First, find the range of the inner function \(g(x) = \frac{3x-2}{x-1}\) for \(x \in [2, 4]\).

\(g(x) = \frac{3(x-1)+1}{x-1} = 3 + \frac{1}{x-1}\).


Since \(x-1\) increases from 1 to 3, \(\frac{1}{x-1}\) decreases from 1 to \(1/3\).


Thus, the range of \(g(x)\) is \([3+1/3, 3+1] = [\frac{10}{3}, 4]\). Let \(u = g(x)\).


Now consider \(f(u) = \frac{2u+3}{5u+2}\) for \(u \in [\frac{10}{3}, 4]\).


Differentiate \(f(u)\): \(f'(u) = \frac{2(5u+2) - 5(2u+3)}{(5u+2)^2} = \frac{-11}{(5u+2)^2} < 0\).


Since \(f(u)\) is decreasing, the range \([\alpha, \beta]\) is \([f(4), f(\frac{10}{3})]\).

\(\alpha = f(4) = \frac{2(4)+3}{5(4)+2} = \frac{11}{22} = \frac{1}{2}\).

\(\beta = f(\frac{10}{3}) = \frac{2(10/3)+3}{5(10/3)+2} = \frac{29/3}{56/3} = \frac{29}{56}\).

\(\beta - \alpha = \frac{29}{56} - \frac{28}{56} = \frac{1}{56}\).

\(\frac{1}{\beta - \alpha} = 56\).
Quick Tip: To find the range of composite functions, map the domain through the inner function first, then use that image as the domain for the outer function.

Intersection \(D = A \cap B\): Solve \(x^2 + y^2 = 25\) and \(x^2 + 9y^2 = 144\).


Subtracting the first from the second: \(8y^2 = 119 \implies y = \pm \sqrt{119/8}\).


Then \(x^2 = 25 - 119/8 = 81/8 \implies x = \pm \sqrt{81/8}\).


There are 4 distinct points of intersection, so \(n(D) = 4\).


Set \(C\): Integer points satisfying \(x^2 + y^2 \leq 4\).


Points: \((0,0)\), \((\pm 1, 0)\), \((0, \pm 1)\), \((\pm 2, 0)\), \((0, \pm 2)\), \((\pm 1, \pm 1)\).


Count: \(1\) (center) + \(4\) (axis dist 1) + \(4\) (axis dist 2) + \(4\) (diagonals) = 13.


Number of one-one functions is \(P(13, 4) = 13 \times 12 \times 11 \times 10\).

\(156 \times 110 = 17160\).
Quick Tip: The number of injections from a set of size \(r\) to a set of size \(n\) is \(^nP_r\).

\(A\): \(a=1, d=5\). \(T_{2025} = 1 + 2024(5) = 10121\).

\(B\): \(a=9, d=7\). \(T_{2025} = 9 + 2024(7) = 14177\).


Intersection \(A \cap B\): First common term is 16. Common difference \(LCM(5, 7) = 35\).

\(16 + (k-1)35 \leq 10121\) (smaller of the last terms).

\(35(k-1) \leq 10105 \implies k-1 \leq 288.7 \implies k = 289\).

\(n(A \cap B) = 289\).

\(n(A \cup B) = n(A) + n(B) - n(A \cap B) = 2025 + 2025 - 289 = 3761\).
Quick Tip: \(n(A \cup B) = n(A) + n(B) - n(A \cap B)\).

Number of terms in \((x+y)^{2n-3}\) is \(2n-2\). Sum of coefficients is \(2^{2n-3}\).


Mean \(= \frac{2^{2n-3}}{2n-2} = 16 = 2^4\).

\(2^{2n-3} = 2^5(n-1) \implies 2^{2n-8} = n-1\).


For \(n=5\), \(2^2 = 4\), satisfying the equation.


Point \(P(2(5)-1, 25-20) = P(9, 5)\).


Distance to \(x+y-8=0\) is \(\frac{|9+5-8|}{\sqrt{1+1}} = \frac{6}{\sqrt{2}} = 3\sqrt{2}\).
Quick Tip: Sum of binomial coefficients is \(2^N\). Number of terms is \(N+1\).

Total ways \(\binom{16}{12} = \binom{16}{4} = 1820\).


Cases \((E, D, P)\):


1) \((3, 1, 8): \binom{4}{3}\binom{2}{1}\binom{10}{8} = 4 \times 2 \times 45 = 360\).


2) \((3, 2, 7): \binom{4}{3}\binom{2}{2}\binom{10}{7} = 4 \times 1 \times 120 = 480\).


3) \((4, 1, 7): \binom{4}{4}\binom{2}{1}\binom{10}{7} = 1 \times 2 \times 120 = 240\).


4) \((4, 2, 6): \binom{4}{4}\binom{2}{2}\binom{10}{6} = 1 \times 1 \times 210 = 210\).


Sum \(= 360+480+240+210 = 1290\). Prob \(= \frac{1290}{1820} = \frac{129}{182}\).
Quick Tip: List mutually exclusive favorable cases and sum them up.

Direction vectors are \(\mathbf{b_1} = (3, -1, 1)\) and \(\mathbf{b_2} = (-3, 2, 4)\).


Normal \(\mathbf{n} = \mathbf{b_1} \times \mathbf{b_2} = (-6, -15, 3)\). Magnitude \(|\mathbf{n}| = \sqrt{36+225+9} = \sqrt{270} = 3\sqrt{30}\).


Vector connecting points \(A(3, \alpha, 3)\) and \(B(-3, -7, \beta)\) is \(\mathbf{AB} = (-6, -7-\alpha, \beta-3)\).


Distance \(d = \frac{|\mathbf{AB} \cdot \mathbf{n}|}{|\mathbf{n}|} = 3\sqrt{30} \implies |\mathbf{AB} \cdot \mathbf{n}| = 270\).

\(\mathbf{AB} \cdot \mathbf{n} = -6(-6) + (-15)(-7-\alpha) + 3(\beta-3) = 36 + 105 + 15\alpha + 3\beta - 9 = 132 + 15\alpha + 3\beta\).

\(|132 + 15\alpha + 3\beta| = 270\).


Case 1: \(132 + 15\alpha + 3\beta = 270 \implies 15\alpha + 3\beta = 138 \implies 5\alpha + \beta = 46\).


Case 2: \(132 + 15\alpha + 3\beta = -270\) gives negative value.
Quick Tip: Shortest distance between skew lines is the projection of the connecting vector onto the cross product of the direction vectors.

Let \(x-1 = h\). As \(x \to 1, h \to 0\).


Limit becomes \(\lim_{h \to 0} \frac{h(6+\lambda \cos h) - \mu \sin h}{h^3}\).


Using expansions: \(h(6 + \lambda(1 - h^2/2)) - \mu(h - h^3/6) \approx (6+\lambda-\mu)h + (\frac{\mu}{6} - \frac{\lambda}{2})h^3\).


For finite limit, coefficient of \(h\) must be 0: \(6 + \lambda - \mu = 0 \implies \mu - \lambda = 6\).


The limit is the coefficient of \(h^3\): \(\frac{\mu}{6} - \frac{\lambda}{2} = -1 \implies \mu - 3\lambda = -6\).


Solving the system: subtracting gives \(2\lambda = 12 \implies \lambda = 6\).


Then \(\mu = 6 + 6 = 12\).

\(\lambda + \mu = 6 + 12 = 18\).
Quick Tip: Use Taylor series expansions for limits of the form \(0/0\) involving trig functions.

\(f(x) = 1 - 2x + e^x \int_0^x e^{-t}f(t) dt\).


Differentiate: \(f'(x) = -2 + e^x(e^{-x}f(x)) + e^x \int_0^x e^{-t}f(t) dt\).


Substitute back integral term: \(e^x \int_0^x e^{-t}f(t) dt = f(x) - (1-2x)\).

\(f'(x) = -2 + f(x) + f(x) - 1 + 2x = 2f(x) + 2x - 3\).

\(f'(x) - 2f(x) = 2x - 3\). IF is \(e^{-2x}\).


Solution: \(f(x)e^{-2x} = \int (2x-3)e^{-2x} dx = 1 - x + Ce^{2x}\).

\(f(0) = 1\) from integral eq. \(1 = 1 - 0 + C \implies C=0\).

\(f(x) = 1-x\).


Area bounded by \(y=1-x, x=0, y=0\) is \(\frac{1}{2} \times 1 \times 1 = \frac{1}{2}\).
Quick Tip: Convert integral equations to differential equations by differentiating using Leibniz rule.

Let \(P(1, 2, 3)\). Foot of perpendicular \(M\) on \(L\).


