Class 11th

Due to high crystallity Be has the highest M.P.

Be = 1560 K

Mg = 925 K

Ca = 1120 K

Sr = 1062 K

During the electrolysis of dilute H₂SO₄

$2H_{2}S{O}_4\stackrel{\left(1\right)electrolysis}{\to }2HS{O}_4^{-}+2H^{+}$  

$H_2{S}_2{O}_8+2H_{2}O\stackrel{hydrolysis}{\to }2H_{2}S{O}_4+\underset{\left(A\right)}{H_2{O}_2}$  

$2H^{+}+2e\to H_2\uparrow$  

In the solid form of  $H_2{O}_2$ dihedral angle is equal to 90.2°.

In boron family lower O.S (+1) is more stabilize down the group due to inert pair effect.

According to question, we can write

$\mathcal{l}=2\pi r\Rightarrow r=\frac{\mathcal{l}}{2\pi }$

$I_1=\frac{m{\mathcal{l}}^2}{3},and{I}_2=\frac{m{r}^2}{2}=\frac{m{\mathcal{l}}^2}{8{\pi }^2}$

$\Rightarrow \frac{I_1}{I_2}=\frac{m{\mathcal{l}}^2}{3}*\frac{8{\pi }^2}{m{\mathcal{l}}^2}=\frac{8{\pi }^2}{3}$

$Bondorder=\frac{e^{-}ofBondingMO-e^{-}ofABMO}{2}$  

Hence Bond order for  $O_2^{+}=\frac{10-5}{2}=2.5$  

$O_2=\frac{10-6}{2}=2.0$

$O_2^{-}=\frac{10-7}{2}=1.5$

$O_2^{--}=\frac{10-8}{2}=1.0$  

Radius of Bohr's orbit $R_n=0.529\stackrel{o}{A}*\frac{n^2}{z}$  

Radius of Bohr's orbit for hydrogen,   $R_n=0.529\stackrel{o}{A}*n^2$

For third orbit (R₃) = $0.529\stackrel{o}{A}*9$ = r₃ and

Fourth orbit (R₄) = $0.59\stackrel{o}{A}*16=r_4$

$\therefore r_4=\frac{16r_3}{9}$  

Wt of liquid = 135 – 40 = 95 gm

Volume of liquid =  $\frac{wt.}{density}=\frac{95}{0.95}=100ml$  

Hence volume of vessel = 100 ml = 0.1 lit from ideal gas equation,

$M=\frac{dRT}{P}=\frac{0.5}{0.1}*\frac{0.0821*250}{0.82}$  

$\therefore M=125g/mol$  

Since velocity does not change, so acceleration will be zero.

mg = FB + Fv $\Rightarrow \frac{4\pi }{3}{r}^3\rho g=\frac{4\pi }{3}{r}^3\sigma g+6\pi \eta rv$

$\Rightarrow v=\frac{2r^2\left(\rho -\sigma \right)g}{9\eta }=\frac{2*0.1*0.1*10^{-6}*\left(10^4-10^3\right)*10}{9*1.0*10^{-5}}$

$\Rightarrow h=\frac{400}{2g}=20m$

Least  count of Vernier = 0.1mm

Reading of Vernier Scale = 5 * 0.1 = 0.5mm

The corrected diameter of sphere = Main Scale Reading + Vernier Scale reading + Zero correction = 1.7 + 0.05 + 0.05 = 1.8cm = 180 * 10^-2 cm.

According to question, we can write

$R=\frac{u^{2}sin2\theta }{g}\Rightarrow u^2=g{R}_{max},where\theta =45°$

$\Rightarrow H_{max}=\frac{u^2}{2g}=\frac{g{R}_{max}}{2g}=\frac{R_{max}}{2}=\frac{100}{2}=50m$