The maximum height reached by the projectile is $h_{max} = \frac{v^2 \sin^2\theta}{2g}$.
Work done by gravity from the point of projection to the highest point is $W = -mgh_{max}$.
$W = -mg \left( \frac{v^2 \sin^2\theta}{2g} \right) = -\frac{mv^2 \sin^2\theta}{2}$.
By applying the work-energy theorem or conservation of energy:
Initial mechanical energy = Final mechanical energy + Work done against friction.
$mgh = mgh' + \mu mgl$
Given $h = 5 \text{ m}$, $\mu = 0.5$, and $l = 8 \text{ m}$.
$5 = h' + 0.5 \times 8 \Rightarrow 5 = h' + 4 \Rightarrow h' = 1 \text{ m}$.
A force of $10 \text{ N}$ in a direction of $30^{\circ}$ east of north means the vector starts from the North direction and deviates $30^{\circ}$ towards the East. This is correctly represented in figure (1).
Both blocks move together. Total mass $M = 3 \text{ kg} + 5 \text{ kg} = 8 \text{ kg}$.
Common acceleration $a = \frac{F}{M} = \frac{8 \text{ N}}{8 \text{ kg}} = 1 \text{ m/s}^2$.
The only horizontal force acting on the upper 3 kg block is the static friction between the two blocks.
Friction force $f = m_{upper} \times a = 3 \text{ kg} \times 1 \text{ m/s}^2 = 3 \text{ N}$.
Given path length $s = t^3 + 5$. Tangential speed $v = \frac{ds}{dt} = 3t^2$.
Tangential acceleration $a_t = \frac{dv}{dt} = 6t$. At $t=2 \text{ s}$, $a_t = 6(2) = 12 \text{ m/s}^2$.
At $t=2 \text{ s}$, $v = 3(2)^2 = 12 \text{ m/s}$.
Radius of curvature from the figure is $R = 20 \text{ m}$.
Centripetal acceleration $a_c = \frac{v^2}{R} = \frac{144}{20} = 7.2 \text{ m/s}^2$.
Total acceleration $a = \sqrt{a_t^2 + a_c^2} = \sqrt{12^2 + 7.2^2} = \sqrt{144 + 51.84} = \sqrt{195.84} \approx 14 \text{ m/s}^2$.
For a perfectly elastic collision in one dimension between two identical masses, the particles simply exchange their velocities.
Therefore, if their initial velocities were $15 \text{ m/s}$ and $10 \text{ m/s}$, their final velocities after the collision will be $10 \text{ m/s}$ and $15 \text{ m/s}$ respectively.
The axis AB passes through the center of the first sphere (P) and is perpendicular to the rod.
Moment of inertia of sphere P about its own center: $I_1 = \frac{2}{5}M(R/2)^2 = \frac{2}{5}M\frac{R^2}{4} = \frac{MR^2}{10}$.
Moment of inertia of sphere Q (at distance 2R) using parallel axis theorem: $I_2 = I_{cm} + Md^2 = \frac{MR^2}{10} + M(2R)^2 = \frac{MR^2}{10} + 4MR^2 = \frac{41MR^2}{10}$.
Total Moment of Inertia $I = I_1 + I_2 = \frac{MR^2}{10} + \frac{41MR^2}{10} = \frac{42MR^2}{10} = \frac{21}{5}MR^2$.
By electrostatic induction, the innermost shell (charge $+Q_1$) induces a charge of $-Q_1$ on the inner surface of the middle shell, and $+Q_1$ on the outer surface of the middle shell.
This $+Q_1$ on the outer surface of the middle shell subsequently induces a charge of $-Q_1$ on the inner surface of the outermost shell.
The charge density on the inner surface of the outermost shell (radius $r_3$) is $\sigma = \frac{\text{Charge}}{\text{Area}} = \frac{-Q_1}{4\pi r_3^2}$.
Assuming the standard configuration of charges $q$, $Q$, and $q$ placed at distances 0, $a$, and $2a$ respectively (as typical for such equilibrium problems despite potential OCR noise like "3q" which leads to mismatch with standard options).
For the charge $q$ at the origin to be in equilibrium, the net force on it must be zero.
Force due to Q + Force due to other q = 0 $\Rightarrow \frac{k q Q}{a^2} + \frac{k q q}{(2a)^2} = 0$.
$\frac{Q}{a^2} + \frac{q}{4a^2} = 0 \Rightarrow Q = -\frac{q}{4}$.
