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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

Consider an electric field \(\mathbf{E}=E_0 \hat{\mathbf{x}}\) where \(\bar{E}_0\) is a constant. The flux through the shaded area ( as shown in the figure) due to this field is

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A small spherical monoatomic ideal gas bubble \(\left(\gamma=\frac{5}{3}\right)\) is trapped inside a liquid of density \(\rho_1\) (see figure). Assume that the bubble does not exchange any heat with the liquid. The bubble contains \(n\) moles of gas. The temperature of the gas when the bubble is at the bottom is \(T_0\), the height of the liquid is \(H\) and the atmospheric pressure is \(p_0\) (Neglect surface tension).

Question:
The buoyancy force acting on the gas bubble is (Assume \(R\) is the universal gas constant)

A composite block is made of slabs \(A, B, C, D\) and \(E\) of different thermal conductivities (given in terms of a constant, \(K\) ) and sizes (given in terms of length, \(L\) ) as shown in the figure. All slabs are of same width. Heat \(Q\) flows only from left to right through the blocks. Then, in steady state

During an experiment with a metre bridge, the galvanometer shows a null point when the jockey is pressed at $40.0\mathrm{cm}$using a standard resistance of $90\mathrm{}\mathrm{\Omega }$ ,as shown in the figure. The least count of the scale used in the metre bridge is $1\mathrm{mm}$. The unknown resistance is

The \(\beta\)-decay process, discovered around 1900 , is basically the decay of a neutron \((n)\). In the laboratory, a proton \((p)\) and an electron \(\left(e^{-}\right)\)are observed as the decay products of the neutron. Therefore, considering the decay of a neutron as a two-body decay process, it was predicted theoretically that the kinetic energy of the electron should be a constant. But experimentally, it was observed that the electron kinetic energy has continuous spectrum. Considering a three-body decay process, i.e.
\(n \rightarrow p+e^{-}+\bar{v}_{e}\), around 1930, Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino \(\left(\bar{v}_{e}\right)\) to be massless and possessing negligible energy, and the neutron to be at rest, momentum and energy conservation principles are applied. From this calculation, the maximum kinetic energy of the electron is \(0.8 \times 10^{6} \mathrm{eV}\). The kinetic energy carried by the proton is only the recoil energy.

Question:
If the anti-neutrino had a mass of \(3 \mathrm{eV} / \mathrm{c}^{2}\) (where \(\mathrm{c}\) is the speed of light) instead of zero mass, what should be the range of the kinetic energy, \(K\), of the electron?

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]

The tangent to the curve \(y=e^x\) drawn at the point \(\left(c, e^c\right)\) intersects the line joining the points \(\left(c-1, e^{c-1}\right)\) and \(\left(c+1, e^{c+1}\right)\)

The electric field E is measured at a point $P(\mathrm{0,0},d)$ , generated due to various charge distributions and the dependence of E on d is found to be different for different charge distributions. List-I contains different relations between E and d. List-II describes different electric charge distributions, along with their locations. Match the functions in List-I with the related charge distributions in List-II.

	LIST -I		LIST -II
A)	is independent of d	P)	A point charge at the origin
B)	\(E \propto \frac{1}{d}\)	Q)	A small dipole with point charges \(Q\) at \((0,0, l)\) and \(-\operatorname{Qat}(0,0, l)\). Take \(2 l \ll d\)
C)	\(E \propto \frac{1}{d^2}\)	R)	An infinite line charge coincident with the \(x\)-axis, with uniform linear charge density \(\lambda\)
D)	\(E \propto \frac{1}{d^3}\)	S)	Two infinite wires carrying uniform linear charge density parallel to the \(x\) - axis. The one along \((y-0, z=l)\) has a charge density \(+\lambda\) and the one along \((y=0, z=-l)\) has a charge density \(-\lambda\). Take \(2 l \ll d\)
		T)	Infinite plane charge coincident with the \(x y\) - plane with uniform surface charge density

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.

Let \(a\) and \(b\) be non-zero and real numbers. Then, the equation \(\left(a x^2+b y^2+c\right)\left(x^2-5 x y+6 y^2\right)=0\) represents

Let $-\frac{\pi }{6}<\theta <-\frac{\pi }{12}$ . Suppose ${\alpha }_1$ and ${\beta }_1$ are the roots of the equation $x^2-2x\sec \theta +1=0$ and ${\alpha }_2$ and ${\beta }_2$ are the roots of the equation $x^2+2x\tan \theta -1=0.$ If ${\alpha }_1>{\beta }_1$ and ${\alpha }_2>{\beta }_2$ , then ${\alpha }_1+{\beta }_2$ equals

Consider an elliptically shaped rail PQ in the vertical plane with $\mathrm{O}\mathrm{P}\mathrm{}=\mathrm{}3\mathrm{}\mathrm{m}$ and $\mathrm{O}\mathrm{Q}\mathrm{}=\mathrm{}4\mathrm{}\mathrm{m}$ . A block of mass 1 kg is pulled along the rail from P to Q with a force of 18 N, which is always parallel to line PQ (see the figure given). Assuming no frictional losses, the kinetic energy of the block when it reaches Q is $(\mathrm{n}\times 10)$ Joules. The value of n is (take acceleration due to gravity = $10\mathrm{}\mathrm{m}{\mathrm{s}}^{–2}$ )