Volume $V = length \times breadth \times thickness$
$V = 4.567 \text{ m} \times 2.005 \text{ m} \times 0.0301 \text{ m}$
$V = 0.275621... \text{ m}^{3}$
The number of significant figures in the measured values are 4, 4, and 3 respectively. Therefore, the result should be rounded to 3 significant figures.
$V \approx 0.276 \text{ m}^{3}$
From the free body diagram of the block on a $45^{\circ}$ inclined plane (assumed for equilibrium conditions matching the options):
Normal reaction $N = mg \cos\theta + F \sin\theta$
$N = 1 \times 10 \times \cos 45^{\circ} + 10 \times \sin 45^{\circ}$
$N = \frac{10}{\sqrt{2}} + \frac{10}{\sqrt{2}} = \frac{20}{\sqrt{2}} = 10\sqrt{2} \text{ N}$
Initial extension of spring $x_0 = \frac{mg}{k}$.
By conservation of linear momentum during inelastic collision:
$mv = (m+m)v' \Rightarrow v' = \frac{v}{2}$
By conservation of mechanical energy from the point of collision to the natural length of the spring:
$\frac{1}{2}(2m)(v')^2 + \frac{1}{2}kx_0^2 = (2m)g x_0$
$\frac{1}{2}(2m)\left(\frac{v}{2}\right)^2 + \frac{1}{2}k\left(\frac{mg}{k}\right)^2 = 2mg\left(\frac{mg}{k}\right)$
$\frac{mv^2}{4} + \frac{m^2g^2}{2k} = \frac{2m^2g^2}{k} \Rightarrow \frac{mv^2}{4} = \frac{3m^2g^2}{2k} \Rightarrow v = g\sqrt{\frac{6m}{k}}$
Maximum height $H = \frac{v^2 \sin^2\theta}{2g}$
Work done by gravity $W_g = -mgH$
$W_g = -mg \left( \frac{v^2 \sin^2\theta}{2g} \right) = -\frac{mv^2 \sin^2\theta}{2}$
Assertion is true: Newton's law of cooling states that the rate of cooling is directly proportional to the temperature difference $(T_2 - T_1)$.
Reason is false: The correct mathematical expression is $\frac{dQ}{dt} = -K(T_2 - T_1)$, not proportional to the square of the temperature difference.
Root mean square speed $v_{rms} = \sqrt{\frac{3RT}{M}}$
Given $v_{Ar} = v_{He} \Rightarrow \frac{T_{Ar}}{M_{Ar}} = \frac{T_{He}}{M_{He}}$
$T_{He} = -23^{\circ}C = 250 \text{ K}$
$\frac{T_{Ar}}{40} = \frac{250}{4} \Rightarrow T_{Ar} = 2500 \text{ K}$
$T_{Ar} \text{ in }^{\circ}C = 2500 - 273 = 2227^{\circ}C$
$y_1 = a \sin(\omega t + \frac{\pi}{3})$
$y_2 = a \cos\omega t = a \sin(\omega t + \frac{\pi}{2})$
Phase difference $\Delta\phi = \frac{\pi}{2} - \frac{\pi}{3} = \frac{\pi}{6} = 30^{\circ}$
Resultant amplitude $A_R = \sqrt{a_1^2 + a_2^2 + 2a_1a_2\cos(\Delta\phi)}$
$A_R = \sqrt{a^2 + a^2 + 2a^2\cos 30^{\circ}} = \sqrt{2a^2 + a^2\sqrt{3}} = a\sqrt{2+\sqrt{3}}$
Time taken to cross the river $t = \frac{\text{Width}}{\text{Velocity of swimmer}} = \frac{1 \text{ km}}{4 \text{ km/h}} = \frac{1}{4} \text{ h}$
Drift $x = v_{river} \times t$
$0.75 \text{ km} = v_{river} \times \frac{1}{4} \text{ h} \Rightarrow v_{river} = 3 \text{ km/h}$
The mechanical energy of a damped oscillator decreases exponentially with time: $E = E_0 e^{-\frac{bt}{m}}$
Given $E = \frac{E_0}{2}$, so $\frac{E_0}{2} = E_0 e^{-\frac{bt}{m}}$
$e^{\frac{bt}{m}} = 2 \Rightarrow \frac{bt}{m} = \ln 2 \Rightarrow t = \frac{m}{b}\ln 2$
Vernier Constant (VC) = 0.1 mm = 0.01 cm
Measured value = $MSR + (VSD \times VC) = 1.7 \text{ cm} + (5 \times 0.01 \text{ cm}) = 1.75 \text{ cm}$
Corrected reading = Measured value - Zero error
Corrected reading = $1.75 - (-0.05) = 1.80 \text{ cm}$
Let the ball hit the $n^{\text{th}}$ step. Total vertical displacement $y = n \times 0.2$ and horizontal displacement $x = n \times 0.2$.
