Class 11th

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

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

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

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.

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

Given    n = 2^x. 3^y. 5^z . (i)

$y+z=5\&\frac{1}{y}+\frac{1}{z}=\frac{5}{6}$

On solving we get y = 3, z = 2

So, n = 2^x. 3³. 5²

So that no. of odd divisor = (3 + 1) (2 + 1) = 12

Hence no. of divisors including 1 = 12

${\displaystyle \sum _{n=1}^{\infty }\frac{n^2+6n+10}{\left(2n+1\right)!}}$

put 2n + 1 = r, r = 3, 5, 7,.

so n = $\frac{r-1}{2}$

$Now{\displaystyle \sum _{r=\left(3,5,7,.....\right)}^{\infty }\frac{n^2+6n+10}{\left(2n+1\right)!}}={\displaystyle \sum _{r=\left(3,5,7.....\right)}^{\infty }\left(\frac{\frac{{\left(r-1\right)}^2}{4}+3r-3+10}{r!}\right)}={\displaystyle \sum _{r=\left(3,5,7,......\right)}^{\infty }\left(\frac{r^2+10r+29}{4.r!}\right)}$

$=\frac{1}{8}\{41e-\frac{19}{e}-80)=\frac{41}{8}e-\frac{19}{8}{e}^{-1}-10$

N > O > Be > B

Due to half filled configuration N has more I.E than oxygen and due to fully filled configuration Be has more I.E than B.

(a) → (ii)                (b) → (iii)            (c) → (iv)            (d) → (i)