Class 11th

Heavy water (D₂O) is used in organic reaction to know the kinetics of reaction

Heavy water D₂O is obtained by repeated electrolysis of H₂O.

Heavy water has b.pt 101.4°C and ordinary water has b.pt 100°C

(b.pt of D₂O > b.pt of H₂O)

D₂O has more molecular mass than H₂O, so greater degree of association and hence greater b.pt and viscosity.

Radial node = n $\mathcal{l}-$ 1 (5d)

= 5 – 2 – 1 = 2

Angular node = $\mathcal{l}$

= 2

Vitamin E is helpful in delaying blood clotting while vitamin K help blood clotting.

Not all dipole – dipole interactions are responsible for hydrogen bonding, so assertion A is false. F is most E. N atom and also hydrogen bonds in HF are symmetrical.

  $F_{1}cos45°+F_{2}cos45°+F_3=\frac{m{v}^2}{r}$

$\Rightarrow \frac{G{m}^2}{{\left(\sqrt[]{2}r\right)}^2}.\frac{1}{\sqrt[]{2}}+\frac{G{m}^2}{{\left(\sqrt[]{2}r\right)}^2}.\frac{1}{\sqrt[]{2}}+\frac{G{m}^2}{{\left(2r\right)}^2}=\frac{m{v}^2}{r}$            

  $\Rightarrow v=\sqrt[]{\frac{Gm}{4r}\left(2\sqrt[]{2}+1\right)}$          

Putting m = 1 kg and r = 1 m,

$v=\frac{1}{2}\sqrt[]{G\left(1+2\sqrt[]{2}\right)}$        

Velocity of block in equilibrium, in first case,

$v=A\omega =A.\sqrt[]{\frac{k}{M}}$                

Velocity of block in equilibrium, is second case,

$v\text{'}=A\text{'}\omega \text{'}=A\text{'}\sqrt[]{\frac{k}{M+m}}$                  

From conservation of momentum,

Mv = (M + m) v'

$\Rightarrow MA\sqrt[]{\frac{k}{M}}=\left(M+m\right)A\text{'}\sqrt[]{\frac{k}{M+m}}\Rightarrow A\text{'}=A\sqrt[]{\frac{M}{M+m}}$            

Since, $\frac{x^2}{\alpha kT}$  should be dimensionless.

So, dimension of   $\alpha ,\left[\alpha \right]=\frac{L^2}{M{L}^2{T}^{-2}}=M^{-1}{T}^2$

Dimension of  $\alpha {\beta }^2$ should be that of W.

So, $\left[\alpha {\beta }^2\right]=M{L}^2{T}^{-2}$  

$\Rightarrow \left[{\beta }^2\right]=\frac{M{L}^2{T}^{-2}}{M^{-1}{T}^2}=M^2{L}^2{T}^{-4}\Rightarrow \left[\beta \right]=ML{T}^{-2}$            

Given $ta{n}^{-1}a+t{a}^{-1}b=\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.$ ${}^{}{}^{}\raisebox{1ex}{$$}\!\left/ \!\raisebox{-1ex}{$$}\right.$

$ta{n}^{-1}\left(\frac{a+b}{1-ab}\right)=\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.$

$\frac{a+b}{1-ab}=1...........\left(i\right)$

OR a + b = 1 – ab .(ii)

Now, $\left(a+b\right)-\left(\frac{a^2+b^2}{2}\right)+\left(\frac{a^3+b^3}{3}\right)-\left(\frac{a^4+b^4}{4}\right)+........$ $\frac{{}^{}{}^{}}{}\frac{{}^{}{}^{}}{}\frac{{}^{}{}^{}}{}$

$\left(a-\frac{a^2}{2}+\frac{a^3}{3}+\frac{a^3}{3}-\frac{a^4}{4}+......\right)+\left(b-\frac{b^2}{2}+\frac{b^3}{3}-\frac{b^4}{4}+.......\right)$

log (1 + a) + log(1 + b) = log (1 + a) (1 + b) = log {1 + a + b + ab} = log_e2

$P{A}^2+P{B}^2=co{s}^2\theta +{\left(sin\theta -3\right)}^2+co{s}^2\theta +{\left(sin\theta +6\right)}^2$

= 2 cos² θ + 2 sin² θ + 6 sin q + 45

= 6 sin θ + 47

for maximum of PA² + PB², sin θ = 1

then P (1, 2)

Hence P, A & B will lie on a straight line.

  $\omega =\frac{2\pi }{T}\Rightarrow \frac{{\omega }_1}{{\omega }_2}=\frac{T_2}{T_1}=8$