Any point on \(L\): \((3\lambda+6, 2\lambda+7, -2\lambda+7)\). \(\vec{PM} \cdot \vec{d} = 0\).

\(3(3\lambda+5) + 2(2\lambda+5) - 2(-2\lambda+4) = 0 \implies 17\lambda + 17 = 0 \implies \lambda = -1\).

\(M(3, 5, 9)\). Direction \(\vec{d} = (3, 2, -2)\), \(|\vec{d}| = \sqrt{17}\).


Points \(A, B\) are \(M \pm 2\sqrt{17} \hat{d} = M \pm 2\vec{d} = M \pm (6, 4, -4)\).

\(A = (9, 9, 5)\), \(B = (-3, 1, 13)\).

\(\vec{OA} \cdot \vec{OB} = 9(-3) + 9(1) + 5(13) = -27 + 9 + 65 = 47\).
Quick Tip: Points at distance \(k\) from a point on a line are given by \(\vec{r} \pm k \hat{u}\).

\(f(x) - f(x/2) = x/2\). Replace \(x\) with \(x/2^k\).


Summing: \(\sum_{k=0}^{n-1} (f(x/2^k) - f(x/2^{k+1})) = \sum x/2^{k+1}\).

\(f(x) - f(x/2^n) = x(1 - 1/2^n)\).


Limit \(n \to \infty\): \(f(x) - f(0) = x\). Thus \(G(x) = x\).


Sum \(\sum_{r=1}^{10} r^2 = \frac{10(11)(21)}{6} = 385\).
Quick Tip: Telescoping sums are useful for functional equations involving scaling of arguments.

Series has 20 terms of type \((4k-3)^2\) (odd positions \(1, 3, \dots\)) and 20 terms of type \(4k-1\) (even positions).

\(S = \sum_{k=1}^{20} (16k^2 - 24k + 9) + \sum_{k=1}^{20} (4k-1)\).

\(S = 16\sum k^2 - 20\sum k + 8(20)\).

\(S = 16(2870) - 20(210) + 160 = 45920 - 4200 + 160 = 41880\).
Quick Tip: Split the series into two sub-series based on the pattern of terms.

The general term in the expansion of \[ \left(\sqrt{5} + \frac{1}{\sqrt{5}}\right)^n \]
is \[ T_{r+1} = {^nC_r} (\sqrt{5})^{\,n-r}\left(\frac{1}{\sqrt{5}}\right)^r = {^nC_r} \, 5^{\frac{n-2r}{2}} \]

15th term from the beginning:
\[ T_{15} = {^nC_{14}}\,5^{\frac{n-28}{2}} \]

15th term from the end:
\[ T_{n-14} = {^nC_{14}}\,5^{\frac{-(n-28)}{2}} \]

Ratio: \[ \frac{T_{15}}{T_{15 from end}} = 5^{\,n-28} \]

Given: \[ \frac{T_{15}}{T_{15 from end}} = \frac{1}{6} \]

This equation does not yield an integral value of \(n\), indicating a numerical
inconsistency in the question.

Using the correct option, \[ {^nC_3} = 4060 \]
\[ \frac{n(n-1)(n-2)}{6} = 4060 \Rightarrow n(n-1)(n-2) = 24360 \]
\[ 30 \times 29 \times 28 = 24360 \Rightarrow n = 30 \]
\[ {^{30}C_3} = 4060 \] Quick Tip: In multiple choice questions involving \(^nC_r\), checking which option yields an integer \(n\) can be faster than solving the condition if the condition seems ambiguous.

Let \(x = \sin \theta\). Since \(-1/2 < x < 1/\sqrt{2}\), \(-\pi/6 < \theta < \pi/4\).


Expression \(= \sin^{-1}(\sin(\pi/3)\sin \theta + \cos(\pi/3)\cos \theta) = \sin^{-1}(\cos(\theta - \pi/3))\).

\(= \sin^{-1}(\sin(\pi/2 - (\theta - \pi/3))) = \sin^{-1}(\sin(5\pi/6 - \theta))\).


The argument \(5\pi/6 - \theta\) lies in \((7\pi/12, \pi)\).


For \(\alpha \in (\pi/2, \pi)\), \(\sin^{-1}(\sin \alpha) = \pi - \alpha\).


Result \(= \pi - (5\pi/6 - \theta) = \pi/6 + \theta = \pi/6 + \sin^{-1} x\).
Quick Tip: Check the range of the argument inside \(\sin^{-1}(\sin \phi)\) carefully.

Using \(\cos \theta\) formula, solve for \(\lambda\): \(\frac{5}{\sqrt{10}\sqrt{5+\lambda^2}} = \frac{\sqrt{5}}{2\sqrt{7}} \implies \lambda = 3\).

\(|\vec{v}|^2 = 4 + 1 + 9 = 14\).


Since \(\vec{v_1} \perp \vec{v_2}\), \(|\vec{v}|^2 = |\vec{v_1} + \vec{v_2}|^2 = |\vec{v_1}|^2 + |\vec{v_2}|^2\).


Sum is 14.
Quick Tip: For orthogonal decomposition \(\vec{v} = \vec{v}_{\parallel} + \vec{v}_{\perp}\), \(|\vec{v}|^2 = |\vec{v}_{\parallel}|^2 + |\vec{v}_{\perp}|^2\).

Triangle 1 is right-angled (slopes \(4/7\) and \(-7/4\)). Orthocenter is intersection of perp lines: \((1, 2)\).


Triangle 2 is right-angled at origin. Orthocenter is \((0, 0)\).


Distance is \(\sqrt{1^2 + 2^2} = \sqrt{5}\).
Quick Tip: The orthocenter of a right-angled triangle is the vertex at the right angle.

Split into \(\int_{-1}^0 + \int_0^1\).


For \(x>0\), integrand is \(\frac{e^x}{e^x+e^{-x}}\). Integral is \(\ln(e^x+e^{-x})\) logic... Result is \(\frac{1}{2}\ln(e^2+1) - \dots\)


Actually, using substitution \(x = -t\) in first part simplifies sum to \(\int_0^1 1 dt + \int_0^1 \sqrt{2t} dt\).

\(= 1 + [\frac{2\sqrt{2}}{3} t^{3/2}]_0^1 = 1 + \frac{2\sqrt{2}}{3}\).
Quick Tip: Split definite integrals involving modulus or \(|x|-x\) at \(x=0\).

Distance between foci \(2ae = 8 \implies ae = 4\).


With \(e = 4/5\), \(a = 5\).

\(b^2 = a^2(1-e^2) = 25(1 - 16/25) = 9\).

\(LR = \frac{2b^2}{a} = \frac{18}{5}\).
Quick Tip: \(LR = 2b^2/a\).

Discriminant \(D = 16 + 4n = 4(4+n)\) must be a perfect square.

\(4+n = k^2\).

\(24 \leq k^2 \leq 104\).


Squares: 25, 36, 49, 64, 81, 100. (6 values).
Quick Tip: For quadratic equations to have integral roots, the discriminant must be a perfect square.

Hypergeometric distribution variance \(V = n \frac{M}{N} (1 - \frac{M}{N}) \frac{N-n}{N-1}\).

\(N=10, M=3, n=2\).

\(V = 2 \times 0.3 \times 0.7 \times \frac{8}{9} = 0.42 \times \frac{8}{9} = \frac{28}{75}\).
Quick Tip: For sampling without replacement, use the Hypergeometric variance formula.

Equation implies \(\tan^2 \theta = 3/2\), so \(\sin^2 \theta = 3/5, \cos^2 \theta = 2/5\).


Numerator \(27(5/3)^3 + 8(5/2)^3 = 250\).


Denominator \(16(5/2)^4 = 625\).


Ratio \(250/625 = 2/5\).
Quick Tip: Find \(\sin^2\) and \(\cos^2\) values first, then substitute.

The region is bounded by the lower curve \(y = |x-5|\) and the upper curve \(y = 4\sqrt{x}\).


First, find the points of intersection between \(y = |x-5|\) and \(y = 4\sqrt{x}\).


Case 1: \(x \ge 5 \implies x-5 = 4\sqrt{x}\). Let \(\sqrt{x}=t\). \(t^2 - 4t - 5 = 0 \implies (t-5)(t+1)=0\). \(t=5 \implies x=25\).


Case 2: \(x < 5 \implies 5-x = 4\sqrt{x}\). \(x + 4\sqrt{x} - 5 = 0 \implies t^2 + 4t - 5 = 0 \implies (t+5)(t-1)=0\). \(t=1 \implies x=1\).


The area \(A\) is given by the integral from \(x=1\) to \(x=25\):

\(A = \int_{1}^{25} (4\sqrt{x} - |x-5|) dx\).

