JEE Advanced Questions & Instant Answers

Let $u = 1 + x^2,\ du = 2x\,dx$. The integral becomes $\tfrac12\int_1^2 \tfrac{du}{u} = \tfrac12[\ln u]_1^2 = \tfrac12\ln 2 \approx 0.347$.

Using $\sum_{n=1}^{\infty} n x^n = \dfrac{x}{(1-x)^2}$ for $|x|<1$, put $x = \tfrac12$: $\dfrac{1/2}{(1/2)^2} = \dfrac{1/2}{1/4} = 2$.

$P = \dfrac{\binom{4}{2}\binom{6}{1}}{\binom{10}{3}} = \dfrac{6 \times 6}{120} = \dfrac{36}{120} = 0.3$.

For $x>0$, $e^x$ grows faster than $x^2$ and $e^x > x^2$ throughout (they never meet). For $x<0$, $x^2$ is a large positive parabola while $e^x$ decays from $1$ to $0$, so they cross exactly once (near $x \approx -0.70$). Total: one real solution.

$z$ lies on a circle of radius $2$ centred at the origin. The farthest point from $3$ (a point $3$ units along the real axis) is diametrically opposite, at distance $|{-3}| + 2$… more precisely $\max|z-3| = |3| + |z| = 3 + 2 = 5$.

Lens formula $\tfrac1f = \tfrac1v - \tfrac1u$ with sign convention $u = -60,\ f = +20$: $\tfrac1v = \tfrac{1}{20} + \tfrac{1}{-60} = \tfrac{3-1}{60} = \tfrac{2}{60} = \tfrac{1}{30}$, so $v = +30$ cm (real image on the far side).

Einstein's photoelectric equation: $KE_{\max} = h\nu - \phi = 5 - 2 = 3\text{ eV}$.

$H = \dfrac{u^2 \sin^2\theta}{2g} = \dfrac{20^2 \times (0.5)^2}{2 \times 10} = \dfrac{400 \times 0.25}{20} = \dfrac{100}{20} = 5$ m.

$F = \dfrac{k q_1 q_2}{r^2} = \dfrac{9\times10^9 \times (2\times10^{-6})^2}{2^2} = \dfrac{9\times10^9 \times 4\times10^{-12}}{4} = 9\times10^{-3}\ \text{N} = 9$ mN.

Acceleration $a = \dfrac{g\sin\theta}{1 + I/mR^2}$. The solid sphere has smaller $I/mR^2$ ($2/5$ vs $2/3$ for the hollow shell), giving larger acceleration — so it arrives first, regardless of mass, radius, or angle.

$\mathrm{Fe}^{3+}$ is $d^5$; with the weak-field ligand $\mathrm{H_2O}$ it is high-spin, giving $5$ unpaired electrons. $\mu = \sqrt{n(n+2)} = \sqrt{5\times7} = \sqrt{35} \approx 5.92$ BM.

At equilibrium the Gibbs free energy change of the actual process is zero, $\Delta G = 0$. The standard value $\Delta G^{\circ} = -RT\ln K$ is generally non-zero (it equals zero only when $K=1$).

$\Delta n = (\text{moles gaseous products}) - (\text{moles gaseous reactants}) = 2 - (1+3) = -2$. Hence $K_p = K_c (RT)^{-2}$.

By the Nernst equation, as reactants are consumed and products build up, the reaction quotient $Q$ rises toward $K$, so $E_{cell} = E^{\circ} - \tfrac{RT}{nF}\ln Q$ falls, reaching $0$ at equilibrium (a dead battery).

Acid-catalysed hydration with $\mathrm{HgSO_4}$ adds water with Markovnikov orientation; the enol tautomerizes to the methyl ketone. Hydroboration–oxidation (B) gives anti-Markovnikov addition → an aldehyde.

Moles $= M \times V = 0.5 \times 0.5 = 0.25\ \text{mol}$. Mass $= 0.25 \times 40 = 10$ g.

For a skew-symmetric matrix, $\det(A) = \det(A^T) = \det(-A) = (-1)^n\det(A)$. For odd $n=3$ this forces $\det(A) = -\det(A)$, so $\det(A) = 0$.

KE $= \tfrac12 k(A^2 - x^2)$, PE $= \tfrac12 kx^2$. Setting them equal: $A^2 - x^2 = x^2 \Rightarrow x^2 = A^2/2 \Rightarrow x = A/\sqrt2$.

$l=2$ is the $d$ subshell, with $m_l \in \{-2,-1,0,1,2\}$ — five orbitals, each holding $2$ electrons, so $10$ electrons maximum.

$\cos\theta = \dfrac{\vec a\cdot\vec b}{|\vec a||\vec b|} = \dfrac{0+1+0}{\sqrt2 \cdot \sqrt2} = \dfrac{1}{2}$, so $\theta = 60^{\circ}$.