For a uniform rod, the mass is evenly distributed along its entire length. Due to spatial symmetry, the center of mass will lie exactly at its geometric center, which is its middle point.
When the elevator is at rest or moving with uniform velocity, the tension in the string is $T = mg$.
When the elevator accelerates upwards with acceleration $a$, the pseudo-force acting on the body relative to the elevator is $ma$ downwards. Thus, the new tension is $T' = m(g+a)$.
Given $T' = 2T \Rightarrow m(g+a) = 2mg$
$g+a = 2g \Rightarrow a = g = 9.8 \text{ m/s}^{2}$.
The conservative force is the negative gradient of potential energy:
$F = -\frac{dU}{dx}$
Given $U(x) = -6x^{2} + 2x$, differentiating it yields:
$\frac{dU}{dx} = -12x + 2$
$F = -(-12x + 2) = 12x - 2$
At $x=1$ m, $F = 12(1) - 2 = 10 \text{ N}$.
For a mass tied to a string to just complete a full vertical circle without the string becoming slack, the minimum speed at the lowest point is given by:
$v_{min} = \sqrt{5gR}$
Given $R = 1 \text{ m}$ and $g = 9.8 \text{ m/s}^{2}$.
$v_{min} = \sqrt{5 \times 9.8 \times 1} = \sqrt{49} = 7 \text{ m/s}$.
Assertion is False: A body traveling in a circular path at a constant speed is continually changing its direction, which means its velocity is changing. Thus, it undergoes acceleration (centripetal acceleration).
Reason is True: To traverse a curved path, a centripetal force is mandatory to change the direction of velocity. Therefore, it is impossible to round a curve with zero acceleration.
The escape velocity from a planet is given by $v_{e} = \sqrt{2gR}$.
Given $\frac{R_{A}}{R_{B}} = k_{1}$ and $\frac{g_{A}}{g_{B}} = k_{2}$.
The ratio of escape velocities is $\frac{v_{eA}}{v_{eB}} = \sqrt{\frac{g_{A}R_{A}}{g_{B}R_{B}}} = \sqrt{k_{2}k_{1}}$.
Excess pressure inside a liquid drop is $P = \frac{2T}{R}$. So, $P \propto \frac{1}{R}$.
Given $P_{1} = 4P_{2}$, this means $R_{2} = 4R_{1}$, or $\frac{R_{1}}{R_{2}} = \frac{1}{4}$.
Mass of a drop $M = \frac{4}{3}\pi R^{3}\rho$. Therefore, $M \propto R^{3}$.
Ratio of masses $\frac{M_{1}}{M_{2}} = (\frac{R_{1}}{R_{2}})^{3} = (\frac{1}{4})^{3} = \frac{1}{64}$, or $1:64$.
Terminal velocity of a drop $v \propto r^{2}$.
When two equal drops coalesce, the volume of the new drop is twice the original volume: $\frac{4}{3}\pi R^{3} = 2 \times \frac{4}{3}\pi r^{3} \Rightarrow R = 2^{1/3}r$.
The new terminal velocity $v' \propto R^{2} \propto (2^{1/3}r)^{2} \propto 2^{2/3}r^{2}$.
Therefore, $v' = 2^{2/3}v = 4^{1/3} \times 5 \text{ cm/sec} = 5 \times (4)^{1/3} \text{ cm/sec}$.
In steady (or streamline) flow of a fluid, the velocity of fluid particles at any given fixed point in space remains constant over time. The path traced by a fluid particle in such a flow is called a streamline.
Linear thermal expansion is given by $\Delta L = L \alpha \Delta T$.
Percentage increase in length $= \frac{\Delta L}{L} \times 100 = (\alpha \Delta T) \times 100$.
Given $\alpha = 10^{-5} {}^{\circ}C^{-1}$ and $\Delta T = 500^{\circ}C$..
$\% \text{ increase} = (10^{-5} \times 500) \times 100 = 5 \times 10^{-3} \times 100 = 0.5\%$.
Assertion is Correct: Magnetic susceptibility $\chi$ is defined as the ratio of the intensity of magnetization (I) induced in the material to the magnetizing field intensity (H), i.e., $\chi = \frac{I}{H}$.
Reason is Incorrect: According to the formula, $\chi$ is directly proportional to I (for a given H). Therefore, a greater value of susceptibility implies a larger intensity of magnetization, not a smaller one.
Let's analyze the statements using the electromagnetic spectrum wavelength order: Gamma rays < X-rays < UV < Visible < IR < Microwaves < Radio waves.
