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The nuclear charge \((\mathrm{Ze})\) is non-uniformly distributed within a nucleus of radius \(R\). The charge density \(\rho(r)\) [charge per unit volume] is dependent only on the radial distance \(r\) from the centre of the nucleus as shown in figure. The electric field is only along the radial direction.

Question:
The electric field within the nucleus is generally observed to be linearly dependent on \(r\). This implies

A circular disc of radius $R$ carries surface charge density $\sigma \left(r\right)={\sigma }_0\left(1-\frac{r}{R}\right)$, where ${\sigma }_0$ is a constant and $r$ is the distance from the center of the disc. Electric flux through a large spherical surface that encloses the charged disc completely is ${\varphi }_0$. Electric flux through another spherical surface of radius $\frac{R}{4}$ and concentric with the disc is $\varphi$. Then the ratio $\frac{{\varphi }_0}{\varphi }$ is

A thermally isolated cylindrical closed vessel of height $8\mathrm{m}$ is kept vertically. It is divided into two equal parts by a diathermic (perfect thermal conductor) frictionless partition of mass $8.3\mathrm{kg}$. Thus the partition is held initially at a distance of $4\mathrm{m}$ from the top, as shown in the schematic figure below. Each of the two parts of the vessel contains $0.1$ mole of an ideal gas at temperature $300\mathrm{K}$. The partition is now released and moves without any gas leaking from one part of the vessel to the other. When equilibrium is reached, the distance of the partition from the top (in $\mathrm{m}$) will be ____________ (take the acceleration due to gravity $=10\mathrm{m}{\mathrm{s}}^{-2}$ and the universal gas constant$=8.3\mathrm{J}{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1}$).

An ideal gas is undergoing a cyclic thermodynamic process in different ways as shown in the corresponding P-V diagrams in column 3 of the table. Consider only the path from state 1 to state 2. W denotes the corresponding work done on the system. The equations and plots in the table have standard notations as used in thermodynamic process. Here $\gamma$ is the ratio of heat capacities at constant pressure and constant volume. The number of moles in the gas is n.

Column – 1	Column – 2	Column – 3
(I) \(W_{1 \rightarrow 2}=\frac{1}{\gamma-1}\) \(\left(P_2 V_2-P_1 V_1\right)\)	(i) Isothermal	(P)
(II) \(W_{1 \rightarrow 2}=\) \(-P V_2+P V_1\)	(ii) Isochoric	(Q)
(III) \(W_{1 \rightarrow 2}=0\)	(iii) Isobaric	(R)
(IV) \(W_{1 \rightarrow 2}=\) \(-n R T \ln \left(\frac{V_2}{V_1}\right)\)	(iv) Adiabatic	(S)

Which one of the following options is the correct combination?

In the following circuit $C_1=12 \mathrm{\mu F},C_2=C_3=4 \mathrm{\mu F}$ and $C_4=C_5=2 \mathrm{\mu F}$. The charge stored in $C_3$ is $\mathrm{\mu C}.$

Half-life of a radio active substance ' \(A\) ' is 4 days. The probability that a nucleus will decay in two half-lives is

Two ideal batteries of emf ${\mathrm{V}}_1$ and ${\mathrm{V}}_2$ and three resistances ${\mathrm{R}}_1,\mathrm{}{\mathrm{R}}_2$ and ${\mathrm{R}}_3$ are connected as shown in the figure. The current in resistance ${\mathrm{R}}_2$ would be zero if

Two soap bubbles \(A\) and \(B\) are kept in a closed chamber where the air is maintained at pressure \(8 \mathrm{Nm}^{-2}\). The radii of bubbles \(A\) and \(B\) are \(2 \mathrm{~cm}\), respectively. Surface tension of the soap-water used to make bubbles is \(0.04\) \(\mathrm{Nm}^{-1}\). Find the ratio \(\frac{n_B}{n_A}\), where \(n_A\) and \(n_B\) are the number of moles of air in bubbles \(A\) and \(B\), respectively. [Neglect the effect of gravity]

A water cooler of storage capacity 120 litres can cool water at constant rate of P watts. In a closed circulation system (as shown schematically in the figure), the water from the cooler is used to cool an external device that generates constantly 3 kW of heat (thermal load). The temperature of water fed into the device cannot exceed ${30}^{o}C$ and the entire stored 120 litres of water is initially cooled to ${10}^{o}C$ . The entire system is thermally insulated. The minimum value of P (in watts) for which the device can be operated for 3 hours is:

(Specific heat of water is 4.2 kJ $k{g}^{-1}{K}^{-1}$ and the density of water is 1000 kg $m^{-3}$ )

Let the functions $f:R\to R$ and $g:R\to R$ be defined by
$f\left(x\right)=e^{x-1}-e^{-\left|x-1\right|}$ and $g\left(x\right)=\frac{1}{2}\left(e^{x-1}+e^{1-x}\right)$.
Then the area of the region in the first quadrant bounded by the curves $y=f\left(x\right), y=g\left(x\right)$ and $x=0$ is

A spherical body of radius $\mathrm{R}$ consists of a fluid of constant density and is in equilibrium under its own gravity. If $\mathrm{P}\left(\mathrm{r}\right)$ is the pressure at $\mathrm{r}(\mathrm{r}<\mathrm{R})$ , then the correct option(s) is (are)

The value of \(\frac{(5050) \int_0^1\left(1-x^{50}\right)^{100} d x}{\int_0^1\left(1-x^{50}\right)^{101} d x}\)

If $\alpha =3{\sin }^{-1}\left(\frac{6}{11}\right) and \beta =3{\cos }^{-1}\left(\frac{4}{9}\right),$ where the inverse trigonometric functions take only the principal values, then the correct options(s) is(are)