JEE Main 2025 Jan 28 shift 1

Three infinitely long wires with linear charge density $\lambda$ are placed along the x-axis, y-axis, and z-axis respectively. We need to find the equation of an equipotential surface. The electric potential at a perpendicular distance $r$ from an infinitely long line charge with linear charge density $\lambda$ is: $V = -\frac{\lambda}{2\pi\epsilon_0} \ln r + C$ For a point $(x, y, z)$: - Distance from the x-axis: $r_x = \sqrt{y^2 + z^2}$ - Distance from the y-axis: $r_y = \sqrt{x^2 + z^2}$ - Distance from the z-axis: $r_z = \sqrt{x^2 + y^2}$ $V_{total} = -\frac{\lambda}{2\pi\epsilon_0}\$[\ln\sqrt{y^{2+}z^2} + \ln\sqrt{x^{2+}z^2} + \ln\sqrt{x^{2+}y^2}]\$ + C'$ $= -\frac{\lambda}{4\pi\epsilon_0}\ln\$[(y^{2+}z^2)(x^{2+}z^2)(x^{2+}y^2)]\$ + C'$ For an equipotential surface, $V_{total} = \text{constant}$, which requires: $(y^2 + z^2)(x^2 + z^2)(x^2 + y^2) = \text{constant}$ This can be rewritten as: $(x^2 + y^2)(y^2 + z^2)(z^2 + x^2) = \text{constant}$ The correct answer is Option 3 : $(x^2 + y^2)(y^2 + z^2)(z^2 + x^2) = \text{constant}$.

We need to evaluate the Assertion and Reason about the speed of sound in different media. Assertion A: A sound wave has higher speed in solids than in gases. The speed of sound in a medium is given by $v = \sqrt{\frac{E}{\rho}}$, where $E$ is the appropriate elastic modulus and $\rho$ is the density. In solids, the elastic modulus (Young's modulus) is extremely high compared to gases. Although solids have higher density, the much larger elastic modulus dominates. For example, the speed of sound \in steel is approximately 5960 m/s, while \in air it is approximately 343 m/s. Assertion A is TRUE. Reason R: Gases have higher value of bulk modulus than solids. This is incorrect . Solids have a much higher bulk modulus than gases. For example, the bulk modulus of steel is about $1.6 \times 10^{11}$ Pa, while for air at atmospheric pressure it is approximately $1.42 \times 10^{5}$ Pa. Solids resist compression far more effectively than gases. Reason R is FALSE. Since Assertion A is true but Reason R is false, the correct answer is Option B : A is true but R is false.

Both capacitors are in parallel ⇒ same voltage at all times $q=CV\Rightarrow q\propto C$ From the graph: At final (steady state), $q_2>q_1$ \Rightarrow $C_2>C_1$ Energy stored: $U=\frac{1}{2}CV^2\Rightarrow U\propto C$ (since V is same) So: $U_2>U_1$

We need to find the angle of prism $P_2$ such that the combination of $P_1$ and $P_2$ produces dispersion without deviation. - Prism $P_1$: angle $A_1 = 4°$, refractive index $n_1 = 1.54$ - Prism $P_2$: angle $A_2 = ?$, refractive index $n_2 = 1.72$ For a thin prism, the deviation is: $\delta = (n - 1)A$ For dispersion without deviation, the two prisms must be arranged to cancel each other's deviation. The prisms are placed with opposite orientations: $\delta_1 = \delta_2$ $(n_1 - 1)A_1 = (n_2 - 1)A_2$ $A_2 = \frac{(n_1 - 1)}{(n_2 - 1)} \times A_1 = \frac{(1.54 - 1)}{(1.72 - 1)} \times 4° = \frac{0.54}{0.72} \times 4°$ $= \frac{3}{4} \times 4° = 3°$ The angle of prism $P_2$ is 3 degrees. The correct answer is Option 1 : 3.

