{7/10}} = \sqrt{\frac{50}{84}} = \sqrt{\frac{25}{42}} = \frac{5}{\sqrt{42}}$.
Potential energy $U = \frac{1}{2}kx^2$.
Total energy $E = \frac{1}{2}kA^2$.
When displacement $x = A/2$, $U = \frac{1}{2}k(A/2)^2 = \frac{1}{4}(\frac{1}{2}kA^2) = \frac{E}{4}$.
Ratio of Potential energy to Total energy is $1 : 4$.
Assertion is Correct: Magnetic susceptibility ($\chi$) is defined as the ratio of the intensity of magnetization ($M$) to the magnetic intensity ($H$), $\chi = \frac{M}{H}$.
Reason is Incorrect: According to the formula, $M = \chi H$. For a given $H$, a greater value of susceptibility means a greater value of magnetization intensity, not smaller.
For a regular polygon with $N$ sides, the magnetic field at the center is $B = \frac{\mu_0 I N}{\pi a_0} \sin(\frac{\pi}{N}) \tan(\frac{\pi}{N})$ where $a_0$ is the side length.
Here, the number of sides is $N = n/2$, and the side length is $a$.
Substitute $N = n/2$ into the formula:
$B = \frac{\mu_0 I (n/2)}{\pi a} \sin\left(\frac{\pi}{n/2}\right) \tan\left(\frac{\pi}{n/2}\right) = \frac{n\mu_0 I}{2\pi a} \sin\left(\frac{2\pi}{n}\right) \tan\left(\frac{2\pi}{n}\right)$.
The electromagnetic spectrum order by increasing wavelength is: Gamma rays, X-rays, UV, Visible, IR, Microwaves, Radio waves.
(I) Wavelength of microwaves > UV rays (Correct).
(II) Wavelength of IR < UV (Incorrect).
(III) Wavelength of microwaves < IR (Incorrect).
(IV) Gamma rays have the shortest wavelength (Correct).
Thus, statements I and IV are correct.
Resonance in an AC circuit occurs when the inductive reactance and capacitive reactance cancel each other out ($X_L = X_C$), which fundamentally requires both an inductor (L) and a capacitor (C) to be present. At this point, the voltages across L and C are equal and opposite, cancelling each other, leaving the total source voltage to appear entirely across the resistor R, maximizing the current to $V_m/R$. Both assertion and reason are correct and explanatory.
Resistance of voltmeter $R_v = 10 \text{ k}\Omega = 10000 \Omega$. It is in parallel with the $400 \Omega$ resistor.
Equivalent parallel resistance $R_p = \frac{400 \times 10000}{400 + 10000} = \frac{4000000}{10400} = \frac{5000}{13} \Omega \approx 384.6 \Omega$.
Total circuit resistance $R_{total} = R_p + 800 = 384.6 + 800 = 1184.6 \Omega$.
Main current $I = \frac{V}{R_{total}} = \frac{6}{1184.6} \approx 0.005065 \text{ A}$.
Voltage read by voltmeter $V_p = I \times R_p = 0.005065 \times 384.6 \approx 1.948 \text{ V} \approx 1.95 \text{ V}$.
By simplifying the capacitor ladder network from left to right: The 2C, 2C, C, C network on the left simplifies progressively. The inner loop resolves to C, which in parallel with the adjacent C gives 2C. This repeats, eventually making the entire left block's equivalent capacitance $3C$.
This $3C$ equivalent block sits in series between the top $3C$ and bottom $3C$ capacitors connected to P and Q.
Equivalent capacitance $\frac{1}{C_{eq}} = \frac{1}{3C} + \frac{1}{3C} + \frac{1}{3C} = \frac{3}{3C} = \frac{1}{C} \Rightarrow C_{eq} = C$.
During charging, the battery acts as a load. Current $I = \frac{V_{supply} - E}{R + r}$.
$I = \frac{120 - 8}{15.5 + 0.5} = \frac{112}{16} = 7 \text{ A}$.
Terminal voltage during charging is $V = E + Ir$
$V = 8 + (7 \times 0.5) = 8 + 3.5 = 11.5 \text{ V}$.
The force required to keep a wire moving at a constant velocity is equal and opposite to the magnetic Lorentz force $F = BIl$. However, the diagram shows the wire sliding on rails with no electrical connection completing the circuit (the ends are open). Since the circuit is open, the induced current $I = 0$. Therefore, the magnetic force is zero, and the external force required to maintain constant velocity is zero.
The magnetic energy density (energy per unit volume) in a magnetic field is $u = \frac{B^2}{2\mu_0}$.
Total magnetic energy stored $U = u \times \text{Volume}$.
For a solenoid of cross-sectional area $A$ and length $l$, Volume $= Al$.
$U = \frac{B^2}{2\mu_0} Al$.
The motional emf developed across a conducting rod rotating about one end in a uniform magnetic field is $E = \frac{1}{2}B\omega L^2$.
Given $B = 0.5 \text{ T}$, $\omega = 200 \text{ rad/s}$, and $L = 1 \text{ m}$.
$E = \frac{1}{2}(0.5)(200)(1)^2 = 0.25 \times 200 = 50 \text{ V}$.
Let the midpoint of base BC be origin (0,0).
Field at O due to B (+2q) is $E_B$ directed towards +x. Field at O due to C (+q) is $E_C$ directed towards -x. Since $E_B > E_C$, the net horizontal field is in the +x direction.
Field at O due to A (-q) is $E_A$ directed towards A (which is along the +y direction).
The resultant electric field has both a positive x-component and a positive y-component. Thus, the resultant vector points into the first quadrant, as shown in diagram (1).
