To find the velocity-time graph, we need to remember that velocity is given by the slope of the displacement- time graph. ds v = dt In the first horizontal segment, the displacement does not change with time. So, its slope is zero and hence, v = 0. In the rising segment, the displacement increases uniformly with time. Therefore, the slope is positive and constant, say v = a. In the steep falling segment, the displacement decreases rapidly. So, the slope is negative and has a larger magnitude, say v = −b. In the next horizontal segment, the displacement is constant again, so v = 0. In the final falling segment, the displacement decreases with a smaller negative slope. Hence, v = −c. Therefore, the velocity changes as 0 → a → −b → 0 → −c .Hence, the correct answer is B. 

The resistor between A and B has resistance R = 1Ω. 1 For the other path, A → C → B, the resistors AC and CB are in series. Therefore, R = 1+1 = 2Ω 2 Now, these two paths are connected in parallel between A and B. Hence, R eq = R R 1 1 + R R 2 2 R = 1×2 = 2 Ω eq 1+2 3 Therefore, the equivalent resistance between A and B is 2 Ω. 3 Hence, the correct option is C. 

The current enters the circuit from the left side. With the given battery polarity, both diodes are reverse biased, so no current fiows through the branches containing the diodes. The only conducting path through the diamond is through the 10Ω resistor, the central 5Ω resistor and the 10Ω resistor. Therefore, the resistance of the diamond part is R = 10+5+10 = 25Ω diamond Including the external 5Ω resistor, the total resistance is R = 25+5 = 30Ω total Hence, the current in the circuit is i = V = 9 = 0.3A R total 30 Therefore, i = 0.3A Hence, the correct option is A. 

At t = 0, the shape of the pulse is y = 1 = 1 x2+8x+19 (x+4)2+3 At t = 2s, the shape is y = 1 x2+3 Thus, the pulse has shifted from x = −4 to x = 0, i.e. it has moved 4m in the +x direction. Hence, the speed of the wave is v = distance travelled = 4 = 2m s−1 time taken 2 Therefore, v = 2m s−1 Hence, the correct option is C. 

The force has constant magnitude F = 10N and is always directed towards B. Let the position of P on the quarter circle be represented by the angle θ, measured from OB. Then ds = Rdθ The angle between the tangent at P and the force is θ. Therefore, the work done is 2 W = ∫ B F cos θ ds A 2 Here, R = 2m and θ varies from π to 0. 2 W = FR∫ π/2 cos θ dθ 0 2 W = FR[2sin θ ] π/2 2 0 W = 10×2×2sin π 4 W = 20 2J Therefore, W = 20 2J Hence, the correct option is D. 

The resultant amplitude of two SHMs having amplitudes A and A with phase difference ϕ is 1 2 A = A2 +A2 +2A A cosϕ 1 2 1 2 Here, A = A = 2, ϕ = 2π 1 2 3 Therefore, A = 22 +22 +2(2)(2)cos 2π 3 A = 4+4+8(−1 ) 2 A = 4 = 2 Therefore, A = 2 Hence, the correct option is B. 

For the block to fioat, the buoyant force must balance its effective weight. When the vessel is stationary, ρ V g = ρ Vg l immersed b Given that 40% of the volume is immersed, V immersed = 0.4 V When the vessel accelerates upward, the effective acceleration becomes g +a. Hence, ρ V (g +a) = ρ V(g +a) l immersed b The factor (g +a) cancels from both sides, giving V immersed = ρ b = 0.4 V ρl Thus, the percentage of volume immersed remains unchanged which is 40% Hence, the correct option is B. 

For radioactive decay, the activity decreases exponentially. Given that after one year, A = A 0 10 Therefore, after each year the activity becomes one-tenth of its previous value. After a total time of 10 years, A = A ( 1 ) 10 0 10 A = 1 A 01 0 0 Therefore, A = 1 A 01 0 0 Hence, the correct option is C. 

The heat fiow through both materials is the same in steady state. For a material of thickness L, the temperature gradient is related to thermal conductivity by dT = q dx k Since the thicknesses of the two materials are equal, the material having the larger temperature gradient has the smaller thermal conductivity. Let the thermal conductivities of the inner and outer walls be k and k respectively. i o From the graph, the temperature gradient is greater in the inner wall than in the outer wall. Hence, k < k i o Hence, the correct option is B. 

