Class 11th

Block is not slipping, so

f = mrw² = 2 * 1 * 16 = 32

$S=\frac{F}{L}$  

[S] = [MT^–2]

$F=nA\frac{dv}{dx}$  

$\eta =\frac{ML{T}^{-2}\cdot T}{L^2}=M{L}^{-1}{T}^{-1}$

$\overrightarrow{L}=\overrightarrow{r}*\overrightarrow{p}$

L º [ML²T^–1]

$KE=\frac{1}{2}l{\omega }^2$  

KE º ML²T⁻²

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