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Three moles of an ideal gas undergo a cyclic process consisting of the following three reversible steps:
(a) **Isobaric expansion:** The gas expands at a constant pressure of $2.0 \text{ atm}$ from an initial volume of $15.0 \text{ L}$ to a final volume of $30.0 \text{ L}$.
(b) **Isochoric cooling:** The pressure of t... | **Final Answer:** $w = -9.21 \text{ L}\cdot\text{atm}, q = 9.21 \text{ L}\cdot\text{atm}, \Delta U = 0, \Delta H = 0, \Delta S = 0$ Step (a): $w_{a} = -P \Delta V = -2.0 \times (30.0 - 15.0) = -30.0 \text{ L}\cdot\text{atm}$ Step (b): $w_{b} = -P \Delta V = 0 \text{ L}\cdot\text{atm}$ Step (c): $w_{c} = -nRT \ln\left(\... | Chemistry 11th JEE Mains |
Determine the types of chemical bonds (ionic, covalent, or both) present in the following compounds: $NaNO_3$ and $CaCO_3$. Explain the presence of multiple bond types in these specific salts. | **Final Answer:** Both $NaNO_{3}$ and $CaCO_{3}$ contain both ionic and covalent bonds. Analysis of $NaNO_{3}$: $NaNO_{3} \rightarrow Na^{+} + NO_{3}^{-}$. Bond type: Ionic bond due to electrostatic attraction between $Na^{+}$ and $NO_{3}^{-}$. Bonding within $NO_{3}^{-}$: Nitrogen and Oxygen are both non-metals that s... | Chemistry 11th JEE Mains |
Find the value of $pK_{sp}$ ($-\log K_{sp}$) for a saturated solution of $AgCl$ at $298\text{ K}$ if the EMF of the cell $Ag | AgCl(sat) || Ag^+(0.1\text{ M}) | Ag$ is $0.236\text{ V}$. [Take $\frac{2.303 RT}{F} = 0.059\text{ V}$]
Answer: 10 | **Final Answer:** 10 Given: $E_{cell} = 0.236\text{ V}$, $C_{cathode} = 0.1\text{ M}$, $E_{cell}^{\circ} = 0$ Using $E_{cell} = E_{cell}^{\circ} - \frac{0.059}{n} \log \frac{[Ag^{+}]_{anode}}{[Ag^{+}]_{cathode}}$ $0.236 = 0 - \frac{0.059}{1} \log \frac{[Ag^{+}]_{sat}}{0.1}$ $\log \frac{[Ag^{+}]_{sat}}{10^{-1}} = -\frac... | Chemistry 12th JEE Mains |
Distinguish between a foam and an emulsion based on the physical state of the dispersed phase and the dispersion medium in each system. | **Final Answer:** Foam: Gas (DP) in Liquid (DM); Emulsion: Liquid (DP) in Liquid (DM). $\text{For Foam: Dispersed Phase (DP) = Gas; Dispersion Medium (DM) = Liquid}$ $\text{For Emulsion: Dispersed Phase (DP) = Liquid; Dispersion Medium (DM) = Liquid}$ | Chemistry 12th JEE Mains |
Identify which of the following gases—$NH_3$, $H_2$, $N_2$, or $He$—would demonstrate the highest magnitude of enthalpy of physisorption on a solid surface, and provide a brief justification based on the ease of liquefaction. | **Final Answer:** $NH_{3}$ Relate adsorption to intermolecular forces: $\text{Enthalpy of physisorption} \propto \text{Strength of Van der Waals forces}$. Relate forces to liquefaction: $\text{Strength of Van der Waals forces} \propto \text{Ease of liquefaction} \propto \text{Critical temperature } (T_{c})$. Compare $T... | Chemistry 12th JEE Mains |
Analyze the power loss in transmission cables during the transport of electricity. Show that the power wasted as heat in the wires is inversely proportional to $V^2$, where $V$ is the voltage at which the power is transmitted. | **Final Answer:** Since $P$ and $R$ are constants for a specific transmission, $P_{loss} \propto \frac{1}{V^{2}}$ For a power $P$ transmitted at voltage $V$, the current is $I = \frac{P}{V}$ The power dissipated as heat in the line of resistance $R$ is $P_{loss} = I^{2}R$ Substituting the expression for current: $P_{lo... | Physics 12th JEE Mains |
Name an organic functional group that contains a carbonyl $(>C=O)$ bond but does not show acidic or basic properties when tested with litmus paper. | **Final Answer:** Ketone (or Aldehyde) Identification of carbonyl groups: $R-CHO$ (Aldehyde), $R-CO-R$ (Ketone), $R-COOH$ (Carboxylic acid), $R-COOR$ (Ester). Litmus test evaluation: $R-COOH \xrightarrow{\text{litmus}} \text{Red (Acidic)}$. $R-CONH_{2}$ (Amide) $\implies \text{Weakly basic/neutral}$. Neutral carbonyl g... | Chemistry 11th JEE Mains |
Determine the effective capacitance of the arrangement shown in the figure below between the terminals $P$ and $Q$. The individual capacitances are $C_1 = 4\mu\text{F}$, $C_2 = 8\mu\text{F}$, and $C_3 = 12\mu\text{F}$. Show the steps taken to identify the specific grouping of the capacitors. | **Final Answer:** $C_{eq} = 24\mu\text{F}$ Mapping node potentials: $V_{P} = V_{\text{node after } C_{2}}$ and $V_{Q} = V_{\text{node after } C_{1}}$ Identifying parallel grouping: $C_{1} \parallel C_{2} \parallel C_{3}$ $C_{eq} = C_{1} + C_{2} + C_{3}$ $C_{eq} = 4\mu\text{F} + 8\mu\text{F} + 12\mu\text{F} = 24\mu\text... | Physics 12th JEE Mains |
Chromium pentoxide ($CrO_5$) is an unstable blue-colored compound produced during the reaction of dichromate ions with hydrogen peroxide in an acidic medium. Standard algebraic methods for determining oxidation states often lead to incorrect results for such molecules due to the presence of unusual bonding. Using the p... | **Final Answer:** The oxidation state of $Cr$ is $+6$, with $1$ oxide oxygen atom at $-2$ and $4$ peroxide oxygen atoms at $-1$. Using the structure-based equation: $x + 1(O_{\text{oxide}}) + 4(O_{\text{peroxide}}) = 0$ Substitution: $x + 1(-2) + 4(-1) = 0$ Result: $x - 2 - 4 = 0 \implies x = +6$ | Chemistry 11th JEE Mains |
In the circuit branch shown below, point $A$ is maintained at an electric potential of $V_A = 60\text{ V}$ and point $B$ is at a potential of $V_B = 10\text{ V}$. Two resistors with resistances $R_1 = 3\ \Omega$ and $R_2 = 2\ \Omega$ are connected in series between these two points. Determine the electric potential at ... | **Final Answer:** $V_{C} = 30\ \text{V}$ Using $\Delta V = V_{A} - V_{B} = 60 - 10 = 50\ \text{V}$ Using $R_{total} = R_{1} + R_{2} = 3 + 2 = 5\ \Omega$ Using $I = \frac{\Delta V}{R_{total}} = \frac{50}{5} = 10\ \text{A}$ Using $V_{C} = V_{A} - I R_{1} = 60 - (10 \times 3) = 30\ \text{V}$ | Physics 12th JEE Mains |
Based on the relativistic relation between energy, momentum, and rest mass ($E^2 = p^2c^2 + m_0^2c^4$), show that a particle with zero rest mass and non-zero energy must travel at a speed exactly equal to the speed of light $c$. | **Final Answer:** $v = c$ Given $E^{2} = p^{2}c^{2} + m_{0}^{2}c^{4}$ and $m_{0} = 0$ $E = \sqrt{p^{2}c^{2} + 0^{2}c^{4}} = pc$ Using the relation $v = \frac{pc^{2}}{E}$ $v = \frac{pc^{2}}{pc} = c$ | Physics 12th JEE Mains |
A particle of mass $2.5 \text{ kg}$ is executing Simple Harmonic Motion with an amplitude of $50 \text{ cm}$ and a time period of $12 \text{ s}$. Calculate the following:
(a) the angular frequency
(b) the maximum speed of the particle
(c) the magnitude of the maximum acceleration
(d) the maximum restoring force acting ... | **Final Answer:** (a) $\frac{\pi}{6} \text{ rad/s}$, (b) $\frac{\pi}{12} \text{ m/s}$, (c) $\frac{\pi^{2}}{72} \text{ m/s}^{2}$, (d) $\frac{5\pi^{2}}{144} \text{ N}$, (e) $\frac{\pi}{15} \text{ m/s}$, (f) $\frac{\pi}{24} \text{ m/s}$, (g) $1 \text{ s}$, (h) $2 \text{ s}$ $\omega = \frac{2\pi}{T} = \frac{2\pi}{12} = \fr... | Physics 12th JEE Mains |
A body of mass $4\text{ kg}$ is moving with an initial speed of $12\text{ m/s}$ on a smooth horizontal surface. A constant force of $16\text{ N}$ is applied to the body for a duration of $2\text{ s}$. Determine the magnitude of the final velocity of the body in each of the following cases:
(i) if the force is applied i... | **Final Answer:** (i) $20\text{ m/s}$, (ii) $4\text{ m/s}$, (iii) $4\sqrt{13}\text{ m/s}$ Acceleration $a = \frac{F}{m} = \frac{16\text{ N}}{4\text{ kg}} = 4\text{ m/s}^{2}$ Case (i): $v = u + at = 12 + (4 \times 2) = 20\text{ m/s}$ Case (ii): $v = u - at = 12 - (4 \times 2) = 4\text{ m/s}$ Case (iii): $v = \sqrt{u^{2}... | Physics 11th JEE Mains |
Two isolated conducting spheres with radii $R_1$ and $R_2$ are situated at a large distance from one another. They are electrically interconnected through a thin conducting wire. If a net charge $Q$ is provided to the system, determine the expression for the final charge residing on each sphere once the system arrives ... | **Final Answer:** $q_{1} = \frac{R_{1}Q}{R_{1} + R_{2}}, \, q_{2} = \frac{R_{2}Q}{R_{1} + R_{2}}$ At equilibrium, potentials are equal: $V_{1} = V_{2}$ Formula: $\frac{1}{4\pi\epsilon_{0}} \frac{q_{1}}{R_{1}} = \frac{1}{4\pi\epsilon_{0}} \frac{q_{2}}{R_{2}} \implies \frac{q_{1}}{R_{1}} = \frac{q_{2}}{R_{2}}$ Using $q_{... | Physics 12th JEE Mains |
A solution is prepared by dissolving $62\text{ g}$ of ethylene glycol ($C_2H_6O_2$) in $400\text{ g}$ of water. The resulting solution is placed in a freezer maintained at $265\text{ K}$. Calculate the mass of ice that will separate out from the solution at this temperature.