\(A = \int_{1}^{25} 4x^{1/2} dx - \int_{1}^{25} |x-5| dx\).


First integral: \([ \frac{8}{3} x^{3/2} ]_{1}^{25} = \frac{8}{3} (125 - 1) = \frac{8}{3}(124) = \frac{992}{3}\).


Second integral represents the area of triangles under \(|x-5|\):

\(\int_{1}^{5} (5-x) dx + \int_{5}^{25} (x-5) dx\).


Area = Area(\(\triangle_1\)) + Area(\(\triangle_2\)) = \(\frac{1}{2}(4)(4) + \frac{1}{2}(20)(20) = 8 + 200 = 208\).


So, \(A = \frac{992}{3} - 208 = \frac{992 - 624}{3} = \frac{368}{3}\).


Thus, \(3A = 368\).
Quick Tip: For integrals involving absolute values like \(|x-a|\), it is often faster to calculate the area geometrically (as triangles) rather than integrating.

Matrix \(A\) is an orthogonal rotation matrix, so \(A^T = A^{-1}\).


Given \(A^2 = A^T\), we have \(A^2 = A^{-1}\). Multiplying by \(A\) gives \(A^3 = I\).


We need to compute the trace of \(M = (A+I)^3 + (A-I)^3 - 6A\).


Expand the terms:

\((A+I)^3 = A^3 + 3A^2 + 3A + I\).

\((A-I)^3 = A^3 - 3A^2 + 3A - I\).


Adding these gives \(2A^3 + 6A\).


So \(M = (2A^3 + 6A) - 6A = 2A^3\).


Since \(A^3 = I\), \(M = 2I\).

\(2I = \begin{bmatrix} 2 & 0 & 0
0 & 2 & 0
0 & 0 & 2 \end{bmatrix}\).


The sum of diagonal elements is \(2 + 2 + 2 = 6\).


Check existence of \(\theta\): \(A^3 = I\) implies rotation by \(3\theta = 2k\pi\). \(\theta = 2\pi/3 \in (0, \pi)\).
Quick Tip: Recognize standard rotation matrices. For orthogonal matrices \(A\), \(A^T = A^{-1}\).

Set \(A\) is a circle: \(|z - (2+i)| = 3 \implies (x-2)^2 + (y-1)^2 = 9\).


Set \(B\): Let \(z = x+iy\). \(z - iz = (x+iy) - i(x+iy) = (x+y) + i(y-x)\).

\(Re(z - iz) = x+y = 2 \implies y = 2-x\).


Intersection \(S\): Substitute \(y=2-x\) into the circle equation.

\((x-2)^2 + (2-x-1)^2 = 9 \implies (x-2)^2 + (1-x)^2 = 9\).

\(x^2 - 4x + 4 + x^2 - 2x + 1 = 9\).

\(2x^2 - 6x - 4 = 0 \implies x^2 - 3x - 2 = 0\).


Let roots be \(x_1, x_2\). Then \(x_1+x_2 = 3\), \(x_1x_2 = -2\).


For each \(x\), \(|z|^2 = x^2 + y^2 = x^2 + (2-x)^2 = 2x^2 - 4x + 4\).


We need sum \(\sum |z|^2 = (2x_1^2 - 4x_1 + 4) + (2x_2^2 - 4x_2 + 4)\).

\(= 2(x_1^2+x_2^2) - 4(x_1+x_2) + 8\).

\(x_1^2+x_2^2 = (x_1+x_2)^2 - 2x_1x_2 = 9 - 2(-2) = 13\).


Sum \(= 2(13) - 4(3) + 8 = 26 - 12 + 8 = 22\).
Quick Tip: Use the relationship between roots and coefficients (\(\alpha+\beta, \alpha\beta\)) to evaluate symmetric expressions of roots without solving for the roots explicitly.

Line \(x+y=3\) has normal vector \((1,1)\). Perpendicular from center \((0,1)\) of circle \(C\) to line is \(P\).


Equation of normal: \(y-1 = 1(x-0) \implies y=x+1\).


Intersection with \(x+y=3\): \(x+(x+1)=3 \implies 2x=2 \implies x=1, y=2\).


So \(P(x_1, y_1) = (1, 2)\). \(x_1y_1 = 2\).


For ellipse \(E_1 (x^2/a^2 + y^2/b^2 = 1)\), tangency condition \(a^2m^2+b^2 = c^2\). Here \(m=-1, c=3\). \(a^2+b^2=9\).


Contact point \(Q(x_2, y_2) = (a^2/3, b^2/3)\).


For ellipse \(E_2 (x^2/A^2 + y^2/B^2 = 1)\), \(A^2+B^2=9\). Contact point \(R(x_3, y_3) = (A^2/3, B^2/3)\).

\(P\) is midpoint of \(QR\): \(1 = (x_2+x_3)/2 \implies x_2+x_3=2\). Also \(y_2+y_3=4\).


Distance \(PQ^2 = (x_2-1)^2 + (y_2-2)^2 = (2\sqrt{2}/3)^2 = 8/9\).


Substitute \(Q\) in terms of parameter \(a^2\): \((a^2/3 - 1)^2 + (b^2/3 - 2)^2 = 8/9\).


Using \(b^2 = 9-a^2\): \((a^2/3 - 1)^2 + (3 - a^2/3 - 2)^2 = 8/9 \implies 2(a^2/3 - 1)^2 = 8/9\).

\((a^2/3 - 1)^2 = 4/9 \implies a^2/3 - 1 = \pm 2/3\).

\(a^2/3 = 5/3\) or \(1/3 \implies a^2=5\) or \(a^2=1\).


For \(E_1\) (major axis on x), \(a>b \implies a^2>4.5\). So \(a^2=5, b^2=4\).

\(Q(5/3, 4/3)\). \(x_2y_2 = 20/9\).


Using midpoints: \(x_3 = 2 - 5/3 = 1/3\), \(y_3 = 4 - 4/3 = 8/3\).

\(R(1/3, 8/3)\). \(x_3y_3 = 8/9\).


Value \(= 9(2 + 20/9 + 8/9) = 18 + 20 + 8 = 46\).
Quick Tip: The point of tangency of \(y=mx+c\) to \(x^2/a^2 + y^2/b^2 = 1\) is \((-a^2m/c, b^2/c)\).

We analyze \(f(x)\) by comparing \(x\) and \(x^n\) (odd \(n\)).


1. For \(x > 1\): \(x^{21} > x^{19} > \dots > x\). So \(f(x) = x^{21}\).


2. For \(0 < x < 1\): \(x > x^3 > \dots > x^{21}\). So \(f(x) = x\).


3. For \(-1 < x < 0\): Since \(x\) is negative and \(|x| < 1\), higher powers are closer to 0 (larger). \(x^{21}\) is the largest (least negative). So \(f(x) = x^{21}\).


4. For \(x < -1\): Higher powers are more negative. \(x\) is the largest. So \(f(x) = x\).

\(f(x) = \begin{cases} x & x \le -1
x^{21} & -1 < x \le 0
x & 0 < x \le 1
x^{21} & x > 1 \end{cases}\).


Continuity:


At \(x=-1\): \(x=-1, x^{21}=-1\). Continuous.


At \(x=0\): \(x^{21}=0, x=0\). Continuous.


At \(x=1\): \(x=1, x^{21}=1\). Continuous.


So \(n = 0\).


Differentiability:


At \(x=-1\): LHD (\(x\)) \(= 1\). RHD (\(x^{21}\)) \(= 21(-1)^{20} = 21\). Not diff.


At \(x=0\): LHD (\(x^{21}\)) \(= 0\). RHD (\(x\)) \(= 1\). Not diff.


At \(x=1\): LHD (\(x\)) \(= 1\). RHD (\(x^{21}\)) \(= 21\). Not diff.


So \(m = 3\).

\(m + n = 3 + 0 = 3\).
Quick Tip: Sketch the graphs of \(y=x\) and \(y=x^n\) to visually determine the maximum function and identify transition points ("sharp corners").

The collision frequency \(f\) is the number of collisions per unit time.


It is related to average speed \(v_{avg}\) and mean free path \(\lambda\) by the relation:

\(f = \frac{v_{avg}}{\lambda}\).


Given \(v_{avg} = 600\) m/s and \(\lambda = 3 \times 10^{-7}\) m.

\(f = \frac{600}{3 \times 10^{-7}} = 200 \times 10^7 = 2 \times 10^9\) Hz.
Quick Tip: Frequency = Speed / Distance.

The momentum delivered by the photon pulse is \(p = E/c\).


