Initial velocity $\vec{u} = 30\hat{i} + 40\hat{j} \text{ m/s}$.
Force $\vec{F} = -(\hat{i} + 5\hat{j}) \text{ N}$ and Mass $m = 5 \text{ kg}$.
Acceleration $\vec{a} = \frac{\vec{F}}{m} = -0.2\hat{i} - 1\hat{j} \text{ m/s}^{2}$.
The y-component of velocity becomes zero when $v_y = 0$.
Using $v_y = u_y + a_y t \Rightarrow 0 = 40 + (-1)t$
$t = 40 \text{ seconds}$.
In an LCR series circuit, the total voltage of the source is the phasor sum of the voltages across the resistor, inductor, and capacitor.
$V = \sqrt{V_R^2 + (V_C - V_L)^2}$
Given $V_R = 40 \text{ V}, V_L = 30 \text{ V}, V_C = 60 \text{ V}$.
$V = \sqrt{40^2 + (60 - 30)^2} = \sqrt{1600 + 900} = \sqrt{2500} = 50 \text{ V}$.
Power of the laser $P = \frac{n \cdot hc}{\lambda}$, where $n$ is the number of photons emitted per second.
$n = \frac{P \lambda}{hc} = \frac{(6 \times 10^{-3} \text{ J/s}) \times (663 \times 10^{-9} \text{ m})}{(6.63 \times 10^{-34} \text{ J s}) \times (3 \times 10^8 \text{ m/s})}$
$n = \frac{3978 \times 10^{-12}}{19.89 \times 10^{-26}} = 200 \times 10^{14} = 2 \times 10^{16}$.
The electron enters a uniform magnetic field $\vec{B} = B_0\hat{j}$. The magnetic force $F = -e(\vec{v} \times \vec{B})$ always acts perpendicular to the velocity vector.
Since the force is perpendicular to velocity, it only changes the direction of motion, not the speed. Therefore, the speed $v$ remains constant at $v_0$.
Because speed is constant, the magnitude of momentum $p = mv$ is constant. Thus, the de-Broglie wavelength $\lambda = \frac{h}{p}$ remains constant over time. $\lambda(t) = \lambda_0$.
The instantaneous magnetic energy density is given by $u_B = \frac{B^2}{2\mu_0}$.
Since $B = B_0 \sin(kx - \omega t)$, the energy density is $u_B = \frac{B_0^2 \sin^2(kx - \omega t)}{2\mu_0}$.
The average value of $\sin^2(kx - \omega t)$ over a complete cycle is $\frac{1}{2}$.
Therefore, the average magnetic energy density is $\langle u_B \rangle = \frac{B_0^2}{4\mu_0}$.
Bulk modulus $B = \frac{\Delta P}{\Delta \rho / \rho}$, where $\Delta \rho$ is the change in density.
$\frac{\Delta \rho}{\rho} = \frac{\Delta P}{B} = \frac{80 \times 1.013 \times 10^5 \text{ Pa}}{2.2 \times 10^9 \text{ Pa}} \approx \frac{81.04 \times 10^5}{22 \times 10^8} \approx 0.00368$
Density at depth $\rho' = \rho + \Delta \rho = \rho(1 + \frac{\Delta \rho}{\rho})$
$\rho' = (1.03 \times 10^3) \times (1 + 0.00368) \approx 1.0337 \times 10^3 \text{ kg/m}^3 \approx 1.034 \times 10^3 \text{ kg/m}^3$.
Gravitational force $F = \frac{G M_1 M_2}{r^2}$. Since $M_1 + M_2 = M$, we can write $M_2 = M - M_1$.
$F = \frac{G M_1 (M - M_1)}{r^2}$. To find the maximum force, differentiate with respect to $M_1$ and set to zero:
$\frac{dF}{dM_1} = \frac{G}{r^2} (M - 2M_1) = 0 \Rightarrow M_1 = \frac{M}{2}$.
Therefore, $M_2 = \frac{M}{2}$, making the ratio $\frac{M_1}{M_2} = 1$.
In the given circuit, the diodes are arranged such that their p-sides (anodes) are connected to inputs A and B. The n-sides (cathodes) are tied together to output C, which is also connected to a 5V supply through a pull-up resistor R.
If either input A or B is at 0V (logic 0), the corresponding diode becomes forward-biased. It pulls the output node C down to approximately 0V (logic 0).
Only when both inputs A and B are at 5V (logic 1), both diodes are reverse-biased. No current flows to ground, and the output C remains at 5V (logic 1) due to the pull-up resistor.