(I) $\lambda_{microwave} > \lambda_{UV}$ (Correct)
(II) $\lambda_{IR} < \lambda_{UV}$ (Incorrect, IR is longer)
(III) $\lambda_{microwave} < \lambda_{IR}$ (Incorrect, microwave is longer)
(IV) Gamma rays have the shortest wavelength (Correct)
Thus, statements I and IV are correct.
The resistance of a block is $R = \rho\frac{L}{A}$.
For block A: Length along current direction $= l$. Cross-sectional area perpendicular to current $= l \times t$. So, $R_{A} = \rho\frac{l}{lt} = \frac{\rho}{t}$.
For block B: Length along current direction $= 2l$. Cross-sectional area $= 2l \times t$. So, $R_{B} = \rho\frac{2l}{2lt} = \frac{\rho}{t}$.
Therefore, the ratio $\frac{R_{A}}{R_{B}} = \frac{\rho/t}{\rho/t} = 1/1$.
Manganin and constantan are alloys specially designed to have a very small temperature coefficient of resistivity. Their resistivity is fairly high and remains almost constant with a weak, roughly linear increase over a wide range of temperatures. Graph (2) correctly depicts this slight linear increase. Graph (1) represents a pure metal like copper at low temperatures, and (4) represents a semiconductor.
Let's analyze the nodes in the given circuit diagram from the far left towards nodes P and Q.
Simplifying the network: The leftmost series combination (2C and 2C) gives equivalent C. This is in parallel with the horizontal C, making it 2C. In the next stage, this 2C is in series with another 2C, giving C. This C is in parallel with another horizontal C, giving 2C. Finally, this equivalent 2C is in series with the top 2C, yielding an equivalent of C.
This equivalent capacitance C is in parallel with the directly connected 2C capacitor between terminals P and Q.
Thus, total equivalent capacitance $C_{eq} = C + 2C = 3C$.
Given: Two coils from the same wire. Radius $r_1 = 2r_2$. Assuming the wire is divided or standard case of same total length used for coils: Total length $L_1 = L_2 = L \Rightarrow 2\pi r_1 N_1 = 2\pi r_2 N_2 \Rightarrow N_1 = N_2/2$.
Since they are made from the same wire, their resistances are equal: $R_1 = R_2 = R$.
Magnetic field at center $B = \frac{\mu_0 N i}{2r}$. Since $B_1 = B_2$, we have $\frac{N_1 i_1}{r_1} = \frac{N_2 i_2}{r_2}$.
Substitute $N_1 = N_2/2$ and $r_1 = 2r_2$: $\frac{(N_2/2)i_1}{2r_2} = \frac{N_2 i_2}{r_2} \Rightarrow \frac{i_1}{4} = i_2 \Rightarrow i_1 = 4i_2$.
Voltage ratio $V_1/V_2 = (i_1 R_1) / (i_2 R_2) = (4i_2 \cdot R) / (i_2 \cdot R) = 4/1$.
(A) Electric field due to an infinite plane sheet of charge is $\frac{\sigma}{2\epsilon_0}$. (Matches with 2)
(B) Electric field due to an infinite conducting plane sheet (with charge on both faces) is $\frac{\sigma}{\epsilon_0}$. (Matches with 4)
(C) Electric field at the surface of a uniformly charged solid non-conducting sphere is $E = \frac{kQ}{R^2} = \frac{1}{4\pi\epsilon_0} \frac{\frac{4}{3}\pi R^3 \rho}{R^2} = \frac{R\rho}{3\epsilon_0}$. (Matches with 3)
(D) Electric field at the center of a conducting charged solid sphere is 0. (Matches with 1)
Correct matching is A-2, B-4, C-3, D-1.
In an LCR series resonant circuit, the resonant frequency ($\omega_r$) is the geometric mean of the two half-power frequencies ($\omega_1$ and $\omega_2$).
Therefore, $\omega_r = \sqrt{\omega_1 \omega_2}$, which can be rewritten as $\omega_1 \times \omega_2 = \omega_r^2$.
The diode acts as a simple voltage drop of 0.5 V in the forward-biased condition.
Net voltage available across the resistor = Supply voltage - Barrier potential.
According to the circuit diagram, the battery voltage is 3.5 V.
$V_R = 3.5 \text{ V} - 0.5 \text{ V} = 3.0 \text{ V}$.
Current $I = \frac{V_R}{R} = \frac{3.0}{100} \text{ A} = 0.030 \text{ A} = 30 \text{ mA}$.
