The electrostatic energy of Z protons uniformly distributed throughout a spherical nucleus of radius R is given by $E=\frac{3}{5}\frac{Z\left(Z-1\right)e^2}{4\pi {\epsilon }_{0}R}$ The measured masses of the neutron, ${}_1{}^{1}H, {}_8{}^{15}O$ are 1.008665 u, 1.007825 u, 15.000109 u and 15.003065 u respectively. Given that the radii of both the ${}_7{}^{15}N and {}_8{}^{15}O$ nuclei are same, 1u = 931.5 Me V/ $c^2$ (c is the speed of light) and $\frac{e^2}{\left(4\pi {ϵ}_0\right)}=1.44 MeV fm.$ Assuming that the difference between the binding energies of ${}_7{}^{15}N and {}_8{}^{15}O$ is purely due to the electrostatic energy, the radius of either of the nuclei is $\left(1 fm={10}^{-15}m\right)$

A rod of mass $m$ and length $L,$ pivoted at one of its ends, is hanging vertically. A bullet of the same mass moving at speed $v$ strikes the rod horizontally at a distance $x$ from its pivoted end and gets embedded in it. The combined system now rotates with an angular speed $\omega$ about the pivot. The maximum angular speed ${\omega }_{\mathrm{M}}$ is achieved for $x=x_{\mathrm{M}}$. Then

A thin flexible wire of length \(L\) is connected to two adjacent fixed points and carries a current \(I\) in the clockwise direction, as shown in the figure. When the system is put in a uniform magnetic field of strength \(B\) going into the plane of the paper, the wire takes the shape of a circle. The tension in the wire is

Consider a uniform spherical charge distribution of radius $R_1$ centred at the origin O. In this distribution, a spherical cavity of radius $R_2$ , centred at $P$ with distance $OP=a=R_1-R_2$ (see figure) is made. If the electric field inside the cavity at position $\overrightarrow{r}$ is $\overrightarrow{E}\left(\overrightarrow{r}\right)$ , then the correct statement(s) is (are)

A hydrogen atom, initially at rest in its ground state, absorbs a photon of frequency \(v_1\) and ejects the electron with a kinetic energy of 10 eV . The electron then combines with a positron at rest to form a positronium atom in its ground state and simultaneously emits a photon of frequency \(v_2\). The center of mass of the resulting positronium atom moves with a kinetic energy of 5 eV . It is given that positron has the same mass as that of electron and the positronium atom can be considered as a Bohr atom, in which the electron and the positron orbit around their center of mass. Considering no other energy loss during the whole process, the difference between the two photon energies (in eV ) is \(\qquad\)

\(5.6 \mathrm{~L}\) of helium gas at STP is adiabatically compressed to \(0.7 \mathrm{~L}\). Taking the initial temperature to be \(T_1\), the work done in the process is

In the circuit shown, initially there is no charge on capacitors and keys ${\mathrm{S}}_1$ and ${\mathrm{S}}_2$ are open. The values of the capacitors are ${\mathrm{C}}_1=10\mathrm{\mu }\mathrm{F},{\mathrm{C}}_2=30\mathrm{\mu }\mathrm{F}$ and ${\mathrm{C}}_3={\mathrm{C}}_4=80\mathrm{\mu }\mathrm{F}.$

Which of the statement(s) is/are correct?

Let $x\in R$ and let $P=\left[\begin{array}{ccc}1& 1& 1\\ 0& 2& 2\\ 0& 0& 3\end{array}\right], Q=\left[\begin{array}{ccc}2& x& x\\ 0& 4& 0\\ x& x& 6\end{array}\right]$ and $R=PQ{P}^{-1}.$
Then which of the following options is/are correct?

For any positive integer $n$, define $f_n:\left(0, \infty \right)\to R$ as $f_n\left(x\right)={\sum }_{i=1}^n{\tan }^{-1}\left(\frac{1}{1+\left(x+j\right)\left(x+j-1\right)}\right)$ for all $x\in \left(0,\infty \right)$ . $($Here, the inverse trigonometric function ${\tan }^{-1}x$ assumes values in $\left(-\frac{\pi }{2}, \frac{\pi }{2}\right)$$)$ Then, which of the following statement(s) is (are) TRUE?

For an isosceles prism of angle $A$ and refractive index $\mu$ it is found that, the angle of minimum deviation is ${ \delta }_m=A$. Which of the following options is/are correct?

In the scheme given below, \(\text X\) and \(\text Y\), respectively, are
Metal halide \(\quad \xrightarrow{\text { aq. NaOH} } \quad\) White precipitate \((\text P )+\) Filtrate \((\text Q )\)
\(P \xrightarrow[\text { heat }]{\substack{\text { aq. } H _2 SO _4 \\ PbO _2 \text { (excess) }}}\text X\) (a coloured species in solution)
\(Q \xrightarrow[\text { warm }]{ MnO ( OH )_2, \text { Conc. } H _2 SO _4}\text Y\) (gives blue-coloration with KI-starch paper)

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Consider the polynomial \(f(x)=1+2 x+3 x^2+4 x^3\). Let \(s\) be the sum of all distinct real roots of \(f(x)\) and let \(t=|s|\).Question:
The function \(f^{\prime}(x)\) is

An electron in a hydrogen atom undergoes a transition from an orbit with a quantum number $n_i$ to another quantum number $n_f$ . $V_i \text{and} V_f$ are the initial and final potential energies of the electron. If $\frac{V_i}{V_f}=6.25$ then the smallest $n_f$ is