[Given: $K_f \text{ for water} = 1.86\text{... | **Final Answer:** $167.5 \text{ g}$ Molar mass of $C_{2}H_{6}O_{2} = 2(12) + 6(1) + 2(16) = 62 \text{ g mol}^{-1}$ Using $n = \frac{w}{M}$, $n = \frac{62}{62} = 1 \text{ mol}$ Using $\Delta T_{f} = T_{f}^{\circ} - T$, $\Delta T_{f} = 273 - 265 = 8 \text{ K}$ $$w_{liquid} = \frac{K_{f} \times n \times 1000}{\Delta T_{f}... | Chemistry 12th JEE Mains |
In a specific region of space, the electric potential is given by the function $V(x, y, z) = \alpha x^2 z^2 + \beta x y z$, where $\alpha$ and $\beta$ are dimensional constants. Determine the expression for the resultant electric field vector $\vec{E}$ at a general point $(x, y, z)$ in Cartesian coordinates. | **Final Answer:** $\vec{E} = -(2\alpha x z^{2} + \beta y z)\hat{i} - \beta x z \hat{j} - (2\alpha x^{2} z + \beta x y)\hat{k}$ Using $\vec{E} = -\nabla V = -\left( \frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z}\hat{k} \right)$ $E_{x} = -\frac{\partial}{\parti... | Physics 12th JEE Mains |
Identify the chemical element whose neutral atom in its ground state possesses the following electronic configuration: $1s^2 2s^2 2p^6 3s^2 3p^5$. | **Final Answer:** Chlorine ($Cl$) $n = 2 + 2 + 6 + 2 + 5 = 17$ $Z = n = 17$ $\text{Element} = \text{Chlorine (Cl)}$ | Chemistry 11th JEE Mains |
A small bead $M$ is constrained to slide along a fixed vertical guide rail. The bead is connected to a block $N$ by a light, inextensible string that passes over a small, frictionless fixed pulley as shown in the figure. At a particular instant, the string segment attached to the bead makes an angle $\alpha$ with the v... | **Final Answer:** $v = u \cos \alpha$ Total string length $L = \sqrt{h^{2} + y^{2}} + l_{N}$ where $h$ is horizontal distance and $y$ is vertical distance. $\frac{dL}{dt} = \frac{y}{\sqrt{h^{2} + y^{2}}} \frac{dy}{dt} + \frac{dl_{N}}{dt} = 0$ Substituting $\frac{dy}{dt} = -u$, $\frac{y}{\sqrt{h^{2} + y^{2}}} = \cos \al... | Physics 11th JEE Mains |
Determine the name of the transition metal that is the primary constituent of the basic carbonate mineral known as Azurite, represented by the formula $2CuCO_3 \cdot Cu(OH)_2$. | **Final Answer:** Copper ($Cu$) Chemical Formula of Azurite = $2CuCO_{3} \cdot Cu(OH)_{2}$ Transition Metal Symbol = $Cu$ Name of Metal ($Cu$) = {Copper} | Chemistry 12th JEE Mains |
If the pressure of an ideal gas is maintained at a constant value, determine the temperature (in $^\circ\text{C}$) at which the root mean square speed of the gas molecules will be exactly double the value it had at a temperature of $127^\circ\text{C}$. | **Final Answer:** $1327^{\circ}\text{C}$ Given: $T_{1} = 127 + 273 = 400\text{ K}$, $v_{\text{rms},2} = 2v_{\text{rms},1}$ Using $v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$, so $v_{\text{rms}} \propto \sqrt{T}$ at constant pressure and for same gas $$\frac{v_{\text{rms},2}}{v_{\text{rms},1}} = \sqrt{\frac{T_{2}}{T_{1}}} \i... | Physics 11th JEE Mains |
Calculate the reduction potential of a hydrogen electrode at $298\text{ K}$ when it is immersed in a solution with a $pH = 12.0$ and the pressure of $H_2$ gas is maintained at $1\text{ bar}$. (Use $0.0591$ as the value for $\frac{2.303RT}{F}$) | **Final Answer:** $E = -0.7092\text{ V}$ $2H^{+}(aq) + 2e^{-} \rightarrow H_{2}(g); E^{0} = 0\text{ V}$ $[H^{+}] = 10^{-pH} = 10^{-12}\text{ M}; P_{H_{2}} = 1\text{ bar}$ $E = E^{0} - \frac{0.0591}{2} \log \frac{P_{H_{2}}}{[H^{+}]^{2}}$ $E = 0 - \frac{0.0591}{2} \log \frac{1}{(10^{-12})^{2}} = - \frac{0.0591}{2} \log(1... | Chemistry 12th JEE Mains |
For a specific real gas, arrange the following temperatures in the correct decreasing order:
(A) Boyle's temperature ($T_b$)
(B) Critical temperature ($T_c$)
(C) Inversion temperature ($T_i$) | **Final Answer:** $T_{i} > T_{b} > T_{c}$ $T_{b} = \frac{a}{Rb}$ $T_{c} = \frac{8a}{27Rb}$ $T_{i} = \frac{2a}{Rb}$ $2 > 1 > \frac{8}{27} \approx 0.296$ $T_{i} > T_{b} > T_{c}$ | Chemistry 11th JEE Mains |
Consider the following organic species and evaluate the relative lengths of their $C-N$ bonds. Arrange the compounds in the descending order of their $C-N$ bond lengths (longest to shortest) and justify the sequence based on the concepts of resonance, inductive effect, and hybridization.
(I) $p$-nitroaniline
(II) Anil... | **Final Answer:** $III > II > I > IV$ Bond length $\propto \frac{1}{\text{Bond Order}}$. For (III), bond is $sp^{3}-sp^{3}$ single bond $\approx 1.47 \text{ \AA}$. For (IV), bond is $sp^{2}=sp^{2}$ pure double bond $\approx 1.27 \text{ \AA}$. In (I) and (II), resonance gives partial double bond character: $\text{Bond l... | Chemistry 11th JEE Mains |
A parallel plate capacitor is connected to a battery and fully charged. If a dielectric slab of constant $K$ is then inserted to completely fill the gap between the plates while the battery remains connected, explain how the capacitance and the energy stored in the capacitor change. | **Final Answer:** The capacitance increases to $K C_{0}$ and the energy stored increases to $K U_{0}$. Given: $V = \text{constant}$, initial capacitance $C_{0} = \frac{\epsilon_{0}A}{d}$ New capacitance with dielectric: $C = K \left( \frac{\epsilon_{0}A}{d} \right) = K C_{0}$ Initial energy stored: $U_{0} = \frac{1}{2}... | Physics 12th JEE Mains |
An object cools from $80^{\circ}\text{C}$ to $50^{\circ}\text{C}$ in a time interval of $5$ minutes when placed in a room kept at a constant temperature of $20^{\circ}\text{C}$. Determine the temperature of the object after an additional $5$ minutes has passed, assuming Newton's law of cooling holds. | **Final Answer:** $T = 35^{\circ}\text{C}$ Given: $T_{1} = 80^{\circ}\text{C}$, $T_{2} = 50^{\circ}\text{C}$, $T_{s} = 20^{\circ}\text{C}$, $\Delta t = 5 \text{ min}$ Using $\frac{T_{1} - T_{2}}{\Delta t} = K\left(\frac{T_{1} + T_{2}}{2} - T_{s}\right) \implies \frac{80 - 50}{5} = K\left(\frac{80 + 50}{2} - 20\right)$ ... | Physics 11th JEE Mains |
A crystalline mineral is composed of two elements, $A$ and $B$, arranged in a cubic lattice. The atoms of element $A$ are situated at the corners of the cube, while the atoms of element $B$ are situated at the centers of all the cubic faces. Deduce the empirical formula of this mineral. | **Final Answer:** The empirical formula of the mineral is $\mathbf{AB_3}$.
