The number of \(3 \times 3\) matrices \(A\) whose entries are either 0 or 1 and for which the system \(A\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right]\) has exactly two distinct solutions, is

Let \(\mathbb{R}\) denote the set of all real numbers. Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) and \(g: \mathbb{R} \rightarrow(0,4)\) be functions defined by \(f(x)=\log _e\left(x^2+2 x+4\right)\), and \(g(x)=\frac{4}{1+e^{-2 x}}\)
Define the composite function \(f \circ g^{-1}\) by \(\left(f \circ g^{-1}\right)(x)=\left(g^{-1}(x)\right)\), where \(g^{-1}\) is the inverse of the function \(g\).
Then the value of the derivative of the composite function \(f \circ g^{-1}\) at \(x=2\) is ________ .

Let $\alpha , \beta$ and $\gamma$ be real numbers. Consider the following system of linear equations $x+2y+z=7$ $x+\alpha z=11$ $2x-3y+\beta z=\gamma$ Match each entry in List-I to the correct entries in List-II.

	List-I		List-II
$\left(\mathrm{P}\right)$	If $\beta =\frac{1}{2}\left(7\alpha -3\right)$ and $\gamma =28$ then the system has	$\left(1\right)$	a unique solution
$\left(\mathrm{Q}\right)$	If $\beta =\frac{1}{2}\left(7\alpha -3\right)$ and $\gamma \ne 28$, then the system has	$\left(2\right)$	no solution
$\left(\mathrm{R}\right)$	If $\beta \ne \frac{1}{2}\left(7\alpha -3\right)$ where $\alpha =1$ and $\gamma \ne 28$, then the system has	$\left(3\right)$	infinitely many solutions
$\left(\mathrm{S}\right)$	If $\beta \ne \frac{1}{2}\left(7\alpha -3\right)$ where $\alpha =1$ and $\gamma =28$, then the system has	$\left(4\right)$	$x=11, y=-2$ and $z=0$ as a solution
		$\left(5\right)$	$x=-15, y=4$ and $z=0$ as a solution

The correct option is:

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Let \(p\) be an odd prime number and \(T_p\) be the following set of \(2 \times 2\) matrices
\[
T_p=\left\{A=\left[\begin{array}{ll}
a & b \\
c & a
\end{array}\right] ; a, b, c \in\{0,1,2, \ldots, p-1\}\right\}
\]Question:
The number of \(A\) in \(T_p\) such that \(\operatorname{det}(A)\) is not divisible by \(p\), is

Let $\overrightarrow{\text{PR}}=3\widehat{i}+\widehat{j}-2\widehat{k}\text{and}\overrightarrow{\text{SQ}}=\widehat{i}-3\widehat{j}-4\widehat{k}$ determine diagonals of a parallelogram PQRS and $\overrightarrow{\text{PT}}=\widehat{i}+2\widehat{j}+3\widehat{k}$ be another vector. Then the volume of the parallelepiped determined by the vectors $\overrightarrow{\mathit{\text{PT}}}\text{,}\overrightarrow{\mathit{\text{PQ}}}\text{and}\overrightarrow{\mathit{\text{PS}}}$  is

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Let \(O\) be the origin, and \(\overrightarrow{O X}, \overrightarrow{O Y}, \overrightarrow{O Z}\) be three unit vectors in the directions of the sides \(\overrightarrow{Q R}, \overrightarrow{R P}\), \(\overrightarrow{P Q}\), respectively, of a triangle \(P Q R\).


Question:

\(|\overrightarrow{O X} \times \overrightarrow{O Y}|=\)

Match the following

Which is correct statement if \(\mathrm{N}_2\) is added at equilibrium condition?

Vessel-1 contains \(\mathbf{w}_2 \mathrm{~g}\) of a non-volatile solute \(\mathbf{X}\) dissolved in \(\mathbf{w}_1 \mathrm{~g}\) of water. Vessel- 2 contains \(\mathbf{w}_2 \mathrm{~g}\) of another non-volatile solute \(\mathbf{Y}\) dissolved in \(\mathbf{w}_1 \mathrm{~g}\) of water. Both the vessels are at the same temperature and pressure. The molar mass of \(\mathbf{X}\) is \(80 \%\) of that of \(\mathbf{Y}\). The van't Hoff factor for \(\mathbf{X}\) is 1.2 times of that of \(\mathbf{Y}\) for their respective concentrations.
The elevation of boiling point for solution in Vessel-1 is ________. \(\%\) of the solution in Vessel-2.

Each of the following options contains a set of four molecules. Identify the option(s) where all four molecules possess permanent dipole moment at room temperature.

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A frame of reference that is accelerated with respect to an inertial frame of reference is called a non-inertial frame of reference. A coordinate system fixed on a circular disc rotating about a fixed axis with a constant angular velocity \(\omega\) is an example of a non-inertial frame of reference.

The relationship between the force \(\vec{F}_{\text {rot }}\) experienced by a particle of mass \(m\) moving on the rotating disc and the force \(\vec{F}_{\text {in }}\) experienced by the particle in an inertial frame of reference is

\(\vec{F}_{\mathrm{rot}}=\vec{F}_{\mathrm{in}}+2 m\left(\vec{v}_{\mathrm{rot}} \times \vec{\omega}\right)+m(\vec{\omega} \times \vec{r}) \times \vec{\omega}\),

where \(\vec{v}_{\text {rot }}\) is the velocity of the particle in the rotating frame of reference and \(\vec{r}\) is the position vector of the particle with respect to the centre of the disc.

Now consider a smooth slot along a diameter of a disc of radius \(R\) rotating counter-clockwise with a constant angular speed \(\omega\) about its vertical axis through its center. We assign a coordinate system with the origin at the center of the disc, the \(x\)-axis along the slot, the \(y\)-axis perpendicular to the slot and the \(z\)-axis along the rotation axis \((\vec{\omega}=\omega \hat{k}) .\) A small block of mass \(m\) is gently placed in the slot at \(\vec{r}=(R / 2) \hat{i}\) at \(t=0\) and is constrained to move only along the slot.

Question :
The net reaction of the disc on the block 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:
What is the maximum energy of the anti-neutrino?

Consider the lines \(L_{1}\) and \(L_{2}\) defined by
\[
L_{1}: x \sqrt{2}+y-1=0 \text { and } L_{2}: x \sqrt{2}-y+1=0
\]
For a fixed constant \(\lambda\), let \(C\) be the locus of a point \(P\) such that the product of the distance of \(P\) from \(L_{1}\) and the distance of \(P\) from \(L_{2}\) is \(\lambda^{2}\). The line \(y=2 x+1\) meets \(C\) at two points \(R\) and \(S\), where the distance between \(R\) and \(S\) is \(\sqrt{270}\).
Let the perpendicular bisector of \(R S\) meet \(C\) at two distinct points \(R^{\prime}\) and \(S^{\prime}\). Let \(D\) be the square of the distance between \(R^{\prime}\) and \(S^{\prime}\).
The value of $D$ is