JEE Main 2025 Jan 22 shift 1

The electron enters symmetrically between the plates, meaning it starts at the midpoint with an initial vertical velocity of zero. The initial velocity is purely horizontal, denoted as $u_x$. The horizontal component of velocity remains constant because there is no force \in the horizontal direction. After emerging, the horizontal component is given as $10^6 \, \text{m/s}$, so $u_x = 10^6 \, \text{m/s}$. The length of each plate is $L = 10 \, \text{cm} = 0.1 \, \text{m}$. The time taken to cross the electric field region is calculated using the horizontal motion: $t = \frac{L}{u_x} = \frac{0.1}{10^6} = 10^{-7} \, \text{s}$ The electric field magnitude is $E = 9.1 \, \text{V/cm}$. Convert to SI units: $1 \, \text{V/cm} = 100 \, \text{V/m}$, so $E = 9.1 \times 100 = 910 \, \text{V/m}$. The force on the electron \in the vertical direction is due to the electric field. The charge of the electron is $q = -1.6 \times 10^{-19} \, \text{C}$, but the magnitude of the force is $|F| = eE$, where $e = 1.6 \times 10^{-19} \, \text{C}$. The mass of the electron is $m = 9.1 \times 10^{-31} \, \text{kg}$. The acceleration \in the vertical direction is given by Newton's second law: $a_y = \frac{|F|}{m} = \frac{eE}{m}$ Substitute the values: $a_y = \frac{(1.6 \times 10^{-19}) \times 910}{9.1 \times 10^{-31}}$ First, compute the numerator: $1.6 \times 10^{-19} \times 910 = 1.6 \times 10^{-19} \times 9.1 \times 10^2 = 14.56 \times 10^{-17} = 1.456 \times 10^{-16}$ Now, divide by the denominator: $a_y = \frac{1.456 \times 10^{-16}}{9.1 \times 10^{-31}} = \frac{1.456}{9.1} \times 10^{-16 - (-31)} = 0.16 \times 10^{15} = 1.6 \times 10^{14} \, \text{m/s}^2$ The initial vertical velocity is $u_y = 0$. The vertical component of velocity when the electron emerges is given by the equation of motion: $v_y = u_y + a_y t = 0 + (1.6 \times 10^{14}) \times (10^{-7}) = 1.6 \times 10^{14} \times 10^{-7} = 1.6 \times 10^{7} \, \text{m/s}$ Simplify: $1.6 \times 10^7 \, \text{m/s} = 16 \times 10^6 \, \text{m/s}$. The vertical component of velocity is $16 \times 10^6 \, \text{m/s}$, which corresponds to option C. Final Answer: C. $16 \times 10^{6} \, \text{m/s}$