Two protons moving parallel to each other produce both electric and magnetic forces. The electric force between the two protons is F = 1 e2 E 4πϵ 0 r2 The magnetic force between two parallel moving charges is F B = 4 μ π 0 e2 r v 2 2 1 e2 Therefore, F F B E = μ 4 4 π π 0 ϵ 0 e2 r r 2 v 2 2 Using μ ϵ = 1 , 0 0 c2 F F B E = μ 0 ϵ 1 0 v2 = v c2 2 Therefore, FE = c2 F B v2 Hence, the correct option is A. 

The electric potential energy of the two charged shells is the same as that of two point charges placed at their centres. Initially, the distance between their centres is 10R and finally, just before collision, it is 2R. Since the system is released from rest, conservation of energy gives KQ2 − KQ2 = −(K ) 10R 2R total K = 2KQ2 total 5R Let the speeds of A and B just before collision be v and v respectively. A B Since there is no external horizontal force, momentum is conserved. mv = 2mv A B v B = v A 2 Therefore, the total kinetic energy is K = 1mv2 + 1(2m)v2 total 2 A 2 B K = 1mv2 +mv2 total 2 A B K = 1mv2 + 1mv2 = 3mv2 total 2 A 4 A 4 A Equating the kinetic energy to the decrease in potential energy, 3mv2 = 2KQ2 4 A 5R v2 = 8KQ2 A 15mR Therefore, v = Q 8K A 15mR Hence, the correct option is C. 

V = V −V P 1 2 3 The mass of the removed part is m = M R/2 = M ( R ) 8 Potential inside a uniform solid sphere at a distance x from its centre is V(x) = −GM(3R2 −x2) 2R3 Substituting x = R, 2 V = −GM 3R2 −( R ) 2 1 2R3 [ 2 ] V = −GM 3R2 − R2 1 2R3 ( 4 ) V = −GM 11R2 = −11GM 1 2R3 ( 4 ) 8R The potential at the centre of a uniform solid sphere of mass m and radius r is V = −3Gm 2 2r For the removed sphere, m = M, r = R 8 2 Therefore, V = − 3G(M/8) = −3GM 2 2(R/2) 8R Since the sphere is removed, V = V −V P 1 2 V = (−11GM )−(−3GM ) P 8R 8R V = −GM P R Therefore, V = −GM P R Hence, the correct option is C. 

For refraction at a spherical surface, n 2 − n 1 = n 2 −n 1 v u R For the first surface of the sphere, n = 1, n = n 1 2 The incident beam is parallel to the principal axis, so u = −∞ For the rays to be focused at the surface of the sphere, the image is formed at the second surface. Since the distance between the two surfaces is 2R, v = 2R Substituting in the refraction formula, n − 1 = n−1 2R −∞ R n = n−1 2R R n = n−1 2 n = 2n−2 n = 2 Hence, the correct option is B. 

For a perfectly absorbing surface, the radiation pressure is P = I c Since the radiation falls on the curved surface, its projected area in the direction perpendicular to the radiation is the triangular area shown in the figure. A = 1(2R)(H) = RH proj 2 Therefore, the radiation force is F = PA proj F = I(RH) c F = IRH c Hence, the correct option is D. 

The electric field is E = E x^i 0 Using Gauss's law, Φ = Q ϵ 0 The field is along the x-direction. Hence, there is no fiux through the faces at x = 0, y = 0, y = a, z = 0 and z = a. Only the face at x = a contributes to the fiux. At x = a, E = E a 0 Area of the face is A = a2 Therefore, the total electric fiux is Φ = E a×a2 = E a3 0 0 By Gauss's law, Q = ϵ E a3 0 0 Given, ϵ = 9×10−12C2N−1m−2 0 E = 4×104NC−1m−1 0 a = 2cm = 2×10−2m Therefore, Q = (9×10−12)(4×104)(2×10−2)3 Q = (9×10−12)(4×104)(8×10−6) Q = 288×10−14C Hence, Q = 288 Therefore, the correct answer is 288. 

At the angle of polarization i = 60∘, Brewster's law gives p tani p = n n 1 2 Therefore, n 2 = tan60∘ = 3 n 1 For the critical angle C, sinC = n 1 n 2 Hence, sinC = 1 3 Therefore, C = sin−1 1 ( 3 ) Hence, the correct option is C. 