This truth table corresponds to an AND Gate.
The Zener diode can withstand a maximum current of $I_{Z,max} = 4 \times I_{load} = 4 \times 5 \text{ mA} = 20 \text{ mA}$.
The series resistor $R_s$ must drop the excess voltage safely when the input voltage is at its maximum ($V_{in} = 25\text{V}$) and the load might be disconnected (meaning the full current goes through the Zener to protect it).
The maximum safe current from the source is $I_s = I_{Z,max} = 20 \text{ mA} = 0.02 \text{ A}$.
$R_s = \frac{V_{in} - V_Z}{I_s} = \frac{25 \text{ V} - 5 \text{ V}}{0.02 \text{ A}} = \frac{20}{0.02} = 1000 \Omega$.
The closest approach distance ($r_0$) provides an estimate of the maximum size of the nucleus.
At closest approach, Kinetic Energy = Electric Potential Energy.
$K = \frac{1}{4\pi\epsilon_0} \frac{(2e)(Ze)}{r_0} \Rightarrow r_0 = \frac{1}{4\pi\epsilon_0} \frac{2Ze^2}{K}$
$r_0 = (9 \times 10^9) \times \frac{2 \times 79 \times (1.6 \times 10^{-19})^2}{7.9 \times 10^6 \times 1.6 \times 10^{-19}} = 9 \times 10^9 \times \frac{158 \times 1.6 \times 10^{-19}}{7.9 \times 10^6}$
$r_0 = 9 \times 10^9 \times 20 \times 1.6 \times 10^{-25} = 288 \times 10^{-16} = 2.88 \times 10^{-14} \text{ m}$.
Using the equation of motion: $S = ut + \frac{1}{2}at^2$.
Taking upward direction as positive: displacement $S = -25 \text{ m}$, initial velocity $u = +20 \text{ m/s}$, acceleration $a = -g = -10 \text{ m/s}^2$.
$-25 = 20t - 5t^2 \Rightarrow 5t^2 - 20t - 25 = 0$
Dividing by 5: $t^2 - 4t - 5 = 0 \Rightarrow (t - 5)(t + 1) = 0$.
Since time cannot be negative, $t = 5 \text{ s}$.
The work done in the PV graph is the area under the curve. The given curve is the lower half of an ellipse (or circle in properly scaled coordinates) from $V=2$ to $V=4$.
The area under the curve = Area of the bounding rectangle - Area of the semi-ellipse.
Bounding rectangle area $= \text{width} \times \text{height} = (4-2)\text{ L} \times (2-0)\text{ atm} = 4 \text{ atm L}$.
Semi-ellipse horizontal radius $a = 1 \text{ L}$, vertical radius $b = 1 \text{ atm}$.
Area of semi-ellipse $= \frac{1}{2}\pi a b = \frac{\pi}{2}(1)(1) = \frac{\pi}{2} \text{ atm L}$.
Total Work Done $= 4 - \frac{\pi}{2} \text{ atm L}$.
Number density of molecules $n = \frac{P}{kT} = \frac{2 \times 1.013 \times 10^5}{1.38 \times 10^{-23} \times 290} \approx 5 \times 10^{25} \text{ m}^{-3}$.
Average speed $v = \sqrt{\frac{8RT}{\pi M}} \approx \sqrt{\frac{8 \times 8.31 \times 290}{3.14 \times 0.028}} \approx 470 \text{ m/s}$.
Collision frequency $f = \sqrt{2} \pi d^2 n v$, where $d = 2r = 2 \mathring{A} = 2 \times 10^{-10} \text{ m}$.
$f \approx 1.414 \times 3.14 \times (4 \times 10^{-20}) \times (5 \times 10^{25}) \times 470 \approx 4.2 \times 10^9 \text{ s}^{-1}$.
The closest order of magnitude is $10^9 \text{ s}^{-1}$.
When the suction pump is removed, the pressure difference causing the initial displacement vanishes. Gravity acts as a restoring force trying to level the mercury columns. Due to inertia, the liquid overshoots the equilibrium position, resulting in simple harmonic motion (SHM) around the equilibrium level.
Statement (2) is correct. Bats use echolocation, which involves emitting ultrasonic waves and interpreting the reflected waves (echoes) to determine the distance, direction, and size of surrounding objects.