**Step-by-step solution:**
- $A$ at all 8 corners: $8 \times \frac{1}{8} = 1$
- $B$ at all 6 face centres: $6 \times \frac{1}{2} = 3$
Ratio $A:B = 1:3$ $\Rightarrow$ **Empirical formula = $AB_3$** | Chemistry 12th JEE Mains |
When a solution is either diluted by adding solvent or concentrated by evaporating solvent, which property of the solute remains unchanged? | **Final Answer:** The number of moles (or mass) of the solute remains unchanged. Using $n = M \times V$ $M_{1}V_{1} = M_{2}V_{2}$ $n_{initial} = n_{final}$ | Chemistry 12th JEE Mains |
For a hydrogen half-cell at $298\text{ K}$, determine whether the reduction potential will be positive, negative, or zero if the partial pressure of hydrogen gas $P_{H_2}$ is maintained at $2\text{ bar}$ and the concentration of hydrogen ions $[H^+]$ is $1.0\text{ M}$. Justify your answer using the Nernst equation. | **Final Answer:** Negative Half-reaction: $2H^{+}(aq) + 2e^{-} \rightarrow H_{2}(g)$ Using $E = E^{\circ} - \frac{0.0591}{n} \log \frac{P_{H_{2}}}{[H^{+}]^{2}}$ $E = 0 - \frac{0.0591}{2} \log \frac{2}{1^{2}} = -0.0089\text{ V}$ | Chemistry 12th JEE Mains |
Determine the possible range of oxidation states for the element Phosphorus ($Z=15$). Provide the formula of one compound as an example for both the minimum and maximum oxidation states in this range. | **Final Answer:** Range: $-3$ to $+5$; Minimum compound: $PH_{3}$, Maximum compound: $PCl_{5}$ Electronic configuration of $P (Z=15)$: $[Ne] 3s^{2} 3p^{3}$ Number of valence electrons $(n) = 5$ Minimum oxidation state $= n - 8 = 5 - 8 = -3$ Maximum oxidation state $= +n = +5$ Example for $-3$: $PH_{3}$ (Phosphine) Exam... | Chemistry 11th JEE Mains |
A block of mass $M$ is kept at rest on a rough inclined plane that makes an angle $\phi$ with the horizontal. The coefficient of static friction between the block and the plane is $\mu$. To prevent the block from sliding down, it is tied to a support at the top of the incline using a light, inextensible string parallel... | **Final Answer:** $T = Mg(\sin \phi - \mu \cos \phi)$ Normal force perpendicular to the incline: $N = Mg \cos \phi$ Maximum limiting static friction acting up the plane: $f_{s} = \mu N = \mu Mg \cos \phi$ Force balance along the incline (upward forces = downward forces): $T + f_{s} = Mg \sin \phi$ $T = Mg \sin \phi - \... | Physics 11th JEE Mains |
(i) Four point charges are placed at the corners of a square with side length $10\sqrt{2} \text{ cm}$ as shown in the figure. The magnitudes of the charges are $q_1 = +12 \mu\text{C}$, $q_2 = -15 \mu\text{C}$, $q_3 = +20 \mu\text{C}$, and $q_4 = -7 \mu\text{C}$. Calculate the net electrostatic potential at the geometri... | **Final Answer:** (i) $9 \times 10^{5} \text{ V}$, (ii) $-2 \text{ V}$ Distance $r = \frac{a\sqrt{2}}{2} = \frac{10\sqrt{2} \times \sqrt{2}}{2} = 10 \text{ cm} = 0.1 \text{ m}$ $V_{O} = \frac{k}{r} \sum q_{i} = \frac{9 \times 10^{9}}{0.1} (12 - 15 + 20 - 7) \times 10^{-6} = 9 \times 10^{5} \text{ V}$ $\Delta V = -\vec{... | Physics 12th JEE Mains |
A point object $P$ is moving with a constant speed of $6\text{ cm s}^{-1}$ along the positive x-axis. It is located in front of a system of two plane mirrors, $M_1$ and $M_2$, as depicted in the figure. Mirror $M_1$ lies along the x-axis, and mirror $M_2$ is inclined at an angle of $60^\circ$ to $M_1$. Calculate the ve... | **Final Answer:** $\vec{v}_{rel} = (9\hat{i} - 3\sqrt{3}\hat{j}) \text{ cm s}^{-1}$ $\vec{v}_{O} = 6\hat{i} \text{ cm s}^{-1}$, Mirror $M_{1}$ unit normal $\hat{n}_{1} = \hat{j}$ $\vec{v}_{I1} = \vec{v}_{O} - 2(\vec{v}_{O} \cdot \hat{n}_{1})\hat{n}_{1} = 6\hat{i} - 2(6\hat{i} \cdot \hat{j})\hat{j} = 6\hat{i} \text{ cm ... | Physics 12th JEE Mains |
Explain why chemical adsorption (chemisorption) results in the formation of a unilayer (monomolecular layer) whereas physical adsorption (physisorption) can result in the formation of multilayers on the surface of the adsorbent. | **Final Answer:** Physical adsorption results in multilayers due to weak, non-specific van der Waals forces, while chemical adsorption forms only a unilayer because it involves specific chemical bond formation that saturates the adsorbent's surface sites. $\text{Force of Attraction (Physisorption)} = \text{Weak van der... | Chemistry 12th JEE Mains |
A point charge is released from rest at a position in an electric field where the field lines are curved. Determine whether the particle will follow the path of the electric field line passing through that point. Provide a physical reason for your conclusion. | **Final Answer:** No, the particle will not follow the curved field line because the electric force is always tangential to the field lines and cannot provide the centripetal force required to change the direction of the velocity vector to match the curvature of the line. Initial force: $\vec{F} = q\vec{E}$ Initial acc... | Physics 12th JEE Mains |
Gay-Lussac's Law of Combining Volumes is specifically applicable to gaseous reactions. Explain why this law cannot be directly applied to a reaction where one of the reactants is a solid, such as the combustion of solid carbon to form carbon dioxide gas. | **Final Answer:** Gay-Lussac's Law is not applicable because solids do not follow the $V \propto n$ relationship that gases do; the volume of solid carbon is negligible and does not form a simple whole-number ratio with the volumes of the gases involved. $V_{gas} \propto n$ (at constant $T$ and $P$) $C(s) + O_{2}(g) \r... | Chemistry 11th JEE Mains |
Describe the process of "Reverse Osmosis" (RO). Draw a neat labeled diagram illustrating its application in the desalination of sea water, clearly showing the direction of solvent flow and the role of the semi-permeable membrane. | **Final Answer:** Reverse Osmosis occurs when applied pressure $P$ exceeds osmotic pressure $\pi$ ($P > \pi$), causing solvent molecules to move from the concentrated solution (seawater) through a semi-permeable membrane to the pure solvent (fresh water) side. $P_{applied} > \pi$ $\pi = iCRT$ $\text{Net Flow: Solution ... | Chemistry 12th JEE Mains |
A spherical capacitor consists of two concentric thin conducting shells of radii $R_1$ and $R_2$ ($R_2 > R_1$). Determine the capacitance of the system if the inner shell is connected to the ground (earthed) and a charge $+Q$ is given to the outer shell. | **Final Answer:** $C = \frac{4\pi\epsilon_{0} R_{2}^{2}}{R_{2} - R_{1}}$ $V_{1} = \frac{1}{4\pi\epsilon_{0}} \left( \frac{q_{1}}{R_{1}} + \frac{Q}{R_{2}} \right) = 0$ $q_{1} = -Q \frac{R_{1}}{R_{2}}$ $V_{2} = \frac{1}{4\pi\epsilon_{0}} \left( \frac{q_{1}}{R_{2}} + \frac{Q}{R_{2}} \right) = \frac{1}{4\pi\epsilon_{0} R_{... | Physics 12th JEE Mains |