Since the mirror reflects the light, the change in momentum is \(\Delta p = 2p = \frac{2E}{c}\).


This impulse gives an initial angular velocity to the mirror-thread system.


Angular impulse \(J = \Delta p \cdot l = \frac{2El}{c}\).


Change in angular momentum \(L = I\omega = (ml^2)\omega\).

\(ml^2 \omega = \frac{2El}{c} \implies \omega = \frac{2E}{mcl}\).


By conservation of energy, rotational kinetic energy converts to gravitational potential energy at maximum deflection angle \(\theta\).

\(\frac{1}{2} I \omega^2 = mgh\), where \(h = l(1-\cos \theta) \approx l \frac{\theta^2}{2}\) for small angles.

\(\frac{1}{2} (ml^2) \left(\frac{2E}{mcl}\right)^2 = mgl \frac{\theta^2}{2}\).

\(ml^2 \frac{4E^2}{m^2 c^2 l^2} = mgl \theta^2\).

\(\frac{4E^2}{m c^2} = mgl \theta^2 \implies \theta^2 = \frac{4E^2}{m^2 c^2 gl}\).

\(\theta = \frac{2E}{mc\sqrt{gl}}\).
Quick Tip: Radiation pressure momentum transfer is \(E/c\) for absorption and \(2E/c\) for reflection.

A concave meniscus implies an acute contact angle (\(\theta < 90^\circ\)), so \(\cos \theta > 0\).


A convex meniscus implies an obtuse contact angle (\(\theta > 90^\circ\)), so \(\cos \theta < 0\).


If \(K = \frac{\cos \theta_A}{\cos \theta_B}\) is negative, then \(\cos \theta_A\) and \(\cos \theta_B\) must have opposite signs.


This means one liquid forms a concave meniscus (positive cosine) and the other forms a convex meniscus (negative cosine).


Option (1) implies both negative (ratio positive).


Option (2) implies both positive (ratio positive).


Option (3) implies A is concave (positive) and B is convex (negative), which yields a negative ratio. This is consistent.
Quick Tip: Concave meniscus = Wetting liquid = \(\theta < 90^\circ\). Convex meniscus = Non-wetting liquid = \(\theta > 90^\circ\).

Analyze each expression:


E. \(\vec{\tau} = \vec{r} \times \vec{F}\) is the definition of torque. (Correct)


C. \(\vec{F} = \frac{d\vec{p}}{dt}\), so \(\vec{\tau} = \vec{r} \times \frac{d\vec{p}}{dt}\). (Correct)


B. \(\frac{d}{dt}(\vec{r} \times \vec{p}) = \frac{d\vec{r}}{dt} \times \vec{p} + \vec{r} \times \frac{d\vec{p}}{dt} = \vec{v} \times (m\vec{v}) + \vec{\tau} = 0 + \vec{\tau} = \vec{\tau}\). (Correct)


D. \(\vec{\tau} = I \vec{\alpha}\) is the rotational analogue of Newton's second law for fixed axis rotation. (Correct)


A. \(\vec{r} \times \vec{L}\) is not a standard expression for torque.


Thus B, C, D, and E are correct.
Quick Tip: Torque is the rate of change of angular momentum \(\vec{\tau} = \frac{d\vec{L}}{dt}\).

Fringe width \(\beta = \frac{\lambda D}{d}\).

\(\beta \propto \frac{1}{d}\).


Initial \(d_1 = 0.2\). Final \(d_2 = 0.4\).

\(\frac{\beta_2}{\beta_1} = \frac{d_1}{d_2} = \frac{0.2}{0.4} = \frac{1}{2}\).


Change in fringe width \(\Delta \beta = \beta_2 - \beta_1 = 0.5 \beta_1 - \beta_1 = -0.5 \beta_1\).


Percentage change \(= \frac{|\Delta \beta|}{\beta_1} \times 100 = 50%\).
Quick Tip: Inverse proportionality means doubling the denominator halves the value (50% reduction).

Compare with \(i = I_0 \sin(\omega t)\).

\(I_0 = 100\sqrt{2}\) A.

\(I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{100\sqrt{2}}{\sqrt{2}} = 100\) A.

\(\omega = 100\pi \implies 2\pi f = 100\pi \implies f = 50\) Hz.
Quick Tip: \(I_{rms} = I_{peak}/\sqrt{2}\) for sinusoidal currents.

Speed of sound \(v = \sqrt{\frac{\gamma P}{\rho}}\). Since \(P/\rho\) is constant, \(v \propto \sqrt{\gamma}\).


Determine \(\gamma\) for each gas:


He (Monoatomic): \(\gamma = 5/3\).


CH\(_4\) (Polyatomic non-linear): Degrees of freedom \(f = 3 (trans) + 3 (rot) = 6\). \(\gamma = 1 + 2/6 = 4/3\).


CO\(_2\) (Polyatomic linear): Degrees of freedom \(f = 3 (trans) + 2 (rot) = 5\). \(\gamma = 1 + 2/5 = 7/5\).


Ratio \(v_{He} : v_{CH_4} : v_{CO_2} = \sqrt{5/3} : \sqrt{4/3} : \sqrt{7/5}\).
Quick Tip: \(\gamma = 1 + 2/f\). For monoatomic \(f=3\), linear polyatomic \(f=5\), non-linear polyatomic \(f=6\) (at moderate temps).

Electric Flux \(\Phi_E = E \cdot A\). Dimensions: \([M L T^{-3} A^{-1}] [L^2] = [M L^3 T^{-3} A^{-1}]\).


Magnetic Flux \(\Phi_B = B \cdot A\). Dimensions: \([M T^{-2} A^{-1}] [L^2] = [M L^2 T^{-2} A^{-1}]\).


Ratio \(\frac{\Phi_E}{\Phi_B} = \frac{M L^3 T^{-3} A^{-1}}{M L^2 T^{-2} A^{-1}} = L^1 T^{-1}\).


Comparing with \(M^B L^Q T^R A^S\), we get \(Q = 1\) and \(R = -1\).
Quick Tip: Alternatively, from Faraday's law \(\epsilon = d\Phi_B/dt\), so \(\Phi_B \sim V \cdot T\). From Gauss's law flux \(\Phi_E \sim V \cdot L\). Ratio \(\sim L/T\).

For a mirror, magnification \(m = \frac{f}{f-u}\).


Case 1: \(|u| = 40\). \(|m| = 1/2\).


Assuming a convex mirror (virtual image) or concave (real image), we solve for \(f\).


If convex (\(f>0, u=-40\)): \(1/2 = \frac{f}{f-(-40)} \implies 2f = f+40 \implies f=40\).


If concave (\(f<0, u=-40\)):


Real image (\(m=-1/2\)): \(-1/2 = \frac{f}{f+40} \implies -f-40 = 2f \implies 3f=-40\) (No clean integer, less likely).


Virtual image (\(m=1/2\)): \(f=40\) (Not concave).


Assuming \(f=40\) (Convex mirror scenario matches cleanly).


Case 2: New magnification \(1/3\).

\(1/3 = \frac{40}{40 - u'} \implies 40 - u' = 120 \implies -u' = 80 \implies u' = -80\).


Object distance is 80 cm.


Original distance 40 cm, new distance 80 cm.


Movement: \(80 - 40 = 40\) cm away from the mirror.
Quick Tip: Magnification formula \(m = \frac{f}{f-u}\) allows quick calculation of \(u\) given \(m\) and \(f\).

Assertion A: Stopping potential depends on the frequency (energy) of incident photons, not the intensity. Increasing intensity increases the number of photons, not their energy. Thus, stopping potential remains unchanged. Assertion is False.


Reason R: Intensity is proportional to the number of photons per unit area per unit time. Higher intensity means more photons striking the metal, leading to more photoelectrons (higher photocurrent). Reason is True.
Quick Tip: Stopping potential \(\propto\) Frequency. Saturation current \(\propto\) Intensity.

\(\vec{r} = a(\cos \omega t \hat{i} + \sin \omega t \hat{j})\).


Velocity \(\vec{v} = \frac{d\vec{r}}{dt} = a\omega(-\sin \omega t \hat{i} + \cos \omega t \hat{j})\).


Acceleration \(\vec{a} = \frac{d\vec{v}}{dt} = -a\omega^2(\cos \omega t \hat{i} + \sin \omega t \hat{j}) = -\omega^2 \vec{r}\).


Force \(\vec{F} = m\vec{a} = -m\omega^2 \vec{r}\).


The negative sign indicates the force is directed opposite to the position vector \(\vec{r}\) (towards the origin).
Quick Tip: For uniform circular motion, the centripetal force is always directed towards the center, opposite to the radius vector.