Electric field acts along +x, so force on charge is F=qE Initially particle is at angle $60^{\circ}$ with x-axis and string length l. Initial coordinates: $x_i=l\cos60^{\circ}=\frac{l}{2}$ When it crosses x-axis, particle is at $x_f=l$ (since still constrained on circle radius lll). Displacement along field: $\Delta x=l-\frac{l}{2}=\frac{l}{2}$ Work done by electric field: $W=qE\Delta x$ $=\frac{qEl}{2}$ Tension does no work. Using work-energy theorem: $\frac{1}{2}mv^2=\frac{qEl}{2}$ So $mv^2=qEl$ so $\sqrt{\frac{qEl}{m}}$

We need to evaluate the Assertion and Reason about central forces and potential energy. Assertion (A): "In a central force field, the work done is independent of the path chosen." Analysis: A central force is always directed along the line joining the particle to a fixed center, with magnitude depending only on the distance: $\vec{F} = f(r)\hat{r}$. Central forces are conservative forces -- the work done depends only on the initial and final positions, not on the path. This is because the curl of a central force is zero. Assertion (A) is TRUE. Reason (R): "Every force encountered in mechanics does not have an associated potential energy." Analysis: This statement says that not every mechanical force has an associated potential energy. This is TRUE -- non-conservative forces like friction and viscous drag do not have an associated potential energy function. Only conservative forces (like gravity, spring force, and central forces) have well-defined potential energy functions. Relationship between A and R: While both statements are true, R does not explain A. The fact that "some forces don't have potential energy" does not explain why "central forces are path-independent." The reason A is true is that central forces are conservative (they have a potential energy function), which is essentially the opposite situation from what R describes. The correct answer is Option 2 : Both A and R are true but R is NOT the correct explanation of A.

We need to identify the correct nuclear process for beta-minus decay of a neutron. Key Concept: Beta-minus ($\beta^-$) Decay In \beta-minus decay, a neutron inside a nucleus (or a free neutron) transforms into a proton by emitting an electron and an antineutrino: $n \rightarrow p + e^- + \bar{\nu}$ Conservation Laws: Charge conservation: Neutron (charge 0) $\rightarrow$ proton (charge +1) + electron (charge -1) + antineutrino (charge 0). Total: $0 = +1 - 1 + 0$ $\checkmark$ Lepton number conservation: Left side: lepton number = 0. Right side: electron (lepton number +1) + antineutrino (lepton number -1) = 0. $\checkmark$ Baryon number conservation: Left side: 1. Right side: 1 + 0 + 0 = 1. $\checkmark$ Why other options are wrong: - Option 1: $n \rightarrow p + e^+ + \bar{\nu}$ -- The positron ($e^+$) has charge +1, giving total charge +2 on the right. Charge not conserved. - Option 2: $n \rightarrow p + e^+ + \nu$ -- Same charge violation as above. - Option 3: $n \rightarrow p + e^- + \nu$ -- Lepton number: right side = +1 + 1 = +2 $\neq$ 0. Lepton number not conserved. The correct answer is Option 4 : $n \rightarrow p + e^- + \bar{\nu}$.

A light ray from the coin at $O$ travels to the edge $B$ of the hemispherical vessel.
In the right-angled triangle $\Delta OCB$, where $OC=CB=R$ (radius), the angle of incidence $i$ at the surface relative to the normal is equal to $\angle OBC = 45^\circ$.
For the ray to be visible from the surface level at $E$, the light must graze the liquid surface, meaning the angle of refraction $r = 90^\circ$.
Applying Snell's Law at point $B$: $\mu \sin(i) = 1 \cdot \sin(r)$.
Substituting the values: $\mu \sin(45^\circ) = \sin(90^\circ)$.
$\mu \left( \frac{1}{\sqrt{2}} \right) = 1 \implies \mu = \sqrt{2}$.