The standard enthalpy of formation of sulfur dioxide gas, $SO_{2}(g)$, is $-296.8\text{ kJ mol}^{-1}$. Determine the enthalpy change ($\Delta H$) in kJ for the decomposition of two moles of sulfur dioxide into solid sulfur and oxygen gas according to the following equation:
$2SO_{2}(g) \rightarrow 2S(s) + 2O_{2}(g)$ | **Final Answer:** $\Delta H = 593.6\text{ kJ}$ Given: $\Delta_{f}H^{\circ}[SO_{2}(g)] = -296.8\text{ kJ mol}^{-1}$ For $2SO_{2}(g) \rightarrow 2S(s) + 2O_{2}(g)$, $\Delta H = \sum n\Delta_{f}H^{\circ}(\text{products}) - \sum m\Delta_{f}H^{\circ}(\text{reactants})$ $\Delta H = [2\Delta_{f}H^{\circ}(S,s) + 2\Delta_{f}H^{... | Chemistry 11th JEE Mains |
Identify the official IUPAC designation for the cyclic alkene structure represented in the diagram below. | **Final Answer:** 3-methylcyclohexene Parent ring identification: $6$-carbon cyclic ring with $1$ double bond $\rightarrow \text{cyclohexene}$ Priority rule: Double bond carbons are assigned locants $C_{1}$ and $C_{2}$ Numbering path: From bottom-right carbon towards top-right carbon to minimize methyl locant $\rightar... | Chemistry 11th JEE Mains |
Match the following ores with their correct chemical formulas:
(i) Horn silver
(ii) Azurite
Provide the formulas for both based on their mineralogical compositions. | **Final Answer:** (i) $AgCl$, (ii) $2CuCO_{3} \cdot Cu(OH)_{2}$ Identification of Horn silver: $AgCl$ Identification of Azurite: $2CuCO_{3} \cdot Cu(OH)_{2}$ | Chemistry 12th JEE Mains |
Calculate the depth $d$ below the surface of the Earth at which the acceleration due to gravity $g$ is reduced by $2\%$ of its value at the surface. Assume the Earth to be a uniform sphere of radius $R_e = 6400 \text{ km}$. | **Final Answer:** $d = 128 \text{ km}$ $g_{d} = g_{s} \left( 1 - \frac{d}{R_{e}} \right)$ $0.98 g_{s} = g_{s} \left( 1 - \frac{d}{6400} \right)$ $d = 0.02 \times 6400 = 128 \text{ km}$ | Physics 11th JEE Mains |
Define Brownian movement in colloidal sols. Explain its cause and the factors that influence the magnitude of this motion. | **Final Answer:** Brownian movement is the zig-zag motion of colloidal particles caused by unbalanced collisions with medium molecules; its magnitude is inversely proportional to particle size and viscosity, and directly proportional to temperature. $\text{Brownian movement} = \text{Continuous zig-zag motion of colloid... | Chemistry 12th JEE Mains |
A particle starts from the mean position at $t=0$ and executes SHM with a time period of $12\text{ s}$. The ratio of the displacement covered by the particle in the $1^{\text{st}}$ second to the displacement covered in the $2^{\text{nd}}$ second is $\frac{1}{\sqrt{n}-1}$. Find the value of $n$.
Answer: 3 | **Final Answer:** $n = 3$ $x(t) = A \sin(\omega t) \implies \omega = \frac{2\pi}{T} = \frac{2\pi}{12} = \frac{\pi}{6} \text{ rad/s}$ $d_{1} = x(1) - x(0) = A \sin\left(\frac{\pi}{6} \cdot 1\right) - 0 = \frac{A}{2}$ $d_{2} = x(2) - x(1) = A \sin\left(\frac{\pi}{6} \cdot 2\right) - \frac{A}{2} = \frac{\sqrt{3}A}{2} - \f... | Physics 11th JEE Mains |
The electrical resistance of a conductivity cell filled with $0.02 \ \text{M}$ $\text{KCl}$ solution is $2.456 \ \Omega$. If the electrodes in the cell are $1.5 \ \text{cm}$ apart and have a cross-sectional area of $60 \ \text{cm}^2$, calculate the resistivity of the solution in $\Omega \cdot \text{cm}$ units. | **Final Answer:** $\rho = 98.24 \ \Omega \cdot \text{cm}$ Given: $R = 2.456 \ \Omega, l = 1.5 \ \text{cm}, A = 60 \ \text{cm}^{2}$ $\rho = R \left( \frac{A}{l} \right)$ $$\rho = 2.456 \times \left( \frac{60}{1.5} \right) = 98.24 \ \Omega \cdot \text{cm}$$ | Chemistry 12th JEE Mains |
A comet moves in a highly elliptical orbit around the Sun with an eccentricity denoted by $e$. Determine the ratio of the comet's speed at its point of closest approach (perihelion) to its speed at its point of farthest distance (aphelion) in terms of $e$. | **Final Answer:** $\frac{v_{p}}{v_{a}} = \frac{1 + e}{1 - e}$ By conservation of angular momentum: $m v_{p} r_{p} = m v_{a} r_{a} \implies \frac{v_{p}}{v_{a}} = \frac{r_{a}}{r_{p}}$ Using distance relations: $r_{p} = a(1 - e)$ and $r_{a} = a(1 + e)$ $\frac{v_{p}}{v_{a}} = \frac{a(1 + e)}{a(1 - e)} = \frac{1 + e}{1 - e}... | Physics 11th JEE Mains |
Two identical short bar magnets, each possessing a magnetic moment $M$, are arranged as shown in the figure. The point $P$ is located at the midpoint of the line segment joining the centers of the two magnets, such that the distance of each magnet's center from $P$ is $d$. If point $P$ lies on the axial line of the hor... | **Final Answer:** $B_{net} = \frac{\sqrt{5}\mu_{0}M}{4\pi d^{3}}$ Field due to axial magnet: $B_{1} = \frac{\mu_{0}}{4\pi} \frac{2M}{d^{3}}$ Field due to equatorial magnet: $B_{2} = \frac{\mu_{0}}{4\pi} \frac{M}{d^{3}}$ Resultant field (since $B_{1} \perp B_{2}$): $B_{net} = \sqrt{B_{1}^{2} + B_{2}^{2}}$ $B_{net} = \sq... | Physics 12th JEE Mains |
A space probe is launched from the surface of a planet towards its moon. The distance between the centers of the planet and the moon is $D$. If the mass of the planet is $144$ times the mass of the moon, calculate the distance from the center of the moon where the net gravitational force acting on the probe is zero. | **Final Answer:** $x = \frac{D}{13}$ Condition for zero net force: $F_{planet} = F_{moon}$ $\frac{G M_{p} m}{(D - x)^{2}} = \frac{G M_{m} m}{x^{2}}$ $\frac{144 M_{m}}{(D - x)^{2}} = \frac{M_{m}}{x^{2}} \implies \frac{144}{(D - x)^{2}} = \frac{1}{x^{2}}$ Taking square root: $\frac{12}{D - x} = \frac{1}{x}$ $12x = D - x ... | Physics 11th JEE Mains |
In the electrolysis of acidified water, the decomposition reaction can be represented as $2H_2O(l) \rightarrow 2H_2(g) + O_2(g)$. If the total volume of gases collected at both the anode and cathode combined is such that it corresponds to $1.5\text{ moles}$ of gas at STP, determine the number of moles of water that hav... | **Final Answer:** $1.0\text{ mole}$ Balanced reaction: $2H_{2}O(l) \rightarrow 2H_{2}(g) + O_{2}(g)$ Moles of gas produced: $n_{total} = n_{H_{2}} + n_{O_{2}} = 1.5\text{ mol}$ Stoichiometric ratio: $n_{H_{2}O} : n_{total} = 2 : (2 + 1) = 2 : 3$ $n_{H_{2}O} = \frac{2}{3} \times n_{total} = \frac{2}{3} \times 1.5 = 1.0\... | Chemistry 12th JEE Mains |