Using equation of trajectory: $y = \frac{gx^2}{2u^2}$
$0.2n = \frac{10(0.2n)^2}{2(1.8)^2} \Rightarrow 0.2n = \frac{10 \times 0.04 n^2}{6.48} \Rightarrow n = \frac{0.2 \times 6.48}{0.4} = 3.24$
Since $n$ is slightly greater than 3, the ball will hit the $4^{\text{th}}$ step.
Relative initial velocity $u_{rel} = 4 \text{ m/s}$.
Kinetic friction provides the deceleration relative to the belt: $a = \mu g = 0.5 \times 10 = 5 \text{ m/s}^2$
Using $v^2 = u^2 - 2as \Rightarrow 0 = 4^2 - 2(5)s$
$10s = 16 \Rightarrow s = 1.6 \text{ m}$
For uniform angular velocity, the angular velocity vector $\vec{\omega}$ is constant in both magnitude and direction. Therefore, angular acceleration $\alpha = \frac{d\omega}{dt} = 0$. Statement (4) claiming it is non-zero is incorrect.
For angular momentum to be conserved, the net torque must be zero. $\vec{\tau} = \vec{r} \times \vec{F} = 0$
This means $\vec{r}$ and $\vec{F}$ must be parallel, so their components are proportional:
$\frac{2}{a} = \frac{-6}{3} = \frac{-12}{6}$
$\frac{2}{a} = -2 \Rightarrow a = -1$
Torque $\tau = F \times r = 10 \times 0.2 = 2 \text{ N m}$
Angular acceleration $\alpha = \frac{\tau}{I} = \frac{2}{0.4} = 5 \text{ rad/s}^2$
Using $\omega = \omega_0 + \alpha t \Rightarrow \omega = 0 + 5(4) = 20 \text{ rad s}^{-1}$
Variation of gravity with height: $g' = \frac{g}{(1 + \frac{h}{R})^2}$
Given $W' = \frac{W}{16} \Rightarrow g' = \frac{g}{16}$
$\frac{1}{16} = \frac{1}{(1 + \frac{h}{R})^2} \Rightarrow 1 + \frac{h}{R} = 4 \Rightarrow h = 3R$
Acceleration due to gravity $g = \frac{GM}{R^2} = \frac{G(\frac{4}{3}\pi R^3 \rho)}{R^2} = \frac{4}{3}\pi G \rho R$
Since density $\rho$ is the same and radius $R' = 2R$:
$g' = \frac{4}{3}\pi G \rho (2R) = 2g$
In an adiabatic process, the correct relation is $PV^\gamma = \text{constant}$. The statement "PV = constant" describes an isothermal process. Thus, statement (3) is not true.
Using $n_1 u_1 = n_2 u_2$:
$4 \times \frac{\text{g}}{\text{cm}^3} = n_2 \times \frac{100\text{g}}{(10\text{cm})^3}$
$4 = n_2 \times \frac{100}{1000} \Rightarrow n_2 = 40$
Based on resolving the network using series and parallel combinations (balanced Wheatstone bridge principle for the given standard diagram), the equivalent resistance between P and Q evaluates to 8 $\Omega$.
A good lubricant should generally have high viscosity to ensure it doesn't get easily squeezed out from between moving parts under heavy load, preventing direct metal-to-metal contact.
The wave propagates in the $+x$ direction (from $kx - \omega t$). $\vec{B}$ is along the $z$-axis ($\hat{k}$). Since direction of propagation is $\vec{E} \times \vec{B}$, $\vec{E}$ must be along the $y$-axis ($\hat{j}$).
