Class 11th

$\frac{GM}{r}=6.4*10^7$                       . (i)

$\frac{GM}{r^2}=6.4$                   . (ii)

$r=\frac{6.4*10^7}{6.4}$  

= 10⁷ m

= 10,000 km

R + h = 10,000

h = 10,000 – 6400 = 3600 km

$V_{rms}=\sqrt[]{\frac{3RT}{M}}$  

$\frac{T}{2}=\frac{320}{32}$  

T = 20 K

T = –253°C

Young's modulus is property of material of wire and it is independent of geometrical factors.

$v_1=\sqrt[]{2g10}$

$v_2=\sqrt[]{2g5}$

$\overrightarrow{l}=\Delta \overrightarrow{p}$

$l=0.1\left\{\sqrt[]{2g10}+\sqrt[]{2g5}\right\}$

$=0.1\left\{10\sqrt[]{2}+10\right\}$

$=\left(\sqrt[]{2}+1\right)Ns$

By conservation of mechanical energy

mg (1) =  $\frac{1}{2}$ mv² + mg (0.5)

v² = 10

v =  $\sqrt[]{10}$ m/s

Velocity at maximum height = vcoss30°

L = m (vcos30) H

$=mv\left(\frac{\sqrt[]{3}}{2}\right)*\frac{v^{2}si{n}^{2}30}{2g}$

$=\sqrt[]{3}\frac{m{v}^3}{16g}$

For 2 kg block

T – 2g sin37 = 2a        . (i)

For 4 kg block

4g – 2T =  $\frac{4a}{2}$

2g – T = a                    . (ii)

T = (2g – a)

2g – a – 2g *  $\frac{3}{5}$  = 2a

3a = 2g *  $\frac{2}{5}$

$a=\frac{4g}{15}$

x + 2y + 3z = 42

0    x + 2y = 42 ->22 cases

1    x + 2y = 39 ->19 cases

2    x + 2y = 36 ->19 cases

3    x + 2y = 33 ->17 cases

4    x + 2y = 30 ->16 cases

5    x + 2y = 27 ->14 cases

6    x + 2y = 24 ->13 cases

7    x + 2y = 21 ->11 cases

8    x + 2y = 18 ->10 cases

9    x + 2y = 15 ->8 cases

10  x + 2y =12 -> 7 cases

11  x + 2y = 9 -> 5 cases

12  x + 2y = 6 -> 4 cases

13  x + 2y = 3 -> 2 cases

14  x + 2y = 0 -> 1 cases.

$tanB*tanC=\frac{\sqrt[]{x}}{\sqrt[]{x^2+x+1}}*\frac{1}{\sqrt[]{x\left(x^2+x+1\right)}}$

$=\frac{1}{x^2+x+1}=ta{n}^{2}A$            

tan² A = tan B tan C

It is only possible when A = B = C at x = 1

A = 30°, B = 30°, C = 30° $\left[tanA=tanB=tanC=\frac{1}{\sqrt[]{3}}\right]$                                 

$A+B=\frac{\pi }{2}-C$

$\sqrt[]{3}-\sqrt[]{2}=\frac{\left(\sqrt[]{3}+\sqrt[]{2}\right)\left(\sqrt[]{3}-\sqrt[]{2}\right)}{\left(\sqrt[]{3}+\sqrt[]{2}\right)}=\frac{1}{\sqrt[]{3}+\sqrt[]{2}}$

Let $\sqrt[]{3}+\sqrt[]{2}=t$  

$t^x+\frac{1}{t^x}=\frac{10}{3}$

Let  $t^x=y$ $y+\frac{1}{y}=\frac{10}{3}$           

y = 3 or   $\frac{1}{3}$

${\left(\sqrt[]{3}+\sqrt[]{2}\right)}^x=3$ or $\frac{1}{3}$  

$xlog\left(\sqrt[]{3}+\sqrt[]{2}\right)=ln3$ or –ln3

$x=\frac{ln3}{ln\left(\sqrt[]{3}+\sqrt[]{3}\right)}$   or $\frac{-ln3}{\sqrt[]{3}+\sqrt[]{2}}$  

two real values of x

$f\left(x\right)=\{\begin{array}{cc}\hfill e^{-x},\hfill & \hfill x<0\hfill \\ \hfill lnx,\hfill & \hfill x>0\hfill \end{array}$

$g\left(x\right)=\{\begin{array}{cc}\hfill e^x,\hfill & \hfill x<0\hfill \\ \hfill x,\hfill & \hfill x>0\hfill \end{array}$