The area (in square units) of the triangle formed by the lines \(x=0, y=0\) and \(3 x+4 y=12\), is

If \(f(x)=k x^3-3 x^2-12 x+8\) is strictly decreasing for all \(x \in \mathbf{R}\) then

If $\cos \theta =-\frac{1}{\sqrt{2}}$ and $\tan \theta =1,$ then the general value of $\theta$ is

If the circles $(x-2{)}^2+(y-3{)}^2=25$ and $25x^2+25y^2-40x-70y-160=0$ touch internally at $\left(\alpha ,\beta \right)$ then $\alpha +\beta =$

\(\lim _{x \rightarrow 0} \frac{\tan x-\sin x}{x^2}\) is equal to

Numerically greatest term in the expansion of \((2 x-3 y)^n\) when \(x=\frac{7}{2}, y=\frac{3}{7}\) and \(n=13\) is

If \(m\) and \(M\) are respectively the absolute minimum and absolute maximum values of a function \(f(x)=2 x^3+9 x^2+12 x+1\) defined on \([-3,0]\), then \(m+M=\)

One end each of a resistance \(r\) capacitor \(C\) and resistance \(2 r\) are connected together. The other ends are respectively connected to the positive terminals of batteries, \(P, Q, R\) having respectively emf's \(E, E\) and \(2 E\). The negative terminals of the batteries are then connected together. In this circuit, with steady current the potential drop across the capacitor is :

For the formation of \(\mathrm{NH}_3\) from \(\mathrm{N}_2\) and \(\mathrm{H}_2\) at \(500 \mathrm{~K}\), the concentration of \(\mathrm{N}_2, \mathrm{H}_2\) and \(\mathrm{NH}_3\) at equilibrium are \(1.5 \times 10^{-2} \mathrm{M}, 3.0 \times 10^{-2} \mathrm{M}\) and \(1.2 \times 10^{-2} \mathrm{M}\), respectively. The equilibrium constant for the reverse reaction is

A cell of emf \(10 \mathrm{~V}\) and internal resistance \(3 \Omega\) is connected in parallel with another cell of emf \(7 \mathrm{~V}\) and internal resistance \(\frac{3}{5} \Omega\), such that their positive terminals are joined together and so their are negative terminals. Their positive terminals are joined with the negative terminal and their negative terminal is joined with the positive terminal of a third cell of emf \(20 \mathrm{~V}\) with internal resistance \(2 \Omega\). The combination can be replaced by a battery of emf \(E\) and internal resistance \(r\), then the values of \(E\) and \(R\) are respectively

Calculate the depression in the freezing point of a solution containing \(0.1 \mathrm{~g}\) of \(\mathrm{K}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]\) in \(100 \mathrm{~g}\) of \(\mathrm{H}_2 \mathrm{O}\) ? (molecular weight of \(\mathrm{K}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]=329 ; K_f=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^1\) )

The mean and standard deviation of 100 observations were calculated as 40 and 5.1 respectively. Later on it was found that one of the observations was taken as 50 in the place of 40 . If the wrong entry is replaced by the correct one, then the sum of the squares of all the observations is