The force on a test charge is $F = qE$. This means the electric field magnitude $E$ is the slope of the $|F|$ versus $q$ graph.
From the graph, the slope is $\tan 30^{\circ}$.
$E = \tan 30^{\circ} = \frac{1}{\sqrt{3}} \text{ N/C}$.
Electric field on the axis of a ring: $E = \frac{kQz}{(r^2 + z^2)^{3/2}}$.
For net field to be zero: $E_1 + E_2 = 0 \Rightarrow \frac{k(Q)z}{(R^2 + z^2)^{3/2}} = \frac{k(\sqrt{8}Q)z}{((2R)^2 + z^2)^{3/2}}$.
$\frac{1}{(R^2 + z^2)^{3/2}} = \frac{8^{1/2}}{(4R^2 + z^2)^{3/2}}$. Raising both sides to power $2/3$:
$\frac{1}{R^2 + z^2} = \frac{(8^{1/2})^{2/3}}{4R^2 + z^2} = \frac{2}{4R^2 + z^2}$.
$4R^2 + z^2 = 2R^2 + 2z^2 \Rightarrow z^2 = 2R^2 \Rightarrow z = \sqrt{2}R$.
The total electrostatic potential energy of the system is the sum of energies of all three pairs.
$U = \frac{k(q)(q)}{a} + \frac{k(Q)(q)}{a} + \frac{k(Q)(q)}{a\sqrt{2}} = 0$
$q + Q + \frac{Q}{\sqrt{2}} = 0 \Rightarrow q + Q\left(\frac{\sqrt{2}+1}{\sqrt{2}}\right) = 0$
$Q = -\frac{q\sqrt{2}}{\sqrt{2}+1}$. Multiplying numerator and denominator by $(\sqrt{2}-1)$:
$Q = -q\sqrt{2}(\sqrt{2}-1) = -q(2-\sqrt{2})$.
Let's evaluate Option (2): $\frac{-2q}{2+\sqrt{2}} = \frac{-2q}{\sqrt{2}(\sqrt{2}+1)} = \frac{-\sqrt{2}q}{\sqrt{2}+1}$. This matches perfectly.
By observing the circuit diagram, all four capacitors ($10 \mu F, 20 \mu F, 30 \mu F, 60 \mu F$) are connected across the same two common nodes (A and B). Therefore, they are in parallel.
Equivalent capacitance $C_{eq} = C_1 + C_2 + C_3 + C_4 = 10 + 20 + 30 + 60 = 120 \mu\text{F}$.
Applying Kirchhoff's Current Law (KCL) at the central junction O:
$\frac{V_A - V_O}{R_1} + \frac{V_B - V_O}{R_2} + \frac{V_C - V_O}{R_3} = 0$
$\frac{30 - V_O}{10} + \frac{12 - V_O}{20} + \frac{2 - V_O}{30} = 0$. Multiplying by 60:
$6(30 - V_O) + 3(12 - V_O) + 2(2 - V_O) = 0$
$180 - 6V_O + 36 - 3V_O + 4 - 2V_O = 0 \Rightarrow 220 = 11V_O \Rightarrow V_O = 20 \text{ V}$.
Current through $R_1$: $I_1 = \frac{V_A - V_O}{R_1} = \frac{30 - 20}{10} = 1 \text{ A}$.
By applying the Right-Hand Grip Rule for both current-carrying wires (along +x and +y axes):
In the 1st quadrant, the wire on x-axis produces a field out of the plane (+z), and the wire on the y-axis produces a field into the plane (-z). They cancel along the line $y=x$.
In the 3rd quadrant, the wire on x-axis produces a field into the plane (-z), and the wire on the y-axis produces a field out of the plane (+z). They cancel along the line $y=x$.
Based on the standard quadrant labels for such problems, the resultant magnetic field is zero in regions matching A and C.
For the rod to slide down with a constant velocity, the net force along the incline must be zero. The downward component of gravity must be balanced by the upward magnetic force.
Force of gravity down the incline $= mg \sin\theta$.
Magnetic force on the current-carrying rod $\vec{F}_m = i(\vec{L} \times \vec{B})$. Since $\vec{B}$ is perpendicular to the incline, the force is parallel to the incline and points upwards, magnitude $= i l B$.
Equating them: $mg \sin\theta = i l B \Rightarrow B = \frac{mg \sin\theta}{il}$.
The standard wave equation is $E = E_0 \cos(kz - \omega t)$.