When reflected and refracted rays are mutually perpendicular, Brewster's law applies. The incident angle is Brewster's angle, $\theta_p$.
The problem states "incident at an angle of $30^{\circ}$". If this refers to the glancing angle (angle with the surface), the angle of incidence $i = 90^{\circ} - 30^{\circ} = 60^{\circ}$.
Using Brewster's law: $\mu = \tan \theta_p = \tan 60^{\circ} = \sqrt{3}$.
*(Note: If $i = 30^{\circ}$, $\mu = \tan 30^{\circ} = 1/\sqrt{3}$, but the refractive index of glass from air must be > 1. Thus $\mu = \sqrt{3}$ is the physically meaningful answer for this context.)*
The de Broglie wavelength for an electron accelerated through a potential difference $V$ is given by the standard formula:
$\lambda = \frac{1.227}{\sqrt{V}} \text{ nm}$
Given $V = 121 \text{ V}$.
$\lambda = \frac{1.227}{\sqrt{121}} \text{ nm} = \frac{1.227}{11} \text{ nm} \approx 0.111 \text{ nm}$.
The total number of spectral lines emitted when an atom transitions from the $n^{\text{th}}$ excited state to the ground state is given by $N = \frac{n(n-1)}{2}$.
Given $N = 15$:
$\frac{n(n-1)}{2} = 15 \Rightarrow n^2 - n = 30 \Rightarrow n^2 - n - 30 = 0$
$(n-6)(n+5) = 0$. Since $n$ must be a positive integer, $n = 6$.
Nuclear density is approximately constant for all nuclei and is exceptionally high. It can be calculated using $\rho = \frac{3m}{4\pi R_0^3}$, where $m$ is the average mass of a nucleon. The calculated value is on the order of $2 \times 10^{17} \text{ kg/m}^3$.
By Rydberg's formula for Hydrogen returning to the ground state ($n_1 = 1$) from an excited state $n$:
$\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{n^2} \right) = R \left( 1 - \frac{1}{n^2} \right)$
$\frac{1}{\lambda R} = 1 - \frac{1}{n^2} \Rightarrow \frac{1}{n^2} = 1 - \frac{1}{\lambda R} = \frac{\lambda R - 1}{\lambda R}$
$n^2 = \frac{\lambda R}{\lambda R - 1} \Rightarrow n = \sqrt{\frac{\lambda R}{\lambda R - 1}}$
For a perfectly absorbing surface, the total momentum transferred by electromagnetic radiation is given by $p = \frac{E}{c}$.
Given total energy $E = 6.48 \times 10^5 \text{ J}$ and speed of light $c = 3 \times 10^8 \text{ m/s}$.
$p = \frac{6.48 \times 10^5}{3 \times 10^8} = 2.16 \times 10^{-3} \text{ kg m/s}$.
From Einstein's photoelectric equation, the stopping potential $V_0$ varies with frequency $\nu$ as: $eV_0 = h\nu - \Phi \Rightarrow V_0 = (\frac{h}{e})\nu - \frac{\Phi}{e}$.
The slope of the $V_0$ versus $\nu$ graph is $\frac{h}{e}$. Since both $h$ (Planck's constant) and $e$ (charge of an electron) are universal constants, the slope is independent of the nature of the material. Therefore, the ratio of the slopes is 1.
In Bohr's model of the atom, the radius of the $n^{\text{th}}$ stationary orbit is proportional to the square of the principal quantum number ($r \propto n^2$).
Let $r_3 = R$. So, $R \propto 3^2 = 9$.
For the fourth orbit, $r_4 \propto 4^2 = 16$.
$\frac{r_4}{R} = \frac{16}{9} \Rightarrow r_4 = \frac{16}{9}R$.
The time period of a simple pendulum is given by $T = 2\pi\sqrt{\frac{L}{g}}$, so $T \propto \sqrt{L}$.
If the length is increased to 4 times its original value ($L' = 4L$), the new time period is:
$T' \propto \sqrt{4L} = 2\sqrt{L} = 2T$.
Given $T = 2 \text{ sec}$, the new time period $T' = 2 \times 2 = 4 \text{ sec}$.
The velocity of sound in a gas is determined by Newton-Laplace's formula: $v = \sqrt{\frac{\gamma P}{\rho}} = \sqrt{\frac{B}{\rho}}$, where $B$ is the bulk modulus (a measure of elasticity) and $\rho$ is the density of the gas. Thus, it depends upon both density and elasticity.