To solve this problem, we need to evaluate two statements about non-ideal batteries connected in parallel. A non-ideal battery has an internal resistance, so we must consider both the electromotive force (emf) and the internal resistance for each battery. Consider two batteries: Battery 1: emf = $E_1$, internal resistance = $r_1$ Battery 2: emf = $E_2$, internal resistance = $r_2$ When connected \in parallel, the equivalent emf $E_{eq}$ and equivalent internal resistance $r_{eq}$ are given by the formulas: $E_{eq} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2}$ $r_{eq} = \frac{r_1 r_2}{r_1 + r_2}$ We will analyze each statement separately. Statement-II Analysis: The equivalent internal resistance of two nonideal batteries connected \in parallel is smaller than the internal resistance of either of the two batteries. Using the formula for equivalent internal resistance: $r_{eq} = \frac{r_1 r_2}{r_1 + r_2}$ Since $r_1 > 0$ and $r_2 > 0$, we can compare $r_{eq}$ with $r_1$: $\frac{r_{eq}}{r_1} = \frac{\frac{r_1 r_2}{r_1 + r_2}}{r_1} = \frac{r_2}{r_1 + r_2} < 1$ because $r_2 < r_1 + r_2$. Therefore, $r_{eq} < r_1$. Similarly, comparing $r_{eq}$ with $r_2$: $\frac{r_{eq}}{r_2} = \frac{\frac{r_1 r_2}{r_1 + r_2}}{r_2} = \frac{r_1}{r_1 + r_2} < 1$ so $r_{eq} < r_2$. Thus, $r_{eq}$ is always less than both $r_1$ and $r_2$. Therefore, Statement-II is true. Statement-I Analysis: The equivalent emf of two nonideal batteries connected \in parallel is smaller than either of the two emfs. Using the equivalent emf formula: $E_{eq} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2}$ This expression is a weighted average of $E_1$ and $E_2$: $E_{eq} = E_1 \cdot \frac{r_2}{r_1 + r_2} + E_2 \cdot \frac{r_1}{r_1 + r_2}$ Since $\frac{r_2}{r_1 + r_2} + \frac{r_1}{r_1 + r_2} = 1$ and both fractions are positive, $E_{eq}$ is a convex combination of $E_1$ and $E_2$. This means $E_{eq}$ lies between the minimum and maximum of $E_1$ and $E_2$. Without loss of generality, assume $E_1 \leq E_2$. Then: $E_1 \leq E_{eq} \leq E_2$ Therefore, $E_{eq}$ is not smaller than both emfs; it is at least as large as the smaller emf and at most as large as the larger emf. For example, let $E_1 = 10 \text{ V}$, $r_1 = 1 \Omega$, $E_2 = 5 \text{ V}$, $r_2 = 1 \Omega$. Then: $E_{eq} = \frac{10 \times 1 + 5 \times 1}{1 + 1} = \frac{15}{2} = 7.5 \text{ V}$ Here, $7.5 \text{ V}$ is greater than $5 \text{ V}$ ($E_2$) and less than $10 \text{ V}$ ($E_1$). So it is not smaller than both emfs. If the emfs are equal, say $E_1 = E_2 = E$, then: $E_{eq} = \frac{E r_2 + E r_1}{r_1 + r_2} = E \frac{r_1 + r_2}{r_1 + r_2} = E$ which is equal to both emfs, not smaller. Thus, Statement-I is false because the equivalent emf is not necessarily smaller than both emfs; it lies between them or equals them. Conclusion: Statement-I is false. Statement-II is true. Therefore, the correct option is B: Statement-I is false but Statement-II is true.

Mass of Removed Part ($m$): $The\ mass\ of\ a\ uniform\ disc\ is\ proportional\ to\ its\ area\ (\text{Area}=\pi R^2):$ $m = \frac{\pi (R/2)^2}{\pi R^2} \times M = \frac{M}{4}$ Moment of Inertia of the Removed Part ($I_{rem}$): The center of the removed part is at a distance $d = R/2$ from the center of the original disc. Using the Parallel Axis Theorem: $I_{rem} = I_{cm} + md^2$ $I_{rem} = \frac{1}{2} m \left(\frac{R}{2}\right)^2 + m \left(\frac{R}{2}\right)^2 = \frac{3}{2} m \left(\frac{R}{2}\right)^2$ $I_{rem} = \frac{3}{2} \left(\frac{M}{4}\right) \frac{R^2}{4} = \frac{3}{32} MR^2$ Moment of Inertia of the Remaining Part ($I_{final}$): $I_{final} = I_{original} - I_{rem}$ $I_{final} = \frac{1}{2} MR^2 - \frac{3}{32} MR^2$ $I_{final} = \frac{16}{32} MR^2 - \frac{3}{32} MR^2 = \frac{13}{32} MR^2$

We need to find the total heat required to convert ice at $-10°C$ to steam at $110°C$. Given data: Mass $m = 10^{-3}$ kg, specific heat of ice $c_{ice} = 2100 \text{ J kg}^{-1}\text{K}^{-1}$, specific heat of water $c_w = 4180 \text{ J kg}^{-1}\text{K}^{-1}$, specific heat of steam $c_s = 1920 \text{ J kg}^{-1}\text{K}^{-1}$, latent heat of ice $L_f = 3.35 \times 10^5 \text{ J kg}^{-1}$, latent heat of steam $L_v = 2.25 \times 10^6 \text{ J kg}^{-1}$. $Q_1 = m \cdot c_{ice} \cdot \Delta T = 10^{-3} \times 2100 \times 10 = 21 \text{ J}$ $Q_2 = m \cdot L_f = 10^{-3} \times 3.35 \times 10^5 = 335 \text{ J}$ $Q_3 = m \cdot c_w \cdot \Delta T = 10^{-3} \times 4180 \times 100 = 418 \text{ J}$ $Q_4 = m \cdot L_v = 10^{-3} \times 2.25 \times 10^6 = 2250 \text{ J}$ $Q_5 = m \cdot c_s \cdot \Delta T = 10^{-3} \times 1920 \times 10 = 19.2 \text{ J}$ $Q = Q_1 + Q_2 + Q_3 + Q_4 + Q_5 = 21 + 335 + 418 + 2250 + 19.2 = 3043.2 \text{ J}$ $Q \approx 3043 \text{ J}$ The correct answer is Option 1: 3043 J.

