Your Answer : (Not Answered)

Correct Answer : A

The major product is (A).

Step (i) is a Paterno-Buchi [2+2] photocycloaddition of acetaldehyde to one C(CH₃)=CH bond of the diene, forming an oxetane on the ring with a quaternary C-CH₃. Step (ii), ozonolysis of the remaining C(CH₃)=CH bond, opens the five-membered ring, leaving a methyl ketone on the quaternary carbon and an aldehyde at the other end of the chain.

Step (iii) is an intramolecular aldol condensation. The methyl ketone's enolate attacks the aldehyde, closing a six-membered ring, and dehydration gives a cyclohexenone whose carbonyl is next to the quaternary methyl-bearing carbon, as in (A). (B) puts the carbonyl on the other side of the ring, and (C) and (D) keep a five-membered ring, which the ozonolysis destroys.

Your Answer : (Not Answered)

Correct Answer : D

X is (D), the spiro ketone with a cyclohexanone ring spiro-fused to a cyclopentane.

Step 1, TiCl₃ with Zn-Cu, is the McMurry coupling: two cyclopentanone molecules give the alkene bi(cyclopentylidene). Step 2, cold alkaline KMnO₄, adds two OH groups across the alkene to give the 1,2-diol (a pinacol). Step 3, H₂SO₄, is the pinacol rearrangement: loss of water gives a tertiary cation and a ring carbon of the neighbouring ring migrates, expanding that ring from five to six carbons and leaving a ketone.

(A) would have no ring expansion, (B) has the ketone in the five-membered ring, and (C) would need both rings to grow, so only (D) matches the one ring expansion.

Your Answer : (Not Answered)

Correct Answer : A

The major product is (A), cis-cyclohexane-1,2-diol.

TiCl₃(DME)₂ reduced by the zinc-copper couple gives low-valent titanium, the McMurry reagent. It transfers electrons to both aldehyde groups of hexane-1,6-dial, and the two ketyl carbons, held on the same titanium surface, bond to each other: a pinacol coupling that closes a six-membered ring between C1 and C6. Because both oxygens stay bound to titanium during the bond formation, the two OH groups end up on the same face, giving the cis-1,2-diol.

At room temperature the reaction stops at the diol; deoxygenation to cyclohexene needs heating. The epoxide (B) is not formed by this reagent, and the cyclopentane products (C) and (D), with a CH₂OH group, would need a different C-C bond (an aldol-type closure), not the coupling of the two carbonyl carbons.

Your Answer : (Not Answered)

Correct Answer : C

The terminal velocity is proportional to \(r^{2}\).

At terminal velocity, weight minus buoyancy equals Stokes' viscous drag \(6\pi\eta r v\). The weight and buoyancy both depend on the volume, \(\frac{4}{3}\pi r^{3}\).

\[ \begin{aligned} \tfrac{4}{3}\pi r^{3}(\rho - \sigma)g &= 6\pi\eta r v_t \\ v_t &= \frac{2r^{2}(\rho - \sigma)g}{9\eta} \end{aligned} \]

The \(r^{3}\) from the volume is divided by the \(r\) in the drag, leaving \(r^{2}\). So \(r^{3}\) ignores the drag's dependence on radius, and the inverse options have the trend backwards: bigger balls fall faster.

Your Answer : (Not Answered)

Correct Answer : A

The field is zero at \(\frac{x}{3}\) from m.

Between the two masses the fields point in opposite directions. At distance d from m they balance when

\[ \begin{aligned} \frac{Gm}{d^{2}} &= \frac{G(4m)}{(x - d)^{2}} \\ x - d &= 2d \Rightarrow d = \frac{x}{3} \end{aligned} \]

The neutral point lies nearer the smaller mass. \(\frac{x}{2}\) would be the midpoint, correct only for equal masses, and \(\frac{x}{5}\) comes from equating \(m/d\) with \(4m/(x - d)\) without squaring.

Your Answer : (Not Answered)

Correct Answer : D

One can use a blade.

A circuit needs a conducting path. A blade is made of metal (steel), which conducts electricity, so it can bridge the gap left by a missing wire, as in the simple tester activities of the NCERT science books.

Paper, rubber and wood are insulators: they do not allow current to pass, so a paper strip, a rubber belt or a wooden stick would leave the circuit open.

Your Answer : (Not Answered)

Correct Answer : D

Statements (a) and (d) are true.

Water pressure increases with depth, because pressure in a liquid is proportional to the height of liquid above the point. The lowest hole has the greatest depth of water above it, so water comes out there with the greatest force (a) and is pushed out fastest, travelling the greatest distance (d).