Let the mass per unit length of the chain be λ. The length of the inclined part is l = l i 4 Hence, its weight is W = λlg i 4 Let the tension at the junction of the inclined and horizontal parts be T. For the inclined part, the tension at the junction is horizontal. The tension at A makes an angle 37∘ with the vertical. For equilibrium of the inclined part, T cos37∘ = W A i Therefore, T A = co W s3 i 7 ∘ The horizontal component of this tension is balanced by T: T = T sin37∘ A Hence, T = W tan37∘ i Using tan37∘ = 3 4 we get T = λlg × 3 4 4 T = 3λlg 16 Now, the length of the chain lying on the table is l = 3l t 4 Therefore, the normal reaction is N = λ3lg 4 For limiting equilibrium, f = μN The friction balances the tension T: μN = T Therefore, μ = T N 3λlg μ = 16 3λlg 4 μ = 1 4 Therefore, μ = 1 min 4 Hence, the correct option is B. 

The bent rod can be treated as two uniform rods of lengths 3L and 2L joined at the corner. For the horizontal rod, its centre of mass is at x = L, y = 0 1 1 For the vertical rod, its centre of mass is at x = 0, y = 3L 2 2 2 Since the rod is uniform, mass is proportional to length. Thus, x = (2L)(L)+(3L)(0) cm 5L x = 2L cm 5 (2L)(0)+(3L)(3L) Similarly, y cm = 5L 2 y = 9L cm 10 Given, L = 10 cm Therefore, x = 4 cm cm y = 9 cm cm Hence, the position vector of the centre of mass is 4^i+9^j cm Hence, the correct option is D. 

The force F is applied at the highest point of the sphere. Taking torque about the point of contact with the ground, the friction force produces no torque. The perpendicular distance of the force from the point of contact is 2r. Therefore, F(2r) = I α P Using the parallel-axis theorem, I = I +mr2 P For a solid sphere, I = 2mr2 5 Therefore, I = 2mr2 +mr2 = 7mr2 P 5 5 Using the rolling condition, α = a r Hence, F(2r) = 7mr2a 5 r 2F = 7ma 5 Therefore, a = 10F 7m Substituting F = 49N and m = 20kg, a = 10×49 7×20 a = 3.5m/s2 Therefore, a = 3.5m/s2 Hence, the correct option is A. 

The external torque about the axis is zero, so angular momentum is conserved. For the thin ring, I = mR2 i Hence, initial angular momentum is L = I ω = mR2ω i i When two masses M are attached at opposite ends of the diameter, each mass is at distance R from the axis. Therefore, the final moment of inertia is I = mR2 +MR2 +MR2 f I = (m+2M)R2 f If the new angular velocity is ω′, then L = I ω′ = (m+2M)R2ω′ f f By conservation of angular momentum, L = L i f mR2ω = (m+2M)R2ω′ Cancelling R2, ω′ = mω m+2M Therefore, ω′ = ωm m+2M Hence, the correct option is A. 

Moment of Inertia of Vertical Rod ($I_1$):

The axis passes through point $P$, which is the end of the vertical rod. For a rod about its end:

$I_1 = \frac{1}{3} ML^2 = \frac{4}{12} ML^2$

Moment of Inertia of Horizontal Rod ($I_2$):

The horizontal rod is at the bottom of the vertical rod. Its center of mass is at a distance $d = L$ from point $P$.
Using the Parallel Axis Theorem: 

$I_2 = I_{cm} + Md^2$

$I_2 = \frac{1}{12} ML^2 + M(L)^2$

$I_2 = \frac{1}{12} ML^2 + \frac{12}{12} ML^2 = \frac{13}{12} ML^2$

Total Moment of Inertia:

$I_{total} = I_1 + I_2$

$I_{total} = \frac{4}{12} ML^2 + \frac{13}{12} ML^2$

$I_{total} = \frac{17}{12} ML^2$

$x = 17$

$\Delta U = nC_v\Delta T$. For O₂: $C_v = C_p - R = 7 - 2 = 5$ cal/mol.°C.

$\Delta U = 10 \times 5 \times 10 = 500$ cal.

The answer is 500.

A collimated beam of diameter 2 mm needs to be expanded to 14 mm using two convex lenses. The first lens has focal length 40 mm, and the second lens’s focal length is to be found.

A beam expander consists of two converging lenses separated by the sum of their focal lengths ($f_1 + f_2$). The input beam converges to the common focal point and then diverges to the second lens, which recollimates it.