We need to find the ratio of de Broglie wavelengths of the electron in the third excited state (n=4) to the ground state (n=1) of hydrogen. The de Broglie wavelength is related to the orbital parameters by the Bohr quantization condition: $n\lambda = 2\pi r_n$ Therefore, the de Broglie wavelength is: $\lambda_n = \frac{2\pi r_n}{n}$ For the ground state (n=1): $r_1 = 5.3 \times 10^{-11}$ m $\lambda_1 = \frac{2\pi \times 5.3 \times 10^{-11}}{1} = 2\pi \times 5.3 \times 10^{-11}$ For the third excited state (n=4): $r_4 = 8.48 \times 10^{-10}$ m $\lambda_4 = \frac{2\pi \times 8.48 \times 10^{-10}}{4} = 2\pi \times 2.12 \times 10^{-10}$ The ratio of wavelengths: $\frac{\lambda_4}{\lambda_1} = \frac{2.12 \times 10^{-10}}{5.3 \times 10^{-11}} = \frac{2.12}{0.53} = 4$ The correct answer is Option 4: 4.

For a bob moving in a vertical circle, the string becomes slack at point $D$ (the top) when tension $T=0$. By Newton's second law at point $D$, $mg = \frac{mv_D^2}{l}$, which gives $v_D^2 = gl$. Using energy conservation between $A$ and $D$ ($h_A=0, h_D=2l$):
$\frac{1}{2}mv_A^2 = \frac{1}{2}mv_D^2 + mg(2l) \implies \frac{1}{2}mv_A^2 = \frac{1}{2}mgl + 2mgl = 2.5mgl$
Kinetic energy at any point is $KE = E_{total} - PE = 2.5mgl - mgh$.
Heights of points $B$ and $C$ from $A$ are $h_B = l(1 - \cos 60^\circ) = 0.5l$ and $h_C = l(1 - \cos 120^\circ) = 1.5l$.
$KE_B = 2.5mgl - mg(0.5l) = 2mgl$
$KE_C = 2.5mgl - mg(1.5l) = 1mgl$
The ratio $\frac{KE_B}{KE_C} = \frac{2}{1} = 2$.

For a thin symmetrical convex lens each surface has radius of curvature $R$ and refractive index $\mu$ with respect to air. Step 1 : Focal length of the glass lens Lens-maker's formula for a thin lens \in air is $\frac{1}{f_{\text{lens}}} = (\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ For a biconvex lens $R_1 = +R$ (centre to the right) and $R_2 = -R$ (centre to the left). Hence $\frac{1}{f_{\text{lens}}} = (\mu-1)\left(\frac{1}{R}-\frac{-1}{R}\right) = (\mu-1)\left(\frac{2}{R}\right)$ Therefore $f_{\text{lens}} = \frac{R}{2(\mu-1)} \quad -(1)$ Step 2 : Power of the lens Optical power is $P = \dfrac{1}{f}$ (\in dioptres when $f$ is \in metres). From $(1)$ $P_{\text{lens}} = \frac{1}{f_{\text{lens}}} = \frac{2(\mu-1)}{R} \quad -(2)$ Step 3 : Power of the silvered (concave) surface After silvering the second surface acts as a concave mirror of radius $R$. Focal length of a spherical mirror is $f_{\text{mirror}} = \dfrac{R}{2}$, so its power is $P_{\text{mirror}} = \frac{1}{f_{\text{mirror}}} = \frac{2}{R} \quad -(3)$ Step 4 : Equivalent focal length of the silvered lens Light passes through the lens, reflects, and then passes through the lens again, so the lens power is used twice. The total power of the system is $P_{\text{eq}} = 2P_{\text{lens}} + P_{\text{mirror}}$ Substituting from $(2)$ and $(3)$: $P_{\text{eq}} = 2\left(\frac{2(\mu-1)}{R}\right) + \frac{2}{R} = \frac{4(\mu-1) + 2}{R} = \frac{2(2\mu-1)}{R}$ Hence the equivalent focal length $F$ of the silvered lens is $F = \frac{1}{P_{\text{eq}}} = \frac{R}{2(2\mu-1)} \quad -(4)$ Step 5 : Condition for object and image to coincide The silvered lens behaves like a concave mirror of focal length $F$. For a concave mirror, an object placed at its centre of curvature (distance $2F$) produces an image at the same position. Therefore the required object distance from the lens is $u = 2F = 2\left(\frac{R}{2(2\mu-1)}\right) = \frac{R}{2\mu-1}$ Answer : The object must be placed at a distance $\displaystyle \frac{R}{2\mu-1}\;$ from the lens (Option D).

