Ask & Answer: India's Largest Education Community

Case I :- $2?-?=\frac{1}{2}m{v}_1^2........\left(i\right)$  

Case II :-     $10?-?=\frac{1}{2}m{v}_2^2........\left(ii\right)$

$\Rightarrow \frac{1}{9}=\frac{v_1^2}{v_2^2}$                                        

$\Rightarrow v_1:v_2=1:3$                            

->x = 1

Kindly consider the following figure

$V=k{T}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$  

-> $T{V}^{-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=K$

$\Rightarrow \gamma -1=-3/2$

$\Rightarrow \gamma =-1/2$            

Work done =  $\frac{nR\Delta T}{-y+1}=\frac{1\times R\times 90}{3/2}=6R$

->n = 60

No, the exam body doesn't send the VSAT result through post or any other offline means. To download the VSAT result, candidates must visit the official website only.

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

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

Total oscillation =   $\frac{1}{2}+\frac{1}{8}=\frac{5}{8}$

$\Rightarrow timetaken=\frac{T}{2}+\frac{T}{12}=\frac{7T}{12}$            

->a = 7

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

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

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

Kindly go through the solution

 $y=\overline{\overline{A}+B}=A.\overline{B}$

The placement record for the Bachelor of Engineering in Computer Science and Engineering with a specialisation in Mobile Computing at Chandigarh University is generally good. Many reputed national and international companies visit the campus for recruitment, offering roles in software development, mobile application development, and related fields. However, the highest packages are offered to a limited number of students, and the average placement depends on individual skills, academic performance, and project work.

ax² + bx + c = 0

D = b² – 4ac

D = 0

b² – 4ac = 0

b² = 4ac

(i) AC = 1, b = 2 (1, 2, 1) is one way

(ii) AC = 4, b = 4

$\begin{array}{l}a=4c=1\\ a=2c=2\\ a=1c=4\end{array}\}3ways$

(iii) AC = 9, b = 6, a = 3, c = 3 is one way

1 + 3 + 1 = 5 way

Required probability = $\frac{5}{216}$   

All the charge given to a conducting sphere resides on outer surface.

Let sin = t

${\displaystyle \frac{}{}}$ ${\displaystyle \int \frac{sin\theta \left(2sin\theta .cos\theta \right)\left(si{n}^6\theta +si{n}^4\theta +si{n}^2\theta \right)\sqrt[]{2si{n}^4\theta +3si{n}^2\theta +6}}{2si{n}^2\theta }}$  d

sin = t

cos . d = dt

${\displaystyle \int \frac{u^{1/2}}{12}}du=\frac{u^{3/2}}{18}+C=\frac{{\left(2t^6+3t^4+6t^2\right)}^{3/2}}{18}+C=\frac{{\left(2si{n}^6\theta +3si{n}^4\theta +6si{n}^2\theta \right)}^{3/2}}{18}+C$

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 a = $\frac{1}{2}$              

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

  $\Delta E=13.6\left(\frac{1}{1^2}-\frac{1}{5^2}\right)=13.6\times \frac{24}{25}eV$

$\Rightarrow \frac{hc}{\lambda }=13.6\times \frac{24}{25}eV..........\left(1\right)$          

With the help of conservation of linear momentum, we can write

$\frac{h}{\lambda }=m_H{v}_H\Rightarrow \frac{hc}{\lambda }=c{m}_H{v}_H\Rightarrow v_H=\frac{\frac{hc}{\lambda }}{c{m}_H}=\frac{13.6\times \frac{24}{25}\times 1.6\times 10^{-19}}{3\times 10^8\times 1.67\times 10^{-27}}=4.17m/s$