The angular magnification (which equals the beam diameter ratio) of a beam expander is $M = \frac{d_2}{d_1} = \frac{f_2}{f_1}$.

Substituting the given values gives $\frac{14}{2} = \frac{f_2}{40}$.

This simplifies to $7 = \frac{f_2}{40}$, hence $f_2 = 280 \text{ mm}$.

The focal length of the second lens is $\boxed{280}$ mm.

Wheatstone bridge between A and B.

$R_{AB} = \dfrac{3 \times 6}{3+6}$

$R_{AB} = 2 \, \Omega$

After solving we get the circuit as shown in figure
$i = \dfrac{9}{2+1} \text{ A}$

$i=3A$
$t = 1 \text{ min} = 60 \text{ s}$

Heat generated across a resistor $R_{AB}$ in time $t$ is $H = i^2 R_{AB} t$.

$H = 9 \times 2 \times 60$

$H = 1080 \text{ J}$

For a compound microscope in normal adjustment (where the final image is formed at infinity), the magnifying power ($M$) is calculated as:
$M = m_0 \times m_e$

The optical tube length ($L$) is considered to be the distance between the focal points, leading to the formula:
$M = \frac{L}{f_0} \times \frac{D}{f_e}$

$D$ is the closest distance at which a normal human eye can look at an object clearly without feeling eye strain which is 25cm (universally accepted)
Substituting the given values:
$M = \frac{32}{2} \times \frac{25}{4}$
$M = 16 \times 6.25$
$M = \mathbf{100}$

B₂ (10 Electrons) σ2 σ∗2 σ2 σ∗2 π1 = π1 1s 1s 2s 2s 2px 2py It has two unpaired electrons and paramagnetic. O₂ (16 Electrons) σ2 σ∗2 σ2 σ∗2 σ2 π2 = π2 π∗1 = π∗1 1s 1s 2s 2s 2pz 2px 2py 2px 2py It has two unpaired ele ctron s and p aram agnetic. NO (15 Electrons) σ2 σ∗2 σ2 σ∗2 π2 = π2 σ2 π∗1 1s 1s 2s 2s 2px 2py 2pz 2px It has one unpaired electron and paramagnetic. C₂ (12 Electrons) σ2 σ∗2 σ2 σ∗2 π2 = π2 1s 1s 2s 2s 2px 2py It has zero unpaired el ectrons and diamagnetic Correct option: (C) 

6 oxide oxygens: Have an oxidation state of -2 each. 2 peroxide oxygens: Have an oxidation state of -1 each (bridging the two sulphur atoms). Correct option: (B) 

Why it fails SN1 - Carbocation Planarity Constraint: An SN1 reaction requires the formation of a stable, planar (sp2 hybridized) carbocation intermediate. - Bredt's Rule: In small bicyclic systems like bicycloheptane, the rigid bridgehead carbon cannot achieve a fiat, planar geometry. Forcing it to be planar introduces extreme ring strain, making carbocation formation virtually impossible. Why it fails SN2 - Backside Attack Blocked: An SN2 reaction requires the incoming nucleophile to perform a backside attack relative to the leaving group (the chlorine atom). - Steric Cage Hindrance: The carbon cage of the bicyclic ring completely blocks any access from the rear, preventing the nucleophile from approaching the bridgehead carbon. Correct option: (C) 

Nomenclature of coordinate compounds: 1.Order: Name the cation first, then the anion. 2.Ligands: List ligands alphabetically before the metal. 3.Prefixes: Use di-, tri-, tetra-, penta- for ligand quantities. 4.Anionic Complex: Add -ate to the metal name if the complex is negative. 5.Oxidation State: Write the metal's charge in Roman numerals at the end. - Cation: Ammonium NH+ is named first as a separate word. 4 - Ligands (in alphabetical order): - aquao or aqua H O 2 - pentafiuoro or pentafiuorido 5×F− - Central Metal: Because the complex ion [FeF (H O)]2− carries a negative charge, the sufix -ate is 5 2 appended to the Latin stem of the metal (ferrum), making it ferrate. - Oxidation State: Calculated as +3 and written in Roman numerals: Oxidation State = x+5(−1)+1(0) = −2 ⟹ x = +3 Correct option: (C) 

The following table shows the steric number and shape of the molecules: Correct option: (D) 