By comparing the given equation $\vec{E} = \hat{i} 40 \cos(kz - 6\times 10^8 t)$, we get $\omega = 6 \times 10^8 \text{ rad/s}$.
The speed of electromagnetic waves in a vacuum is $c = \frac{\omega}{k}$.
$k = \frac{\omega}{c} = \frac{6 \times 10^8}{3 \times 10^8} = 2 \text{ m}^{-1}$.
As the loop falls away from the wire, the inward magnetic flux passing through it decreases. By Lenz's law, the induced current will try to oppose this decrease by creating more inward flux, which requires a clockwise current.
The upper arm of the loop is closer to the wire, so the upward repulsive magnetic force it experiences is stronger than the downward attractive force on the lower arm. The net magnetic force is upward, reducing the net downward acceleration to $a < g$.
The sudden release of the piston represents an adiabatic expansion. The relation between temperature and volume is $TV^{\gamma-1} = \text{constant}$.
$T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1} \Rightarrow \frac{T_1}{T_2} = \left(\frac{V_2}{V_1}\right)^{\gamma-1}$.
Since $V \propto L$ (length of gas column), $\frac{T_1}{T_2} = \left(\frac{L_2}{L_1}\right)^{\gamma-1}$.
For a monoatomic gas, $\gamma = \frac{5}{3}$, so $\gamma - 1 = \frac{2}{3}$. Thus, $\frac{T_1}{T_2} = \left(\frac{L_2}{L_1}\right)^{2/3}$.
The first law of thermodynamics ($\Delta Q = \Delta U + \Delta W$) is fundamentally a restatement of the law of conservation of energy applied to thermodynamic systems.
Amount of substance in moles $n = \frac{m}{M} = \frac{1 \text{ g}}{4 \text{ g/mol}} = \frac{1}{4} \text{ mol}$.
For a monoatomic gas (Helium), the molar specific heat at constant volume $C_v = \frac{3}{2}R$.
Heat energy required $Q = n C_v \Delta T = \frac{1}{4} \times \left(\frac{3}{2}R\right) (T_2 - T_1) = \frac{3}{8}R (T_2 - T_1)$.
Since $R = N_A k_B$, the heat required is $\frac{3}{8} N_A k_B (T_2 - T_1)$.
The intensity of a sound wave is proportional to the square of its amplitude and the square of its frequency: $I \propto A^2 f^2$.
If $A' = 2A$ and $f' = f/4$, then the new intensity is:
$I' \propto (2A)^2 \left(\frac{f}{4}\right)^2 = 4A^2 \frac{f^2}{16} = \frac{1}{4} (A^2 f^2) = \frac{1}{4} I$.
Thus, the intensity will decrease by a factor of 4.
The amplitude of a damped oscillator decays exponentially: $A(t) = A_0 e^{-\lambda t}$.
At $t=2$, $A = A_0 / 3$, so $e^{-2\lambda} = \frac{1}{3}$.
At $t=6$, $A(6) = A_0 e^{-6\lambda} = A_0 (e^{-2\lambda})^3 = A_0 \left(\frac{1}{3}\right)^3 = \frac{A_0}{27}$.
Thus $n = 27$. (Note: $3^3 = 27$, which matches option (4) where $3^3$ is misread as 33 due to OCR/typographical errors).
With the first tuning fork (440 Hz) and 5 Hz beats, the string frequency is either $440 + 5 = 445 \text{ Hz}$ or $440 - 5 = 435 \text{ Hz}$.
With the second tuning fork (437 Hz) and 8 Hz beats, the string frequency is either $437 + 8 = 445 \text{ Hz}$ or $437 - 8 = 429 \text{ Hz}$.
The common frequency in both cases is 445 Hz.
Radius of the raindrop $R = \frac{0.02}{2} \text{ cm} = 0.01 \text{ cm} = 10^{-4} \text{ m}$.
Excess pressure $P = \frac{2T}{R} = \frac{2 \times 72 \times 10^{-3} \text{ N/m}}{10^{-4} \text{ m}} = 1440 \text{ N/m}^2$.
To convert to CGS units: $1 \text{ N/m}^2 = 10 \text{ dyne/cm}^2$.
$1440 \text{ N/m}^2 = 14400 \text{ dyne/cm}^2 = 1.44 \times 10^4 \text{ dyne/cm}^2$.
According to Einstein's photoelectric equation: $K_{max} = \frac{hc}{\lambda} - \phi$.