A $12 \, \mu\text{F}$ capacitor is charged by a $200 \, \text{V}$ battery. It is then disconnected from the source and connected in parallel to another identical uncharged capacitor. Calculate the total electrostatic energy stored in the system after the connection is made. | **Final Answer:** $0.12 \, \text{J}$ Given: $C_{1} = 12 \times 10^{-6} \, \text{F}, V_{1} = 200 \, \text{V}, C_{2} = 12 \times 10^{-6} \, \text{F}, V_{2} = 0 \, \text{V}$ $V_{c} = \frac{C_{1}V_{1} + C_{2}V_{2}}{C_{1} + C_{2}} = \frac{(12 \times 200) + 0}{12 + 12} = 100 \, \text{V}$ $U_{f} = \frac{1}{2} (C_{1} + C_{2}) ... | Physics 12th JEE Mains |
A $127 \text{ g}$ sample of a metal chloride ($MCl_2$) contains exactly $71 \text{ g}$ of chlorine. Evaluate the equivalent weight of this metal chloride. | **Final Answer:** $63.5$ $W_{M} = 127 \text{ g} - 71 \text{ g} = 56 \text{ g}$ $E_{M} = \frac{W_{M}}{W_{Cl}} \times 35.5 = \frac{56}{71} \times 35.5 = 28$ $E_{MCl_{2}} = E_{M} + E_{Cl} = 28 + 35.5 = 63.5$ | Chemistry 11th JEE Mains |
A projectile has the same horizontal range $R$ for two different angles of projection. If $T_1$ and $T_2$ are the times of flight corresponding to these two angles, show that the horizontal range is given by $R = \frac{1}{2} g T_1 T_2$. | **Final Answer:** $R = \frac{1}{2} g T_{1} T_{2}$ For the same range $R$ and velocity $u$, the two angles of projection are $\theta$ and $(90^{\circ} - \theta)$. $T_{1} = \frac{2u \sin \theta}{g}$ and $T_{2} = \frac{2u \sin(90^{\circ} - \theta)}{g} = \frac{2u \cos \theta}{g}$ $$T_{1} T_{2} = \left( \frac{2u \sin \theta... | Physics 11th JEE Mains |
Describe the metallurgical process for the extraction of zinc from zinc blende ($ZnS$), detailing the stages of roasting and reduction with carbon. Provide all relevant chemical equations. | **Final Answer:** Zinc is extracted via roasting zinc blende ($ZnS$) to zinc oxide ($ZnO$), followed by reduction with carbon (coke) at $1673 K$ to obtain metallic zinc. $2ZnS + 3O_{2} \xrightarrow{1200 K} 2ZnO + 2SO_{2}$ $ZnO + C \xrightarrow{1673 K} Zn + CO$ $Zn(g) \xrightarrow{\text{condense}} Zn(l)$ | Chemistry 12th JEE Mains |
Determine the maximum speed of photoelectrons emitted from a metallic surface having a work function of $3.8 \text{ eV}$ when it is illuminated by radiation of frequency $2.4 \times 10^{15} \text{ Hz}$. (Take $h = 6.63 \times 10^{-34} \text{ J s}$, $m_e = 9.1 \times 10^{-31} \text{ kg}$, and $1 \text{ eV} = 1.6 \times ... | **Final Answer:** $v_{max} \approx 1.47 \times 10^{6} \text{ m/s}$ Given: $\phi = 3.8 \text{ eV}$, $\nu = 2.4 \times 10^{15} \text{ Hz}$, $h = 6.63 \times 10^{-34} \text{ J s}$, $m_{e} = 9.1 \times 10^{-31} \text{ kg}$ $E = h\nu = (6.63 \times 10^{-34}) \times (2.4 \times 10^{15}) = 1.5912 \times 10^{-18} \text{ J}$ $E... | Physics 12th JEE Mains |
"The choice of a catalyst is crucial for the efficiency of industrial chemical processes." Justify this statement by taking the example of the manufacture of ammonia via the Haber's process. | **Final Answer:** Iron catalyst lowers the activation energy, enabling the Haber process to reach equilibrium rapidly at moderate temperatures, ensuring industrial efficiency. Chemical Equilibrium: $N_{2}(g) + 3H_{2}(g) \rightleftharpoons 2NH_{3}(g); \Delta H = -92.4 \text{ kJ mol}^{-1}$ Arrhenius Equation: $k = A e^{-... | Chemistry 12th JEE Mains |
The diagram below illustrates a composite rod formed by joining an aluminum rod and an iron rod end-to-end. Both rods are of the same length and have identical cross-sectional areas. The external end of the aluminum rod is maintained at a constant temperature of $10^\circ\text{C}$, while the external end of the iron ro... | **Final Answer:** $T_{j} = 30^{\circ}\text{C}$ $H_{Fe} = H_{Al}$ $$\frac{k_{Fe} A (110 - T_{j})}{L} = \frac{k_{Al} A (T_{j} - 10)}{L}$$ $50(110 - T_{j}) = 200(T_{j} - 10)$ $110 - T_{j} = 4(T_{j} - 10) \Rightarrow 110 - T_{j} = 4T_{j} - 40$ $$5T_{j} = 150 \Rightarrow T_{j} = 30^{\circ}\text{C}$$ | Physics 11th JEE Mains |
A circular conducting loop of radius $R$ carries a steady current $I$. Let $B_{center}$ be the magnitude of the magnetic field at the center of the loop. On the axis of the loop, at a distance $x$ from the center, the magnetic field magnitude is $B_{axis}$. If the ratio of the magnetic fields is given by $\frac{B_{cent... | **Final Answer:** $\frac{x}{R} = 2$ $B_{center} = \frac{\mu_{0} I}{2R}$ and $B_{axis} = \frac{\mu_{0} I R^{2}}{2(R^{2} + x^{2})^{3/2}}$ $\frac{B_{center}}{B_{axis}} = \frac{\mu_{0} I / (2R)}{\mu_{0} I R^{2} / (2(R^{2} + x^{2})^{3/2})} = \frac{(R^{2} + x^{2})^{3/2}}{R^{3}}$ $\frac{B_{center}}{B_{axis}} = \left( 1 + \fra... | Physics 12th JEE Mains |
A rescue aircraft is flying horizontally at a constant speed of $270 \text{ km/h}$ at an altitude of $490 \text{ m}$ above the ground. Determine the horizontal distance from a stationary target on the ground at which a relief package should be released so that it hits the target. (Take $g = 9.8 \text{ m/s}^2$). | **Final Answer:** $750 \text{ m}$ $u_{x} = 270 \times \frac{5}{18} = 75 \text{ m/s}$ $t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 490}{9.8}} = 10 \text{ s}$ $x = u_{x} \times t = 75 \times 10 = 750 \text{ m}$ | Physics 11th JEE Mains |
A block of mass $M$ is attached to the lower end of a vertical spring of spring constant $k$, the other end of which is fixed to a ceiling. Initially, the block is held such that the spring is at its natural length. If the block is suddenly released, determine the maximum extension produced in the spring. | **Final Answer:** $x = \frac{2Mg}{k}$ Using Conservation of Mechanical Energy: $E_{i} = E_{f}$ $K_{i} + U_{gi} + U_{si} = K_{f} + U_{gf} + U_{sf}$ $0 + 0 + 0 = 0 - Mgx + \frac{1}{2}kx^{2}$ $Mgx = \frac{1}{2}kx^{2} \implies x = \frac{2Mg}{k}$ | Physics 11th JEE Mains |
A sample of hydrogen atoms in the ground state is exposed to monochromatic electromagnetic radiation of wavelength $972.5$ $\text{\AA}$, which excites the atoms to the $n=4$ state. Calculate the total number of distinct spectral lines (induced radiations) that can be observed in the emission spectrum as the atoms retur... | **Final Answer:** 6 Using $E = \frac{12400}{\lambda}$, $E = \frac{12400}{972.5} \approx 12.75\text{ eV}$ $\Delta E = 13.6\left(1 - \frac{1}{n^{2}}\right) = 12.75 \implies n=4$ Total lines $N = \frac{n(n-1)}{2} = \frac{4(4-1)}{2} = 6$ | Physics 12th JEE Mains |
Analyze the nature of work done by the force of friction. State whether the work done by friction on a body can be positive, negative, or zero, and briefly justify your answer with a conceptual explanation. | **Final Answer:** Work done by friction can be positive, negative, or zero depending on the displacement relative to the force. Using $W = \vec{f} \cdot \vec{s} = fs \cos \theta$ Negative Work: $\theta = 180^{\circ} \implies W = fs \cos 180^{\circ} = -fs$ (e.g., kinetic friction on a sliding block) Positive Work: $\the... | Physics 11th JEE Mains |