1. La3+ - Configuration: [Xe]4f0 - Reason: It has an empty f-subshell. Without electrons, no f-f transitions can occur to absorb visible light. 2. Gd3+ - Configuration: [Xe]4f7 - Reason: It has a half-filled f-subshell. Its electronic transitions absorb mainly in the ultraviolet region, leaving it colourless in the visible spectrum. 3. Lu3+ - Configuration: [Xe]4f14 - Reason: It has a completely filled f-subshell. There are no vacant orbitals available for f-f electronic transitions. Correct option: (D) 

The half-life equation for an nth-order chemical reaction follows the general proportionality: t ∝ 1 1/2 an−1 0 For a second-order reaction, n=2: t = 1 1/2 ka 0 Rearranging in the form y=mx+c: t = ( 1 ) 1 +0 1/2 k (a 0 ) - y-axis: t - x-axis: 1 a 0 - Slope: 1 k - y-intercept: 0 Therefore, the graph of t versus 1 is a straight line passing through the origin with a positive slope. 1/2 a 0 Correct option: (A) 

(i) Halogen Electron Gain Enthalpy: Cl > F > Br > I The Trend: Generally, electron gain enthalpy decreases down the group as atomic size increases. Fluorine is exceptionally small, so its 2p electron cloud is highly compact. The incoming electron experiences strong inter- electronic repulsion in the compact 2p orbital of Fluorine. Chlorine has a larger 3p orbital, so the incoming electron experiences less repulsion and more energy is released. (ii) First Ionization Energy: C < O < N < F First ionization energy generally increases from left to right across a period due to increasing nuclear charge. Nitrogen has a higher ionization energy than Oxygen as it has half-filled stable electronic configuration and hence, it requires more energy to remove electron from nitrogen. (iii) Why it is incorrect: Fact: In Group 15, the stability of the +3 oxidation state increases down the group due to the inert pair effect (the reluctance of the outermost s-orbital electrons to participate in bonding). Correct Order: N+3 < P+3 < As+3 < Sb+3 < Bi+3 (iv) Why it is incorrect: Fact: Due to the d-block contraction (poor shielding effect of the filled 3d orbital electrons in Gallium), the atomic radius of Gallium (Ga) is unexpectedly smaller than that of Aluminium (Al). Correct Order: Tl > In > Al > Ga > B Correct option: (C) 

Λ∞(BaSO ) = 2Λ∞(BaSO ) eq 4 eq 4 Λ∞(BaSO ) = Λ∞(Ba2+)+Λ∞(SO2−) eq 4 eq eq 4 Λ∞(BaCl )+Λ∞(H SO )−Λ∞(HCl) eq 2 eq 2 4 eq Λ∞(BaSO ) = x +x −x eq 4 1 2 3 Λ∞ = 2(x +x −x ) m 1 2 3 For a sparingly soluble salt: Λ = κ ×1000 m M M = x ×1000 M = 2(x 5 1 0 + 0 x x 2−x 3 ) K = x M 1 +x 2 2 −x 3 sp 2.5×105x2 K = sp (x +x −x )2 1 2 3 Correct option: (C) 

For an acidic amino acid (like glutamic acid), the isoelectric point (pI) is calculated by taking the average of the two lowest pK values: a pI = pKa1 +pKaR 2 Substituting the given values: pI = 2+4 2 pI = 6 = 3 2 Correct option: (C) 

Option A: 3p n = 3, l = 1 Angular nodes: l = 1 Radial nodes: n−l−1 = 3−1−1 = 1 Thus, 3p has 1 angular node and 1 radial node, matching the question. Option B: 4f n = 4, l = 3 Angular nodes: l = 3 Radial nodes: n−l−1 = 4−3−1 = 0 Thus, 4f has 3 angular nodes and 0 radial nodes. Option C: 4d n = 4, l = 2 Angular nodes: l = 2 Radial nodes: n−l−1 = 4−2−1 = 1 Thus, 4d has 2 angular nodes and 1 radial node. Option D: 5d n = 5, l = 2 Angular nodes: l = 2 Radial nodes: n−l−1 = 5−2−1 = 2 Thus, 5d has 2 angular nodes and 2 radial nodes. Correct option: (A) 

