The entropy versus temperature plot for phases $\alpha$ and $\beta$ at $1$ bar pressure is given. ${\mathrm{S}}_{\mathrm{T}}$ and ${\mathrm{S}}_0$ are entropies of the phases at temperatures $\mathrm{T}$ and $0 \mathrm{K}$, respectively. The transition temperature for $\alpha$ to $\beta$ phase change is $600 \mathrm{K}$ and ${\mathrm{C}}_{\mathrm{p},\mathrm{\beta }}-{\mathrm{C}}_{\mathrm{p},\mathrm{\alpha }}=1 \mathrm{J} {\mathrm{mol}}^{-1} {\mathrm{K}}^{-1}$. Assume $\left({\mathrm{C}}_{\mathrm{p},\mathrm{\beta }}-{\mathrm{C}}_{\mathrm{p},\mathrm{\alpha }}\right)$ is independent of temperature in the range of $200$ to $700 \mathrm{K}.{\mathrm{C}}_{\mathrm{p},\mathrm{\alpha }}$ and ${\mathrm{C}}_{\mathrm{p},\mathrm{\beta }}$ are heat capacities of $\alpha$ and $\beta$ phases, respectively.
The value of entropy change, ${\mathrm{S}}_{\beta }-{\mathrm{S}}_{\alpha }$ (in ${\mathrm{Jmol}}^{-1} {\mathrm{K}}^{-1}$ ), at $300 \mathrm{K}$ is [Use: $\ln 2=0.69$0. Given: ${\mathrm{S}}_{\beta }-{\mathrm{S}}_{\alpha }=0$ at $0 \mathrm{K}$]

The term that corrects for the attractive forces present in a real gas in the van der Waals' equation is

A plot of the number of neutrons $(\text{N})$ against the number of protons $(\text{P})$ of the stable nuclei exhibits upwards deviation from linearity for atomic number, $\text{Z}>20$. For an unstable nucleus having $\frac{\text{N}}{\text{P}}$ ratio less than $1$, the possible mode(s) of decay is (are):

Among the triatomic molecules/ions, $\mathrm{B}\mathrm{e}\mathrm{C}{\mathrm{l}}_2,\mathrm{ }{\mathrm{N}}_3^{-},\mathrm{ }{\mathrm{N}}_2\mathrm{O},\mathrm{ }{\mathrm{N}\mathrm{O}}_2^{+},\mathrm{ }{\mathrm{O}}_3,\mathrm{S}\mathrm{C}{\mathrm{l}}_2,\mathrm{ }{\mathrm{I}\mathrm{C}\mathrm{l}}_2^{-},\mathrm{ }{\mathrm{I}}_3^{-}$ and $\mathrm{X}\mathrm{e}{\mathrm{F}}_2$ , the total number of linear molecule(s)/ion(s) where the hybridization of the central atom does not have contribution from the d-orbital(s) is
[Atomic number: S = 16, Cl = 17, I = 53 and Xe = 54]

For ${\mathrm{He}}^{+}$, a transition takes place from the orbit of radius $105.8\mathrm{pm}$ to the orbit of radius $26.45\mathrm{pm}$. The wavelength (in $\mathrm{nm}$ ) of the emitted photon during the transition is
Bohr radius, $\mathrm{a}=52.9 \mathrm{pm}$
Rydberg constant, ${\mathrm{R}}_{\mathrm{H}}=2.2\times {10}^{-18} \mathrm{J}$
Planck's constant, $\mathrm{h}=6.6\times {10}^{-34}\mathrm{Js}$
Speed of light, $\mathrm{c}=3\times {10}^8 {\mathrm{ms}}^{-1}$ ]

${\mathit{\text{EDTA}}}^{4-}$ is ethylenediaminetetraacetate ion. The total number of N - Co - O bond angles in ${\left[\mathit{\text{Co}}\left(\mathit{\text{EDTA}}\right)\right]}^{1-}$ complex ion is

Paragraph:
The hydrogen-like species \(\mathrm{Li}^{2+}\) is in a spherically symmetric state \(S_1\) with one radial node. Upon absorbing light the ion undergoes transition to a state \(S_2\). The state \(S_2\) has one radial node and its energy is equal to the ground state energy of the hydrogen atom.
Question:
The orbtial angular momentum quantum number of the state \(S_2\) is

Most materials have the refractive index, \(n>1\). So, when a light ray from air enters a naturally occurring material, then by Snell's

law, \(\frac{\sin \theta_{1}}{\sin \theta_{2}}=\frac{n_{2}}{n_{1}}\), it is understood that the refracted ray bends towards the normal. But it never emerges on the same side of the normal as the incident ray. According to electromagnetism, the refractive index of the medium is given by the relation, \(\mathrm{n}=c / v=\pm \sqrt{\varepsilon_{r} \mu_{r}}\), where \(c\) is the speed of electromagnetic waves in vacuum, \(v\) its speed in the medium, \(\varepsilon_{r}\) and \(\mu_{r}\) are the relative permittivity and permeability of the medium respectively. In normal materials, both \(\varepsilon_{r}\) and \(\mu_{r}\), are positive, implying positive \(n\) for the medium. When both \(\varepsilon_{r}\) and \(\mu_{r}\) are negative, one must choose the negative root of \(n\). Such negative refractive index materials can now be artificially prepared and are called metamaterials. They exhibit significantly different optical behavior, without violating any physical laws. Since \(n\) is negative, it results in a change in the direction of propagation of the refracted light. However, similar to normal materials, the frequency of light remains unchanged upon refraction even in meta-materials.
Question : Choose the correct statement.

A student performs an experiment for determination of \(g\left(=\frac{4 \pi^2 l}{T^2}\right), l \approx 1 \mathrm{~m}\), and he commits an error of \(\Delta l\). For \(T\) he takes the time of \(n\) oscillations with the stop watch of least count \(\Delta T\) and he commits a human error of \(0.1 \mathrm{~s}\). For which of the following data, the measurement of \(g\) will be most accurate?

The correct statements about the oxoacids, ${\mathrm{HClO}}_4\mathrm{ }\mathrm{and}\mathrm{ }\mathrm{HClO}$ are,

Column I contains a list of processes involving expansion of an ideal gas. Match this with Column II describing the thermodynamic change during this process. Indicate your answer by darkening the appropriate bubbles of the \(4 \times 4\) matrix given in the ORS.

Let ${\text{S}}_{\text{n}}=\underset{\text{k}=1}{\overset{4\text{n}}{\Sigma }}{\left(-1\right)}^{\frac{\text{k}\left(\text{k}+1\right)}{2}}{\text{k}}^2$. Then S_n can take value(s)

Concentration of ${\mathrm{H}}_2{\mathrm{SO}}_4$ and ${\mathrm{Na}}_2{\mathrm{SO}}_4$ in a solution is $1\mathrm{M}$ and $1.8\times {10}^{-2}\mathrm{M}$, respectively. Molar solubility of ${\mathrm{PbSO}}_4$ in the same solution is $\mathrm{X}\times {10}^{-\mathrm{Y}}\mathrm{M}$ (expressed in scientific notation). The value of $\mathrm{Y}$ is
[Given: Solubility product of ${\mathrm{PbSO}}_4\left({\mathrm{K}}_{\mathrm{sp}}\right)=1.6\times {10}^{-8}$. For ${\mathrm{H}}_2{\mathrm{SO}}_4, {\mathrm{K}}_{\mathrm{a}1}$ is very large and ${\mathrm{K}}_{\mathrm{a}2}=1.2\times {10}^{-2}$]
If the numerical value has more than two decimal places, truncate/round-off the value to TWO decimal places.