Twelve identical resistors, each with a resistance of $R$, are interconnected to form the framework of a cube. The circuit diagram below shows a two-dimensional projection of this cubic network. Determine the net equivalent resistance of the entire arrangement between terminals $A$ and $B$, which are located at the opp... | **Final Answer:** $R_{eq} = \frac{3}{4} R$ $V_{AB} = \sum I_{i}R_{i}$ $V_{AB} = I \times \frac{3}{4} R$ $R_{eq} = \frac{V_{AB}}{I} = \frac{3}{4} R$ | Physics 12th JEE Mains |
Describe the principle and the chemical reactions involved in the estimation of nitrogen in an organic compound using the Dumas method. | **Final Answer:** The principle involves converting nitrogen to $N_{2}$ gas using $CuO$, and the percentage is given by $\%N = \frac{V_{0}}{8m}$ where $V_{0}$ is volume at STP and $m$ is mass of compound. $C_{x}H_{y}N_{z} + (2x + \frac{y}{2})CuO \rightarrow xCO_{2} + \frac{y}{2}H_{2}O + \frac{z}{2}N_{2} + (2x + \frac{y... | Chemistry 11th JEE Mains |
Describe the variation in the surface tension of water as soap is added gradually. Specifically, discuss the behavior of surface tension before reaching the Critical Micelle Concentration (CMC) and after the CMC is reached. | **Final Answer:** Surface tension decreases linearly with the logarithm of concentration until the CMC is reached, after which it remains constant. Adding soap (surfactant) molecules reduces the surface tension $\gamma$ of water by populating the air-water interface and disrupting hydrogen bonds. For concentration rang... | Chemistry 12th JEE Mains |
In the illustrated mechanical arrangement, a block of mass $M_0$ is connected to a light, inextensible string. This string is anchored to a rigid floor at one end, passes over an ideal (massless and frictionless) movable pulley, and is attached to the block at the other end. The movable pulley is suspended from a rigid... | **Final Answer:** $T = 4\pi\sqrt{\frac{M_{0}}{K}}$ Constraint Relation: $x = 2y$ where $x$ is block displacement and $y$ is pulley displacement. Force Balance on Pulley: $2T = F_{spring} = K(y_{0} + y) \implies T = \frac{1}{2}K(y_{0} + y)$ Equation of Motion for $M_{0}$: $M_{0}\ddot{x} = M_{0}g - T$ Substitution: $M_{0... | Physics 11th JEE Mains |
Sodium hydroxide ($NaOH$) is known to be a deliquescent substance. Explain what happens to the physical state of $NaOH$ pellets when they are exposed to the atmosphere for a prolonged duration. | **Final Answer:** The $NaOH$ pellets first absorb moisture from the air to form a concentrated solution and then react with atmospheric $CO_{2}$ to form a white solid crust of sodium carbonate ($Na_{2}CO_{3}$). $NaOH_{(s)} + xH_{2}O_{(g)} \rightarrow NaOH_{(aq)}$ $2NaOH_{(aq)} + CO_{2(g)} \rightarrow Na_{2}CO_{3(s)} + ... | Chemistry 11th JEE Mains |
The vapour pressure of pure water at $100^\circ\text{C}$ is $760\text{ mm Hg}$. Calculate the vapour pressure of an aqueous solution of urea (a non-volatile solute) that is $2.0$ molal. (Assume the solution behaves ideally). | **Final Answer:** $P_{s} = 733.59\text{ mm Hg}$ Given: $P_{0} = 760\text{ mm Hg}$, $m = 2.0\text{ mol/kg}$ $n_{\text{H}_{2}\text{O}} = \frac{1000}{18} = 55.56\text{ mol}$ $x_{\text{urea}} = \frac{m}{m + 55.56} = \frac{2.0}{2.0 + 55.56} = 0.03475$ $P_{s} = P_{0}(1 - x_{\text{urea}}) = 760(1 - 0.03475) = 733.59\text{ mm ... | Chemistry 12th JEE Mains |
Explain the following categories of redox reactions, providing a balanced chemical equation for each to illustrate the transfer of electrons:
(i) Combination reaction
(ii) Decomposition reaction | **Final Answer:** Redox reactions are categorized as Combination ($C + O_{2} \rightarrow CO_{2}$) and Decomposition ($2NaH \rightarrow 2Na + H_{2}$) when they involve change in oxidation states of elemental species. (i) Combination: $A + B \rightarrow C$ (where reactants are in elemental state) $C(s) + O_{2}(g) \righta... | Chemistry 11th JEE Mains |
Using the half-reaction (ion-electron) method, balance the following redox reaction occurring in an acidic medium:
$$I_2 + H_2O_2 \longrightarrow IO_3^- + H_2O$$
Provide the balanced equation and clearly show the separated oxidation and reduction half-reactions. | **Final Answer:** $I_{2} + 5H_{2}O_{2} \longrightarrow 2IO_{3}^{-} + 2H^{+} + 4H_{2}O$ Oxidation half-reaction: $I_{2} + 6H_{2}O \longrightarrow 2IO_{3}^{-} + 12H^{+} + 10e^{-}$ Reduction half-reaction: $H_{2}O_{2} + 2H^{+} + 2e^{-} \longrightarrow 2H_{2}O$ Multiplying reduction half by 5: $5H_{2}O_{2} + 10H^{+} + 10e^... | Chemistry 11th JEE Mains |
A small particle of mass $m$ is placed at the topmost point of a smooth hemispherical shell of mass $M = 2m$. The shell is kept on a smooth horizontal floor and is free to move. The particle is given a negligible push and begins to slide down the outer surface of the shell. At the instant the particle is at an angular ... | **Final Answer:** $2 \csc \theta$ Horizontal force on the shell: $F_{x} = N \sin \theta$ Acceleration of the shell: $A = \frac{F_{x}}{M} = \frac{N \sin \theta}{2m}$ Magnitude of the pseudo force: $F_{p} = m A = m \left( \frac{N \sin \theta}{2m} \right) = \frac{N \sin \theta}{2}$ Ratio of forces: $\frac{N}{F_{p}} = \fra... | Physics 11th JEE Mains |
The pH of a $0.2\text{ M}$ monobasic acid is measured to be $1.0$. Determine the osmotic pressure of this solution at a given temperature $T\text{ K}$ in terms of $R$ and $T$. | **Final Answer:** $\Pi = 0.3RT$ Given: $C = 0.2\text{ M}$, $pH = 1.0$ $[H^{+}] = 10^{-pH} = 10^{-1.0} = 0.1\text{ M}$ $[H^{+}] = C\alpha \implies 0.1 = 0.2\alpha \implies \alpha = 0.5$ Using $i = 1 + (n-1)\alpha = 1 + (2-1)0.5 = 1.5$ $\Pi = iCRT = 1.5 \times 0.2 \times R \times T = 0.3RT$ | Chemistry 12th JEE Mains |
A block of mass $10 \text{ kg}$ is resting on a rough horizontal floor. The coefficients of static and kinetic friction between the block and the floor are $\mu_s = 0.5$ and $\mu_k = 0.3$ respectively. A pull of $100 \text{ N}$ is applied to the block at an angle of $30^\circ$ with the horizontal as shown in the figure... | **Final Answer:** $a = 7.16\text{ m/s}^{2}$ $N = mg - F\sin 30^{\circ} = 10(10) - 100(0.5) = 50\text{ N}$ $F_{x} = F\cos 30^{\circ} = 100(0.866) = 86.6\text{ N} > f_{s,\max} = \mu_{s}N = 25\text{ N}$ $a = \frac{F_{x} - \mu_{k}N}{m} = \frac{86.6 - 0.3(50)}{10} = 7.16\text{ m/s}^{2}$ | Physics 11th JEE Mains |
A uniform electric field $\vec{E}$ exists within a cylindrical region and is oriented parallel to the cylinder's longitudinal axis. The magnitude of this electric field varies over time $t$ as shown in the provided graph for four different cases (labeled 1, 2, 3, and 4).