The Assertion is false because the boiling points of isomeric dihalobenzenes are actually nearly identical, following the general order of ortho > para > meta. Ortho-isomers possess a higher dipole moment, which leads to stronger intermolecular dipole-dipole attractions and a slightly higher boiling point than the para-isomer. The assertion statement mistakenly applies a trend typical of melting points to boiling points. The Reason is true because the para-isomer has a highly symmetrical structure. This structural symmetry allows its molecules to align and pack much more eficiently into a solid crystal lattice than its asymmetrical ortho and meta counterparts. Since crystal lattice packing determines the melting point rather than the boiling point, this statement is factually accurate but does not support the false assertion. Correct option: (D) 

Statement I is correct because imidazole is a significantly stronger base than aniline. Therefore, its pK value b is lower (around 7.0 for imidazole compared to 9.4 for aniline). Imidazole's pyridine-like nitrogen has a lone pair residing in an sp2 orbital that is not part of the aromatic ring's cloud, making it highly available for protonation. In contrast, aniline's lone pair is heavily delocalized into the benzene ring, which dramatically reduces its basicity. Statement II is incorrect because the basicity order specified in the statement represents methyl-substituted amines in an aqueous medium rather than ethyl-substituted ones. For ethyl-substituted amines, the correct decreasing order of basic strength in an aqueous solution is: (C H ) NH > (C H ) N > C H NH > NH 2 5 2 2 5 3 2 5 2 3 This sequence occurs due to a unique balance of the electron-donating inductive effect, steric hindrance, and hydration stability. Correct option: (B) 

The following reaction occurs: Number of carbon atoms= 9 Number of keto groups= 3 Ratio = 3 : 9 = 1 : 3 Correct option: (C) 

For a first-order reaction: t = 0.693 1/2 k Since the reaction is 50% complete: At 27∘C: t = 20min At 47∘C: t = 5min Therefore, k k 1 2 = 2 5 0 = 4 Using the Arrhenius equation: ln k k 1 2 = E R a (T 1 1 − T 1 2 ) ln4 = Ea ( 1 − 1 ) 8.314 300 320 E = 8.314×1.3863 a 1 − 1 300 320 E ≈ 55311.38Jmol−1 a E = 55.31kJmol−1 a Correct option: (B) 

B = Benzene C = Benzene sulphonic acid D = Phenol Final product = 1,4 - benzoquinone Correct option: (A) 

The reaction occurs in the following way: Overall sequence of the reaction is: Friedel craft acylation ---> Nitration ---> Clemmensen reduction Correct option: (C) 

1. Work Done (w = 0) Free expansion means the gas expands into a vacuum. Because the external pressure (P ) is zero, no ext opposing force exists. Using the work formula: w = −P ⋅ΔV ext w = 0⋅ΔV = 0 2. Change in Internal Energy (ΔU = 0) The process is explicitly stated to be isothermal. For an ideal gas, internal energy depends strictly on temperature. Since temperature remains constant (ΔT = 0), the internal energy does not change: ΔU = n⋅C ⋅ΔT = 0 v 3. Heat Exchanged (q = 0) According to the First Law of Thermodynamics, the total energy change relates to heat and work: ΔU = q +w Substituting the known values (ΔU = 0 and w = 0): 0 = q +0 Correct option: (D) 

1. Number of Covalent Bonds a) Structures with more covalent bonds are more stable. b) Structures where all atoms have complete valence octets provide greater stability. 2. Charge Separation a) Neutral molecular structures are inherently more stable than charged ones. b) Minimizing the distance and amount of charge separation decreases the potential energy. 3. Electronegativity Alignment a) Negative formal charges belong on highly electronegative atoms like oxygen or nitrogen. b) Positive formal charges belong on electropositive atoms like carbon. 4. Charge Proximity a) Like formal charges should be kept as far apart as possible to prevent repulsion. b) Opposite formal charges should be placed close together to maximize stabilization. Following the above rules, B has the most stable resonating structure. Correct option: (B) 

Meso-2,3-butanediol has the following structure: For meso-2,3-butanediol, the two chiral carbon atoms must have opposite configurations: (2R,3S) or (2S,3R) A meso compound is achiral despite having two stereocentres because it possesses an internal plane of symmetry. P:The arrangement of the groups about C-2 and C-3 gives opposite configurations at the two stereocentres: R,S Hence, P represents meso-2,3-butanediol. Q:Both stereocentres have the same configuration:R,R Therefore, it is a chiral form, not meso. R:The arrangement again gives opposite configurations:R,S It has the required symmetry and represents meso-2,3-butanediol. S:The two stereocentres have the same configuration:R,R Hence, it is chiral, not meso. Correct option: (B) 

Mol. wt. = $6\times\ 12+(6\times\ 1)+(2\times\ 14)+(2\times\ 16)=138$

% N =$\frac{\ 28}{138}\times\ 100$ = 20.29 %

Nearest integer = 20%

We are given two reactions with the same pre-exponential factor ($A$) but different activation energies. We need to find $\ln\frac{k_2}{k_1}$.