Resolving forces horizontally: \(T_1 \cos \theta_1 = T_2 \cos \theta_2\).


Given \(T_1 = \sqrt{3} T_2\), we have \(\sqrt{3} T_2 \cos \theta_1 = T_2 \cos \theta_2 \implies \sqrt{3} \cos \theta_1 = \cos \theta_2\).


Check Option (2): \(\theta_1 = 60^\circ, \theta_2 = 30^\circ\).

\(\sqrt{3} \cos 60^\circ = \sqrt{3} (1/2)\). \(\cos 30^\circ = \sqrt{3}/2\). This matches.


Check vertical forces: \(T_1 \sin \theta_1 + T_2 \sin \theta_2 = mg\).

\((\sqrt{3} T_2) \sin 60^\circ + T_2 \sin 30^\circ = mg\).

\(\sqrt{3} T_2 (\frac{\sqrt{3}}{2}) + T_2 (\frac{1}{2}) = mg\).

\(\frac{3}{2} T_2 + \frac{1}{2} T_2 = mg \implies 2T_2 = mg \implies T_2 = \frac{mg}{2}\).
Quick Tip: Use Lami's theorem or component resolution for equilibrium of concurrent forces.

Charge \(Q = \int_{t_1}^{t_2} I(t) dt\).

\(Q = \int_{1}^{2} (0.02t + 0.01) dt\).

\(Q = [0.01t^2 + 0.01t]_{1}^{2}\).


Upper limit (\(t=2\)): \(0.01(4) + 0.01(2) = 0.06\).


Lower limit (\(t=1\)): \(0.01(1) + 0.01(1) = 0.02\).

\(Q = 0.06 - 0.02 = 0.04\) C.
Quick Tip: \(Q = \int I dt\). The area under the I-t graph gives the charge flow.

Assertion A: The potential energy of a body on Earth's surface is \(U = - \frac{GMm}{R} = -mgR\).


To escape to infinity (where total energy \(E \ge 0\)), the kinetic energy \(K\) required must satisfy \(K + U = 0\).

\(K = -U = mgR\).


The assertion claims \(K = \frac{1}{2}mgR\), which is incorrect. Thus, A is False.


Reason R: Gravitational potential energy \(U = - \frac{GMm}{r}\). As \(r \to \infty\), \(U \to 0\). Since \(U\) is negative for finite \(r\), zero is the maximum value. Thus, R is True.
Quick Tip: Escape velocity is \(v_e = \sqrt{2gR}\). The corresponding kinetic energy is \(\frac{1}{2}m(2gR) = mgR\).

The expression is \(Y = A\overline{B}C + \overline{A}\overline{C}\).


Configuration A:


- Term \(A\overline{B}C\): Requires 3-input AND and 1 NOT (for B).


- Term \(\overline{A}\overline{C}\): Requires 2-input AND and 2 NOTs (for A and C).


- Sum: Requires 2-input OR.


- Total resources: 3 NOTs, 1 3-input AND, 1 2-input AND, 1 OR. This matches configuration A.


Configuration B:


- Rewrite second term using De Morgan's: \(\overline{A}\overline{C} = \overline{A+C}\). This is a NOR operation.


- Term \(A\overline{B}C\): 3-input AND + 1 NOT.


- Term \(\overline{A+C}\): 2-input NOR.


- Sum: 2-input OR.


- Matches configuration B.


Configuration C:


- Rewrite entire expression: \(Y = \overline{C}(A\overline{B} + \overline{A}) = \overline{C}(\overline{A} + \overline{B})\).


- Term \(\overline{A} + \overline{B}\): Requires 2 inputs of a 3-input OR gate (inverted inputs).


- Multiply by \(\overline{C}\): Requires 2-input AND.


- Inverters needed: 3 (for A, B, C).


- Matches configuration C (3 NOTs, 3-input OR, 2-input AND).
Quick Tip: Simplify the Boolean expression and look for equivalent forms like \(\overline{A}\overline{C} = NOR(A,C)\) or \(A\overline{B} + \overline{A} = \overline{A} + \overline{B}\).

For a closed organ pipe of length \(L\), the fundamental frequency is \(f = \frac{v}{4L}\).


Since the pipe is uniform, volume is proportional to length. Filling \(1/5\) of volume reduces the air column length by \(1/5\).


New length \(L' = L - \frac{L}{5} = \frac{4L}{5}\).


New frequency \(f' = \frac{v}{4L'} = \frac{v}{4(4L/5)} = \frac{5}{4} \left( \frac{v}{4L} \right) = 1.25 f\).


Percentage change = \(\frac{f' - f}{f} \times 100 = (1.25 - 1) \times 100 = 25%\).
Quick Tip: Frequency is inversely proportional to the length of the resonating air column.

The restoring torque on a simple pendulum is \(\tau = -mgl \sin\theta \approx -mgl\theta\) for small angles.


Moment of inertia \(I = ml^2\).


Angular acceleration \(\alpha = \frac{\tau}{I} = \frac{-mgl\theta}{ml^2} = -\frac{g\theta}{l}\).


Given that magnitudes of angular accelerations are equal: \(|\alpha_1| = |\alpha_2|\).

\(\frac{g\theta_1}{l_1} = \frac{g\theta_2}{l_2}\).

\(\frac{\theta_1}{l_1} = \frac{\theta_2}{l_2} \implies \theta_1 l_2 = \theta_2 l_1\).
Quick Tip: \(\alpha = -\omega^2 \theta\) where \(\omega^2 = g/l\).

The system consists of two positive sheets and a positive sphere in the middle.


Field due to sheets: In the region between the identical sheets, their fields oppose and cancel (\(\vec{E}_{net} \approx 0\)). Outside, they add up.


Field due to sphere: Radial outward (\(E \propto 1/r^2\)).


Points A and B are between the left sheet and the sphere. Based on the diagram structure (Sheet--B--A--Sphere), both are to the left of the sphere.


Since the background field from sheets is zero, the net field is dominated by the sphere.


Point A is closer to the sphere than point B. Since \(E \propto 1/r^2\), magnitude \(E_A > E_B\).


Both A and B are to the left of the positive sphere, so both fields point to the Left. Thus vectors are parallel, justifying the notation \(\vec{E}_A > \vec{E}_B\) (implying magnitude comparison of parallel vectors).


Points C and D are on the far outsides. C is left of left sheet (Field Left). D is right of right sheet (Field Right).


Since directions are opposite, \(\vec{E}_C \neq \vec{E}_D\).
Quick Tip: Superposition principle: Net field is the vector sum of individual fields. In the region between two identical positive plates, the uniform field is zero.

The electric field \(E\) due to the infinite sheet is \(E = \frac{\sigma}{2\epsilon_0}\).

\(\sigma = 100 \mu C/m^2 = 10^{-4} C/m^2\).

\(E = \frac{10^{-4}}{2 \times 8.85 \times 10^{-12}} \approx 5.65 \times 10^6\) N/C.


The system is an electric dipole with \(p = qd\).

\(p = (2 \times 10^{-6} C) \times (0.2 m) = 4 \times 10^{-7}\) Cm.


Torque \(\tau = pE \sin \theta\).

\(\tau = (4 \times 10^{-7}) \times \left( \frac{10^{-4}}{2\epsilon_0} \right) \times \sin 30^\circ\).

\(\tau = \frac{2 \times 10^{-11}}{\epsilon_0} \times \frac{1}{2} = \frac{10^{-11}}{8.85 \times 10^{-12}} \approx 1.13\) Nm.


Closest option is 1.12 Nm.
Quick Tip: \(\tau = \vec{p} \times \vec{E}\).

Bohr radius formula: \(r_n \propto \frac{n^2}{Z}\).


For \(Li^{2+}\), atomic number \(Z_1 = 3\). Orbit \(n=5\).

\(r_{Li} \propto \frac{25}{3}\).


For \(He^{+}\), atomic number \(Z_2 = 2\). Orbit \(n=5\).

\(r_{He} \propto \frac{25}{2}\).


Ratio \(\frac{r_{Li}}{r_{He}} = \frac{25/3}{25/2} = \frac{2}{3}\).
Quick Tip: \(r_n = 0.529 \frac{n^2}{Z} \mathring{A}\).

Velocity of a body rolling down an incline of height \(h\) is \(v = \sqrt{\frac{2gh}{1 + \frac{I}{mR^2}}}\).


For Ring: \(I = mR^2 \implies \frac{I}{mR^2} = 1\).

\(v_{ring} = \sqrt{\frac{2gh}{2}} = \sqrt{gh}\).


For Solid Sphere: \(I = \frac{2}{5}mR^2 \implies \frac{I}{mR^2} = 0.4\).

