If the circle \(\mathrm{S}=0\) cuts the circles \(x^2+y^2-2 x+6 y=0\). \(x^2+y^2-4 x-2 y+6=0\) and \(x^2+y^2-12 x+2 y+3=0\) orthogonally, then equation of the tangent at \((0,3)\) on \(S=\) 0 is

If \(\omega\) is a complex cube root of unity, then \(\sum_{r=1}^9 r(r+1-\omega)\left(r+1-\omega^2\right)=\)

The substitution \(x=v y\) converts which one of the following differential equation to an equation solvabie by variable separable method?

If \(\frac{3 x}{(x-a)(x-b)}=\frac{2}{x-a}+\frac{1}{x-b}\), then \(a: b\) is equal to

If \(f: R \rightarrow R\) is defined by \(f(x)=[2 x]-2[x]\) for \(x \in R\), where \([x]\) is the greatest integer not exceeding \(x\), then the range of \(f\) is :

The angle between the tangents drawn from a point \((4,3)\) to the circle \(x^2+y^2-2 x-4 y=0\) is

Match the items of List - I with those of the entires of List - II
\(List - I\)
\(\begin{aligned} & \text { (I) } \sin ^2 5^{\circ}+\sin ^2 10^{\circ}+ \\ & \sin ^2 15^{\circ}+\ldots+\sin ^2 90^{\circ}=\end{aligned}\)
\(\begin{aligned} & \text { (II) } \tan ^2 5^{\circ} \cdot \tan ^2 10^{\circ} \text {. } \\ & \tan ^2 15^{\circ} \ldots \tan ^2 85^{\circ}= \\ & \end{aligned}\)
\(\begin{aligned} & \text { (III) } \cos ^2 5^{\circ}+\cos ^2 10^{\circ} \\ & +\cos ^2 15^{\circ}+\ldots+\cos ^2 180^{\circ}=\end{aligned}\)
\(\begin{gathered}\text { (IV) } \cot 5^{\circ}+\cot 10^{\circ}+\cot 15^{\circ} \\ +\ldots .+\cot 175^{\circ}=\end{gathered}\)
\(List - II\)
(A) \(0\)
(B) \(\frac{19}{2}\)
(C) \(18\)
(D) \(1\)
(E) \(-1\)

Which among the following can be explained by adsorption theory?

If the values \(x=\alpha, y=\beta, z=\gamma\) satisfy all the 3 equations \(x+2 y+3 z=4\), \(3 x+y+z=3\) and \(x+3 y+3 z=2\), then \(3 \alpha+\gamma=\)

A projectile is thrown straight upward from the earth's surface with an initial speed $v=\alpha v_E$, where $\alpha$ is a constant and $v_E$ is the escape speed. The projectile travels upto a height $800 \mathrm{km}$ from earth's surface, before it comes to rest. The value of the constant $\alpha$ is,
(Radius of the earth $=6400 \mathrm{km}$)

The variance of the first 50 even natural numbers is

Match the following
\(\begin{array}{ll}
\text{List - I} & \text{List - II} \\
\text{i) Isothermal process} & \text{a)} 0 \\
\text{ii) Isobaric process} & \text{b)} \frac{1}{\gamma-1}\left[\mathrm{P}_2 \mathrm{~V}_2-\mathrm{P}_1 \mathrm{~V}_1\right] \\
\text{iii) Isochoric process} & \text{c)} \mu \mathrm{RT} \ln \left(\frac{V_2}{V_1}\right) \\
\text{iv) Adiabatic process} & \text{d)} \mathrm{P}\left(\mathrm{V}_2-\mathrm{V}_1\right) \\
\end{array}\)
The correct answer is

\(6 \mathrm{~g}\) of a non-volatile, non-electrolyte \(X\) dissolved in \(100 \mathrm{~g}\) of water freezes at \(-0.93^{\circ} \mathrm{C}\). The molar mass of \(X\) in \(\mathrm{g} \mathrm{mol}^{-1}\) is \(\left(K_f\right.\) of \(\left.\mathrm{H}_2 \mathrm{O}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)\)