*(Note: Statement 1 is false because a displacement node is a pressure antinode. Statement 3 is false due to differences in overtones/timbre. Statement 4 is false because dispersive mediums change the pulse shape.)*
The rubbing of synthetic clothes against the skin or other garments generates static electricity due to friction. When you take them off, the sudden separation of the charged surfaces leads to a rapid electric discharge through the air, producing the visible spark and audible crackle.
The net polarisation of a dielectric material is a result of the competition between two opposing factors: the external electric field (which provides dipole potential energy tending to align the molecular dipoles) and thermal agitation (thermal energy which tends to randomize and disrupt this alignment). Thus, both factors are involved.
Kepler's Third Law states that the square of the time period ($T^2$) is directly proportional to the cube of the semi-major axis or average distance ($r^3$). Therefore, $T^2 = k \cdot r^3$.
If we plot $T$ on the y-axis and $r^3$ on the x-axis, the equation becomes $y^2 = k \cdot x$, which mathematically represents a parabola.
In a closed organ pipe (closed at one end, open at the other), the fundamental mode ($0^{\text{th}}$ overtone) has exactly 1 node and 1 antinode.
The $1^{\text{st}}$ overtone has 2 nodes, the $2^{\text{nd}}$ overtone has 3 nodes, and generally, the $n^{\text{th}}$ overtone contains $(n+1)$ nodes.
Since the pipe is vibrating with 10 nodes, $n+1 = 10 \Rightarrow n = 9$. It is vibrating in its $9^{\text{th}}$ overtone.
In the half-deflection method for finding galvanometer resistance, the resistance of the galvanometer $G$ is given by the formula:
$G = \frac{RS}{R - S}$
Substituting the given values $R = 20 \Omega$ and $S = 12 \Omega$:
$G = \frac{20 \times 12}{20 - 12} = \frac{240}{8} = 30 \Omega$.
The dot (scalar) product of two vectors is commutative, meaning the order of the vectors does not change the result. Therefore, $\vec{A} \cdot \vec{B} = \vec{B} \cdot \vec{A}$ is a true statement.
*(Note: The cross product is anti-commutative, $\vec{A} \times \vec{B} = -(\vec{B} \times \vec{A})$).*
Centripetal acceleration is $a = \frac{v^2}{R}$.
If the velocity is halved ($v' = v/2$) while the radius $R$ remains the same:
$a' = \frac{(v/2)^2}{R} = \frac{v^2 / 4}{R} = \frac{1}{4} \left(\frac{v^2}{R}\right) = \frac{a}{4}$.
Magnetic flux $\Phi = B \cdot A = B \cdot \pi r^2$.
According to Faraday's law of induction, $E = -\frac{d\Phi}{dt} = -B \cdot \pi \cdot 2r \frac{dr}{dt}$.
Given $B = 0.025 \text{ T}$, rate of shrinking $\frac{dr}{dt} = -1 \text{ mm/s} = -10^{-3} \text{ m/s}$, and $r = 2 \text{ cm} = 0.02 \text{ m}$.
$E = -0.025 \times \pi \times 2(0.02) \times (-10^{-3}) = 0.001 \times 10^{-3} \times \pi = \pi \times 10^{-6} \text{ V} = \pi \mu\text{V}$.
The work done in rotating a magnetic dipole in a magnetic field from angle $\theta_1$ to $\theta_2$ is $W = -MB(\cos\theta_2 - \cos\theta_1)$.
Work to rotate from $0^{\circ}$ to $90^{\circ}$: $W_{90} = -MB(\cos 90^{\circ} - \cos 0^{\circ}) = -MB(0 - 1) = MB$.
Work to rotate from $0^{\circ}$ to $60^{\circ}$: $W_{60} = -MB(\cos 60^{\circ} - \cos 0^{\circ}) = -MB(0.5 - 1) = 0.5MB$.
Given $W_{90} = n W_{60} \Rightarrow MB = n(0.5MB) \Rightarrow n = 2$.
Fringe width in Young's double slit experiment is given by $\beta = \frac{\lambda D}{d}$.
The fringe width ($\beta$) is directly proportional to the distance ($D$) between the slits and the screen. Therefore, if the separation $D$ increases, the fringe width will also increase.
At the lowest point of a vertical circular path, both tension ($T$) acting upwards toward the center and weight ($mg$) acting downwards are aligned along the vertical axis.
The net force providing the required centripetal acceleration is towards the center: $T - mg = \frac{mv^2}{r}$.
Therefore, $T = \frac{mv^2}{r} + mg$.
A negative zero error occurs when the zero mark of the Vernier scale lies to the left of the zero mark of the main scale.