Using the Ampere-Maxwell law, rank these four ... | **Final Answer:** Increasing order: 4, 3, 2, 1 Using Ampere-Maxwell Law: $\oint \vec{B} \cdot d\vec{l} = \mu_{0}\epsilon_{0} \frac{d\Phi_{E}}{dt}$ Substitution for cylindrical symmetry: $B(2\pi R) = \mu_{0}\epsilon_{0} \frac{d(E \cdot \pi R^{2})}{dt}$ Simplifying for $B$: $B = \frac{\mu_{0}\epsilon_{0}R}{2} \left| \fra... | Physics 12th JEE Mains |
In an aqueous solution, the observed molar mass of potassium nitrate ($KNO_3$) is determined to be $63.125$ g/mol. Calculate the degree of dissociation of potassium nitrate in this solution. (Atomic masses: $K = 39, N = 14, O = 16$) | **Final Answer:** $\alpha = 0.6$ $M_{\text{calculated}} = 39 + 14 + 3(16) = 101 \text{ g/mol}$ $i = \frac{M_{\text{calculated}}}{M_{\text{observed}}} = \frac{101}{63.125} = 1.6$ $i = 1 + (n - 1)\alpha \implies 1.6 = 1 + (2 - 1)\alpha \implies \alpha = 0.6$ | Chemistry 12th JEE Mains |
A snapshot of a sinusoidal progressive wave travelling on a string is captured at a certain time $t$, as depicted in the figure. Several elemental sections of the string are identified by numbers $1, 2, 3,$ and $4$. Based on the principles of wave energy distribution in a travelling wave, identify the specific element(... | **Final Answer:** Point 2 $u_{p} = \frac{1}{2} T \left( \frac{\partial y}{\partial x} \right)^{2}$ $\text{For } y = A \sin(kx - \omega t), \text{ slope } \frac{\partial y}{\partial x} = Ak \cos(kx - \omega t)$ $\text{At Point 2, } y = 0 \implies \cos(kx - \omega t) = \pm 1 \implies \left| \frac{\partial y}{\partial x} ... | Physics 11th JEE Mains |
A $25\text{ mL}$ sample of $H_2O_2$ solution requires $40\text{ mL}$ of $0.05\text{ M } KMnO_4$ for complete titration in an acidic medium. Calculate the volume strength of the $H_2O_2$ solution. | **Final Answer:** Volume Strength = $2.24 \text{ V}$ $n_{KMnO_{4}} = 5, \quad n_{H_{2}O_{2}} = 2$ $N_{KMnO_{4}} = M \times n = 0.05 \times 5 = 0.25 \text{ N}$ $N_{1}V_{1} = N_{2}V_{2} \implies N_{H_{2}O_{2}} \times 25 = 0.25 \times 40$ $N_{H_{2}O_{2}} = \frac{10}{25} = 0.4 \text{ N}$ $\text{Volume Strength} = 5.6 \time... | Chemistry 11th JEE Mains |
Three point charges $q$, $2q$, and $-q$ coulomb are placed at the points $(\hat{i} + \hat{j} - \hat{k})$, $(\hat{i} - \hat{j} + \hat{k})$, and $(-\hat{i} + \hat{j} + \hat{k})$ respectively. Determine the $y$-component of the resultant electric field at the origin. | **Final Answer:** $\frac{q}{6\sqrt{3}\pi\epsilon_{0}}$ Given: $q_{1} = q, \vec{r}_{1} = (1, 1, -1)$; $q_{2} = 2q, \vec{r}_{2} = (1, -1, 1)$; $q_{3} = -q, \vec{r}_{3} = (-1, 1, 1)$ Distance $r = \sqrt{1^{2} + 1^{2} + (-1)^{2}} = \sqrt{3}$ for all point charges. Using $E_{y} = \sum \frac{1}{4\pi\epsilon_{0}} \frac{q_{i}(... | Physics 12th JEE Mains |
Calculate the mole fraction of methanol ($\text{CH}_3\text{OH}$) and water ($\text{H}_2\text{O}$) in a binary solution prepared by mixing $16\text{ g}$ of methanol with $54\text{ g}$ of water. | **Final Answer:** $\chi_{\text{CH}_{3}\text{OH}} = 0.143, \chi_{\text{H}_{2}\text{O}} = 0.857$ $M_{\text{CH}_{3}\text{OH}} = 12 + 4(1) + 16 = 32\text{ g mol}^{-1}$ $n_{\text{CH}_{3}\text{OH}} = \frac{16}{32} = 0.5\text{ mol}$ $M_{\text{H}_{2}\text{O}} = 2(1) + 16 = 18\text{ g mol}^{-1}$ $n_{\text{H}_{2}\text{O}} = \fra... | Chemistry 11th JEE Mains |
Identify the atomic number and provide the official IUPAC symbol for the element temporarily named "Ununquadium" ($Uuq$). | **Final Answer:** Atomic number: $114$, Symbol: $Fl$ IUPAC roots: $\text{un} = 1, \text{un} = 1, \text{quad} = 4$ $$Z = (1 \times 10^{2}) + (1 \times 10^{1}) + (4 \times 10^{0}) = 114$$ Official symbol mapping: $Z = 114 \rightarrow Fl$ | Chemistry 11th JEE Mains |
Provide the chemical formulas and the common mineral names of two important carbonate ores used in the extraction of metals. | **Final Answer:** Calamine ($ZnCO_{3}$) and Siderite ($FeCO_{3}$) Identifying Zinc carbonate ore: Mineral Name = Calamine, Formula = $ZnCO_{3}$ Identifying Iron carbonate ore: Mineral Name = Siderite, Formula = $FeCO_{3}$ | Chemistry 12th JEE Mains |
An aqueous solution contains $0.1 \text{ M}$ sucrose. Describe the changes observed in the freezing point and the boiling point of this solution if a small amount of solid potassium sulfate ($K_2SO_4$) is added and dissolved in it. | **Final Answer:** The boiling point increases and the freezing point decreases. Dissociation of $K_{2}SO_{4} \rightarrow 2K^{+} + SO_{4}^{2-}$ Van't Hoff factor $i = 1 + (3-1)1 = 3$ $M_{total} = M_{sucrose} + i \cdot M_{K_{2}SO_{4}} > 0.1 \text{ M}$ $\Delta T_{b} = K_{b} \cdot M_{total} \implies T_{b} = T_{b}^{o} + \De... | Chemistry 12th JEE Mains |
Why is it recommended to spread sand or grit on a slippery, oil-covered floor to prevent people from falling? Explain in terms of the coefficients of friction. | **Final Answer:** Spreading sand increases the coefficient of static friction $\mu_{s}$, which significantly increases the limiting friction $f_{s} = \mu_{s} N$ required to provide necessary grip and prevent falling. $f_{s} \le \mu_{s} N$ $\mu_{\text{oil}} \approx 0.05 \implies f_{\text{oil}} = 0.05 N$ $\mu_{\text{sand... | Physics 11th JEE Mains |
(i) Define the term soil pollution.
(ii) State two major environmental disadvantages of using organochlorines like DDT.
(iii) Why are modern herbicides often preferred over traditional broad-spectrum pesticides? | **Final Answer:** (i) Soil pollution is the buildup of toxic substances that change the soil profile. (ii) DDT is persistent and biomagnifies in the food chain. (iii) Herbicides are preferred because they are biodegradable and target-specific. $\text{Soil Pollution} = \text{Contamination of soil by pollutants} \rightar... | Chemistry 11th JEE Mains |
A solid cylinder begins to rotate about its longitudinal axis with a time-dependent angular acceleration given by the expression $\alpha = 3pt^2 - 2qt$, where $p$ and $q$ are constant coefficients and $t$ is the time in seconds. Assuming the cylinder has an initial angular velocity $\omega_0$ at $t = 0$ and starts from... | **Final Answer:** (a) $\omega = \omega_{0} + pt^{3} - qt^{2}$ and (b) $\theta = \omega_{0}t + \frac{pt^{4}}{4} - \frac{qt^{3}}{3}$ $\frac{d\omega}{dt} = \alpha = 3pt^{2} - 2qt$ $\int_{\omega_{0}}^{\omega} d\omega = \int_{0}^{t} (3pt^{2} - 2qt) dt \implies \omega = \omega_{0} + pt^{3} - qt^{2}$ $\frac{d\theta}{dt} = \om... | Physics 11th JEE Mains |
Determine the dimensional formula of the Modulus of Rigidity ($G$). Compare its dimensions with those of pressure and clarify if they are identical. | **Final Answer:** Dimensional formula: $G = [ML^{-1}T^{-2}]$. Yes, its dimensions are identical to those of pressure. Formula: $G = \frac{\text{Shear Stress}}{\text{Shear Strain}}$ $[G] = \frac{[MLT^{-2}] / [L^{2}]}{[M^{0}L^{0}T^{0}]} = [ML^{-1}T^{-2}]$ Formula: $P = \frac{\text{Force}}{\text{Area}}$ $[P] = \frac{[MLT^... | Physics 11th JEE Mains |
Define the term "vapour pressure" of a liquid. How is the boiling point of a liquid related to its vapour pressure and the external atmospheric pressure? | **Final Answer:** The boiling point is the temperature at which the vapour pressure of a liquid equals the external atmospheric pressure ($P_{vap} = P_{ext}$). Definition: $P_{vap} = \text{Pressure exerted by vapours in equilibrium with the liquid phase at a constant temperature}$ Boiling Condition: $P_{vap}(T) = P_{ex... | Chemistry 12th JEE Mains |
A uniform thin rod of mass $M$ and length $L$ is moving on a frictionless horizontal floor. At a particular instant, the rod is aligned horizontally, and the velocities of its endpoints are perpendicular to its length. The left end is moving with a speed of $3v$ in the upward direction, while the right end is moving wi... | **Final Answer:** $\frac{7}{6} Mv^{2}$ Using $v_{cm} = \frac{v_{left} + v_{right}}{2}$ with upward as positive: $v_{cm} = \frac{3v + (-v)}{2} = v$ Using $\omega = \frac{v_{left} - v_{right}}{L}$ with upward as positive: $\omega = \frac{3v - (-v)}{L} = \frac{4v}{L}$ Using $K = \frac{1}{2}Mv_{cm}^{2} + \frac{1}{2}I_{cm}\... | Physics 11th JEE Mains |