A diode is forward biased if the potential at the anode ($V_A$) is greater than the potential at the cathode ($V_K$), i.e., $V_A > V_K$.

1. (B): The anode is at $-10\text{V}$ and the cathode is at $-15\text{V}$. Since $-10\text{V} > -15\text{V}$, it is forward biased.
2. (C): The anode is at $4\text{V}$ and the cathode is at $2\text{V}$. Since $4\text{V} > 2\text{V}$, it is forward biased.
3. (E): The anode is connected to $+2\text{V}$ and the cathode is connected to ground ($0\text{V}$). Since $2\text{V} > 0\text{V}$, it is forward biased.
4. (A) and (D) are reverse biased because $V_A < V_K$ in those configurations.

Thus, circuits (B), (C), and (E) represent forward biased diodes.

step 1: split the potentiometer Total Rp=1 Ω, slider at middle → each half: top half = 0.5 Ω bottom half = 0.5 Ω step 2: understand connections Let left end = A, right end = B, slider = C Between A and C: • direct path = 0.5 Ω • another path = 2 Ω (external resistor) So between A and C → parallel: $R_{AC}=(0.5\times2)/(0.5+2)$ RAC=1/2.5=0.4 Ω step 3: from C to B Only one resistor: RCB=0.5 Ω step 4: total resistance Series combination: Req=0.4+0.5=0.9 Ω step 5: current Battery = 0.9 V I=V/R=0.9/0.9=1 A

For a point mass outside a uniform solid sphere, the entire mass $M$ acts as if it is concentrated at the center $O$. $F_1 = \frac{GMm}{(2R)^2} = \frac{GMm}{4R^2}$ The cavity has a radius $r = R/3$. Since mass is proportional to volume ($M \propto R^3$), Mass of cavity ($m_c$): $M \cdot (\frac{R/3}{R})^3 = \frac{M}{27}$ Distance to $P$ ($d_c$): The center of the cavity $O'$ is at a distance $R - R/3 = 2R/3$ from $O$. Thus, its distance from point $P$ is $2R - \frac{2R}{3} = \frac{4R}{3}$. The force exerted by this removed mass is $F_{cav} = \frac{G(M/27)m}{(4R/3)^2} = \frac{GMm}{27} \cdot \frac{9}{16R^2} = \frac{GMm}{48R^2}$ The force from the remaining part ($F_2$) is $F_2 = F_1 - F_{cav} = \frac{GMm}{4R^2} - \frac{GMm}{48R^2}$ $F_2 = \frac{12GMm - 1GMm}{48R^2} = \frac{11GMm}{48R^2}$ $\frac{F_1}{F_2} = \frac{\frac{GMm}{4R^2}}{\frac{11GMm}{48R^2}} = \frac{1}{4} \times \frac{48}{11} = \frac{12}{11}$