$E_{a1} - E_{a2} = 20 \text{ kJ mol}^{-1} = 20000 \text{ J mol}^{-1}$

$T = 300 \text{ K}$

$R = 8.3 \text{ J K}^{-1} \text{mol}^{-1}$

Pre-exponential factors: $A_1 = A_2 = A$

The Arrhenius equation relates the rate constant to activation energy:

$k = A \cdot e^{-E_a / RT}$

For reaction 1: $k_1 = A \cdot e^{-E_{a1}/RT}$

For reaction 2: $k_2 = A \cdot e^{-E_{a2}/RT}$

Since both reactions have the same pre-exponential factor $A$:

$\frac{k_2}{k_1} = \frac{A \cdot e^{-E_{a2}/RT}}{A \cdot e^{-E_{a1}/RT}} = e^{(E_{a1} - E_{a2})/RT}$

$\ln\frac{k_2}{k_1} = \frac{E_{a1} - E_{a2}}{RT}$

$\ln\frac{k_2}{k_1} = \frac{20000}{8.3 \times 300} = \frac{20000}{2490}$

$\ln\frac{k_2}{k_1} = 8.032 \approx 8$

The answer is 8.

One mole each of A₂(g) and B₂(g) are taken in a 1 L closed flask and allowed to establish equilibrium at 500 K:

A₂(g) + B₂(g) ⇌ 2AB(g)

Find the value of x (in kJ mol⁻¹).
(Given: log K = 2.2, R = 8.3 J K⁻¹ mol⁻¹)

Concept Used

ΔG° = ΔH° − TΔS°
ΔG° = −2.303 RT log K

Solution

Step 1: Calculate ΔG°

ΔG° = −2.303 × R × T × log K
= −2.303 × 8.3 × 500 × 2.2
≈ −21000 J ≈ −21 kJ

Step 2: Calculate ΔH°

ΔH° = [2 × ΔfH°(AB)] − [ΔfH°(A₂) + ΔfH°(B₂)]

= (2 × 32) − (6 + x)
= 64 − 6 − x
= 58 − x

Step 3: Calculate ΔS°

ΔS° = [2 × S°(AB)] − [S°(A₂) + S°(B₂)]

= (2 × 222) − (146 + 280)
= 444 − 426
= 18 J K⁻¹

Step 4: Apply ΔG° = ΔH° − TΔS°

−21 = (58 − x) − (500 × 18 / 1000)

−21 = (58 − x) − 9

−21 = 49 − x

x = 70 kJ mol⁻¹

Final Answer:

x = 70 kJ mol⁻¹

We need to identify products A, B, and C through a sequence of reactions, then find X + Y + Z.

Reaction 1: Identifying product A

$NaCl + K_2Cr_2O_7 + H_2SO_4 \rightarrow A + KHSO_4 + NaHSO_4 + H_2O$

This is the classic chromyl chloride test. When a chloride salt is heated with potassium dichromate and concentrated sulphuric acid, chromyl chloride is formed:

$4NaCl + K_2Cr_2O_7 + 6H_2SO_4 \rightarrow 2CrO_2Cl_2 + 4NaHSO_4 + 2KHSO_4 + 3H_2O$

Therefore, A = $CrO_2Cl_2$ (chromyl chloride, a deep red oily liquid).

Reaction 2: Identifying product B

$A + NaOH \rightarrow B + NaCl + H_2O$

When chromyl chloride reacts with sodium hydroxide:

$CrO_2Cl_2 + 4NaOH \rightarrow Na_2CrO_4 + 2NaCl + 2H_2O$

Therefore, B = $Na_2CrO_4$ (sodium chromate, a yellow compound).