Define the term "node" in the context of atomic orbitals and specify the mathematical value of the probability density function $(\psi^2)$ at any nodal point. | **Final Answer:** At a node, the probability density $\psi^{2} = 0$ At any nodal point, the wave function $\psi = 0$ $\text{Probability density} = |\psi|^{2} = (0)^{2}$ $|\psi|^{2} = 0$ | Chemistry 11th JEE Mains |
An experiment determines the actual speed of sound in a specific gas to be $340 \text{ m/s}$. At the time of the measurement, the gas pressure is $1.0 \times 10^5 \text{ Pa}$ and its density is $1.21 \text{ kg/m}^3$. Calculate the percentage error in the speed of sound as predicted by Newton's formula relative to the e... | **Final Answer:** $15.45\%$ $v_{N} = \sqrt{\frac{P}{\rho}}$ $v_{N} = \sqrt{\frac{1.0 \times 10^{5}}{1.21}} = 287.48 \text{ m/s}$ $\text{Percentage Error} = \frac{|v_{exp} - v_{N}|}{v_{exp}} \times 100$ $\text{Percentage Error} = \frac{340 - 287.48}{340} \times 100 = 15.45\%$ | Physics 11th JEE Mains |
A thin convex lens of focal length $F$ and aperture diameter $D$ forms a real image of intensity $I$ of a distant object. If the central part of the lens, corresponding to a diameter of $D/2$, is covered with an opaque black disc, determine the new focal length of the lens and the intensity of the image formed. | **Final Answer:** Focal length remains $F$; Image intensity becomes $\frac{3}{4}I$ $F_{new} = F$ (Focal length depends on lens curvature and refractive index, not aperture area) Initial Area $A = \pi \left(\frac{D}{2}\right)^{2} = \frac{\pi D^{2}}{4}$ Blocked Area $A_{b} = \pi \left(\frac{D/2}{2}\right)^{2} = \frac{\pi... | Physics 12th JEE Mains |
A cylindrical vessel filled with a liquid of density $\rho$ is being accelerated vertically upwards with a constant acceleration of $a = 15 \text{ m/s}^2$. Consider two points, $A$ and $B$, within the liquid that are separated by a vertical distance $h$, where point $B$ is located directly below point $A$. If the magni... | **Final Answer:** 25 Given: $a = 15 \text{ m/s}^{2}$, $g = 10 \text{ m/s}^{2}$ $$g_{\text{eff}} = g + a = 10 + 15 = 25 \text{ m/s}^{2}$$ $$|P_{B} - P_{A}| = \rho g_{\text{eff}} h = 25 \rho h$$ $$k \rho h = 25 \rho h \implies k = 25$$ | Physics 11th JEE Mains |
Although $[10]$-annulene (cyclodeca-$1,3,5,7,9$-pentaene) possesses $10\pi$ electrons, which corresponds to the $(4n + 2)$ rule for $n=2$, it is found to be non-aromatic. Evaluate the structural constraints, specifically concerning the internal hydrogen atoms, that prevent the molecule from achieving the planar geometr... | **Final Answer:** Due to the steric repulsion between internal hydrogen atoms, the molecule is forced into a non-planar geometry, rendering it non-aromatic. $N_{\pi} = 10 \implies 4n + 2 = 10$ $n = 2$ $d(\text{H}_{int} \cdots \text{H}_{int}) < 2.4 \text{ \AA} \implies \text{Steric Repulsion}$ $\text{Steric Energy} > \t... | Chemistry 11th JEE Mains |
A laboratory experiment is conducted to analyze how the magnetic susceptibility ($\chi$) of various substances changes with absolute temperature ($T$). The resulting data for four different samples are plotted in the graphs labeled (a), (b), (c), and (d) below.
(i) Based on the fundamental properties of magnetic mater... | **Final Answer:** (i) Graph (d) represents the diamagnetic substance. (ii) $\chi = -1.4 \times 10^{-5}$ Properties of diamagnetic substances: $\chi < 0$ and $\frac{d\chi}{dT} = 0$ From the given plots, only graph (d) shows $\chi$ as a constant negative value. Given $\chi_{280} = -1.4 \times 10^{-5}$ Since $\chi$ is ind... | Physics 12th JEE Mains |
What are real gases? Under which specific conditions of temperature and pressure do they most closely approximate the behavior of an ideal gas? | **Final Answer:** Real gases behave ideally under conditions of Low Pressure and High Temperature. $\left(P + \frac{an^{2}}{V^{2}}\right)(V - nb) = nRT$ $\text{For } P \rightarrow 0, V \rightarrow \infty \implies \frac{an^{2}}{V^{2}} \rightarrow 0$ $\text{For } V \gg nb \implies (V - nb) \approx V$ $$PV = nRT$$ | Chemistry 11th JEE Mains |
In the case of chemisorption, explain why the extent of adsorption ($x/m$) initially increases with a rise in temperature before it eventually decreases. | **Final Answer:** Chemisorption requires activation energy ($E_{a}$) for bond formation causing an initial increase, but because it is exothermic ($\Delta H < 0$), the extent of adsorption eventually decreases at higher temperatures. $x/m = f(T) \cdot e^{-E_{a}/RT}$ $\text{Initially, } T \uparrow \implies e^{-E_{a}/RT}... | Chemistry 12th JEE Mains |
Determine the effect on the electrode potential of a copper electrode ($E_{Cu^{2+}/Cu}$) if the concentration of $Cu^{2+}$ ions in the half-cell is decreased at a constant temperature. | **Final Answer:** The electrode potential ($E_{Cu^{2+}/Cu}$) decreases. Reduction half-reaction: $Cu^{2+}(aq) + 2e^{-} \rightarrow Cu(s)$ $E_{Cu^{2+}/Cu} = E^{\circ}_{Cu^{2+}/Cu} - \frac{2.303RT}{nF} \log \left(\frac{1}{[Cu^{2+}]}\right)$ At $298 \, K$: $E_{Cu^{2+}/Cu} = E^{\circ}_{Cu^{2+}/Cu} + \frac{0.0591}{2} \log [... | Chemistry 12th JEE Mains |
A uniform solid cylindrical block of length $L$ and mass density $d$ is floating vertically in a deep, non-viscous liquid of density $\sigma$ (where $\sigma > d$). Initially, the block is in equilibrium. It is then depressed vertically by a small distance $y_0$ and released.
(a) Show that the subsequent motion of the c... | **Final Answer:** $T = 2\pi \sqrt{\frac{Ld}{\sigma g}}$ At equilibrium: $mg = F_{B} \implies (A \cdot L \cdot d)g = (A \cdot h \cdot \sigma)g$ where $h$ is submerged depth. Equilibrium relation: $Ld = h\sigma \implies h = \frac{Ld}{\sigma}$ Restoring force when depressed by distance $y$: $F_{restoring} = mg - A(h+y)\si... | Physics 11th JEE Mains |
Explain the physical significance of 'cyclotron frequency' for a charged particle moving in a uniform magnetic field. Show that this frequency does not depend on the particle's speed. | **Final Answer:** The cyclotron frequency $f = \frac{qB}{2\pi m}$ is independent of the particle's speed $v$. $qvB = \frac{mv^{2}}{r}$ $r = \frac{mv}{qB}$ $T = \frac{2\pi r}{v} = \frac{2\pi m}{qB}$ $f = \frac{1}{T} = \frac{qB}{2\pi m}$ | Physics 12th JEE Mains |
A truck is accelerating horizontally along a straight road with a constant acceleration $A$. An observer standing on the floor of the truck releases a small marble from his hand. Calculate the magnitude of the acceleration of the marble relative to the observer and determine the angle that this acceleration vector make... | **Final Answer:** Magnitude: $\sqrt{A^{2} + g^{2}}$, Angle: $\theta = \tan^{-1}\left(\frac{g}{A}\right)$ with the horizontal direction $\vec{a}_{M} = -g\hat{j}, \vec{a}_{O} = A\hat{i}$ $\vec{a}_{MO} = \vec{a}_{M} - \vec{a}_{O} = -A\hat{i} - g\hat{j}$ $a_{MO} = \sqrt{(-A)^{2} + (-g)^{2}}$ $a_{MO} = \sqrt{A^{2} + g^{2}}$... | Physics 11th JEE Mains |
Determine the change in the electrode potential (in mV) of a $Zn^{2+}/Zn$ half-cell at $298\text{ K}$ if the concentration of $Zn^{2+}$ ions in the solution is increased by a factor of $100$. Use the value of $\frac{2.303RT}{F} = 0.059\text{ V}$. | **Final Answer:** 59 { mV} $Zn^{2+}(aq) + 2e^{-} \rightarrow Zn(s) \implies n = 2$ $E = E^{\circ} + \frac{0.059}{2} \log [Zn^{2+}]$ $\Delta E = E_{2} - E_{1} = \frac{0.059}{2} \log \left( \frac{[Zn^{2+}]_{final}}{[Zn^{2+}]_{initial}} \right)$ $$\Delta E = \frac{0.059}{2} \log(100) = 0.059 \text{ V}$$ $\Delta E = 0.059 ... | Chemistry 12th JEE Mains |
At a specific geographic location, the true value of the angle of dip is $45^\circ$. Calculate the apparent angle of dip recorded by a magnetic needle if the dip circle is rotated such that its plane makes an angle of $60^\circ$ with the magnetic meridian. | **Final Answer:** $\delta' = \tan^{-1}(2) \approx 63.4^{\circ}$ Given: True dip $\delta = 45^{\circ}$ and angle with meridian $\alpha = 60^{\circ}$ Using formula $\tan \delta' = \frac{\tan \delta}{\cos \alpha}$ Substitution: $\tan \delta' = \frac{\tan 45^{\circ}}{\cos 60^{\circ}}$ $\tan \delta' = \frac{1}{1/2} = 2$ $$\... | Physics 12th JEE Mains |
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