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Ofcourse, you should join Chandigarh University. The university is really growing fast and it has always

been the first choice of students. Students dream to join here, not just because of the curriculum, but

because the university gives more than that. It gives top- class education, many specialization, and also a

lot of practical learning. The programs are designed as per the latest market demand and industry

trends. Along with this, the university keeps arranging workshops, seminars, and guest lectures. These

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future jobs.

Maxwell simply wanted to generate a magnetic field using electric current, which displacement current had the capability to offer. Since both current flows were able to produce an electric field, both of them could be simultaneously used for using in the equation.

According to question,

$\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{1}{4}$

$\frac{1}{a}+\frac{1}{b}=\frac{1}{2}........(i)$

Equation of required line is $\frac{x}{a}+\frac{y}{b}=1$ $\frac{}{}\frac{}{}$

Obviously B (2, 2) satisfying condition (i)

$\widehat{r}-\widehat{i}=2cos\theta \widehat{n}......\left(1\right)$

$\Rightarrow \widehat{r}=\widehat{i}-2\left(\widehat{i}.\widehat{n}\right)\widehat{n}$

$\Rightarrow \overrightarrow{b}=\overrightarrow{a}-2\left(\overrightarrow{a}.\overrightarrow{c}\right)\overrightarrow{c}$

No. There is no provision for the candidates to pay the WB ANM GNM application fee offline. The WB ANM GNM application fee payment mode is online only. Make use of Net Banking, Debit Card, or Credit Card only to pay the application fee.

$\Delta -\left|\begin{array}{ccc}\hfill 3\hfill & \hfill -2\hfill & \hfill -k\hfill \\ \hfill 2\hfill & \hfill -4\hfill & \hfill -2\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill -1\hfill \end{array}\right|=\left|\begin{array}{ccc}\hfill 3\hfill & \hfill -2\hfill & \hfill -k\hfill \\ \hfill 0\hfill & \hfill -8\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill -1\hfill \end{array}\right|,\left[R_2-R_1-2R_3\right]$

= 8 (3 + k)

For inconsistent $\Delta =0\Rightarrow k=3$

${\Delta }_x=\left|\begin{array}{ccc}\hfill 10\hfill & \hfill -2\hfill & \hfill -k\hfill \\ \hfill 6\hfill & \hfill -4\hfill & \hfill -2\hfill \\ \hfill 5m\hfill & \hfill 2\hfill & \hfill -1\hfill \end{array}\right|=32-40m\ne 0\Rightarrow m\ne \frac{4}{5}$

UG students must submit score of at least 65% along with necessary documents like LORs SOP English proficiency scores etc. After submitting documents students can pay the application fee and submit the application.

  $N=mgcos30°+qEsin30°$

$a=\frac{mgsin\theta -qEcos\theta -\mu N}{m}=2.30m/s^2$

$S=ut+\frac{1}{2}a{t}^2$

$\Rightarrow t=\sqrt[]{\frac{2\mathcal{l}}{a}}=1.31sec$             

y = x³

${\frac{dy}{dx}=3x^2\Rightarrow \frac{dy}{dx}|}_{\left(t,t^3\right)}=3t^2$

Equation of tangent y – t³ = 3t² (x – t) 

Let again meet the curve at $Q\left(t_1,t_1^3\right)$

$\Rightarrow t_1^3-t^3=3t^2\left(t_1-t\right)$

$t_1^2+t{t}_1+t^2=3t^2\left[?t_1\ne t\right]$

$t_1^2+t{t}_1-2t^2=0$

=> t₁ = -2t

Required ordinate = $\frac{2t^3+t_1^3}{3}=\frac{2t^3-8t^3}{3}=-2t^3$   

Frequency increases on filing. So, initial frequency of A is 335 Hz.

f = 340 – 5 = 335 Hz

The key difference between these two terms is that displacement current is deprived of electrons which flow in the case of an actual current. It's role is to help generate magnetic field which is produced when the medium changes between two different bodies.

-> $Z=\sqrt[]{R^2+{\left(X_L-X_C\right)}^2}$

$Z=\sqrt[]{{\left(120\right)}^2+{\left(10-100\right)}^2}=150\Omega$

$\omega =\frac{1}{\sqrt[]{LC}}=\frac{1}{\sqrt[]{10^{-1}\times 10^{-4}}}=\sqrt[]{10^5}$

$?\omega =2\pi f$

$\Rightarrow f=\frac{10^3}{2\pi \sqrt[]{10}}=50Hz$

$\frac{dp}{dt}=0.5p-450andP\left(0\right)=850$

$\Rightarrow \frac{dp}{P-900}=0.5dt$

${\displaystyle \underset{850}{\overset{0}{\int }}\frac{dp}{P-900}={\displaystyle \underset{0}{\overset{T}{\int }}0.5dt}}$

$\Rightarrow ln\left(P-900\right){|}_{805}^0=0.5T$

$\frac{T}{2}=ln\left|\frac{900}{50}\right|=ln18$

T = 2 ln 18

$A=A_0{e}^{-\lambda t_1}$     [Radio active decay law]

$\frac{A}{5}=A_0{e}^{-\lambda \left(t_2-t_1\right)}$

$\Rightarrow Averagelife=\frac{1}{\lambda }=\frac{\left(t_2-t_1\right)}{\mathcal{l}n5}$  

The exam pattern of VSAT consists of the subjects, topics and sub-topics on which the exam question paper will be based. As per the exam pattern, there are a total of 120 questions asked in the exam and the time duration is 2 hours and 30 minutes.

Stress = Y × strain

-> $\frac{T_1}{A}=Y\times -\frac{\left({\mathcal{l}}_1-\mathcal{l}\right)}{\mathcal{l}}\left(i\right)$  

  $\frac{T_2}{A}=Y\times \frac{\left({\mathcal{l}}_2-\mathcal{l}\right)}{\mathcal{l}}\left(ii\right)$                          

$\Rightarrow$ $\frac{T_1}{T_2}=\frac{{\mathcal{l}}_1-\mathcal{l}}{{\mathcal{l}}_2-\mathcal{l}}$                      

$\Rightarrow \mathcal{l}=\frac{T_1{\mathcal{l}}_2-T_2{\mathcal{l}}_2}{T_1-T_2}=\frac{T_2{\mathcal{l}}_1-T_1{\mathcal{l}}_2}{T_2-T_1}$            

TS PECET admit card will be visible for candidates some days before the exam. It is important to carry the hard copy of the exam if you wish to appear for the exam.

$y=\alpha x-\beta x^2$

$\Rightarrow \frac{dy}{dx}=\alpha -2\beta x=0$  

$\Rightarrow x=\frac{\alpha }{2\beta }$           

$y_{max}=\alpha \times \frac{\alpha }{2\beta }-\beta \times {\left(\frac{\alpha }{2\beta }\right)}^2=\frac{{\alpha }^2}{4\beta }$           

Range = $2x=\frac{\alpha }{\beta }=\frac{2u^{2}sin\theta .cos\theta }{g}$  

On comparing with

y = x tan θ - $\frac{g{x}^2}{2u^{2}co{s}^2\theta }$