Assuming the standard alignment case where the 7th division coincides on the leftward shifted Vernier scale (which aligns with typical -0.03 error problems in this context):
Negative Zero Error $= -(\text{Total divisions} - \text{Coinciding division}) \times \text{Least Count}$
Error $= -(10 - 7) \times 0.1 \text{ mm} = -0.3 \text{ mm} = -0.03 \text{ cm}$.
For an orbiting satellite, Kinetic Energy $KE = \frac{GMm}{2r}$ and Potential Energy $PE = -\frac{GMm}{r}$.
If the radius $r$ is increased, the $KE$ clearly decreases due to the larger denominator.
The $PE$ is negative; increasing $r$ makes the magnitude $\frac{GMm}{r}$ smaller, meaning the value becomes less negative. Moving from a larger negative number to a smaller negative number constitutes an increase in potential energy.
When $n$ drops combine to form a big drop, charge is conserved: $Q_B = n q_S$.
Volume is conserved: $\frac{4}{3}\pi R_B^3 = n \frac{4}{3}\pi r_S^3 \Rightarrow R_B = n^{1/3} r_S$.
Capacitance of a spherical drop is proportional to its radius, so $C_B = n^{1/3} C_S$.
Energy stored in a drop is $U = \frac{Q^2}{2C}$.
$\frac{U_B}{U_S} = \frac{(n q_S)^2 / (2 n^{1/3} C_S)}{q_S^2 / (2 C_S)} = \frac{n^2}{n^{1/3}} = n^{2 - 1/3} = n^{5/3}$.
Applying Kirchhoff's voltage path rule from point A to point B in the direction of the 2A current:
$V_A - I(2\Omega) - E - I(1\Omega) = V_B$
$V_A - 2(2) - 3 - 2(1) = V_B$
$V_A - 4 - 3 - 2 = V_B \Rightarrow V_A - 9 = V_B$
$V_A - V_B = +9 \text{ V}$.
The circuit features two identical resistors $R$ connected in parallel across a 15V source.
Equivalent resistance $R_{eq} = \frac{R}{2}$.
Power dissipated $P = \frac{V^2}{R_{eq}} \Rightarrow 150 = \frac{15^2}{R/2} = \frac{225 \times 2}{R} = \frac{450}{R}$.
$R = \frac{450}{150} = 3 \Omega$.
Let the mass of both balls be $m$. Initial velocity of first ball $= u_1$, second ball $u_2 = 0$.
After collision, first ball stops ($v_1 = 0$). By conservation of momentum: $mu_1 + 0 = m(0) + mv_2 \Rightarrow v_2 = u_1$.
The coefficient of restitution $e = \frac{\text{Relative velocity of separation}}{\text{Relative velocity of approach}} = \frac{v_2 - v_1}{u_1 - u_2}$.
$e = \frac{u_1 - 0}{u_1 - 0} = 1$. This indicates a perfectly elastic collision.
The radius of the helical path is determined by the perpendicular component of velocity relative to the magnetic field.
$R = \frac{m v_{\perp}}{qB} = \frac{v \sin\theta}{(q/m)B}$
$R = \frac{3 \times 10^5 \times \sin 30^{\circ}}{10^8 \times 0.3} = \frac{3 \times 10^5 \times 0.5}{3 \times 10^7} = \frac{1.5 \times 10^5}{3 \times 10^7} = 0.5 \times 10^{-2} \text{ m}$.
$R = 0.5 \text{ cm}$.
The magnetic field magnitude at a distance $r=1\text{m}$ from a long straight wire is $B = \frac{\mu_0 I}{2\pi r}$.
$B = \frac{4\pi \times 10^{-7} \times 10}{2\pi \times 1} = 2 \times 10^{-6} \text{ T} = 2 \mu\text{T}$.
Using the right-hand grip rule, for a current along the +z-axis, the magnetic field lines form counterclockwise circles in the xy-plane. At a point on the +x-axis, the tangent points in the +y direction.
The force exerted on a test charge $q$ placed on the x-axis by two symmetric charges $Q$ at $(0, d/2)$ and $(0, -d/2)$ is directed along the x-axis.
$F_x = \frac{2kQq x}{(x^2 + (d/2)^2)^{3/2}}$.
At the origin ($x=0$), the force is zero. As $x \to \infty$, the force also approaches zero. The force reaches a maximum magnitude at $x = \frac{d}{2\sqrt{2}}$. This corresponds to the standard curve rising from origin to a peak and then gradually dropping toward the x-axis.