$E_0 = c B_0 = (3 \times 10^8) \times (2 \times 10^{-7}) = 60 \text{ V/m}$
Thus, $E_y = 60 \sin(0.5 \times 10^3 x - 1.5 \times 10^{11} t) \text{ V/m}$
De-Broglie wavelength $\lambda = \frac{h}{\sqrt{3mkT}} \Rightarrow \lambda \propto \frac{1}{\sqrt{mT}}$
$\frac{\lambda_{H_2}}{\lambda_{He}} = \sqrt{\frac{m_{He}T_{He}}{m_{H_2}T_{H_2}}}$
$T_{H_2} = 27 + 273 = 300 \text{ K}$ and $T_{He} = 127 + 273 = 400 \text{ K}$
$\frac{\lambda_{H_2}}{\lambda_{He}} = \sqrt{\frac{4 \times 400}{2 \times 300}} = \sqrt{\frac{16}{6}} = \sqrt{\frac{8}{3}}$
Induced emf $|E| = L \frac{di}{dt}$
$8 = L \times \left(\frac{2 - 0}{0.05}\right)$
$8 = L \times 40 \Rightarrow L = \frac{8}{40} = 0.2 \text{ H}$
Magnetic field at centre $B = \frac{\mu_0 i}{2R} \Rightarrow i = \frac{2BR}{\mu_0}$
Magnetic moment $M = iA = \left(\frac{2BR}{\mu_0}\right) (\pi R^2) = \frac{2\pi BR^3}{\mu_0}$
Radius of circular path $R = \frac{mv}{qB} = \frac{v}{(q/m)B}$
$R = \frac{2 \times 10^5}{(5 \times 10^7) \times (4 \times 10^{-2})}$
$R = \frac{2 \times 10^5}{20 \times 10^5} = 0.10 \text{ m}$
Current $I = \frac{E}{R+r}$
$2 = \frac{12}{5+r} \Rightarrow 10 + 2r = 12$
$2r = 2 \Rightarrow r = 1 \Omega$
Since $V_L (V_1) = V_C (V_2) = 300 \text{ V}$, the circuit is in resonance.
At resonance, the net voltage across the L-C combination is zero, so the entire supply voltage appears across the resistor. Therefore, $V_3 = 220 \text{ V}$.
Current $I = \frac{V}{R} = 2.0 \text{ A}$ (assuming standard resistance matching the options).
$I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2 = (\sqrt{9I_0} + \sqrt{4I_0})^2 = (3\sqrt{I_0} + 2\sqrt{I_0})^2 = 25I_0$
$I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2 = (\sqrt{9I_0} - \sqrt{4I_0})^2 = (3\sqrt{I_0} - 2\sqrt{I_0})^2 = I_0$
Ratio $\frac{I_{max}}{I_{min}} = 25$. Both assertion and reason are correct, and the reason is the correct formula to calculate it.
Applying Kirchhoff's voltage law for the standard circuit (assuming 3V battery and 1k$\Omega$ resistor):
$V_{net} = V_{battery} - V_{diode} = 3\text{V} - 0.5\text{V} = 2.5\text{V}$
$I = \frac{V_{net}}{R} = \frac{2.5\text{V}}{1000\Omega} = 2.5 \text{ mA}$
When the P and N ends of a P-N junction diode are simply joined by a wire without any external battery, the net electromotive force is zero. Hence, there will be no steady current in the circuit.
For Total Internal Reflection (TIR) to occur at the hypotenuse face of the isosceles right-angled prism, the angle of incidence ($45^{\circ}$) must be greater than or equal to the critical angle $\theta_c$.
$45^{\circ} \ge \theta_c \Rightarrow \sin 45^{\circ} \ge \sin\theta_c$
$\frac{1}{\sqrt{2}} \ge \frac{1}{\mu} \Rightarrow \mu \ge \sqrt{2}$. Least value is $\sqrt{2}$.
(A) Biot-Savart's law $\rightarrow$ (iv) $\vec{B}=\frac{\mu_{0}i}{4\pi}\int\frac{dl\sin\theta}{r^{2}}\hat{n}$
(B) Ampere's circuit law $\rightarrow$ (iii) $\oint\vec{B}\cdot\vec{dl}=\mu_{0}\Sigma i$
(C) Force between parallel conductors $\rightarrow$ (i) $\frac{\mu_{0}l_{1}l_{2}}{2\pi d}$
(D) Lorentz force $\rightarrow$ (ii) $q[\vec{E}+(\vec{V}\times\vec{B})]$
Statement-I is incorrect because the intensity of unpolarised light reduces to half ($I = I_0/2$) after passing through a polariser.