$\frac{1}{2}mv^2 = \frac{hc - \lambda\phi}{\lambda} \Rightarrow v^2 = \frac{2(hc - \lambda\phi)}{m\lambda}$.
Taking the square root gives $v = \left[ \frac{2(hc - \lambda\phi)}{m\lambda} \right]^{1/2}$.
When phosphorus (a pentavalent donor) is added, the electron concentration becomes approximately equal to the donor concentration: $n \approx N_D = 10^{21} / \text{m}^3$.
By the Law of Mass Action, $n \times p = n_i^2$.
The new hole concentration is $p = \frac{n_i^2}{n} = \frac{(10^{16})^2}{10^{21}} = \frac{10^{32}}{10^{21}} = 10^{11} \text{ per m}^3$.
Power $P = 1000 \text{ kW} = 10^6 \text{ W} = 10^6 \text{ J/s}$.
Energy produced per hour $E = P \times t = 10^6 \times 3600 = 3.6 \times 10^9 \text{ J}$.
Using mass-energy equivalence $E = \Delta m c^2$:
$\Delta m = \frac{E}{c^2} = \frac{3.6 \times 10^9}{(3 \times 10^8)^2} = \frac{3.6 \times 10^9}{9 \times 10^{16}} = 0.4 \times 10^{-7} \text{ kg}$.
$\Delta m = 4 \times 10^{-8} \text{ kg} = 40 \times 10^{-6} \text{ g} = 40 \mu\text{g}$.
The diode has its P-side connected to 10V and N-side to 5V. Since $10V > 5V$, the diode is forward-biased. As it is an ideal diode, it acts as a short circuit with 0 resistance.
The current is $I = \frac{\Delta V}{R} = \frac{10\text{V} - 5\text{V}}{100\Omega} = \frac{5}{100} \text{ A} = \frac{1}{20} \text{ A}$.
A wavefront is defined as the locus of all points having the same phase of oscillation. For a parallel beam of light (light coming from an infinite distance), the locus of points with the same phase forms a plane perpendicular to the direction of propagation. Thus, it is a plane wavefront.
The core of a transformer is laminated (made of thin insulated sheets instead of a solid block) to break the circulating paths of eddy currents. This significantly increases resistance to these currents, thereby minimizing energy losses in the form of heat.
In the Binding Energy per Nucleon curve, stability is highest in the middle mass region (near Iron).
Lighter elements (like P and Q) can undergo nuclear fusion to form heavier, more stable nuclei, releasing energy.
Heavier elements (like R and S) can undergo nuclear fission to split into intermediate-mass fragments, also becoming more stable and releasing energy.
Therefore, statements (1) and (3) are both correct.
Air is supplied at a constant rate, meaning the volume $V$ increases linearly with time: $V \propto t \Rightarrow R^3 \propto t \Rightarrow R \propto t^{1/3}$.
The excess pressure inside a soap bubble is $P = \frac{4T}{R}$.
Substituting the relation for R: $P \propto \frac{1}{t^{1/3}} = t^{-1/3}$.
This represents a curve that decreases with time and is concave upwards, which matches graph (2).
The silvered plano-convex lens acts as a concave mirror. Equivalent power $P = 2P_{lens} + P_{mirror}$.
$P_{lens} = \frac{1}{f} = \frac{1}{10}$. Since plane mirror power $P_{mirror} = 0$, $P = \frac{2}{10} = \frac{1}{5}$.
Equivalent focal length $F_{eq} = 5 \text{ cm}$ (concave mirror).
For the image to form on the object itself, the object must be placed at the center of curvature of this equivalent mirror. $d = 2 \times F_{eq} = 2 \times 5 = 10 \text{ cm}$.
Terminal velocity is reached when the effective downward acceleration becomes zero.
$a = g - kv^2 \Rightarrow 0 = g - kv_t^2$
$v_t^2 = \frac{g}{k} = \frac{9.8}{0.002} = \frac{9800}{2} = 4900$
$v_t = \sqrt{4900} = 70 \text{ m/s}$.
Induced emf $E = \frac{\Delta\phi}{\Delta t}$. Induced current $I = \frac{E}{R} = \frac{\Delta\phi}{R\Delta t}$.
Total charge passing through the circuit $Q = I \times \Delta t = \left( \frac{\Delta\phi}{R\Delta t} \right) \times \Delta t = \frac{\Delta\phi}{R}$.
This shows that the total charge $Q$ depends only on the net change in magnetic flux and the resistance, and is completely independent of the time interval $\Delta t$. Therefore, $Q \propto \Delta t^0$.