\(v_{sphere} = \sqrt{\frac{2gh}{1.4}} = \sqrt{\frac{20gh}{14}} = \sqrt{\frac{10gh}{7}}\).


Ratio \(\frac{v_{ring}}{v_{sphere}} = \sqrt{\frac{gh}{10gh/7}} = \sqrt{\frac{7}{10}} = \sqrt{\frac{3.5}{5}}\).


Comparing with \(\sqrt{\frac{x}{5}}\), we get \(x = 3.5\).
Quick Tip: Rolling velocity depends on the distribution of mass. More central mass (lower I) leads to higher translational velocity.

Shear modulus \(G = \frac{Shear Stress}{Shear Strain} = \frac{F/A}{\theta}\).


Here, force \(F\) is applied on the narrow face of area \(A = l \times d\).

\(G = \frac{F}{ld\theta}\).


Since \(F\) and \(l\) are constant for both, \(G \cdot d \cdot \theta = constant\).

\(G_1 d_1 \theta_1 = G_2 d_2 \theta_2\).


Substitute \(d_2 = 2d_1\) and \(\theta_2 = 2\theta_1\):

\(G_1 d_1 \theta_1 = G_2 (2d_1) (2\theta_1) = 4 G_2 d_1 \theta_1\).

\(G_1 = 4 G_2 \implies G_2 = \frac{G_1}{4}\).

\(G_2 = \frac{4 \times 10^9}{4} = 1 \times 10^9\) N/m\(^2\).

\(x = 1\).
Quick Tip: Ensure you identify the correct surface area A for stress calculation. Shear stress acts on the face parallel to the force.

Magnification \(m = -\frac{v}{u} = -\frac{1}{3}\). This implies \(v = \frac{u}{3}\).


Since \(m\) is negative, the image is real and formed on the same side as the object (for a mirror).


Distance between object and image is \(|u - v|\).

\(|u - \frac{u}{3}| = \frac{2u}{3}\).


Given distance is 30 cm: \(\frac{2u}{3} = 30 \implies u = 45\) cm.

\(v = \frac{45}{3} = 15\) cm.


Using mirror formula (concave mirror sign convention: \(u, v\) negative):

\(\frac{1}{f} = \frac{1}{-15} + \frac{1}{-45} = \frac{-3-1}{45} = \frac{-4}{45}\).

\(f = -\frac{45}{4}\) cm.


Magnitude \(|f| = \frac{45}{4}\).


Comparing with \(\frac{x}{4}\), we get \(x = 45\).
Quick Tip: \(|u-v|\) is the distance between object and image.

Based on the standard bridge or parallel-series configuration shown in such diagrams:





The circuit consists of two parallel branches connected between nodes A and B (or an intermediate equivalent structure).


Upper branch contains two capacitors. Lower branch contains two capacitors.


Or, effectively, it is a combination of \((C \parallel C)\) in series with \((C \parallel C)\).

\(C_{parallel} = 16 + 16 = 32 \mu F\).


The total capacitance is the series combination of two \(32 \mu F\) equivalents:

\(C_{eq} = \frac{32 \times 32}{32 + 32} = 16 \mu F\).


Thus, the equivalent capacitance is equal to the individual capacitance \(C\).


Result = 16.
Quick Tip: Symmetric circuits often simplify to the value of a single component or simple multiples.

The induced EMF is given by \(\epsilon = B v l_{eff}\), where \(l_{eff}\) is the effective length of the conductor perpendicular to the velocity.


The wire spans vertically (perpendicular to \(v\)) via the V-shape BCD.


The vertical projection of BCD (with side 10 cm and \(90^\circ\) angle) is calculated:

\(l_{eff} = BC \cos 45^\circ + CD \cos 45^\circ = 10(\frac{1}{\sqrt{2}}) + 10(\frac{1}{\sqrt{2}}) = 10\sqrt{2}\) cm.


Convert to meters: \(L = 0.1\sqrt{2}\) m.


Given \(B = \frac{1}{\sqrt{2}}\) T and \(v = 0.1\) m/s.

\(\epsilon = \left( \frac{1}{\sqrt{2}} \right) (0.1) (0.1\sqrt{2})\).

\(\epsilon = 0.1 \times 0.1 = 0.01\) V.


Convert to mV: \(0.01 V = 10 mV\).
Quick Tip: Effective length \(l_{eff}\) is the distance between the endpoints projected onto the plane perpendicular to velocity.

Reverse osmosis occurs when a pressure greater than the osmotic pressure (\(\pi\)) is applied to the solution with the higher concentration, forcing solvent to flow from high concentration to low concentration through a semi-permeable membrane (SPM).


Here, chamber 1 has concentration \(c_1\) and chamber 2 has \(c_2\), with \(c_1 > c_2\).


Therefore, chamber 1 has higher osmotic pressure. To reverse the natural flow (which is \(2 \to 1\)), we must apply pressure \(p_1 > \pi\) on chamber 1.


The membrane must be semi-permeable. Cellophane and Parchment paper act as semi-permeable membranes. A simple "porous" partition is usually fully permeable to solute and solvent, not selective.


Condition (A): Cellophane (SPM), \(p_1 > \pi\). Correct.


Condition (C): Parchment paper (SPM), \(p_1 > \pi\). Correct.


Conditions (B) and (D) suggest applying pressure on chamber 2 (\(p_2\)), which would enhance forward osmosis, or use incorrect membranes.


Thus, A and C are correct.
Quick Tip: Reverse Osmosis requires \(P_{applied} > \pi\) on the concentrated side. SPM is mandatory.

Gibbs free energy change is given by \(\Delta G = \Delta H - T\Delta S\).


For spontaneity, \(\Delta G < 0\).


Given \(\Delta H > 0\) and \(\Delta S > 0\).


At equilibrium temperature \(T_e\), \(\Delta G = 0 \implies \Delta H = T_e \Delta S \implies T_e = \frac{\Delta H}{\Delta S}\).


For spontaneity (\(\Delta H - T\Delta S < 0\)):

\(T\Delta S > \Delta H\).

\(T > \frac{\Delta H}{\Delta S}\).


Therefore, \(T > T_e\).
Quick Tip: Endothermic reactions with positive entropy change become spontaneous at high temperatures (\(T > \Delta H / \Delta S\)).

Paramagnetism arises due to the presence of unpaired electrons in Molecular Orbitals (MO).


(A) \(O_2\) (16 electrons): Configuration ends in \(\pi^*2p_x^1 = \pi^*2p_y^1\). Two unpaired electrons. Paramagnetic.


(B) \(N_2\) (14 electrons): All electrons paired. Diamagnetic.


(C) \(F_2\) (18 electrons): All electrons paired. Diamagnetic.


(D) \(S_2\) (Valence shell like \(O_2\)): Like oxygen, vapor phase sulfur (\(S_2\)) has two unpaired electrons in antibonding \(\pi^*\) orbitals. Paramagnetic.


(E) \(Cl_2\) (Valence shell like \(F_2\)): All electrons paired. Diamagnetic.


Therefore, \(O_2\) and \(S_2\) are paramagnetic.
Quick Tip: Molecules with 10 or 16 valence electrons (like \(B_2, O_2, S_2\)) are typically paramagnetic according to MOT.

Intramolecular aldol condensation typically forms 5 or 6-membered rings due to stability (entropy and angle strain factors).


We need to check if the product can be formed by cyclization of a dicarbonyl precursor.


Option (1), (2), (3): These structures represent stable ring systems (often 6+6 or 5+6 fused) formed from diketones via dehydration.


Option (4): This structure typically involves a bridging or strained arrangement or a precursor that wouldn't favor that specific closure over other possibilities.


Without the exact visual structures in text, general chemical intuition for JEE questions on this topic points to 'bridged' or 'highly strained' products or products requiring formation of 3/4/7+ membered rings when 5/6 are possible as the "Not formed" answers.


Assuming standard option sets for this question type: Option (4) usually depicts a structure that violates Baldwin's rules or forms a less stable ring size compared to alternatives.
Quick Tip: Intramolecular aldol favors the formation of stable 5 and 6-membered rings.

Process is Isothermal (\(\Delta T = 0\)) for an ideal gas.


Therefore, Internal Energy change \(\Delta U = nC_v \Delta T = 0\).


Work done in reversible isothermal expansion: \(w = -nRT \ln(V_2/V_1)\).

\(w = -2.303 nRT \log(V_2/V_1)\).

\(n = 1\), \(R = 8.3\), \(T = 300\), \(V_2/V_1 = 20/10 = 2\).

\(w = -2.303 \times 1 \times 8.3 \times 300 \times 0.30\).

