A planet of radius $\mathrm{R}=\frac{1}{10}\times$ (radius of Earth) has the same mass density as Earth. Scientists dig a well of depth $\frac{\mathrm{R}}{5}$ on it and lower a wire of the same length and of linear mass density ${10}^{-3}\mathrm{ }\mathrm{k}\mathrm{g}{\mathrm{m}}^{-1}$ into it. If the wire is not touching anywhere, the force applied at the top of the wire by a person holding it in place is (take the radius of Earth $=6\times {10}^6\mathrm{ }\mathrm{m}$ and the acceleration due to gravity of Earth is $10\mathrm{ }\mathrm{m}{\mathrm{s}}^{-2}$ )

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Consider a simple \(R C\) circuit as shown in Figure \(1\).

Process 1: In the circuit the switch \(S\) is closed at \(t=0\) and the capacitor is fully charged to voltage \(V_{0}\) (i.e., charging continues for time \(T>>R C\) ). In the process some dissipation \(\left(E_{D}\right)\) occurs across the resistance \(R\). The amount of energy finally stored in the fully charged capacitor is \(E_{C}\).

Process 2: In a different process the voltage is first set to \(\frac{V_{0}}{3}\) and maintained for a charging time \(T>>R C\). Then the voltage is raised to \(\frac{2 V_{0}}{3}\) without discharging the capacitor and again maintained for a time \(T>>R C\). The process is repeated one more time by raising the voltage to \(V_{0}\) and the capacitor is charged to the same final voltage \(V_{0}\) as in Process 1.

These two processes are depicted in Figure \(2 .\)







Question:

In Process 1 , the energy stored in the capacitor \(E_{C}\) and heat dissipated across resistance \(E_{D}\) are related by:

Using the expression $2d \sin \theta =\lambda$, one calculates the values of $d$ by measuring the corresponding angle $\text{θ}$ in the range $0$ to $90°$. The wavelength $\lambda$ is exactly known and the error in $\theta$ is constant for all the values of $\theta$. As $\text{θ}$ increases from $0°$,

A point mass is subjected to two simultaneous sinusoidal displacements in \(x\)-direction, \(x_1(t)=A \sin \omega t\) and \(x_2(t)=A \sin \left(\omega t+\frac{2 \pi}{3}\right)\). Adding a third sinusoidal displacement \(x_3(t)=B \sin (\omega t+\phi)\) bring the mass to a complete rest. The values of \(B\) and \(\phi\) are

Four solid spheres each of diameter \(\sqrt{5} \mathrm{~cm}\) and mass \(0.5 \mathrm{~kg}\) ar placed with their centres at the corners of a square of side \(4 \mathrm{~cm}\). The moment of inertia of the system about the diagonal of the square is \(N \times 10^{-4} \mathrm{~kg}-\mathrm{m}^2\), then \(N\) is

A meter bridge is set up as shown in figure, to determine an unknown resistance \(X\) using a standard \(10 \Omega\) resistor. The galvanometer shows null point when tapping key is at \(52 \mathrm{~cm}\) mark. The end-corrections are \(1 \mathrm{~cm}\) and \(2 \mathrm{~cm}\) respectively for the ends \(A\) and \(B\). The determined value of \(X\) is

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The mass of a nucleus \({ }_{Z}^{A} X\) is less than the sum of the masses of \((A-Z)\) number of neutrons and \(Z\) number of protons in the nucleus. The energy equivalent to the corresponding mass difference is known as the binding energy of the nucleus. A heavy nucleus of mass \(M\) can break into two light nuclei of masses \(m_{1}\) and \(m_{2}\) only if \(\left(m_{1}+m_{2}\right) \lt M\). Also two light nuclei of masses \(m_{3}\) and \(m_{4}\) can undergo complete fusion and form a heavy nucleus of mass \(M^{\prime}\) only if \(\left(m_{3}+m_{4}\right) \gt M^{\prime}\). The masses of some neutral atoms are given in the table below:


\(\left(1 u=932 M e V / c^{2}\right)\)

Question:

The correct statement is

Two spheres $P$ and $Q$ of equal radii have densities ${\rho }_1$ and ${\rho }_2$ , respectively. The spheres are connected by a massless string and placed in liquids $L_1$ and $L_2$ of densities ${\sigma }_1$ and ${\mathrm{\sigma }}_2$ and viscosities ${\eta }_1$ and ${\eta }_2$ , respectively. They float in equilibrium with the sphere $P$ in $L_1$ and sphere $Q$ in $L_2$ and the string being taut (see figure). If sphere $P$ alone in $L_2$ has terminal velocity ${\overrightarrow{V}}_P$ and $Q$ alone in $L_1$ has terminal velocity ${\overrightarrow{V}}_Q$ , then

The electrochemical extraction of aluminium from bauxite ore involves

Three boys and two girls stand in a queue. The probability that the number of boys ahead of every girl is atleast one more than the number of girls ahead of her is

Match the reactions (in the given stoichiometry of the reactants) in List-I with one of their products given in List-II and choose the correct option.

List - I	List - II
(P) \(P _2 O _3\) \(+~3 H _2 O \rightarrow\)	(1) \(P ( O )\left( OCH _3\right)\) \(Cl _2\)
(Q) \(P _4+3 NaOH\) \(+~3 H _2 O \rightarrow\)	(2) \(H _3 PO _3\)
(R) \(PCl _5+ CH _3\) \(COOH \rightarrow\)	(3) \(PH _3\)
(S) \(H _3 PO _2\) \(+~2 H _2 O ~+\) \(4 AgNO _3 \rightarrow\)	(4) \(POCL _3\)
	(5) \(H _3 PO _4\)

To check the principle of multiple proportions, a series of pure binary compounds $\left({\mathrm{P}}_{\mathrm{m}}{\mathrm{Q}}_{\mathrm{n}}\right)$ were analysed and their composition is tabulated below. The correct option(s) is(are)

Compound	Weight $\%$ of $\mathrm{P}$	Weight $\%$ of $\mathrm{Q}$
$1$	$50$	$50$
$2$	$44.4$	$55.6$
$3$	$40$	$60$

A thin uniform rod of length \(L\) and certain mass is kept on a frictionless horizontal table with a massless string of length \(L\) fixed to one end (top view is shown in the figure). The other end of the string is pivoted to a point \(\mathrm{O}\). If a horizontal impulse \(P\) is imparted to the rod at a distance \(x=L / n\) from the mid-point of the rod (see figure), then the rod and string revolve together around the point \(\mathrm{O}\), with the rod remaining aligned with the string. In such a case, the value of \(n\) is _______ .