Let the magnitudes of the two vectors be $A$ and $B$ (with $A > B$).
Maximum resultant $R_{max} = A + B = 10$.
Minimum resultant $R_{min} = A - B = 6$.
Adding the two equations: $2A = 16 \Rightarrow A = 8 \text{ unit}$.
Substituting $A$ back: $8 + B = 10 \Rightarrow B = 2 \text{ unit}$.
Resistance of wire is $R = \rho\frac{L}{A}$. Its volume $V = A \times L$ remains constant upon melting.
Given new radius $r' = 4r$. New area $A' = \pi(4r)^2 = 16\pi r^2 = 16A$.
Since volume is constant, $A'L' = AL \Rightarrow (16A)L' = AL \Rightarrow L' = L/16$.
New resistance $R' = \rho\frac{L'}{A'} = \rho\frac{L/16}{16A} = \frac{1}{256} (\rho\frac{L}{A}) = \frac{R}{256}$.
$R' = \frac{256}{256} = 1 \Omega$.
Force on the particle $F = Q E = Q E_0 \sin\omega t$.
Acceleration $a = \frac{F}{m} = (\frac{Q E_0}{m}) \sin\omega t$.
For Simple Harmonic Motion, the maximum acceleration is related to amplitude $A$ by $a_{max} = \omega^2 A$.
Equating the maximum acceleration: $\frac{Q E_0}{m} = \omega^2 A \Rightarrow A = \frac{Q E_0}{m\omega^2}$.
For a concavo-convex lens, both surfaces curve in the same direction, so $R_1$ and $R_2$ have the same sign convention. Let $R_1 = +20 \text{ cm}$ and $R_2 = +30 \text{ cm}$.
By Lens Maker's Formula: $\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$
$\frac{1}{f} = (1.5 - 1)\left(\frac{1}{20} - \frac{1}{30}\right) = 0.5 \left(\frac{3 - 2}{60}\right) = 0.5 \left(\frac{1}{60}\right) = \frac{1}{120}$
Therefore, $f = +120 \text{ cm}$.
For the block to slide with constant velocity, the component of gravity down the incline must perfectly balance the viscous drag force.
Force of gravity component $F_g = mg \sin\theta = (\rho \cdot a^3) g \sin\theta$.
Viscous force $F_v = \eta A \frac{dv}{dx} = \eta (a^2) (\frac{v}{t})$.
Equating them: $\rho a^3 g \sin\theta = \eta a^2 \frac{v}{t} \Rightarrow \eta = \frac{\rho a^3 g \sin\theta \cdot t}{a^2 v} = \frac{\rho a g t \sin\theta}{v}$.
Given path difference $\Delta x = 171.5 \lambda$.
We are also given the numerical value: $\Delta x = 0.01372 \text{ cm} = 0.01372 \times 10^{-2} \text{ m}$.
$171.5 \lambda = 1.372 \times 10^{-4} \text{ m}$
$\lambda = \frac{1.372 \times 10^{-4}}{171.5} = 0.008 \times 10^{-4} \text{ m} = 8 \times 10^{-7} \text{ m} = 8000 \mathring{A}$.
In an equilateral prism (or any symmetric prism), the condition for minimum deviation is achieved when the ray passes symmetrically through the prism. This means the angle of incidence equals the angle of emergence, and the refracted ray inside the prism (QR) travels parallel to the base of the prism (BC). Thus, QR || BC.
Assertion is False: If an external point charge is placed in an existing electric field, the net electric field at nearby points is the vector sum of the existing field and the field of the point charge. Depending on the sign of the charge and the point chosen, the resultant field can either increase or decrease.
Reason is False: Electric field created by a point charge is $\vec{E} = \frac{kq}{r^2} \hat{r}$. The direction fundamentally depends on the nature (sign) of the charge.
Given equation: $y = 8 \sin[\pi(\frac{t}{10} - \frac{x}{4}) + \frac{\pi}{3}]$
Distributing $\pi$: $y = 8 \sin[\frac{\pi}{10}t - \frac{\pi}{4}x + \frac{\pi}{3}]$.
Comparing this with the standard wave equation $y = A \sin(\omega t - kx + \phi)$, we get the wave number $k = \frac{\pi}{4}$.
Since $k = \frac{2\pi}{\lambda}$, we have $\frac{2\pi}{\lambda} = \frac{\pi}{4} \Rightarrow \lambda = 8 \text{ m}$.
Using Rydberg's formula for frequencies: $f \propto (\frac{1}{n_1^2} - \frac{1}{n_2^2})$. Let constant be $C$.