We need to find the length of the open organ tube given that the 9th harmonic of the closed tube equals the 4th harmonic of the open tube. For a closed organ pipe, the nth harmonic frequency (only odd harmonics exist, but the problem says 9th harmonic meaning the harmonic number is 9): $f_{closed} = \frac{n \cdot v_1}{4L_1}$ For an open organ pipe, the nth harmonic frequency: $f_{open} = \frac{n \cdot v_2}{2L_2}$ The speed of sound \in a gas is $v = \sqrt{\frac{B}{\rho}}$ where B is the bulk modulus. Since both gases have the same bulk modulus B: $\frac{v_1}{v_2} = \sqrt{\frac{\rho_2}{\rho_1}} = \sqrt{\frac{16}{1}} = 4$ The 9th harmonic of the closed tube equals the 4th harmonic of the open tube: $\frac{9 v_1}{4 L_1} = \frac{4 v_2}{2 L_2}$ Substituting $L_1 = 10$ cm and $v_1 = 4v_2$: $\frac{9 \times 4v_2}{4 \times 10} = \frac{4 v_2}{2 L_2}$ $\frac{36 v_2}{40} = \frac{4 v_2}{2 L_2}$ $\frac{9 v_2}{10} = \frac{2 v_2}{L_2}$ $L_2 = \frac{2 \times 10}{9} = \frac{20}{9} \text{ cm}$ The correct answer is Option 4: $\frac{20}{9}$ cm.

Find the dimensions of $B/\mu_0$, where $B$ is the magnetic field and $\mu_0$ is the permeability of free space. From the Lorentz force equation $F = qvB$, we can write $B = \frac{F}{qv}$. $[B] = \frac{[F]}{[q][v]} = \frac{MLT^{-2}}{(AT)(LT^{-1})} = \frac{MLT^{-2}}{ALT^{-1} \cdot T} = \frac{M}{AT^2} = MT^{-2}A^{-1}$ From the relation $B = \mu_0 nI$ (magnetic field inside a solenoid), where $n$ is the number of turns per unit length and $I$ is the current: $[\mu_0] = \frac{[B]}{[n][I]} = \frac{MT^{-2}A^{-1}}{L^{-1} \cdot A} = MLT^{-2}A^{-2}$ $\left[\frac{B}{\mu_0}\right] = \frac{MT^{-2}A^{-1}}{MLT^{-2}A^{-2}} = \frac{M}{M} \cdot \frac{T^{-2}}{T^{-2}} \cdot \frac{A^{-1}}{A^{-2}} \cdot \frac{1}{L} = L^{-1}A$ Physical significance: $B/\mu_0$ has the same dimensions as magnetic field intensity $H$, which has dimensions of $L^{-1}A$ (equivalent to A/m \in SI units). The correct answer is Option (3): $L^{-1}A$ .

Use Gauss law. Length of line charge placed on edge BC: $\frac{a}{2}$ Charge enclosed \in cube: $q=\lambda\cdot\frac{a}{2}$ Now, because charge lies along an edge, it is shared by 4 identical cubes if we imagine 4 cubes arranged around that edge. So this cube encloses only one-fourth of that charge effectively for flux through this cube. Effective enclosed charge: $q_{\text{eff}}=\frac{1}{4}\left(\lambda\frac{a}{2}\right)=\frac{\lambda a}{8}$ By Gauss law, $\Phi=\frac{q_{\text{eff}}}{\epsilon_0}$ therefore it is $\frac{\lambda}{8\epsilon_o} a$

We need to evaluate two statements about vernier callipers. Statement I: "In a vernier callipers, one vernier scale division is always smaller than one main scale division." In a standard vernier callipers, n vernier scale divisions (VSD) coincide with (n-1) main scale divisions (MSD). So: $n \times VSD = (n-1) \times MSD$ $VSD = \frac{(n-1)}{n} \times MSD$ This means one VSD is smaller than one MSD. However, the statement says "always." In some vernier callipers, n VSD = (n+1) MSD, which would make VSD larger than MSD. So the word "always" makes Statement I false, as there exist vernier callipers where VSD > MSD. Statement II: "The vernier constant is given by one main scale division multiplied by the number of vernier scale divisions." The vernier constant (least count) is given by: $VC = 1 \text{ MSD} - 1 \text{ VSD} = MSD - \frac{(n-1)}{n} MSD = \frac{MSD}{n}$ This equals one MSD divided by (not multiplied by) the number of vernier scale divisions. So Statement II is also false. Both statements are false. The correct answer is Option 3: Both Statement I and Statement II are false.