The highest hole has the least pressure, so its stream is weakest and shortest, which makes (b) and (e) false; and the force and distance are not equal for all holes, which makes (c) and (f) false.

Your Answer : (Not Answered)

Correct Answer : B

The integrating factor is \(e^{-x/2}\).

Put the equation in the standard linear form \(\frac{dy}{dx} + Py = Q\):

\[ \begin{aligned} 2\frac{dy}{dx} &= y + \cos x \\ \frac{dy}{dx} - \tfrac{1}{2}y &= \tfrac{1}{2}\cos x \end{aligned} \]

Here \(P = -\frac{1}{2}\), so the integrating factor is \(e^{\int P\,dx} = e^{-x/2}\). The option \(e^{x/2}\) misses the minus sign that appears when \(y\) is moved to the left, and \(e^{-2x}\) forgets to divide by 2 first.

Your Answer : (Not Answered)

Correct Answer : C

The product is \(\frac{\sqrt{2}}{8}\).

Since \(\sin 75^{\circ} = \cos 15^{\circ}\):

\[ \begin{aligned} \sin 15^{\circ}\cos 15^{\circ} &= \tfrac{1}{2}\sin 30^{\circ} = \tfrac{1}{4} \\ \text{product} &= \tfrac{1}{4} \times \tfrac{\sqrt{2}}{2} = \frac{\sqrt{2}}{8} \end{aligned} \]

All three sines are positive, so the negative options are impossible. \(\frac{\sqrt{2}}{4}\) forgets the \(\frac{1}{2}\) in the double-angle step.

Your Answer : (Not Answered)

Correct Answer : B

The integral is \(4x^{4}\log_{e}(x) - x^{4} + C\).

Integrate by parts with \(u = \ln x\) and \(dv = 16x^{3}dx\), so \(v = 4x^{4}\):

\[ \begin{aligned} \int 16x^{3}\ln x\,dx &= 4x^{4}\ln x - \int 4x^{4} \cdot \tfrac{1}{x}\,dx \\ &= 4x^{4}\ln x - x^{4} + C \end{aligned} \]

Differentiating gives \(16x^{3}\ln x + 4x^{3} - 4x^{3}\), which checks. The + sign and the \(-4x^{4}\) options mishandle \(\int 4x^{3}dx = x^{4}\).

Your Answer : (Not Answered)

Correct Answer : D

The number of ways is 441.

With at least three women among six members, count each possible split of women and men (7 women, 4 men).

\[ \begin{aligned} 3W, 3M &: \tbinom{7}{3}\tbinom{4}{3} = 35 \times 4 = 140 \\ 4W, 2M &: \tbinom{7}{4}\tbinom{4}{2} = 35 \times 6 = 210 \\ 5W, 1M &: \tbinom{7}{5}\tbinom{4}{1} = 21 \times 4 = 84 \\ 6W &: \tbinom{7}{6} = 7 \end{aligned} \]

The total is \(140 + 210 + 84 + 7 = 441\). Leaving out the all-women committee gives 434, and the larger options count committees with fewer than three women.

Your Answer : (Not Answered)

Correct Answer : B

\(n(B') = \) 13.

Find \(n(B)\) from the union formula, then subtract it from \(n(U)\):

\[ \begin{aligned} n(B) &= n(A \cup B) - n(A) + n(A \cap B) \\ &= 20 - 14 + 6 = 12 \\ n(B') &= n(U) - n(B) = 25 - 12 = 13 \end{aligned} \]

12 is \(n(B)\) itself, the usual slip of stopping one step early; 14 is \(n(A)\), and 15 or 16 come from adding \(n(A \cap B)\) with the wrong sign.

Check: \(n(A \cup B) = 14 + 12 - 6 = 20\), as given.

Your Answer : (Not Answered)

Correct Answer : A

The variation is \(-\text{T}(\partial^{2}\text{V}/\partial\text{T}^{2})_{\text{P}}\).

Start from \(C_{p} = T(\partial S/\partial T)_{P}\) and differentiate with respect to P at constant T, using the Maxwell relation \((\partial S/\partial P)_{T} = -(\partial V/\partial T)_{P}\):

\[ \begin{aligned} \left(\frac{\partial C_{p}}{\partial P}\right)_{T} &= T\frac{\partial}{\partial T}\left(\frac{\partial S}{\partial P}\right)_{T} \\ &= -T\left(\frac{\partial^{2} V}{\partial T^{2}}\right)_{P} \end{aligned} \]

\(T(\partial S/\partial T)_{P}\) is \(C_{p}\) itself, not its pressure dependence. The forms with \(\partial^{2}S/\partial T^{2}\) or \(\partial^{2}P/\partial T^{2}\) at constant V do not follow from the Maxwell relation. For an ideal gas V is linear in T, so the result correctly gives zero.