$f_1$ (Lyman series limit): $n_1=1, n_2=\infty \Rightarrow f_1 = C(1 - 0) = C$.
$f_2$ (Lyman first line): $n_1=1, n_2=2 \Rightarrow f_2 = C(1 - 1/4) = 0.75C$.
$f_3$ (Balmer series limit): $n_1=2, n_2=\infty \Rightarrow f_3 = C(1/4 - 0) = 0.25C$.
Evaluating the column matches:
a. $f_1 = C$. It is greater than both $f_3 (0.25C)$ and $f_2 (0.75C)$. So a $\rightarrow$ p, r.
b. $f_2 = 0.75C$. It is greater than $f_3 (0.25C)$. So b $\rightarrow$ p.
c. $f_2 - f_1 = 0.75C - C = -0.25C$. It is negative. So c $\rightarrow$ q.
d. $f_1 - f_2 = C - 0.75C = 0.25C = f_3$. It is equal to $f_3$. So d $\rightarrow$ s.
Matching sequence: a-p,r; b-p; c-q; d-s.
The fundamental frequency of a stretched string is $f = \frac{1}{2L}\sqrt{\frac{T}{m}}$, where $m$ is linear mass density.
Linear mass density $m = \text{volume} \times \text{density} / \text{length} = (\pi r^2 L \rho)/L = \pi r^2 \rho$. So $m \propto r^2$.
Therefore, $f = \frac{1}{2L}\sqrt{\frac{T}{\pi r^2 \rho}}$. This shows $f \propto \frac{1}{r}$ when T, L, $\rho$ are constant.
$\frac{f_1}{f_2} = \frac{r_2}{r_1} = \frac{2}{1}$.
The speed of light in free space is given by $c = \frac{1}{\sqrt{\mu_0\epsilon_0}}$.
The given expression is $(\frac{2}{\mu_0\epsilon_0})^{1/2} = \sqrt{2} \cdot \frac{1}{\sqrt{\mu_0\epsilon_0}} = \sqrt{2}c$.
Since it evaluates to a constant multiple of speed, its dimensional formula is that of velocity, which is $[M^0 L T^{-1}]$.
According to Gauss's Law, the total electric flux originating from a point charge 'q' enclosed entirely is constant ($\Phi = q/\epsilon_0$).
The total flux is shared among face A, face B, and the curved surface.
When the charge is moved away from the center towards face A, it gets closer to face A and farther from face B. The solid angle subtended by face B at the location of the charge decreases, meaning fewer electric field lines will intercept face B. Thus, the flux related to face 'B' will decrease.
Applying the principle of conservation of mechanical energy between the point of projection (height h) and the point of landing (ground, height 0):
$K.E_{initial} + P.E_{initial} = K.E_{final} + P.E_{final}$
$\frac{1}{2}mu^2 + mgh = \frac{1}{2}mv^2 + 0$
$mgh = \frac{1}{2}mv^2 - \frac{1}{2}mu^2 \Rightarrow gh = \frac{v^2 - u^2}{2}$
$h = \frac{v^2 - u^2}{2g}$.
Ignoring slight expansion implies zero work done ($\Delta W = 0$). By the first law of thermodynamics, $\Delta U = \Delta Q$.
Heat added $\Delta Q = m c \Delta T$
Mass $m = 100 \text{ g} = 0.1 \text{ kg}$. $\Delta T = 50^{\circ}C - 30^{\circ}C = 20 \text{ K}$.
$\Delta U = 0.1 \text{ kg} \times 4184 \text{ J kg}^{-1}\text{K}^{-1} \times 20 \text{ K} = 8368 \text{ J}$.
$8368 \text{ J} \approx 8.4 \text{ kJ}$.
Helium is a monoatomic gas: $C_{v1} = \frac{3}{2}R$. Moles $n_1 = 2$.
Hydrogen (rigid molecule) is a diatomic gas: $C_{v2} = \frac{5}{2}R$. Moles $n_2 = 3$.
Molar specific heat of the mixture at constant volume is:
$C_{v_{mix}} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2} = \frac{2(\frac{3}{2}R) + 3(\frac{5}{2}R)}{2 + 3} = \frac{3R + 7.5R}{5} = \frac{10.5R}{5} = 2.1R$.
Substituting $R = 8.3 \text{ J/mol K}$:
$C_{v_{mix}} = 2.1 \times 8.3 = 17.43 \text{ J/mol K} \approx 17.4 \text{ J/mol K}$.