The efficiency of a Carnot engine is given by the formula: $\eta = 1 - \frac{T_c}{T_h}$ where $T_c$ is the cold reservoir temperature and $T_h$ is the hot reservoir temperature, both \in Kelvin. For the original engine $E$ working between $T_1 = 473 \, \text{K}$ and $T_2 = 273 \, \text{K}$: $\eta_{12} = 1 - \frac{T_2}{T_1} = 1 - \frac{273}{473} = \frac{473 - 273}{473} = \frac{200}{473}$ For engine $E_1$ working between $T_1 = 473 \, \text{K}$ and $T_3 = 373 \, \text{K}$: $\eta_1 = 1 - \frac{T_3}{T_1} = 1 - \frac{373}{473} = \frac{473 - 373}{473} = \frac{100}{473}$ For engine $E_2$ working between $T_3 = 373 \, \text{K}$ and $T_2 = 273 \, \text{K}$: $\eta_2 = 1 - \frac{T_2}{T_3} = 1 - \frac{273}{373} = \frac{373 - 273}{373} = \frac{100}{373}$ Now, evaluate the options: Option A: $\eta_{12} = \eta_1 \eta_2$ Compute $\eta_1 \eta_2$: $\eta_1 \eta_2 = \left( \frac{100}{473} \right) \times \left( \frac{100}{373} \right) = \frac{10000}{473 \times 373}$ Compare with $\eta_{12} = \frac{200}{473}$: $\frac{200}{473} \neq \frac{10000}{473 \times 373}$ Since they are not equal, option A is incorrect. Option B: $\eta_{12} \geq \eta_1 + \eta_2$ Compute $\eta_1 + \eta_2$: $\eta_1 + \eta_2 = \frac{100}{473} + \frac{100}{373} = 100 \left( \frac{1}{473} + \frac{1}{373} \right) = 100 \left( \frac{373 + 473}{473 \times 373} \right) = 100 \times \frac{846}{473 \times 373} = \frac{84600}{473 \times 373}$ Now, express $\eta_{12}$ with the same denominator: $\eta_{12} = \frac{200}{473} = \frac{200 \times 373}{473 \times 373} = \frac{74600}{473 \times 373}$ Compare $\frac{74600}{473 \times 373}$ and $\frac{84600}{473 \times 373}$: Since $74600 < 84600$, it follows that $\eta_{12} < \eta_1 + \eta_2$. Thus, $\eta_{12} \geq \eta_1 + \eta_2$ is false, so option B is incorrect. Option C: $\eta_{12} = \eta_1 + \eta_2$ From above: $\eta_{12} = \frac{200}{473} \approx 0.4228$ $\eta_1 + \eta_2 = \frac{100}{473} + \frac{100}{373} \approx 0.2114 + 0.2681 = 0.4795$ Since $0.4228 \neq 0.4795$, they are not equal. Thus, option C is incorrect. Option D: $\eta_{12} \leq \eta_1 + \eta_2$ From the comparison \in option B, $\eta_{12} < \eta_1 + \eta_2$, which implies $\eta_{12} \leq \eta_1 + \eta_2$ is true. Therefore, option D is correct. The correct answer is option D.

The radial magnetic force is $F_r = m \frac{dv_r}{dt} = -q v_z B$, and the force in the z-direction is $F_z = m \frac{dv_z}{dt} = q v_r B$, where $B = \frac{\mu_{\circ} I}{2\pi r}$.
Dividing these equations gives $\frac{dv_z}{dv_r} = -\frac{v_r}{v_z}$, which implies $v_r^2 + v_z^2 = v_{\circ}^2$ (conservation of kinetic energy).
We also have $\frac{dv_z}{dr} = \frac{dv_z}{dt} \frac{1}{v_r} = \frac{q v_r B}{m v_r} = \frac{q \mu_{\circ} I}{2\pi m r}$.
Integrating from the initial state $(v_{\circ}, a)$ to the turning point $(v_z, x)$ where $v_r = 0$:
Since $v_r^2 + v_z^2 = v_{\circ}^2$ and $v_r=0$ at the turning point, $v_z$ must reach $-v_{\circ}$.
Integrating the expression for $dv_z$: $\int_{v_{\circ}}^{-v_{\circ}} dv_z = \frac{q \mu_{\circ} I}{2\pi m} \int_a^x \frac{dr}{r} \implies -2v_{\circ} = \frac{q \mu_{\circ} I}{2\pi m} \ln\left(\frac{x}{a}\right)$.
Solving for $x$: $\ln\left(\frac{x}{a}\right) = -\frac{4\pi m v_{\circ}}{q \mu_{\circ} I} \implies x = a e^{-\frac{4\pi m v_{\circ}}{q \mu_{\circ} I}}$.