The coefficient of friction ($\mu$) is the ratio of two forces: the frictional force to the normal reaction force ($\mu = F_f / N$).
Since it is a ratio of identical physical quantities (forces), it is a dimensionless quantity.
Therefore, its dimensional formula has zero power for all fundamental dimensions, represented generally as $[M^0 L^0 T^0]$, or corresponding zero exponents for any substituted base quantities ($\alpha^0 \beta^0 \gamma^0$ or similar purely dimensionless forms).
Variation of $g$ with height (for $h \ll R$): $g_h = g(1 - \frac{2h}{R})$.
Variation of $g$ with depth: $g_d = g(1 - \frac{d}{R})$.
If gravity is equal at these positions, $g_h = g_d \Rightarrow 1 - \frac{2h}{R} = 1 - \frac{d}{R}$.
Canceling terms gives $\frac{d}{R} = \frac{2h}{R} \Rightarrow d = 2h$.
Statement 1 is incorrect because Bernoulli's equation relies on the assumption of streamline (laminar), non-viscous, and steady flow. A rapidly flowing river is highly turbulent, making Bernoulli's principle inapplicable.
Statement 2 is incorrect because while pressure differences are functionally the same, Bernoulli's equation theoretically relates the absolute total pressure states. Mixing gauge and absolute inconsistently causes errors, and the equation is formally established using absolute pressure.
Trade winds are driven by global atmospheric convection cells. The equatorial regions receive more direct solar heating than polar regions, creating thermal pressure gradients (unequal heating). As this air moves, the rotation of the Earth applies the Coriolis effect, deflecting the winds and creating the distinct directional pattern known as the trade winds.
Distance is the integral of velocity over time: $S = \int_{t_1}^{t_2} v \, dt$.
$S = \int_{2}^{3} (2t + 6t^2) \, dt = \left[ t^2 + 2t^3 \right]_{2}^{3}$
$S = (3^2 + 2(3^3)) - (2^2 + 2(2^3)) = (9 + 54) - (4 + 16)$
$S = 63 - 20 = 43 \text{ units}$.
When charges move in an electric field without suffering collisions (which typically dissipate energy as heat), the work done by the electric field entirely converts into the kinetic energy of the charges. Because it is a conservative system with no dissipative losses, the total mechanical energy (Kinetic + Potential) remains conserved or unchanged.
By Lens Maker's formula, the focal length of a lens in a medium is $f_m = f_a \left( \frac{\mu_g - 1}{\frac{\mu_g}{\mu_m} - 1} \right)$.
Given $f_a = 0.15 \text{ m}, \mu_g = \frac{3}{2}, \mu_w = \frac{4}{3}$.
$f_w = 0.15 \times \frac{(\frac{3}{2} - 1)}{(\frac{3/2}{4/3} - 1)} = 0.15 \times \frac{1/2}{9/8 - 1} = 0.15 \times \frac{1/2}{1/8} = 0.15 \times 4 = 0.6 \text{ m}$.
The magnifying power of an astronomical telescope in normal adjustment is $M = \frac{f_o}{f_e}$.
Given $M = 10$ and focal length of eye-piece $f_e = 20 \text{ cm}$.
$10 = \frac{f_o}{20} \Rightarrow f_o = 10 \times 20 = 200 \text{ cm}$.
The maximum acceleration in Simple Harmonic Motion is given by $a_{max} = \omega^2 A$.
Since both SHMs have the same displacement amplitude $A$, the ratio of their maximum accelerations is:
$\frac{a_1}{a_2} = \frac{\omega_1^2 A}{\omega_2^2 A} = \left(\frac{\omega_1}{\omega_2}\right)^2 = \left(\frac{100}{1000}\right)^2 = \left(\frac{1}{10}\right)^2 = \frac{1}{100}$.
Therefore, the ratio is $1 : 10^2$.
Horizontal velocity acquired: $u_x = at = 3 \text{ m/s}^2 \times 30 \text{ s} = 90 \text{ m/s}$.
For a projectile, maximum height $H = \frac{u_y^2}{2g}$.
Given $H = 80 \text{ m} \Rightarrow \frac{u_y^2}{2(10)} = 80 \Rightarrow u_y^2 = 1600 \Rightarrow u_y = 40 \text{ m/s}$.
Angle of projection $\tan\theta = \frac{u_y}{u_x} = \frac{40}{90} = \frac{4}{9}$.
Therefore, $\theta = \tan^{-1}\left(\frac{4}{9}\right)$.