Class 11th

Two same group on any C-atom of double bond ⇒ (compound does not show geometrical isomerism)

Gutta percha is a synthetic rubber and its monomer is isoprene .
Since isoprene has EDG, so it is prepared by cationic addition polymerization

From Reaction

Δn? = 2 – 1 = 1

Kp = Kc (RT)^Δn?

Kp = Kc (RT)¹

Kc = Kp (RT)? ¹

Thermal stability of compound is directly proportional to large anion of s-block elements:

T.S ∝ 1/ (PP of cation)

And p.p. ∝ charge ∝ 1/size, So, [T.S. ∝ size ∝ 1/ (charge)]

So, Sr [NO? ]? is highly stable

But Mg (NO? )? ⇒ poor

All have same molarity, pH ∝ 1/ (Acidic strength) ∝ Basic strength

H? SO? → Acidic

NH? Cl (Salt of weak base and strong acid)

Acidic but less than H? SO?

NaCl → (Neutral solution)

NaOH → Basic (Strong Base)

So NaOH > NaCl > NH? Cl > H? SO?

and basic strength ∝ 1/pK? ∝ K? ∝ +R ∝ 1/ (-R) ∝ +I ∝ 1/ (-I)

So basic strength order ⇒ IV > III > II > I, so pK? order is IV < III < II < I

? Solvent is H? O, which is in excess

So using m ( molality ) = (x? *1000)/ (x? * (M? )? )

? x? = 0.74 (Mol? = 18 g)

x? = 1 – 0.74 = 0.26 ∴ m = (0.26 * 1000)/ (0.74 * 18) = 19.5

If hunds rule is followed

${N_2}^{-2}:\sigma 1s^2{\sigma }^{*}1s^2\sigma 2s^2{\sigma }^{*}2s^2\sigma 2{p_z}^2\pi 2{p^2}_x\pi 2{p_y}^2{\pi }^{*}2{p_x}^1{\pi }^{*}2{p_y}^1$            

If Hunds rule is not followed

${N_2}^{-2}:\sigma 1s^2{\sigma }^{*}1s^2\sigma 2s^2{\sigma }^{*}2s^2\sigma 2{p_z}^2\pi 2{p^2}_x\pi 2{p_y}^2{\pi }^{*}2{p_x}^2$            

The authority has not specified any seggregated weigtage of the SITEEE syllabus for class 11th and 12th.

Acceleration of the combined system along the incline is given by

$a=g\mathrm{s}\mathrm{i}\mathrm{n}?{37}^{?}a=\frac{3g}{5}$

With respect to block M :

$\begin{array}{ll}\therefore \mathrm{}& N=mg-ma\mathrm{s}\mathrm{i}\mathrm{n}?{37}^{?}=mg-\frac{3mg}{5}\cdot \frac{3}{5}\\ & N=\frac{16}{25}mg\\ & \mu N=ma\mathrm{c}\mathrm{o}\mathrm{s}?{37}^{?}\\ \Rightarrow \mathrm{}& \mu \frac{16}{25}mg=\frac{3mg}{5}\cdot \frac{4}{5}\Rightarrow \mu =\frac{3}{4}=\frac{750}{1000}\Rightarrow I=750\end{array}$