We need to find the ratio of resistance between two endpoints of an edge in a triangle vs. a square, both made from identical wires of resistance R. The wire of resistance R is bent into an equilateral triangle. Each side has resistance $R/3$. Between two endpoints of one edge: one path has resistance $R/3$ (direct), and the other path has $2R/3$ (around the other two sides). These are \in parallel: $R_{\triangle} = \frac{(R/3)(2R/3)}{R/3 + 2R/3} = \frac{2R^2/9}{R} = \frac{2R}{9}$ The identical wire of resistance R is bent into a square. Each side has resistance $R/4$. Between two endpoints of one edge: one path has $R/4$ (direct), and the other has $3R/4$ (around three sides). In parallel: $R_{\square} = \frac{(R/4)(3R/4)}{R/4 + 3R/4} = \frac{3R^2/16}{R} = \frac{3R}{16}$ $\frac{R_{\triangle}}{R_{\square}} = \frac{2R/9}{3R/16} = \frac{2}{9} \times \frac{16}{3} = \frac{32}{27}$ The correct answer is Option (3): 32/27.

We analyze each statement about the experiment measuring viscosity $\eta$ using a falling ball \in a liquid. Statement A: Graph between terminal velocity V and R is a parabola. The terminal velocity formula from Stokes' law is: $V_t = \frac{2R^2(\rho_s - \rho_l)g}{9\eta}$ Since $V_t \propto R^2$, the graph of $V_t$ vs $R$ is indeed a parabola. Statement A is CORRECT. Statement B: Terminal velocities of different diameter balls are constant for a given liquid. Since $V_t \propto R^2$, different diameter balls have different terminal velocities. Statement B is INCORRECT. Statement C: Measurement of terminal velocity depends on temperature. Viscosity $\eta$ is temperature-dependent. As temperature changes, the terminal velocity changes. Statement C is CORRECT. Statement D: This experiment can assess the density of a given liquid. From the terminal velocity equation $V_t = \frac{2R^2(\rho_s - \rho_l)g}{9\eta}$, if we know $V_t$, $R$, $\rho_s$, and $\eta$, we can determine $\rho_l$. Statement D is CORRECT. Statement E: If balls are dropped with some initial speed, the value of $\eta$ will change. Viscosity is an intrinsic property of the liquid. Regardless of the initial speed of the ball, it will eventually reach the same terminal velocity. The measured $\eta$ remains unchanged. Statement E is INCORRECT. The correct statements are A, C, and D. The answer is Option C) A, C and D Only .

We are given an EM wave with electric field $E = 100\sin(\omega t - kx)$ NC$^{-1}$ inside a cylinder of length 200 cm. A second cylinder has the same length but half the diameter, and must hold the same amount of EM energy. The average energy density of an EM wave is proportional to $E_0^2$: $u = \frac{1}{2}\epsilon_0 E_0^2$ Total energy stored \in a cylinder of volume $V$ is: $U = u \times V = \frac{1}{2}\epsilon_0 E_0^2 \times V$ For a cylinder with diameter $d$ and length $L$: $V = \frac{\pi d^2}{4} \cdot L$. The second cylinder has half the diameter, so its cross-sectional area is $(1/2)^2 = 1/4$ of the first, giving $V_2 = \frac{V_1}{4}$. Equating the energies, $\frac{1}{2}\epsilon_0 E_1^2 V_1 = \frac{1}{2}\epsilon_0 E_2^2 V_2$ implies $E_1^2 V_1 = E_2^2 \cdot \frac{V_1}{4}$, so $E_2^2 = 4E_1^2$ and hence $E_2 = 2E_1 = 2 \times 100 = 200 \text{ NC}^{-1}$. The modified electric field is $E = 200\sin(\omega t - kx)$ NC$^{-1}$. The correct answer is Option B) $200\sin(\omega t - kx)$ NC$^{-1}$