Work done by an external agent against the electric field is $W = q(V_B - V_A)$.
$24 \text{ J} = 0.01 \text{ C} \times (V_B - V_A)$
$V_B - V_A = \frac{24}{0.01} = 2400 \text{ V} = 2.4 \times 10^3 \text{ V}$.
The electric field is $\vec{E} = E_0\cos(\omega t - kz)\hat{i}$. The wave propagates in the +z direction ($\hat{k}$).
The direction of propagation is given by $\vec{E} \times \vec{B}$. Since $\hat{i} \times \hat{j} = \hat{k}$, the magnetic field must oscillate in the $\hat{j}$ direction.
The magnitude relationship is $B_0 = \frac{E_0}{c}$.
Thus, $\vec{B} = \frac{E_0}{c}\cos(\omega t - kz)\hat{j}$.
De-Broglie wavelength is given by $\lambda = \frac{h}{mv}$.
For particles moving with the same velocity $v$, $\lambda \propto \frac{1}{m}$.
To maximize $\lambda$, the mass $m$ must be minimum. Among the given options (Neutron, Proton, $\beta$-particle, $\alpha$-particle), the $\beta$-particle (electron) has the smallest mass.
Energy difference $\Delta E = h\nu$
Given $\Delta E = 3.3 \text{ eV}$. Planck's constant $h = 4.14 \times 10^{-15} \text{ eV s}$.
$\nu = \frac{\Delta E}{h} = \frac{3.3}{4.14 \times 10^{-15}} \approx 0.797 \times 10^{15} \text{ Hz} \approx 0.8 \times 10^{15} \text{ Hz}$.
The diode is forward-biased. Applying Kirchhoff's Voltage Law:
Net driving voltage $V_{net} = V_{battery} - V_{barrier} = 4.5 \text{ V} - 0.5 \text{ V} = 4.0 \text{ V}$.
Current $I = \frac{V_{net}}{R} = \frac{4.0}{200} \text{ A} = 0.02 \text{ A} = 20 \text{ mA}$.
The logic operation of an OR gate is to output a HIGH (1) state if at least one of its inputs is HIGH (1). Therefore, the output Y will be 1 when input A or B or both are 1.
The energy gap $E_g$ depends on the atomic spacing. Carbon (Diamond) has the smallest atomic size and highest binding, thus the largest energy gap (~5.4 eV), making it an insulator. Silicon has $E_g \approx 1.1 \text{ eV}$, and Germanium has $E_g \approx 0.7 \text{ eV}$.
Thus, $E_g(C) > E_g(Si) > E_g(Ge)$.
Angular width $\theta \propto \lambda$.
When the apparatus is immersed in water, the wavelength of light decreases to $\lambda' = \frac{\lambda}{\mu}$.
Therefore, the new angular width $\theta' = \frac{\theta}{\mu} = \frac{0.4^{\circ}}{4/3} = 0.4 \times \frac{3}{4} = 0.3^{\circ}$.
Initial binding energy $BE_i = 240 \times 9 \text{ MeV} = 2160 \text{ MeV}$.
Final binding energy of the two fragments $BE_f = 2 \times (120 \times 10 \text{ MeV}) = 2400 \text{ MeV}$.
Total gain in binding energy $= BE_f - BE_i = 2400 - 2160 = 240 \text{ MeV}$.
In eV: $240 \times 10^6 \text{ eV} = 2.4 \times 10^8 \text{ eV}$.
Energy of incident photon $E = \frac{hc}{\lambda} = \frac{1240 \text{ eV-nm}}{500 \text{ nm}} = 2.48 \text{ eV}$.
From Einstein's photoelectric equation, Work Function $\Phi = E - K_{max}$.
$\Phi = 2.48 \text{ eV} - 0.8 \text{ eV} = 1.68 \text{ eV}$.
At minimum deviation, the ray passes symmetrically, so the internal angle of refraction $r = A/2$.
Given ray is refracted at $30^{\circ}$ at the first surface, so $r = 30^{\circ} \Rightarrow A/2 = 30^{\circ} \Rightarrow A = 60^{\circ}$.
Minimum deviation $\delta_m = 30^{\circ}$.
Refractive index $n = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin(A/2)} = \frac{\sin\left(\frac{60+30}{2}\right)}{\sin(60/2)} = \frac{\sin 45^{\circ}}{\sin 30^{\circ}} = \frac{1/\sqrt{2}}{1/2} = \sqrt{2}$.
Therefore, $n^2 = (\sqrt{2})^2 = 2$.
Fringe width $\beta = \frac{\lambda D}{d}$.
If $\lambda$ is increased by 10%, $\lambda' = 1.1\lambda$.
If separation $d$ is increased by 10%, $d' = 1.1d$.
New fringe width $\beta' = \frac{1.1\lambda D}{1.1d} = \frac{\lambda D}{d} = \beta$.
The fringe width remains unchanged at 0.24 mm.
A magnified virtual image is formed by a concave mirror. Magnification $m = -\frac{v}{u} = +2 \Rightarrow v = -2u$.
(By sign convention, $u$ is negative, so $v$ is positive, on the other side of the mirror).
Distance between object and virtual image $= v - u = -2u - u = -3u$.
Given distance $= 15 \text{ cm} \Rightarrow -3u = 15 \Rightarrow u = -5 \text{ cm}$.
Then $v = -2(-5) = +10 \text{ cm}$.
Mirror formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{10} + \frac{1}{-5} = \frac{1 - 2}{10} = -\frac{1}{10}$.
Focal length $f = -10 \text{ cm}$.