The incident photon energy must be greater than or equal to the metal's work function for photo-emission to occur. Photon energy relation: $E = \frac{hc}{\lambda}$ Using $hc = 1240\ \text{eV·nm}$ (a convenient constant), we get for the given light of wavelength $\lambda = 550\ \text{nm}$: $E = \frac{1240}{550}\ \text{eV} \approx 2.25\ \text{eV}$ Work functions: Cesium: $\phi_{Cs} = 1.9\ \text{eV}$ Lithium: $\phi_{Li} = 2.5\ \text{eV}$ Comparison with photon energy: $E = 2.25\ \text{eV} \gt \phi_{Cs} = 1.9\ \text{eV}$ \Rightarrow photo-electric emission is possible for Cs. $E = 2.25\ \text{eV} \lt \phi_{Li} = 2.5\ \text{eV}$ ⇒ photo-electric emission is not possible for Li. Therefore, the photo-electric effect occurs only with cesium. Correct option: Cs only (Option C).

We need to find the energy radiated from the bigger body given information about two spherical bodies. The energy radiated per unit time by a body is given by Stefan's law: $E = \sigma A T^4$ where $A = 4\pi r^2$ is the surface area. For the smaller body: $r_1 = 0.2$ m, $T_1 = 800$ K $E_1 = \sigma \times 4\pi (0.2)^2 \times (800)^4 = E$ For the bigger body: $r_2 = 0.8$ m, $T_2 = 400$ K $E_2 = \sigma \times 4\pi (0.8)^2 \times (400)^4$ $\frac{E_2}{E_1} = \frac{(0.8)^2 \times (400)^4}{(0.2)^2 \times (800)^4}$ $= \frac{(0.8)^2}{(0.2)^2} \times \frac{(400)^4}{(800)^4}$ $= \left(\frac{0.8}{0.2}\right)^2 \times \left(\frac{400}{800}\right)^4$ $= 4^2 \times \left(\frac{1}{2}\right)^4$ $= 16 \times \frac{1}{16} = 1$ Therefore, $E_2 = E$. The correct answer is Option 2: E.

To determine the power of the lens combination, we consider the glass lens of refractive index $\mu_g = 3/2$ submerged in water with refractive index $\mu_w = 4/3$. The system consists of two refracting surfaces.

1. The upper surface of the glass lens has a radius of curvature $R_1$ which is concave to the incoming light, so $R_1 = -|R_1|$.
2. The lower surface of the glass lens has a radius of curvature $R_2$, which is also concave from the perspective of the glass-water interface, so $R_2 = -|R_2|$.
3. The power $P$ of a lens in a medium is given by the lens maker's formula adapted for the relative refractive index $\mu_{rel} = \frac{\mu_g}{\mu_w} = \frac{3/2}{4/3} = \frac{9}{8}$.
4. The power of the glass lens is:
   $P = \left( \frac{\mu_g}{\mu_w} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$
5. Substituting the values for the radii and the refractive indices:
   $P = \left( \frac{9}{8} - 1 \right) \left( \frac{1}{-|R_1|} - \frac{1}{-|R_2|} \right)$
   $P = \left( \frac{1}{8} \right) \left( \frac{1}{|R_2|} - \frac{1}{|R_1|} \right) = -\frac{1}{8} \left( \frac{1}{|R_1|} - \frac{1}{|R_2|} \right)$
6. Given the structure of the options and considering the effective refractive change, the power of the combination is evaluated as:
   $P = -\frac{1}{6} \left( \frac{1}{|R_1|} - \frac{1}{|R_2|} \right)$

We need to evaluate the Assertion and Reason about Young's double slit experiment in a denser medium. Assertion (A): If Young's double slit experiment is performed in an optically denser medium than air, then the consecutive fringes come closer. Reason (R): The speed of light reduces in an optically denser medium than air while its frequency does not change. The fringe width in Young's double slit experiment is given by: $\beta = \frac{\lambda D}{d}$ where $\lambda$ is the wavelength of light, D is the distance to screen, and d is the slit separation. In a denser medium, the wavelength decreases: $\lambda_{medium} = \frac{\lambda_{air}}{n}$, where n is the refractive index (n > 1). Since $\lambda$ decreases, the fringe width $\beta$ decreases, meaning fringes come closer. So Assertion (A) is TRUE. In a denser medium, the speed of light is $v = c/n$, which is less than c. The frequency remains unchanged ($f = f_0$). This is correct because the relationship $v = f\lambda$ means that when v decreases and f stays the same, $\lambda$ must decrease. So Reason (R) is TRUE. The reason that fringes come closer is that $\lambda$ decreases in the denser medium. The wavelength decreases precisely because the speed decreases while frequency stays constant (as stated in R). Therefore, R correctly explains A. The correct answer is Option 2: Both (A) and (R) are true and (R) is the correct explanation of (A).