When the length of the spring becomes R, the angle spring will form from the vertical will be $\theta\ =60^{\circ\ }$. Since all forces acting on the body are conservative, total energy remains conserved. Potential energy of the bead at the top due to gravity, $PE_{g}=mg\left(2R\right)=2mgR$ Potential energy of the bead at the top due to the spring, $PE_{sp}=\frac{1}{2}kx^2\ =\frac{1}{2}k\left(2R-R\right)^2=\frac{1}{2}kR^2$ \Rightarrow Initial potential energy, $PE_{initial}=PE_g+PE_{sp}=2mgR+\frac{1}{2}kR^2$ When spring becomes the length R, the extension \in the spring becomes 0. Therefore, Potential energy = gravitational potential energy Final potential energy, $PE_{final}=mgH_{final}\ =\ mg\left(R\cos60^{\circ\ }\right)=\frac{mgR}{2}$ Loss \in potential energy = Gain \in kinetic energy $KE_{final}-KE_{initial}=PE_{initial}-PE_{final}$ \Rightarrow $KE_{final}=PE_{initial}-PE_{final}\ =\ 2mgR+\frac{1}{2}kR^{2-}\frac{mgR}{2}=\frac{3}{2}mgR+\frac{1}{2}kR^2$ $\frac{1}{2}mv^2=\frac{3}{2}mgR+\frac{1}{2}kR^2$ $\therefore\ v=\sqrt{\ 3gR+\frac{kR^2}{m}}$

Label the three resistors each as r/3 and the nodes: Left terminal = A Right terminal = D Between them are three resistors with wires: • top wire connects A to node after second resistor • bottom wire connects node between first and second resistor to D step 1: identify node equivalences Because of wires: A ≡ node after second resistor node between first and second ≡ D So the circuit collapses to just two nodes A and D. step 2: check each resistor Each of the three resistors (r/3) is connected between A and D. So all three are in parallel. step 3: equivalent resistance $R_{eq}=\frac{r/3}{3}=\frac{r}{9}$

Let us analyze each statement about fluid properties. Statement A: Surface tension arises due to extra energy of molecules at the interior compared to molecules at the surface. This is INCORRECT . Surface tension arises because molecules at the surface have higher potential energy than those in the interior (not the other way around). Interior molecules are surrounded by neighbors on all sides and are in a lower energy state. Statement B: As temperature of liquid rises, the coefficient of viscosity increases. This is INCORRECT . For liquids, viscosity decreases with increasing temperature because higher thermal energy helps molecules overcome intermolecular forces more easily. Statement C: As temperature of gas increases, the coefficient of viscosity increases. This is CORRECT . For gases, viscosity increases with temperature. This is because viscosity in gases arises from momentum transfer between molecular layers, and higher temperature means greater molecular speeds and more momentum transfer. The relationship is $\eta \propto \sqrt{T}$. Statement D: The onset of turbulence is determined by Reynolds number. This is CORRECT . The Reynolds number $Re = \frac{\rho v d}{\eta}$ determines whether flow is laminar or turbulent. Turbulence sets \in when $Re$ exceeds a critical value. Statement E: In steady flow, two streamlines never intersect. This is CORRECT . If streamlines intersected, a fluid element at the intersection would have two velocity directions simultaneously, which is impossible in steady flow. The correct statements are C, D, and E. The answer is Option A) C, D, E Only .

$The\ mass\ per\ unit\ area\ is\ \sigma=\frac{\sigma_0x}{ab}.\ We\ integrate\ over\ the\ area\ ofthe\ plate\ (x\in[0,a],\ y\in[0,b]):$ $M = \int_{0}^{a} \int_{0}^{b} \sigma \, dy \, dx = \int_{0}^{a} \int_{0}^{b} \frac{\sigma_0 x}{ab} \, dy \, dx$ $M = \frac{\sigma_0}{ab} \left[ \frac{x^2}{2} \right]_0^a \left[ y \right]_0^b = \frac{\sigma_0}{ab} \left( \frac{a^2}{2} \right) (b) = \frac{\sigma_0 a}{2}$ X-coordinate of Center of Mass: $x_{cm} = \frac{1}{M} \int \int x \sigma \, dA$ $x_{cm} = \frac{1}{M} \int_{0}^{a} \int_{0}^{b} x \left( \frac{\sigma_0 x}{ab} \right) \, dy \, dx = \frac{1}{M} \frac{\sigma_0}{ab} \int_{0}^{a} x^2 \, dx \int_{0}^{b} dy$ $x_{cm} = \frac{2}{\sigma_0 a} \cdot \frac{\sigma_0}{ab} \left( \frac{a^3}{3} \right) (b) = \frac{2}{\sigma_0 a} \cdot \frac{\sigma_0 a^2}{3} = \frac{2}{3}a$ Y-coordinate of Center of Mass: $Since\ the\ density\ \sigma\ depends\ only\ on\ x$ the mass is distributed uniformly \in the y-direction for any given x. Therefore: $y_{cm}=\frac{b}{2}$ The center of mass is at: $(x_{cm},y_{cm})=\left(\frac{2}{3}a,\frac{b}{2}\right)$

