Match the nuclear processes given in column I with the appropriate option(s) in column II.

	Column I		Column II
A.	Nuclear fusion	P	Absorption of thermal neutrons by ${}_{92}{}^{235}\mathrm{U}$
B.	Fission in a nuclear reactor	Q	${}_{27}{}^{60}\mathrm{C}\mathrm{o}$ nucleus
C.	$\mathrm{\beta }$ -decay	R	Energy production in stars via hydrogen conversion to helium
D.	$\mathrm{\gamma }$ -ray emission	S	Heavy water
		T	Neutrino emission

A uniform circular disc of mass $\text{50 kg}$ and radius $\text{0.4 m}$ is rotating with an angular velocity of ${\text{10 rad s}}^{-1}$ about its own axis, which is vertical. Two uniform circular rings, each of mass $\text{6.25 kg}$ and radius $\text{0.2 m}$, are gently placed symmetrically on the disc in such a manner that they are touching each other along the axis of the disc and are horizontal. Assume that the friction is large enough such that the rings are at rest relative to the disc and the system rotates about the original axis. The new angular velocity $\left({in\; rad\; s}^{-1}\right)$ of the system is

A table tennis ball has radius \((3 / 2) \times 10^{-2} \mathrm{~m}\) and mass \((22 / 7) \times 10^{-3} \mathrm{~kg}\). It is slowly pushed down into a swimming pool to a depth of \(d=0.7 \mathrm{~m}\) below the water surface and then released from rest. It emerges from the water surface at speed \(v\), without getting wet, and rises up to a height \(H\). Which of the following option(s) is(are) correct?
[Given: \(\pi=22 / 7, g=10 \mathrm{~m} \mathrm{~s}^{-2}\), density of water \(=1 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}\), viscosity of water \(=1 \times 10^{-3} \mathrm{~Pa}\)-s.]

A conducting solid sphere of radius \(R\) and mass \(M\) carries a charge \(Q\). The sphere is rotating about an axis passing through its center with a uniform angular speed \(\omega\). The ratio of the magnitudes of the magnetic dipole moment to the angular momentum about the same axis is given as \(\alpha \frac{Q}{2 M}\). The value of \(\alpha\) is _______ .

Two solid spheres \(A\) and \(B\) of equal volumes but of different densities \(d_A\) and \(d_B\) are connected by a string. They are fully immersed in a fluid of density \(d_F\). They get arranged into an equilibrium state as shown in the figure with a tension in the string. The arrangement is possible only if

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Light guidance in an optical fiber can be understood by considering a structure comprising of thin solid glass cylinder of refractive index \(n_{1}\) surrounded by a medium of lower refractive index \(n_{2}\). The light guidance in the structure takes place due to successive total internal reflections at the interface of the media \(n_{1}\) and \(n_{2}\) as shown in the figure. All rays with the angle of incidence \(i\) less than a particular value \(i_{m}\) are confined in the medium of refractive index \(n_{1}\). The numerical aperture (NA) of the structure is defined as \(\sin i_{m}\).



Question:

If two structures of same cross-sectional area, but different numerical apertures \(N A_{1}\) and \(N A_{2}\left(N A_{2} < N A_{1}\right)\) are joined longitudinally, the numerical aperture of the combined structure is

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The figure shows a surface \(X Y\) separating two transparent media, medium-1 and medium-2. The lines \(a b\) and \(c d\) represent wavefronts of a light wave travelling in medium-1 and incident on \(X Y\). The lines ef and \(g h\) represent wavefronts of the light wave in medium-2 after refraction.
Question:
Light travels as a

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The noble gases have closed-shell electronic configuration and are monoatomic gases under normal conditions. The low boiling points of the lighter noble gases are due to weak dispersion forces between the atoms and the absence of other interatomic interactions.
The direct reaction of xenon with fluorine leads to a series of compounds with oxidation numbers \(+2,+4\) and \(+6 . \mathrm{XeF}_4\) reacts violently with water to give \(\mathrm{XeO}_3\). The compounds of xenon exhibit rich stereochemistry and their geometries can be deduced considering the total number of electron pairs in the valence shell.
Question:
The structure of \(\mathrm{XeO}_3\) is

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A special metal \(S\) conducts electricity without any resistance. A closed wire loop, made of \(S\), does not allow any change in flux through itself by inducing a suitable current to generate a compensating flux. The induced current in the loop cannot decay due to its zero resistance. This current gives rise to a magnetic moment which in turn repels the source of magnetic field or flux. Consider such a loop, of radius \(a\), with its center at the origin. A magnetic dipole of moment \(m\) is brought along the axis of this loop from infinity to a point at distance \(r(\gg a)\) from the center of the loop with its north pole always facing the loop, as shown in the figure below.

The magnitude of magnetic field of a dipole \(m\), at a point on its axis at distance \(r\), is \(\frac{\mu_{0}}{2 \pi} \frac{m}{r^{3}}\), where \(\mu_{0}\) is the permeability of free space. The magnitude of the force between two magnetic dipoles with moments, \(m_{1}\) and \(m_{2}\), separated by a distance \(r\) on the common axis, with their north poles facing each other, is \(\frac{k m_{1} m_{2}}{r^{4}}\), where \(k\) is a constant of appropriate dimensions. The direction of this force is along the line joining the two dipoles.

Question :
The work done in bringing the dipole from infinity to a distance$r$from the centre of the loop by the given process is proportional to:

Considering only the principal values of the inverse trigonometric functions, the value of $\frac{3}{2}{\cos }^{-1}\sqrt{\frac{2}{2+{\pi }^2}}+\frac{1}{4}{\sin }^{-1}\frac{2\sqrt{2}\pi }{2+{\pi }^2}+{\tan }^{-1}\frac{\sqrt{2}}{\pi }$ is _______. (If the numerical value has more than two decimal places, truncate/round-off the value to TWO decimal places)  

\(\mathrm{B}(\mathrm{OH})_3+\mathrm{NaOH} \rightleftharpoons \mathrm{NaBO}_2+\mathrm{Na}[\mathrm{B}\)\((\mathrm{OH})_4]+\mathrm{H}_2 \mathrm{O}\).
How can this reaction is made to proceed in forward direction?

The value of \(\int \frac{\left(x^2-1\right) d x}{x^3 \sqrt{2 x^4-2 x^2+1}}\) is

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The amount of energy required to break a bond is same as the amount of energy released when the same bond is formed. In gaseous state, the energy required for homolytic cleavage of a bond is called Bond Dissociation Energy (BDE) or Bond Strength. BDE is affected by \(s\)-character of the bond and the stability of the radicals formed. Shorter bonds are typically stronger bonds. BDEs for some bonds are given below:





Question:
\(\mathrm{CH}_{4}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \stackrel{\text { light }}{\longrightarrow} \mathrm{CH}_{3} \mathrm{Cl}(\mathrm{g})+\mathrm{HCl}(\mathrm{g})\) the correct statement is