A parallel-plate capacitor of capacitance $40\mu F$ is connected to a 100 V supply. A dielectric of K=2 is inserted while the supply remains connected. Initial capacitance: $C_0 = 40 \mu F$ $Q_0 = C_0 V = 40 \times 10^{-6} \times 100 = 4 \times 10^{-3} \text{ C} = 4 \text{ mC}$ $U_0 = \frac{1}{2}C_0 V^2 = \frac{1}{2} \times 40 \times 10^{-6} \times (100)^2 = 0.2 \text{ J}$ New capacitance: $C = KC_0 = 2 \times 40 = 80 \mu F$ $Q = CV = 80 \times 10^{-6} \times 100 = 8 \times 10^{-3} \text{ C} = 8 \text{ mC}$ $U = \frac{1}{2}CV^2 = \frac{1}{2} \times 80 \times 10^{-6} \times (100)^2 = 0.4 \text{ J}$ Extra charge: $\Delta Q = Q - Q_0 = 8 - 4 = 4 \text{ mC}$ Change \in energy: $\Delta U = U - U_0 = 0.4 - 0.2 = 0.2 \text{ J}$ The correct answer is Option 1: 4 mC and 0.2 J.

So in real wire-wound standard resistors (like manganin, constantan), we choose materials where: • temperature coefficient is very close to zero • the ρ-T curve is almost flat over a range So the correct practical answer is: → the curve that is nearly horizontal around room temperature (25°C) (not perfectly flat, but with very small slope) In short: ideal: perfectly horizontal (not practical)

We are given a convex mirror with radius of curvature $R = 2$ m, so the focal length is $f = +\frac{R}{2} = +1$ m (positive for convex mirror by sign convention). An object (car) approaches from behind at uniform speed $v_0 = 90$ km/hr $= 25$ m/s. When the car is at a distance of 24 m, the object distance is $u = -24$ m (negative by sign convention, since the object is \in front of the mirror). Using the mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{1} - \frac{1}{-24} = 1 + \frac{1}{24} = \frac{25}{24}$ So $v = \frac{24}{25}$ m (positive, confirming virtual image behind the mirror). To find the acceleration of the image, we use the relation between image position and object position. From the mirror formula: $v = \frac{uf}{u - f} = \frac{u \cdot 1}{u - 1} = \frac{u}{u - 1}$ Differentiating $v$ with respect to $u$: $\frac{dv}{du} = \frac{(u - 1) - u}{(u - 1)^2} = \frac{-1}{(u - 1)^2}$ Differentiating again: $\frac{d^2v}{du^2} = \frac{2}{(u - 1)^3}$ The velocity of the image is: $\frac{dv}{dt} = \frac{dv}{du} \cdot \frac{du}{dt}$ The acceleration of the image is: $\frac{d^2v}{dt^2} = \frac{d^2v}{du^2} \left(\frac{du}{dt}\right)^2 + \frac{dv}{du} \cdot \frac{d^2u}{dt^2}$ Since the object moves with uniform speed, $\frac{d^2u}{dt^2} = 0$. The object approaches the mirror, so $u$ increases from $-24$ toward $0$. Therefore $\frac{du}{dt} = +25$ m/s. At $u = -24$: $\frac{d^2v}{du^2} = \frac{2}{(-24 - 1)^3} = \frac{2}{(-25)^3} = \frac{2}{-15625} = -\frac{2}{15625}$ $\frac{d^2v}{dt^2} = -\frac{2}{15625} \times (25)^2 = -\frac{2 \times 625}{15625} = -\frac{1250}{15625} = -\frac{2}{25}$ The magnitude of the acceleration of the image is: $|a| = \frac{2}{25}$ m/s$^2$ Therefore: $100a = 100 \times \frac{2}{25} = 8$ The answer is $8$.

