${}_{92}{}^{238}\mathrm{U}$ is known to undergo radioactive decay to form ${}_{82}{}^{206}\mathrm{Pb}$ by emitting alpha and bet a particles. A rock initially contained $68\times {10}^{-6}\mathrm{g}$ of ${}_{92}{}^{238}\mathrm{U}$. If the number of alpha particles that it would emit during its radioactive decay of ${}_{92}{}^{238}\mathrm{U}$ to ${}_{82}{}^{206}\mathrm{Pb}$ in three half-lives is $\mathrm{Z}\times {10}^{18}$, then what is the value of $\mathrm{Z}$?

At \(298 \mathrm{~K}\), the limiting molar conductivity of a weak monobasic acid is \(4 \times 10^{2} \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\). At \(298 \mathrm{~K}\), for an aqueous solution of the acid the degree of dissociation is \(\boldsymbol{\alpha}\) and the molar conductivity is \(\mathbf{y} \times 10^{2} \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\). At \(298 \mathrm{~K}\), upon 20 times dilution with water, the molar conductivity of the solution becomes \(3 \mathbf{y} \times 10^{2} \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\)
The value of $\mathrm{\alpha }$ is _______ .

For 'invert sugar', the correct statement(s) is (are)
(Given: specific rotations of $\left(+\right)-$sucrose, $\left(+\right)-$maltose, $\mathrm{L}-\left(-\right)-$glucose and $\mathrm{L}-\left(+\right)-$fructose in aqueous solution are $+\mathrm{ }{66}^{°},\mathrm{ }+{140}^{°},\mathrm{ }–{52}^{°}$ and $+\mathrm{ }{92}^{°}$, respectively)

Dissolving $1.24 \mathrm{g}$ of white phosphorous in boiling $\mathrm{NaOH}$ solution in an inert atmosphere gives a gas $\mathrm{Q}$. The amount of ${\mathrm{CuSO}}_4$ (in $\mathrm{g}$) required to completely consume the gas $\mathrm{Q}$ is____
[Given: Atomic mass of
\(\text H =1,\text O =16, \text{Na} =23,\text P =31,\text S=32,\) \(\text{Cu} =63]\)

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When a metal rod \(M\) is dipped into an aqueous colourless concentrated solution of compound \(\mathrm{N}\), the solution turns light blue. Addition of aqueous \(\mathrm{NaCl}\) to the blue solution gives a white precipitate \(\mathrm{O}\). Addition of aqueous \(\mathrm{NH}_3\) dissolves \(\mathrm{O}\) and gives an intense blue solution.
Question:
The metal \(\operatorname{rod} M\) is

Among the complex ions,
\(\left[\mathrm{Co}\left(\mathrm{NH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{NH}_{2}\right)_{2} \mathrm{Cl}_{2}\right]^{+},\)\(\left[\mathrm{CrCl}_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}\right]^{3-},\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}(\mathrm{OH})_{2}\right]^{+}\), \(\left[\mathrm{Fe}\left(\mathrm{NH}_{3}\right)_{2}(\mathrm{CN})_{4}\right]^{-},\)\(\left[\mathrm{Co}(\mathrm{NH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{NH}_{2})_{2}\left(\mathrm{NH}_{3}\right) \mathrm{Cl}\right]^{2+}\) and \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4}\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{Cl}\right]^{2+}\), the number of complex ion(s) that show(s) cis-trans isomerism is

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\(RCONH _2\) is converted into \(RNH _2\) by means of Hofmann bromamide degradation.

In this reaction, \(R \mathrm{CONHBr}\) is formed from which this reaction has derived its name. Electron donating group at phenyl activates the reaction. Hofmann degradation reaction is an intramolecular reaction.
Question:
How can the conversion of (i) to (ii) be brought about?

At time $t=0$, a disk of radius $1 \mathrm{m}$ starts to roll without slipping on a horizontal plane with an angular acceleration of $\alpha =\frac{2}{3} \mathrm{rad} {\mathrm{s}}^{-2}$. A small stone is stuck to the disk. At $t=0$, it is at the contact point of the disk and the plane. Later, at time $t=\sqrt{\pi } \mathrm{s}$, the stone detaches itself and flies off tangentially from the disk. The maximum height (in $\mathrm{m}$) reached by the stone measured from the plane is $\frac{1}{2}+\frac{x}{10}$. The value of $x$ is [Take $g=10 \mathrm{m} {\mathrm{s}}^{-2}$.] If the numerical value has more than two decimal places, truncate/round-off the value to TWO decimal places.

List-I includes starting materials and reagents of selected chemical reactions. List-II gives structures of compounds that may be formed as intermediate products and/or final products from the reactions of List-I.

List-I	List-II
(I)	$\left(\mathrm{P}\right)$
(II)	$\left(\mathrm{Q}\right)$
(III)	$\left(\mathrm{R}\right)$
(IV)	$\left(\mathrm{S}\right)$
	$\left(\mathrm{T}\right)$
	$\left(\mathrm{U}\right)$

Which of the following options has correct combination considering List-I and List-II?

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A wooden cylinder of diameter \(4 r\), height \(H\) and density \(\rho / 3\) is kept on a hole of diameter \(2 r\) of a tank, filled with liquid of density \(\rho\) as shown in the figure.

Question :
Now level of the liquid starts decreasing slowly. When the level of liquid is at a height \(h_1\) above the cylinder, the block starts moving up. At what value of \(h_1\), will the block rise?

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Consider the lines: \(L_1: \frac{x+1}{3}=\frac{y+2}{1}=\frac{z+1}{2}, L_2: \frac{x-2}{1}=\frac{y+2}{2}=\frac{z-3}{3}\)Question:
The distance of the point \((1,1,1)\) from the plane passing through the point \((-1,-2,-1)\) and whose normal is perpendicular to both the lines \(L_1\) and \(L_2\), is

Column I shows four situations of standard Young's double slit arrangement with the screen placed far away from the slits \(S_1\) and \(S_2\). In each of these cases \(S_1 P_0=S_2 P_0\), \(S_1 P_1-S_2 P_1=\frac{\lambda}{4}\) and \(S_1 P_2-S_2 P_2=\frac{\lambda}{3}\), where \(\lambda\) is the wavelength of the light used. In the cases \(\mathrm{B}, \mathrm{C}\) and \(\mathrm{D}\), a transparent sheet of refractive index \(\mu\) and thickness \(t\) is pasted on slit \(S_2\). The thickness of the sheets are different in different cases. The phase difference between the light waves reaching a point \(P\) on the screen from the two slits is denoted by \(\delta(P)\) and the intensity by \(I(P)\). Match each situation given in Column-I with the statement(s) in Column-II valid for that situation.

The cyanide process of gold extraction involves leaching out gold from its ore with ${\mathrm{C}\mathrm{N}}^{-}$ in the presence of $\mathrm{Q}$ in water to form $\mathrm{R}$ . Subsequently, $\mathrm{R}$ is treated with $\mathrm{T}$ to obtain $\mathrm{A}\mathrm{u}$ and $\mathrm{Z}$ . Choose the correct option(s).