Class 11th

As we go pole to equator, acceleration due to gravity decreases. So, weight of the body will also reduces. Thus, weight on equator will be slightly smaller than 49 N.

$T=2\pi \sqrt[]{\frac{L}{g}}or{T}^2=4{\pi }^2\left(\frac{L}{g}\right)$

$\Rightarrow g=4{\pi }^2\left(\frac{L}{T^2}\right)$

$\Rightarrow \frac{\Delta g}{g}\mathrm{\%}=\left(\frac{\Delta L}{L}+\frac{2\Delta T}{T}\right)\mathrm{\%}=\left[\frac{1mm}{1m}+\frac{2*0.01}{1.95}\right]*100\mathrm{\%}=\left(0.001+0.0102\right)*100=1.13\mathrm{\%}$

Equation of given line x – y + 1 = 0 .(i),

equation of per perpendicular line PP' is

-x – y + $\lambda =0$  

As line passing through (3, 5)

$\therefore -3-5+\lambda =0$   

$\lambda =8$  

Equation of line PP' is –x – y + 8 = 0 .(ii)

Solving (i) and (ii)

$\begin{array}{l}-2y=-9\\ y=\frac{9}{2}\end{array}\}Q=\left(\frac{7}{2},\frac{9}{2}\right)$

$y=\frac{9}{2}$

$\therefore x-\frac{9}{2}+1=0$

$P\text{'}=\left(4,4\right)$           

$\frac{5+y_1}{2}=\frac{9}{2}$           

y₁ = 4

According to KTG, the gas exerts pressure because its molecule :

suffer change in momentum when impinge on the walls of container.

$x^2-2\left(3k-1\right)x+8k^2-7>0\Rightarrow a>0,\&D<0,$       

a = 1 > 0 and D < 0

4 (3k – 1)² – 4 (8k² – 7) < 0

$\left(9k^2+1-6k-8k^2+7<0\right)$            

$k(k-4)-2(k-4)<0$    

$k\in \left(2,4\right)$           

K = 3

  $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{-2}=r$

For 'B', x = 2r + 1, y = 3r – 1, z = -2r + 1

As AB is perpendicular to the line,

$\left\{\left(2r+1\right)-0\right\}2+\left\{\left(3r-1\right)-1\right\}3+\left\{\left(-2r+1\right)-2\right\}$

$r=\frac{2}{17}\Rightarrow B\left(\frac{2}{17},\frac{-11}{7},\frac{13}{17}\right)$ ->

direction ratios of AB

(2r + 1, 3r – 2, -2r – 1)

Equation of AB

$\frac{x}{-3}=\frac{y-1}{4}=\frac{z-2}{3}$  

l + m – n = 0

l + m = n . (i)

l² + m² = n²

Now from (i)

l² + m² = (l + m)²

->2lm = 0

->lm = 0

l = 0 or m = 0

->m = n Þ l = n

if we take direction consine of line

$0,\frac{1}{\sqrt[]{2}},\frac{1}{\sqrt[]{2}}and\frac{1}{\sqrt[]{2}},0,\frac{1}{\sqrt[]{2}}$  

cos α =   $\frac{1}{2}$

$si{n}^4\alpha +co{s}^4\alpha ={\left(\frac{\sqrt[]{3}}{2}\right)}^4={\left(\frac{1}{2}\right)}^4=\frac{9}{16}+\frac{1}{16}=\frac{10}{16}=\frac{5}{8}$           

xyz = 24

24 = 2³ * 3

Let's distribute 2, 3 among 3 variables. No. of positive integral solution =

No. of ways to distribute = 

$\therefore In\Delta AQP$         

$tan30°=\frac{PQ}{AQ}$ $\frac{1}{\sqrt[]{3}}=\frac{h}{x+y}$        

x + y = $\sqrt[]{3}h.....\left(i\right)$  

$\therefore ln\Delta BQP$          

$tan45°=\frac{h}{y}$          

h = y .(ii)

(i) & (ii) x + y = $\sqrt[]{3}y$  

$\Rightarrow x=\left(\sqrt[]{3}-1\right)y.......\left(iii\right)$     

Let the speed be S

$\frac{x}{S}=20$

x = 20.S

from (iii)

$\frac{y}{S}=10\left(\sqrt[]{3}+1\right)$