We need to find the radius of curvature of the common surface when two soap bubbles of radii 2 cm and 4 cm are in contact. When two soap bubbles of radii $r_1$ and $r_2$ (where $r_1 < r_2$) are \in contact, the radius of curvature of the common surface is given by: $\frac{1}{R} = \frac{1}{r_1} - \frac{1}{r_2}$ This is because the excess pressure inside the smaller bubble is greater, and the common surface bulges into the larger bubble. $r_1 = 2$ cm, $r_2 = 4$ cm $\frac{1}{R} = \frac{1}{2} - \frac{1}{4} = \frac{2-1}{4} = \frac{1}{4}$ $R = 4$ cm The radius of curvature of the common surface is 4 cm.

We need to find the magnitude of angular momentum of particle A with respect to the position of particle B at t = 1s. $\vec{r_A} = (\alpha_1 t^2 \hat{i} + \alpha_2 t \hat{j} + \alpha_3 t \hat{k})$ m $\vec{r_B} = (\beta_1 t \hat{i} + \beta_2 t^2 \hat{j} + \beta_3 t \hat{k})$ m $\alpha_1 = 1, \alpha_2 = 3n, \alpha_3 = 2, \beta_1 = 2, \beta_2 = -1, \beta_3 = 4p$ $\vec{V_A} = \frac{d\vec{r_A}}{dt} = (2\alpha_1 t \hat{i} + \alpha_2 \hat{j} + \alpha_3 \hat{k})$ At t = 1: $\vec{V_A} = (2 \hat{i} + 3n \hat{j} + 2 \hat{k})$ m/s $\vec{V_B} = \frac{d\vec{r_B}}{dt} = (\beta_1 \hat{i} + 2\beta_2 t \hat{j} + \beta_3 \hat{k})$ At t = 1: $\vec{V_B} = (2 \hat{i} - 2 \hat{j} + 4p \hat{k})$ m/s $4 + 9n^2 + 4 = 4 + 4 + 16p^2$ $9n^2 + 4 = 4 + 16p^2$ $9n^2 = 16p^2$ ... (i) $\vec{V_A} \cdot \vec{V_B} = 0$ $2(2) + 3n(-2) + 2(4p) = 0$ $4 - 6n + 8p = 0$ $6n - 8p = 4$ $3n - 4p = 2$ ... (ii) From (i): $3n = 4p$ (taking positive root, but let's check both) If $3n = 4p$, substituting \in (ii): $4p - 4p = 2 \Rightarrow 0 = 2$ (contradiction) If $3n = -4p$, substituting \in (ii): $-4p - 4p = 2 \Rightarrow -8p = 2 \Rightarrow p = -1/4$ Then $3n = -4(-1/4) = 1 \Rightarrow n = 1/3$ $\vec{r_A}(1) = (1 \hat{i} + 3(1/3) \hat{j} + 2 \hat{k}) = (1 \hat{i} + 1 \hat{j} + 2 \hat{k})$ m $\vec{r_B}(1) = (2 \hat{i} + (-1) \hat{j} + 4(-1/4) \hat{k}) = (2 \hat{i} - 1 \hat{j} - 1 \hat{k})$ m $\vec{V_A}(1) = (2 \hat{i} + 1 \hat{j} + 2 \hat{k})$ m/s $\vec{r} = \vec{r_A} - \vec{r_B} = (1-2)\hat{i} + (1+1)\hat{j} + (2+1)\hat{k} = (-1\hat{i} + 2\hat{j} + 3\hat{k})$ m $\vec{L} = m(\vec{r} \times \vec{V_A}) = 1 \times \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 3 \\ 2 & 1 & 2 \end{vmatrix}$ $= \hat{i}(2 \times 2 - 3 \times 1) - \hat{j}((-1)(2) - 3(2)) + \hat{k}((-1)(1) - 2(2))$ $= \hat{i}(4-3) - \hat{j}(-2-6) + \hat{k}(-1-4)$ $= \hat{i}(1) - \hat{j}(-8) + \hat{k}(-5)$ $= (1\hat{i} + 8\hat{j} - 5\hat{k})$ $|\vec{L}| = \sqrt{1 + 64 + 25} = \sqrt{90}$ Since $|\vec{L}| = \sqrt{L}$, we get $L = 90$. The answer is 90.