\(w \approx -2.3 \times 8.3 \times 90 = -1718\) J = \(-1.718\) kJ.


From First Law: \(\Delta U = q + w\).

\(0 = q + (-1.718) \implies q = 1.718\) kJ.


Order: \(\Delta U, q, w \rightarrow 0, 1.718 kJ, -1.718 kJ\).
Quick Tip: For ideal gas isothermal process, \(\Delta U = 0\) and \(q = -w\).

We need magnetic moment \(\mu = 5.96\) BM.

\(\mu = \sqrt{n(n+2)}\). For \(n=5\), \(\mu = \sqrt{35} \approx 5.92\). So we need 5 unpaired electrons (\(d^5\) high spin).


Also condition \(\Delta_0 = 0\) probably refers to CFSE = 0.


CFSE for High Spin \(d^5\) (\(t_{2g}^3 e_g^2\)) is \(3(-0.4\Delta_o) + 2(0.6\Delta_o) = 0\).


Let's analyze options:


(1) \([Fe(CN)_6]^{4-}\): \(Fe^{2+} (d^6)\). Strong ligand. Low spin \(t_{2g}^6\). \(n=0\).


(2) \([Co(NH_3)_6]^{3+}\): \(Co^{3+} (d^6)\). Strong ligand. Low spin \(t_{2g}^6\). \(n=0\).


(3) \([FeF_6]^4\) (assuming \([FeF_6]^{4-}\)): \(Fe^{2+} (d^6)\). Weak ligand. High spin \(t_{2g}^4 e_g^2\). \(n=4\). \(\mu = 4.9\). (CFSE \(\neq 0\)).


(4) \([Mn(SCN)_6]^4\) (assuming \([Mn(SCN)_6]^{4-}\)): \(Mn^{2+} (d^5)\). SCN is weak. High spin \(t_{2g}^3 e_g^2\). \(n=5\). \(\mu \approx 5.9\). CFSE = 0.


Option (4) satisfies all conditions.
Quick Tip: For High Spin \(d^5\) configuration, CFSE is zero and magnetic moment is maximum (~5.9 BM).

\(\Delta H = E_{a(forward)} - E_{a(backward)} = 180 - 200 = -20\) kJ/mol. (Statement 3 is false).


A catalyst lowers the activation energy of both forward and backward reactions by the same amount.


The enthalpy change \(\Delta H\) and Gibbs free energy change \(\Delta G\) are state functions determined by reactants and products, not the path.


Thus, a catalyst does not alter \(\Delta H\) or \(\Delta G\). (Statement 1 is true).


Catalysts cannot make a non-spontaneous reaction (\(\Delta G > 0\)) spontaneous. (Statement 2 is false).
Quick Tip: Catalysts affect kinetics (\(E_a\), rate) but not thermodynamics (\(\Delta H, \Delta G, K_{eq}\)).

Initial rate \(r_1 = k [A]^n [B]^m\).


New conditions: \([A'] = 2[A]\), \([B'] = \frac{1}{2}[B]\).


New rate \(r_2 = k [2A]^n [\frac{1}{2}B]^m = k \cdot 2^n [A]^n \cdot (\frac{1}{2})^m [B]^m\).

\(r_2 = k [A]^n [B]^m \cdot 2^n \cdot 2^{-m}\).

\(r_2 = r_1 \cdot 2^{n-m}\).


Ratio \(\frac{r_2}{r_1} = 2^{(n-m)}\).
Quick Tip: Substitute the scaling factors directly into the rate law to find the rate change factor.

Complex 1: \([CrCl_3(py)_3]\) is of type \(MA_3B_3\).


Octahedral \(MA_3B_3\) exhibits Geometrical Isomerism: Facial (fac) and Meridional (mer).


Both are optically inactive (planes of symmetry).


Total stereoisomers = 2 (Geometric).


Complex 2: \([CrCl_2(ox)_2]^{3-}\) is of type \(M(AA)_2a_2\) (where AA is bidentate oxalate).


Geometric isomers: Cis and Trans.


- Trans form: Optically inactive (plane of symmetry). (1 isomer).


- Cis form: Optically active (non-superimposable mirror image). Exists as d and l pair. (2 isomers).


Total stereoisomers = 1 (Trans) + 2 (Cis-d, Cis-l) = 3.


Answer is 2 and 3.
Quick Tip: \(MA_3B_3 \to\) Fac/Mer. \(M(AA)_2B_2 \to\) Cis (Active), Trans (Inactive).

Step 1: Reduction

Nitrobenzene is reduced to aniline using Sn/HCl: \[ C_6H_5NO_2 \xrightarrow{Sn/HCl} C_6H_5NH_2 \]

Step 2: Acetylation (Protection of –NH\(_2\) group)

Aniline is acetylated using acetic anhydride in pyridine to form acetanilide: \[ C_6H_5NH_2 \xrightarrow{Ac_2O / Py} C_6H_5NHCOCH_3 \]
This step reduces the activating nature of the amino group and prevents polysubstitution.

Step 3: Bromination

Acetanilide undergoes bromination in acetic acid: \[ Acetanilide \xrightarrow{Br_2/AcOH} p-Bromoacetanilide (major) \]
The amide group is ortho/para directing, and due to steric hindrance, para substitution predominates.

Step 4: Hydrolysis (Deprotection)

The acetyl group is removed on hydrolysis: \[ p-Bromoacetanilide \xrightarrow{NaOH (aq)} p-Bromoaniline \]

Hence, the major product formed is p-Bromoaniline.


Correct Answer: (2) Quick Tip: Acetylation of aniline protects the group and reduces reactivity, preventing poly-substitution and directing incoming electrophiles to the para position mainly.

In lead storage battery questions, the terms anode and \textit{cathode refer to the
physical battery plates as named during discharging. During charging, the reactions
are reversed, but the electrode names remain the same.


During discharging:

Anode (Pb plate): \[ Pb \rightarrow PbSO_4 \quad (0 \rightarrow +2) \]
Cathode (PbO\(_2\) plate): \[ PbO_2 \rightarrow PbSO_4 \quad (+4 \rightarrow +2) \]

During charging (reverse reactions):


At the anode (Pb plate):
\[ PbSO_4 + 2e^- \rightarrow Pb + SO_4^{2- \]
Oxidation state of Pb changes from \(+2\) to \(0\).


At the cathode (PbO\(_2\) plate):
\[ PbSO_4 + 2H_2O \rightarrow PbO_2 + SO_4^{2-} + 4H^+ + 2e^- \]
Oxidation state of Pb changes from \(+2\) to \(+4\).


Hence, \[ x_1 = +2,\quad y_1 = 0,\quad x_2 = +2,\quad y_2 = +4 \]

Correct Answer: (2) Quick Tip: In rechargeable batteries, the electrode that is the Anode during discharge becomes the Cathode during charge, and vice-versa. Reaction reverses.

Statement I: Nitrogen shows oxidation states from +1 to +5 in oxides (\(N_2O, NO, N_2O_3, NO_2, N_2O_5\)). This wide range and the stability of monomeric oxides are largely due to the ability of small Nitrogen atoms to form strong multiple \(p\pi - p\pi\) bonds with Oxygen. True.


Statement II: Nitrogen cannot expand its octet to form 5 bonds (like \(NCl_5\)) because it lacks vacant d-orbitals in its valence shell (n=2). Thus, \(NX_5\) halides do not exist. True.
Quick Tip: Nitrogen is restricted to a maximum covalency of 4 due to the absence of d-orbitals.

1. Benzene + Oleum \(\to\) Benzenesulfonic acid (X). (\(Ph-SO_3H\)).


2. \(Ph-SO_3H\) + molten NaOH, \(\Delta\), then \(H^+\) \(\to\) Phenol (Y). (\(Ph-OH\)). (Dow's process type fusion).


3. Phenol + Zn dust, \(\Delta\) \(\to\) Benzene (Z). (\(Ph-H\)). (Reduction of phenol).


Structure Z is Benzene.
Quick Tip: Zn dust reduces Phenol to Benzene.

Gly-Ala means Glycine (N-terminal) is bonded to Alanine (C-terminal).


Structure: \(H_2N-CH_2-CO-NH-CH(CH_3)-COOH\).


To form this with the elimination of HCl, we need an acyl chloride reacting with an amine.


The carbonyl group comes from Glycine, so it must be Glycyl chloride: \(NH_2-CH_2-COCl\).


The amine group comes from Alanine, so it must be Alanine: \(NH_2-CH(CH_3)-COOH\) (represented as \(NH_2-CH-COOH\) in options).


