Class 12th

$\left(\frac{-\pi }{2},0\right)$

At $\frac{dy}{\sqrt[]{1-y^2}}+\frac{dx}{\sqrt[]{1-x^2}}=0\Rightarrow {\mathrm{s}\mathrm{i}\mathrm{n}}^{-1}?y+{\mathrm{s}\mathrm{i}\mathrm{n}}^{-1}?x=c$ $\frac{}{\text{exception 25:}}\frac{}{\text{exception 25:}}{}^{}{}^{}$

Hence $x=\frac{1}{2},y=\frac{\sqrt[]{3}}{2}\Rightarrow c=\frac{\pi }{2}\Rightarrow {\mathrm{s}\mathrm{i}\mathrm{n}}^{-1}?y={\mathrm{c}\mathrm{o}\mathrm{s}}^{-1}?x$ $\frac{}{}\frac{\text{exception 25:}}{}\frac{}{}{}^{}{}^{}$

KNO? , HCl and NaCl are strong electrolytes for these electrolytes of Λ_m with √c will be liner which can be given as
Λ_m = Λ? _m - A√c for strong electrolyte
Since given variation is not linear it has to be a weak electrolyte
CH? COOH is a weak electrolyte

Gas + Solid? GS? H = -ve
Adsorbed gas
Adsorption of gas is an exothermic process. Increase in temperature reduces the extent of adsorption.
x/m = Kp¹/? (n > 1)

$\mathrm{}{f}^{\mathrm{\text{'}}}\left(x\right)=x\left(\pi -{\mathrm{c}\mathrm{o}\mathrm{s}}^{-1}?(\mathrm{s}\mathrm{i}\mathrm{n}?|x\left|\right)\right)=x\left(\pi -\left(\frac{\pi }{2}-{\mathrm{s}\mathrm{i}\mathrm{n}}^{-1}?(\mathrm{s}\mathrm{i}\mathrm{n}?|x\left|\right)\right)\right)=x\left(\frac{\pi }{2}+\left|x\right|\right)$

$f\left(x\right)=\left\{\begin{array}{ll}x\left(\frac{\pi }{2}+x\right)& x\ge 0\\ x\left(\frac{\pi }{2}-x\right)& x<0\end{array}\right.$

$f^{\mathrm{\text{'}}}\left(x\right)=\left\{\begin{array}{ll}\frac{\pi }{2}+2x& x\ge 0\\ \frac{\pi }{2}-2x& x<0\end{array}\right.$  is increasing in $f^{\mathrm{\text{'}}}\left(x\right)$ ${}^{}$ and decreasing in $\left(0,\frac{\pi }{2}\right)$

In FCC octahedral voids are present at the edge centers and body center

Consider a diagonal projected form edge centre passing through the body centre
Distance between octahedral voids = √2a/2 = a/√2

δ_min = (µ - 1)A
= (1.5 - 1)1
= 0.5
δ_min = 5/10
N=5

$=\frac{1}{56}*277.85=496$

$\mathrm{}\pm 1=\left|\begin{array}{lll}1& 1& \lambda \\ 1& 1& 3\\ 2& 1& 1\end{array}\right|\Rightarrow =-\lambda +3=\pm 1$ or $\Rightarrow \lambda =2$

For $\lambda =4$ $$

$\lambda =4$

$\mathrm{c}\mathrm{o}\mathrm{s}?\theta =\frac{2+1+4}{\sqrt[]{6}\sqrt[]{18}}=\frac{7}{6\sqrt[]{3}}$

E? = φ + K?
E? = φ + K?
E? - E? = K? - K?
Now V? /V? = 2
K? /K? = 4; K? = 4K?
Now from equation (2)
⇒ 4 - 2.5 = 4K? - K?
1.5 = 3K?
K? = 0.5eV
Now putting this
Value in equation (2)
2.5 = φ + 0.5eV
φ = 2eV

Let us assume the potential at A = V_A = 0.
Now at junction C, according to KCL
i? + i? = i?
1 A + i? = 2 A

i? = 2 A
Now analyse potential along ACDB
V_A + 1 + i? (2) - 2 = V_B
0 + 1 + 2 (1) - 2 = V_B
V_B = 3 - 2
V_B = 1amp.

Figure of Merit = C = i/θ
= C = (6 * 10? ³)/2 = 3 * 10? ³ Am²