Class 11th

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

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

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

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

$9^n-8n-1={\left(1+8\right)}^n-8n-1$  

$=(1+8n{+}^n\text{?}{C}_2\cdot 8^2{+}^n\text{?}{C}_3\cdot 8^3+....]-8n-1$  

So, a = ⁿC₂ + ⁿC₃8 + ⁿC₄8² +….

Similarly, b = ⁿC₂ + ⁿC₃ 5 + ⁿC₄ 5² +….

a - b = ⁿC₃ (8 – 5) + ⁿC₄ (8² – 5²) +….

pva -> $\left(\sim rvp\right)$  

$\sim \left(p\vee q\right)\vee \left(\sim rvp\right)$  

its negation as asked in question

$\left(p\vee q\right)\wedge \left(\sim p\wedge r\right)$  

=  $\left(p\wedge \sim p\wedge r\right)\vee \left(q\wedge r\wedge \sim p\right)$  

=  $\left(p\wedge r\wedge \sim p\right)\left[asp\wedge \sim pisfalse\right]$  

According to Newton's law of motion, we can write

ma = mg – N = mg - $\frac{mg}{4}=\frac{3mg}{4}$

$\Rightarrow a=\frac{3g}{4}$                             

According to conservation of energy, we can write

Gain in kinetic energy = Loss in potential energy

=>K_f – K_in = U_in - U_f

=>K - = mgy – mg (y – y₀) = mgy₀