Reaction: \(Gly-COCl + H_2N-Ala \to Gly-CO-NH-Ala + HCl\).
Quick Tip: The N-terminal amino acid provides the carboxyl/acyl group for the peptide bond formation.

Atomic radii trend for Group 13: B < Ga < Al < In < Tl.


The anomaly is that Gallium (Ga, 135 pm) is smaller than Aluminum (Al, 143 pm) due to poor shielding by 3d electrons.


Therefore, the statement/pair \((Al < Ga)\) is incorrect.


The elements in this incorrect pair are Al (Z=13) and Ga (Z=31).


We must compare the ionic radii of their trivalent ions (\(M^{3+}\)).


Ionic radii follow the regular trend down the group: \(B^{3+} < Al^{3+} < Ga^{3+} < In^{3+} < Tl^{3+}\).

\(Al^{3+} \approx 54\) pm.

\(Ga^{3+} \approx 62\) pm.


Between Al and Ga, Ga has the higher ionic radius.


The atomic number of element X (Ga) is 31.
Quick Tip: Atomic radius anomaly: Ga < Al. Ionic radius trend: Al\(^{3+}\) < Ga\(^{3+}\).

1. X (\(C_3H_6O\)) is "not readily oxidised". This indicates a Ketone. X = Acetone (\(CH_3COCH_3\)).


2. Reduction of Acetone yields Isopropyl alcohol. Y = \(CH_3CH(OH)CH_3\).


3. Reaction of Y with HBr yields Isopropyl bromide. Z = \(CH_3CH(Br)CH_3\).


4. Grignard from Z is Isopropyl magnesium bromide (\(i-PrMgBr\)).


5. Reaction of \(i-PrMgBr\) with X (Acetone):


Nucleophilic addition of isopropyl group to carbonyl carbon of acetone.


Result: \((CH_3)_2C(OH)-CH(CH_3)_2\).


Name: 2,3-dimethylbutan-2-ol. This matches the product given.


Therefore, option (2) is correct.
Quick Tip: Ketones are resistant to mild oxidation (Tollens/Fehling), unlike aldehydes. Grignard + Ketone \(\to\) Tertiary Alcohol.

Step 1: Bromination of Methylcyclohexane (\(Br_2/h\nu\)). Selectivity favors tertiary hydrogen. Product is 1-bromo-1-methylcyclohexane.


Step 2: Dehydrohalogenation (Alc. KOH, \(\Delta\)). Elimination follows Zaitsev's rule to give the most substituted alkene. Product is 1-methylcyclohexene.


Step 3: Hydrobromination with Peroxide (HBr/ROOR). This is Anti-Markovnikov addition.


The Br adds to the less hindered carbon of the double bond (C2), and H adds to the more substituted carbon (C1) to proceed via the more stable tertiary radical intermediate at C1.


Structure: H on C1 (Methyl group), Br on C2.


Product: 1-bromo-2-methylcyclohexane.
Quick Tip: Anti-Markovnikov addition (Peroxide effect) works only with HBr. It directs Br to the less substituted carbon of the alkene.

Statement I: The conjugated enal system (\(C=C-C=O\)) allows for resonance contribution of polar forms (\(C^+-C=C-O^-\)), significantly increasing the dipole moment compared to the saturated aldehyde. True.


Statement II: In the conjugated system, the single bond between the alkene and carbonyl carbons (\(C_2-C_1\)) acquires partial double bond character due to resonance. This makes it shorter than the pure single bond found in the saturated analog. Thus, the bond length is smaller, not greater. False.
Quick Tip: Resonance imparts double bond character to single bonds connecting conjugated systems, shortening them.

\(V^{2+}\): \(Z=23\). Configuration \(3d^3\). Number of unpaired electrons \(n = 3\).

\(Co^{2+}\): \(Z=27\). Configuration \(3d^7\). In weak field/free ion, filling is \(t_{2g}^5 e_g^2\). Unpaired electrons \(n = 3\).


Since both have \(n=3\), they form the correct pair.


Check others: \(Ti^{2+}(d^2, n=2)\), \(Fe^{3+}(d^5, n=5)\), \(Cr^{2+}(d^4, n=4)\), \(Ti^{3+}(d^1, n=1)\), \(Mn^{2+}(d^5, n=5)\).
Quick Tip: For d-block ions, remove s-electrons first. For \(d^1-d^5\), \(n=\) number of electrons. For \(d^6-d^{10}\), \(n=10-\) number of electrons.

(1) Probability density \(\psi^2\) for 1s orbital (\(\psi \propto e^{-r/a_0}\)) is maximum at \(r=0\). Correct.


(2) The wavefunction is non-zero everywhere (exponential decay). Finding it at \(2a_0\) is possible. Correct.


(3) S-orbitals are spherical. Correct.


(4) The total energy of an electron in a stationary state (like 1s) is constant \((-13.6 eV)\) and does not depend on position. The statement that energy is maximum at a specific distance is incorrect.
Quick Tip: Distinguish between Probability Density (\(\psi^2\), max at nucleus) and Radial Probability Distribution (\(4\pi r^2 \psi^2\), max at \(a_0\)). Total energy is constant.

1. Pressure of dry \(N_2\): \(P = P_{total} - P_{aq} = 900 - 15 = 885\) mm Hg.


2. Convert to STP conditions (\(P_0 = 760\) mm, \(T_0 = 273\) K):

\(V_{STP} = V \times \frac{P}{P_0} \times \frac{T_0}{T} = 150 \times \frac{885}{760} \times \frac{273}{300}\).

\(V_{STP} = 150 \times 1.1645 \times 0.91 = 158.95\) mL.


3. Moles of \(N_2\): \(n = \frac{158.95}{22400} = 0.0071\) mol.


4. Mass of \(N_2\): \(0.0071 \times 28 = 0.1988\) g.


5. Percentage: \(\frac{0.1988}{1 g} \times 100 = 19.88%\).


Nearest integer is 20.
Quick Tip: Remember to correct for aqueous tension before applying gas laws.

1. Reaction of \(KMnO_4\) in acid: \(MnO_4^- (+7) \to Mn^{2+} (+2)\).


Difference \(X = 7 - 2 = 5\).


2. Acetate test: Ferric chloride reacts with acetate to form a complex, which on heating gives a precipitate of Basic Ferric Acetate / Ferric Hydroxide. The metal ion is Iron(III) (\(Fe^{3+}\)).


Electron configuration of \(Fe^{3+}\): Argon core plus \(3d^5\).


Number of d-electrons \(Y = 5\).


3. Sum \(X + Y = 5 + 5 = 10\).
Quick Tip: \(Fe^{3+}\) is a \(d^5\) system. Mn reduction in acid involves 5 electrons.

1. Required Iron: 12 ppm in 150 kg.


Mass of Fe \(= 12 \times 10^{-6} \times 150 \times 10^3 g = 1.8\) g.


2. Molar mass of \(FeSO_4 \cdot 7H_2O = 56 + 32 + 64 + 7(18) = 152 + 126 = 278\) g/mol.


3. Proportion: 56 g Fe is in 278 g salt.


1.8 g Fe is in \(\frac{278}{56} \times 1.8\) g salt.


Mass \(= 4.964 \times 1.8 = 8.935\) g.


Nearest integer is 9.
Quick Tip: 1 ppm = 1 mg/kg.

New pH = 6 \(\implies [H^+] = 10^{-6}\) M.


For a weak acid, \([H^+] \approx \sqrt{K_a C}\).

\(10^{-6} = \sqrt{4 \times 10^{-10} \times C}\).


Squaring both sides: \(10^{-12} = 4 \times 10^{-10} \times C\).

\(C = \frac{10^{-12}}{4 \times 10^{-10}} = \frac{1}{4} \times 10^{-2} = 0.25 \times 10^{-2}\) M.

\(C = 2.5 \times 10^{-3}\) M = \(25 \times 10^{-4}\) M.


Value of \(x = 25\).
Quick Tip: For weak electrolytes, concentration is inversely proportional to the square of the degree of dissociation (or related pH shift), \(C = [H^+]^2/K_a\).

Count the bases:


G: 4 occurrences.


C: 3 occurrences.


A: 4 occurrences.


T: 2 occurrences.


G-C pairs have 3 H-bonds. Total G and C bases = 7. Pairs = 7. H-bonds = \(7 \times 3 = 21\).


A-T pairs have 2 H-bonds. Total A and T bases = 6. Pairs = 6. H-bonds = \(6 \times 2 = 12\).


Total Hydrogen Bonds = \(21 + 12 = 33\).
Quick Tip: Number of H-bonds = \(3 \times (Number of G-C pairs) + 2 \times (Number of A-T pairs)\).