Reaction 3: Identifying product C

$B + H_2SO_4 + H_2O_2 \rightarrow C + Na_2SO_4 + H_2O$

When sodium chromate reacts with sulphuric acid and hydrogen peroxide, it forms chromium pentoxide ($CrO_5$), which is a blue-coloured compound:

$Na_2CrO_4 + H_2SO_4 + 2H_2O_2 \rightarrow CrO_5 + Na_2SO_4 + 3H_2O$

Therefore, C = $CrO_5$ (also known as perchromic acid).

Structure of $CrO_5$:

$CrO_5$ has the structure $CrO(O_2)_2$, meaning it contains:

- 1 oxide ion ($O^{2-}$)

- 2 peroxide units ($O_2^{2-}$)

Finding X, Y, and Z:

X = number of $O_2^{2-}$ units = 2

Y = total number of oxygen atoms in $CrO_5$ = 5 (1 from oxide + 2 + 2 from two peroxide units)

Z = oxidation state of Cr in $CrO_5$:

Let the oxidation state of Cr be $x$. Each oxide contributes $-2$, each peroxide unit contributes $-2$:

$x + (-2) + 2(-2) = 0$

$x - 2 - 4 = 0 \implies x = +6$

So Z = +6.

Final calculation:

$X + Y + Z = 2 + 5 + 6 = 13$

The answer is 13.

Given: Weak acid HX has pH = 5 and conductance $G = 4 \times 10^{-5}$ S (standard conditions).

Electrode area $A = 1$ cm$^2 = 10^{-4}$ m$^2$, electrode distance $l = 15$ cm $= 0.15$ m.

Cell constant = $\frac{l}{A} = \frac{0.15}{10^{-4}} = 1500$ m$^{-1}$.

Specific conductivity: $\kappa = G \times \frac{l}{A} = 4 \times 10^{-5} \times 1500 = 0.06$ S m$^{-1}$.

Since pH = 5, $[H^+] = 10^{-5}$ M. For weak acid HX with $\alpha \ll 1$: $[H^+] = c\alpha \approx 10^{-5}$ M.

Molar conductivity at this concentration: $\Lambda_m = \frac{\kappa}{c}$ where $c$ is in mol/m$^3$.

Since $\alpha \ll 1$, the total concentration $c \gg [H^+]$. However, we can use:

$\Lambda_m = \alpha \cdot \Lambda_m^0$ (limiting molar conductivity)

So $\Lambda_m^0 = \frac{\Lambda_m}{\alpha} = \frac{\kappa}{c\alpha}$.

Since $c\alpha = [H^+] = 10^{-5}$ mol/L $= 10^{-2}$ mol/m$^3$:

$\Lambda_m^0 = \frac{\kappa}{c\alpha \text{ (in mol/m}^3\text{)}} = \frac{0.06}{10^{-2}} = 6$ S m$^2$ mol$^{-1}$.

The answer is 6.

To find the trace of the matrix raised to the fifth power, we should first determine the eigenvalues of the original matrix A. The characteristic equation of matrix A is given by ∣A−λI∣ = 0. Let us compute this determinant. ∣−λ 1 0 ∣ ∣∣ ∣∣ ∣ 0 −λ 1 ∣ = 0 ∣ ∣ ∣ 6 −11 6−λ∣ ∣ ∣ Expanding this determinant along the first row yields the following equation. −λ⋅(−λ⋅(6−λ)+11)−1⋅(0−6) = 0 −λ⋅(λ2 −6λ+11)+6 = 0 −λ3 +6λ2 −11λ+6 = 0 Multiplying by negative one gives the standard characteristic polynomial. λ3 −6λ2 +11λ−6 = 0 By simple inspection, we can see that λ = 1 is a root since 1−6+11−6 = 0. Dividing the cubic polynomial by λ−1 gives a quadratic quotient of λ2 −5λ+6 = 0. Factoring the quadratic equation yields roots of 2 and 3. Thus, the eigenvalues of matrix A are 1,2, and 3. A standard property of matrices states that if a matrix has eigenvalues λ ,λ , and λ , then the matrix raised to 1 2 3 the power of k will have eigenvalues λk,λk, and λk. 1 2 3 Therefore, the eigenvalues of A5 are 15,25, and 35. The trace of any square matrix is always equal to the sum of its eigenvalues. Trace(A5) = 15 +25 +35 Trace(A5) = 1+32+243 Trace(A5) = 276 The correct value is 276. 