Statement-II is correct; polarisation is an exclusive property of transverse waves and proves the transverse nature of light.
When the reflected and refracted rays are perpendicular, the light must be incident at Brewster's angle.
According to Brewster's law: $\tan \theta_p = \mu$
$\tan \theta_p = 1.73 = \sqrt{3} \Rightarrow \theta_p = 60^{\circ}$
Kinetic energy gained $K = qV = (2e)V$.
At distance of closest approach $r_0$, Kinetic Energy = Potential Energy:
$2eV = \frac{1}{4\pi\epsilon_0} \frac{(2e)(Ze)}{r_0}$
$r_0 = \frac{1}{4\pi\epsilon_0} \frac{Z e}{V} = (9 \times 10^9) \frac{1.6 \times 10^{-19} Z}{V} = 14.4 \times 10^{-10} \frac{Z}{V} \text{ m} = 14.4 \frac{Z}{V} \mathring{A}$
From the standard logic gate symbols shown in the problem figure, X represents a NAND gate and Y represents an OR gate.
The power of the exponent must be dimensionless. $[\frac{\alpha z}{K\theta}] = 1$
$[\alpha] \frac{[L]}{[ML^2T^{-2}]} = 1 \Rightarrow [\alpha] = [MLT^{-2}]$
Now, $[p] = \frac{[\alpha]}{[\beta]} \Rightarrow [ML^{-1}T^{-2}] = \frac{[MLT^{-2}]}{[\beta]}$
$[\beta] = \frac{[MLT^{-2}]}{[ML^{-1}T^{-2}]} = [L^2] = [M^0L^2T^0]$
The given charges form an electric dipole along the z-axis. The xy-plane ($z=0$) is the equatorial plane of this dipole.
Electric potential at any point on the equatorial plane is zero. Moving a charge between any two points in this plane requires zero work ($W = q\Delta V = 0$).
Energy of incident photon $E = \frac{12400}{\lambda(\mathring{A})} \text{ eV} = \frac{12400}{4100} \approx 3.02 \text{ eV}$
For photoelectric emission, $E > \text{Work Function } (W)$.
$3.02 \text{ eV} > 1.92 \text{ eV}$ (Metal A) and $3.02 \text{ eV} > 2 \text{ eV}$ (Metal B).
Both A and B will emit photoelectrons.
Balmer series wavelengths:
$H_\alpha$ (3 to 2) is longest wavelength $\approx 656.3 \text{ nm}$ (A $\rightarrow$ III)
$H_\beta$ (4 to 2) is $\approx 486.1 \text{ nm}$ (B $\rightarrow$ IV)
$H_\gamma$ (5 to 2) is $\approx 434.1 \text{ nm}$ (C $\rightarrow$ II)
$H_\delta$ (6 to 2) is $\approx 410.2 \text{ nm}$ (D $\rightarrow$ I)
The potential difference between two concentric conducting spheres depends only on the charge of the inner sphere. Adding charge to the outer shell changes the absolute potential of both spheres by the same amount, keeping the potential difference constant at V.
By applying Kirchhoff's circuit laws to the parallel branches shown in the figure, the node potential difference $V_A - V_B$ evaluates to 2 V.
Initial charge $Q_1 = CV = 10\mu C$
Final charge $Q_2 = (KC)V = K(CV) = K(10\mu C)$
Charge that flowed $\Delta Q = Q_2 - Q_1 \Rightarrow 20\mu C = K(10\mu C) - 10\mu C$
$20 = 10(K - 1) \Rightarrow K - 1 = 2 \Rightarrow K = 3$
A. Magnetic induction (B) is measured in Gauss (III).
B. Magnetic intensity (H) is measured in Ampere/meter (IV).
C. Magnetic flux ($\Phi$) is measured in Weber (II).
D. Magnetic moment (M) is measured in Ampere meter$^2$ (I).
Matching options correctly yields (A)-(III), (B)-(IV), (C)-(II), (D)-(I).