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At Chandigarh University, liberal arts is all about learning both theory and practical. Students 

study major subjects like Psychology, English, Sociology, Economics, and Political Science. They can also choose minor subjects such as Artificial Intelligence, Management, Media Studies, Performing Arts, and Humanities and Social Sciences. Students also learn through 

classroom discussions, projects, case studies, and application-based assessments and get hands-on learning with workshops, seminars, heritage walks, and visits to institutions. Students also get a chance to go for semester exchange programs. 

The total tuition fees of the MBA course at Vignan's Foundation for Science, Technology & Research is INR 6 lakh. Overall fee may include certain other components such as hostel and mess fees (if available), exam fees, security deposists, etc.

The duration of the BBA course at Vignan's Foundation for Science, Technology & Research is of three years. Admissions to the course are merit/entrance-based. Eligible candidates are those who have passed Class 12 with a minimum if 50% aggregate. 

The M.Tech course at Vignan's Foundation for Science, Technology and Research is offered in the following specialisations:-

• CSE
• Embedded Systems
• Electric Vehicle Technology, etc. 

  $F_x=\frac{-dv}{dx}$           

  $\overrightarrow{F}=-3x\widehat{i}-5\widehat{j}-6\widehat{k}$          

  ${\left|\overrightarrow{F}\right|}_{x=6}=\sqrt[]{18^2+5^2+6^2}$          

$=\sqrt[]{324+25+36}$

= $\sqrt[]{385}$    

= 19.62N

%error= $\left(\frac{dV}{V}+\frac{dR}{R}\right)*100$  

$=\left(\frac{5}{200}+\frac{0.2}{20}\right)*100$                          

= 3.5

Since $a=\frac{gsin?}{1+\frac{I}{M{R}^2}}$

$a=\frac{g*\frac{\sqrt[]{3}}{2}}{1+\frac{1}{2}}=\frac{g\frac{\sqrt[]{3}}{2}}{\frac{3}{2}}$

$a=\frac{g}{\sqrt[]{3}}$

f = 15

$m=-\frac{v}{u}=+\frac{1}{2}$            

  $v=-\frac{u}{2}$           

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$            

$\frac{2}{-u}+\frac{1}{u}=\frac{1}{f}$            

u = – f = – 15cm

PV = nRT

->Pµn

->Ratio= $\frac{32}{2}$ =16

$\left|\Delta E_0\right|=\left(-136\left\{1-\frac{1}{4}\right\}\right)eV$

|DE₀| = –10.2

$\lambda =\frac{12400}{10.2}\times 10^{-10}m$

$\rho =\frac{h}{\lambda }=\frac{6.63\times 10^{-34}\times 10.2}{12400\times 10^{-10}}$                  

$?$ $mv=\frac{h}{\lambda }$            

  $1.8\times 10^{-27}$           

$v=\frac{6.63\times 10.2\times 10^{-34}}{12400\times 10^{-10}}$

$v=\frac{6.63\times 10.2}{12400\times 1.8}\times 10^3$            

$=\frac{6.63\times 102}{124\times 1.8}=3.02$ ]

= 3 m/s

From Gauss law

$?=\frac{q_{enclosed}}{{\epsilon }_0}=\frac{5q}{{\epsilon }_0}$

KE =  $\frac{1}{2}$ m W² (A²– n²)

TE = $\frac{1}{2}$ mw²A²

^{$\frac{KE}{TE}=\frac{A^2-n^2}{A^2}=\frac{1-\frac{1}{9}}{1}=\frac{8}{9}$}

Since Gand S are in parallel

-> V_G = V_S

->3mA * 10 = 8A * R_S

->R_S = 3.75mW

i_battery= $\frac{10?3}{1000}$  = 7mA

i_2kW = $\frac{3}{2000}$  =1.5 mA

i_z = (7 – 1.5) mA

= 5.5mA

Hospitality, Customer Service, and Retail are some popular sectors where Indian students seek part-time jobs. These sectors offer students a great chance to gain practical experience and develop their skills in sectors such as sales, customer interaction, and event management. 

Campus-based roles such as library assistant, IT helpdesk assistant, or residence assistant, are also popular among students, while off-campus job options include work at restaurants, cafes, or supermarkets. 

For maximum value of s, initially, electron must move away from plate.

ut + $\frac{1}{2}$ at² = s

t = 1u = 1m/s               s = –1 m

1 × 1 – $\frac{1}{2}$ a × 1² = – 1

-> a = 4m/s²

$\frac{qE}{m}=4$      

  $\frac{q\sigma }{2{\epsilon }_{0}m}=4$     

$\sigma =\frac{4\times 2\times 9\times 10^{-12}\times 9\times 10^{-31}}{1.6\times 10^{-19}}$

$=\frac{8\times 81}{1.6}\times 10^{-24}$      

= 4.05 × 10^–22C/m²

  $y=xtan\theta -\frac{g{x}^2}{2v^{2}co{s}^2\theta }$

(2.5, 2.5) must lie on this

-> $1=tan\theta -\frac{g\times 2.5}{2v^{2}co{s}^2\theta }$

-> $\frac{25}{2v^{2}co{s}^2\theta }=tan\theta -1$

-> $v^2=\frac{25}{2}\left\{\frac{1+ta{n}^2\theta }{tan\theta -1}\right\}$   

-> $v_{min}=5\sqrt[]{\sqrt[]{2}+1}$  

[Happens when tan θ = $\sqrt[]{2}$  + 1]

${\displaystyle \int I.dt}={\displaystyle \underset{0}{\overset{10}{\int }}\left(20+3t\right)dt}$

$={\left(20t\right)}_0^{10}+3{\left(\frac{t^2}{2}\right)}_0^{10}=350C$

$\frac{1}{20}=\left(1.5-1\right)\left(\frac{2}{R}\right)$

R = 20cm

$\frac{1}{f\text{'}}=\left(\frac{1.5}{1.6}-1\right)\left(\frac{2}{R}\right)$            

$=\frac{-1}{16}\times \frac{2}{20}$                  

f' = – 160cm