If \(A=\left|\begin{array}{lll}a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3}\end{array}\right|\) and \(B=\left|\begin{array}{ccc}c_{1} & c_{2} & c_{3} \\ a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3}\end{array}\right|\) then

The differential equation of all the \(2-d\) lines are

If \({ }^{56} P_{r+6}:{ }^{54} P_{r+3}=30800: 1\), then

The turning point of the function \(y=\dfrac{a x-b}{(x-1)(x-4)}\) at the point \(P(2,-1)\) is

The solution of the differential equation \(e^{-x}(y+1) d y+\left(\cos ^{2} x-\sin 2 x\right) y(d x)=0\) subjected to the condition that \(y=1\), when \(x=0\) is

If \({ }^{35} \mathrm{C}_{x+7}={ }^{35} \mathrm{C}_{4 x-2}\), then the values of \(x\) are either

The line \(\dfrac{x-3}{4}=\dfrac{y-4}{5}=\dfrac{z-5}{6}\) is parallel to the plane

\(\text { The value of the integral } \int_\limits{\dfrac{1}{3}}^1 \dfrac{\left(x-x^3\right)^{\dfrac{1}{3}}}{x^4} d x \text { is }\)

The conversion of Propyne to Benzene can be brought out in 4 steps.
Choose the reagents to be used, in the proper sequential order, to bring out the conversion.
Reagents provided: alc. \(\mathrm{KOH}, \mathrm{H}_2 / \mathrm{Pd}, \mathrm{Na} /\) dry ether, \(\mathrm{HBr}, \mathrm{HBr} /\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CO}\right)_2 \mathrm{O}_2\),
Soda-lime, \(\mathrm{Cr}_2 \mathrm{O}_3\) (high T and P), Conc. \(\mathrm{H}_2 \mathrm{SO}_4\)

If \(\alpha, \beta\) and \(\gamma\) are the angles between the vectors \(\overrightarrow{\mathrm{P}}, \overrightarrow{\mathrm{Q}}\), and \(\overrightarrow{\mathrm{R}}\) and \(\alpha=90^{\circ}\) as shown in figure. the product of \((\vec{Q} \times \vec{R}) \cdot \vec{Q}\) is equal to

Given below are 4 equations showing Molar conductivities at infinite dilution of various electrolytes. Which one of them represents the correct equation?
\(\begin{aligned}
& \left(\Lambda_m^0\right)_{N a B r}-\left(\Lambda_m^0\right)_{N a C l}=\left(\Lambda_m^0\right)_{\mathrm{KCl}}-\left(\Lambda_m^0\right)_{\mathrm{KBr}} \\
& \left(\Lambda_m^0\right)_{\mathrm{HCl}}+\left(\Lambda_m^0\right)_{\mathrm{KOH}}-\left(\Lambda_m^0\right)_{\mathrm{KCl}}=\left(\Lambda_m^0\right)_{\mathrm{H}_2 \mathrm{O}} \\
& \left(\Lambda_m^0\right)_{\mathrm{KBr}}-\left(\Lambda_m^0\right)_{\mathrm{NaBr}}=\left(\Lambda_m^0\right)_{\mathrm{NaBr}}-\left(\Lambda_m^0\right)_{\mathrm{Nal}} \\
& \left(\Lambda_m^0\right)_{N H_4 \mathrm{Cl}}-\left(\Lambda_m^0\right)_{\mathrm{NH}_4 \mathrm{NO}_3}=\left(\Lambda_m^0\right)_{N H_4 \mathrm{Cl}}-\left(\Lambda_m^0\right)_{N H 4 B r}
\end{aligned}\)

A string of length \(25 \mathrm{~cm}\) and mass \(10^{-3} \mathrm{~kg}\) is clamped at its ends. The tension in the string is \(2.5 \mathrm{~N}\). The identical wave pulses are generated at one end and at regular interval of time, \(\Delta \mathrm{t}\). The minimum value of \(\Delta \mathrm{t}\), so that a constructive interference takes place between successive pulses is

Particles which are not composite and hence truly elementary are