A $\rightarrow$ 2: Max safe speed on unbanked circular road is $v = \sqrt{\mu Rg}$.
B $\rightarrow$ 3: Stopping distance $S = \frac{v^2}{2\mu g}$. Since $p = mv \Rightarrow v = \frac{p}{m}$, $S = \frac{p^2}{2\mu m^2 g}$.
C $\rightarrow$ 1: $t_{rough} = n \times t_{smooth}$. Here $n=2$. $\mu = (1 - \frac{1}{n^2}) \tan\theta = (1 - \frac{1}{4}) \tan 45^{\circ} = \frac{3}{4} = 0.75$.
D $\rightarrow$ 4: Velocity after sliding down rough incline: $v^2 = 2aL = 2(g\sin\theta - \mu g\cos\theta)L \Rightarrow v = \sqrt{2gL(\sin\theta - \mu\cos\theta)}$.
Assertion is true: The shortest wavelength lines of the Balmer series (like the series limit where transition is from $n=\infty$ to $n=2$, $\lambda \approx 364.6 \text{ nm}$) lie in the near-ultraviolet region, not the visible region.
Reason is true: The given Rydberg formula calculates exactly these wavelengths, proving mathematically that higher transitions drop below the $\sim$400 nm threshold of visible light. The reason perfectly explains the assertion.
For a real image to be formed on a screen by a convex lens, the minimum distance between the object and the screen must be $4f$.
Therefore, $D \ge 4f \Rightarrow f \le \frac{D}{4}$.
Given $D = 3 \text{ m}$, the maximum focal length is $f_{max} = \frac{3}{4} = 0.75 \text{ m}$.
In an LCR series circuit at resonance, the inductive reactance perfectly cancels the capacitive reactance ($X_L = X_C$). The circuit behaves purely resistively. As a result, the voltage and current are exactly in the same phase ($\phi = 0$), making the power factor ($\cos\phi$) equal to 1.
According to the Equation of Continuity ($A_1 v_1 = A_2 v_2$), velocity is highest where the cross-sectional area is smallest. Thus, fluid speed is greatest at B.
By Bernoulli's principle ($P + \frac{1}{2}\rho v^2 = \text{const}$ for horizontal pipe), higher velocity means lower static pressure. Therefore, pressure is lowest at B, while the pressures at A and C are equal and higher. The height of liquid in the vertical manometer tubes directly indicates this static pressure, so the levels in A and C will be equal.
The force is $F = 15 - 3x = -3(x - 5) \text{ N}$.
The equilibrium position (mean position) is where $F = 0$, which gives $x = 5 \text{ m}$.
The particle is released from rest at $x = 7 \text{ m}$. In SHM, the point of release from rest is an extreme position.
Amplitude $A = |x_{extreme} - x_{mean}| = |7 - 5| = 2 \text{ m}$.
Young's Modulus is defined as the ratio of Stress to Strain.
Strain is a dimensionless quantity (change in length / original length). Therefore, the dimensions of Young's Modulus are identical to the dimensions of Stress, which is Force/Area. This is the exact same dimensional formula as Pressure.
Center of mass coordinates:
$x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} = \frac{2(1) + 6(2)}{8} = \frac{14}{8} = 1.75 \text{ m}$.
$y_{cm} = \frac{2(1) + 6(2)}{8} = 1.75 \text{ m}$.
$z_{cm} = \frac{2(1) + 6(1)}{8} = 1 \text{ m}$.
Distance from the 2 kg mass at (1, 1, 1) to the COM (1.75, 1.75, 1):
$d = \sqrt{(1.75 - 1)^2 + (1.75 - 1)^2 + (1 - 1)^2} = \sqrt{0.75^2 + 0.75^2} = \sqrt{2 \times (3/4)^2} = \frac{3\sqrt{2}}{4} \text{ m}$.
On a microscopic level, pressure is proportional to the rate of momentum transfer to the walls. Pressure $P \propto (\text{momentum change per collision}) \times (\text{frequency of collisions})$.
Since $v_{rms} \propto \sqrt{T}$, if temperature is doubled, $v_{rms}$ increases by a factor of $\sqrt{2}$.
Momentum change per collision ($\Delta p = 2mv$) increases by $\sqrt{2}$.
Frequency of collision ($f \propto v$) also increases by $\sqrt{2}$.
Thus, both factors increase, contributing to the overall doubling of the pressure ($\sqrt{2} \times \sqrt{2} = 2$).