The output $Y$ of the given circuit is derived from the inputs $A$ and $B$. The circuit consists of two AND gates, one processing $(A, \bar{B})$ and the other processing $(A, B)$, followed by a NOR gate.
The expression for the output is:
$Y = \overline{(A \cdot \bar{B}) + (A \cdot B)}$
Using Boolean algebra, factor out $A$:
$Y = \overline{A \cdot (\bar{B} + B)}$
Since $\bar{B} + B = 1$:
$Y = \overline{A \cdot 1} = \bar{A}$
Option B represents a NAND gate with both inputs tied to $A$, which performs the NOT operation: $\overline{A \cdot A} = \bar{A}$.

For an ideal gas, rms speed is $v_{rms}=\sqrt{\frac{3kT}{m}}$ where k = Boltzmann constant, m = mass of one molecule. Squaring, $v_{rms}^2=\frac{3k}{m}T$ But mean squared velocity is $\overline{v^2}=v_{rms}^2$ So $\overline{v^2}\propto T$ This means mean squared velocity varies linearly with absolute temperature. So the graph is a straight line through origin.

A proton of mass $m_p$ has the same energy $E$ as a photon of wavelength $\lambda$, and we seek the ratio of the proton's de Broglie wavelength to the photon's wavelength. To begin, we express the photon energy \in terms of its wavelength. For a photon, $E = \frac{hc}{\lambda}$, which implies $\lambda = \frac{hc}{E}$. Next, we determine the proton's de Broglie wavelength by first relating its kinetic energy to its momentum. In the non-relativistic regime, the proton's energy is $E = \frac{p^2}{2m_p}$, from which the momentum follows as $p = \sqrt{2m_p E}$ and hence its de Broglie wavelength is $\lambda_p = \frac{h}{p} = \frac{h}{\sqrt{2m_p E}}$ Combining these expressions to form the desired ratio gives $\frac{\lambda_p}{\lambda} = \frac{h/\sqrt{2m_p E}}{hc/E} = \frac{h}{\sqrt{2m_p E}} \cdot \frac{E}{hc} = \frac{E}{c\sqrt{2m_p E}} = \frac{\sqrt{E}}{c\sqrt{2m_p}} = \frac{1}{c}\sqrt{\frac{E}{2m_p}}$ The correct answer is Option D) $\frac{1}{c}\sqrt{\frac{E}{2m_p}}$ .

In Young's double-slit interference experiment, the position of the nth bright fringe from the central maximum is given by the formula: $x = \frac{n \lambda D}{d}$ where: - $n$ is the fringe order, - $\lambda$ is the wavelength of light, - $D$ is the distance from the slits to the screen, - $d$ is the separation between the slits. For the first light source: Wavelength, $\lambda_1 = 600 \text{ nm} = 600 \times 10^{-9} \text{ m}$, Position of the 10th bright fringe, $x_1 = 10 \text{ mm} = 0.01 \text{ m}$, Fringe order, $n = 10$. Substituting into the formula: $x_1 = \frac{10 \lambda_1 D}{d} = 0.01 \quad ...(1)$ For the second light source: Wavelength, $\lambda_2 = 660 \text{ nm} = 660 \times 10^{-9} \text{ m}$, Fringe order, $n = 10$ (same as before), We need to find the new position $x_2$. The formula becomes: $x_2 = \frac{10 \lambda_2 D}{d} \quad ...(2)$ Since the experimental setup ($D$ and $d$) remains unchanged, we can take the ratio of equations (2) and (1): $\frac{x_2}{x_1} = \frac{\frac{10 \lambda_2 D}{d}}{\frac{10 \lambda_1 D}{d}} = \frac{\lambda_2}{\lambda_1}$ Therefore: $x_2 = x_1 \cdot \frac{\lambda_2}{\lambda_1}$ Substituting the given values (keeping units consistent \in mm and nm for simplicity): $x_1 = 10 \text{ mm}$, $\lambda_1 = 600 \text{ nm}$, $\lambda_2 = 660 \text{ nm}$. So: $x_2 = 10 \times \frac{660}{600}$ Simplifying the fraction: $\frac{660}{600} = \frac{66}{60} = \frac{11}{10} = 1.1$ Thus: $x_2 = 10 \times 1.1 = 11 \text{ mm}$ The distance of the 10th bright fringe from the central maximum for the second wavelength is 11 mm.