Your Answer : (Not Answered)

Correct Answer : D

The EPR spectrum shows 8 hyperfine lines.

VO²⁺ contains V(IV), d¹, with one unpaired electron. Vanadium-51, nearly 100 percent abundant, has nuclear spin I = 7/2, and the electron couples to it to give 2I + 1 lines:

\[ 2\left(\tfrac{7}{2}\right) + 1 = 8 \]

In solution at room temperature the anisotropy averages out and the eight-line pattern of the vanadyl ion is seen. 3 and 5 would correspond to nuclei with I = 1 or 2, and 7 to I = 3.

Your Answer : (Not Answered)

Correct Answer : C

\(\Delta G^{\circ}\) is +0.41 kJ mol^-1.

Use \(\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\), converting \(\Delta S^{\circ}\) to kJ:

\[ \begin{aligned} \Delta G^{\circ} &= -13 - 298 \times (-0.045) \\ &= -13 + 13.41 = +0.41\ \text{kJ mol}^{-1} \end{aligned} \]

The negative entropy change makes \(-T\Delta S^{\circ}\) positive, and at 298 K it just outweighs the favourable enthalpy, so the reaction is slightly non-spontaneous. Getting the sign of \(T\Delta S\) wrong gives -0.41 kJ mol^-1.

Your Answer : (Not Answered)

Correct Answer : C

The unit is \(mol^{-1} L\ s^{-1}\).

For an nth order reaction, the units of k are \((\text{mol L}^{-1})^{1-n}\,\text{s}^{-1}\). For \(n = 2\):

\[ \begin{aligned} k &= \frac{\text{rate}}{[A]^{2}} = \frac{\text{mol L}^{-1}\text{s}^{-1}}{(\text{mol L}^{-1})^{2}} \\ &= \text{L mol}^{-1}\text{s}^{-1} \end{aligned} \]

\(mol\ L^{-1} s^{-1}\) is the unit for a zero order reaction (and of the rate itself). The other options have the powers of mol or L wrong.

Your Answer : (Not Answered)

Correct Answer : B

The answer is same number of d-electrons, with \([vo_{2}]^{+}\) bent and \([uo_{2}]^{2+}\) linear.

In VO₂⁺ vanadium is +5, which removes all five valence electrons (3d³4s²): d⁰. In UO₂²⁺ uranium is +6, which removes all six (5f³6d¹7s²): also no d electrons. So both have zero d electrons.

Their shapes differ. The dioxovanadium(V) ion is bent (cis-dioxo), as usual for d⁰ transition-metal dioxo species, which use d orbitals that favour a cis arrangement. The uranyl ion is strictly linear (O=U=O), because the 5f orbitals let uranium form strong pi bonds to two trans oxygens. That rules out 'bent for both' and 'linear for both', and the oxidation states are irrelevant to the comparison asked.

Your Answer : (Not Answered)

Correct Answer : B

The oxidation state of chromium is +3.

Water is a neutral ligand (charge 0) and each chloride ligand carries −1. With the overall charge of the complex ion +1:

\[ \begin{aligned} x + 4(0) + 2(-1) &= +1 \\ x &= +3 \end{aligned} \]

+2 comes from forgetting the charge on the ion, and +5 and +7 do not fit the charges at all.

Your Answer : (Not Answered)

Correct Answer : D

The low moment is explained by antiferromagnetic coupling.

Copper(II) acetate monohydrate is a dimer in which four bridging acetates hold two Cu²⁺ ions (each d⁹, one unpaired electron) about 2.6 Å apart. The two spins couple antiparallel, giving a singlet ground state and a low-lying triplet. At room temperature the triplet is only partly populated, so the moment per copper, 1.41 B.M., is well below the spin-only value of 1.73 B.M., and it falls further on cooling.

Ferromagnetic coupling would raise the moment above 1.73 B.M. Spin-orbit coupling in Cu(II) usually raises the moment slightly (to about 1.8-2.0 B.M.), and temperature independent paramagnetism is a small additive effect that cannot account for a moment this low.

Your Answer : (Not Answered)

Correct Answer : D

The radioactive isotope of hydrogen is tritium.

Hydrogen has three natural isotopes: protium (one proton, no neutron), deuterium (one proton and one neutron) and tritium (one proton and two neutrons). Protium and deuterium are stable, but tritium is radioactive: it decays by beta emission to helium-3 with a half-life of about 12.3 years.

Helium is a different element, not an isotope of hydrogen, and it is also the product of tritium's decay. Protium and deuterium are hydrogen isotopes, but stable ones.