We need to find the dimension of the ratio $\frac{\text{Modulus of Elasticity}}{\text{Torque}}$ and determine the value of $c$ given that $b = -3$. To begin, we write the dimension of Modulus of Elasticity ($E$). Modulus of elasticity (Young's modulus) is defined as stress/strain. Since strain is dimensionless: $[E] = [\text{Stress}] = \frac{[\text{Force}]}{[\text{Area}]} = \frac{[M L T^{-2}]\$}{[L^2]\$} = \$[M L^{-1} T^{-2}]\$$ Next, we write the dimension of Torque ($\tau$). Torque is given by Force $\times$ distance (lever arm): $[\tau] = \$[M L T^{-2}]\$ \times [L] = \$[M L^2 T^{-2}]\$$ We then compute the dimension of $E/\tau$: $\left[\frac{E}{\tau}\right] = \frac{[M L^{-1} T^{-2}]\$}{[M L^2 T^{-2}]\$}$ Dividing each base dimension: $M: M^1 / M^1 = M^0$ $L: L^{-1} / L^2 = L^{-1-2} = L^{-3}$ $T: T^{-2} / T^{-2} = T^0$ Therefore, $\left[\frac{E}{\tau}\right] = \$[M^0 L^{-3} T^0]\$$ Comparing this result with $[M^a L^b T^c]$ shows that $a = 0$, $b = -3$, and $c = 0$. This confirms the given value $b = -3$ and yields the value of $c = 0$ . The answer is 0.

A rectangular metallic sheet has length $L = 5$ mm and breadth $B = 2.5$ mm, and the screw gauge used has pitch = 0.75 mm with 15 divisions on the circular scale. We first find the least count (LC) of the screw gauge. $LC = \frac{\text{Pitch}}{\text{Number of divisions}} = \frac{0.75}{15} = 0.05 \text{ mm}$ Next, we calculate the area of the sheet. $A = L \times B = 5 \times 2.5 = 12.5 \text{ mm}^2$ We then determine the maximum fractional error \in this area. $\frac{\Delta A}{A} = \frac{\Delta L}{L} + \frac{\Delta B}{B}$ Here, we take $\Delta L = \Delta B = LC = 0.05$ mm (since the maximum absolute error equals the least count). $\frac{\Delta A}{A} = \frac{0.05}{5} + \frac{0.05}{2.5} = 0.01 + 0.02 = 0.03 = \frac{3}{100}$ So $\frac{x}{100} = \frac{3}{100}$, giving $x = 3$. Hence the answer is 3.

Two iron solid discs have radii $R_1$ and $R_2$ with $R_2 = 2R_1$. We first determine the ratio $I_1/I_2$. Next, we recall that the moment of inertia of a solid disc is given by $I = \frac{1}{2}MR^2$ Since both discs are made of iron (same density $\rho$) and have negligible thickness $t$, their mass can be expressed as $M = \rho \cdot \pi R^2 \cdot t$ so $M \propto R^2$. Substituting into the inertia formula gives $I = \frac{1}{2}MR^2 = \frac{1}{2}(\rho \pi R^2 t)R^2 = \frac{\rho \pi t}{2} R^4$ and hence $I \propto R^4$. It follows that $\frac{I_1}{I_2} = \frac{R_1^4}{R_2^4} = \frac{R_1^4}{(2R_1)^4} = \frac{R_1^4}{16R_1^4} = \frac{1}{16}\,.$ Therefore, $I_1/I_2 = 1/x$ where $x = \boxed{16}$.

To compare the moments of inertia of a disc, ring, and sphere (all of mass $M$ and radius $R$) rotating about a diameter, we first recall the standard formulas for each. For a ring about its diameter we have $I_{\text{ring}} = \frac{1}{2}MR^2$, for a disc about its diameter $I_{\text{disc}} = \frac{1}{4}MR^2$, and for a sphere about its diameter $I_{\text{sphere}} = \frac{2}{5}MR^2$. Next, we check the condition given \in the problem by computing the ratio $\frac{I_{\text{disc}}}{I_{\text{ring}}} = \frac{MR^2/4}{MR^2/2} = \frac{1}{2}$. Since the problem states that $I_{\text{disc}} = 2.5 \times I_{\text{ring}}$, this standard result does not hold unless the disc and ring have different radii. We are told, however, that the sphere has the same radius as the disc. Let us denote the ring's radius by $r$ and the disc's (and sphere's) radius by $R$. Imposing $I_{\text{disc}} = 2.5 \times I_{\text{ring}}$ gives $\frac{1}{4}MR^2 = 2.5 \times \frac{1}{2}Mr^2 = \frac{5}{4}Mr^2$. From this we find $R^2 = 5r^2$. Since the sphere has the same radius as the disc, its moment of inertia is $I_{\text{sphere}} = \frac{2}{5}MR^2 = \frac{2}{5}M(5r^2) = 2Mr^2$. Consequently, the ratio of the sphere's moment of inertia to the ring's is $n = \frac{I_{\text{sphere}}}{I_{\text{ring}}} = \frac{2Mr^2}{\frac{1}{2}Mr^2} = 4$. Hence the answer is $n = 4$ .

We analyze both statements about glucose derivatives and starch. Statement I: D-glucose pentaacetate reacts with 2,4-dinitrophenylhydrazine (2,4-DNP). 2,4-DNP reacts with the carbonyl group ($C=O$) of aldehydes and ketones. In D-glucose pentaacetate, all five hydroxyl groups (including the one at C-1) are acetylated. The C-1 hydroxyl is converted to an acetate ester, meaning the hemiacetal/open-chain aldehyde form is blocked. Since there is no free aldehyde group available, D-glucose pentaacetate does NOT react with 2,4-DNP. Statement I is FALSE. Statement II: Starch, on heating with concentrated sulfuric acid at 100 degrees C and 2-3 atm pressure, produces glucose. Starch undergoes acid-catalyzed hydrolysis. When heated with dilute or concentrated sulfuric acid under the given conditions, the glycosidic bonds in starch break down to yield glucose (the monomer unit). Statement II is TRUE. The answer is Option A) Statement I is false but Statement II is true .

We need to identify reactions that both acetaldehyde ($CH_3CHO$) and acetone ($CH_3COCH_3$) undergo individually. A. Iodoform Reaction: Both acetaldehyde and acetone have the $CH_3CO-$ group (methyl ketone/methyl aldehyde). Both undergo the iodoform reaction to give $CHI_3$. YES for both. B. Cannizzaro Reaction: Cannizzaro reaction requires aldehydes without \alpha-hydrogen. Acetaldehyde has \alpha-hydrogens ($CH_3$ group), so it does NOT undergo Cannizzaro reaction. Instead, it undergoes aldol condensation. Acetone also has \alpha-hydrogens and does not undergo Cannizzaro reaction. NO for both. C. Aldol Condensation: Both acetaldehyde and acetone have \alpha-hydrogens and undergo aldol condensation \in the presence of dilute base. YES for both. D. Tollen's Test: Tollen's test (silver mirror test) is specific to aldehydes. Acetaldehyde gives a positive test, but acetone does not. NO (not both). E. Clemmensen Reduction: Clemmensen reduction ($Zn-Hg/HCl$) reduces the carbonyl group to $CH_2$. Both aldehydes and ketones undergo this reduction. YES for both. The reactions common to both are A, C, and E. The answer is Option C) A, C and E Only .