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Yes, Computer Science Engineering is one of the most in demand programs at Chandigarh University. Due to strong placement records, industry collaborations, and specialisation options like AI, Data Science, and Cybersecurity, many students prefer CSE. The university manages large enrollments efficiently through advanced labs and structured academic planning. Despite high demand, the teaching methodology ensures quality learning and project based exposure for every student.

Galvanometer is a device having low resistance which is used to detect tiny bits of electric current scattered among different directions in the circuit. Whereas, a voltmeter is used to calculate the potential difference between two different points in the circuit. A galvanometer can also be used as a voltmeter if connected to high resistance in a series.

The allotment of the WB ANM/GNM exam centre is based on the choices of zones given on the application form. In case a candidate is not allocated a zone of their choice, request to change it will not be entertained.

 As per unofficial sources, the Georgetown Law acceptance rate stands at around 13%. This suggests the institute is very selective at admitting students to its law courses. 
 
Interested candidates must perform really well in school-level & entrance-level exams to have a chance of securing seat at Georgetown Law.

WBJEEB, the official exam conducting body, released a list of official test center zones where the West Bengal ANM GNM 2025 exam was conducted on October 19, 2025. At the time of filling out the West Bengal ANM GNM application form, candidates had to choose the preferred test center zone out of all the choices available in the vicinity of their home.

They core difference between these two terms lies in the fact that self induction involves only one coil as compared to mutual induction involving two. In mutual induction, the change of electric current in one coil produces a magnetic field which results in generating EMF in the second coil. Whereas there is no concept of neighbouring coil in self induction and everything happens within the one coil itself.

kx + y + 2z = 1. (i)

3x – y – 2z = 2 . (ii)

2x – 2y – 4z = 3. (iii)

(ii) × 5 (i) $\equiv$ $$  (iii) × 3 (15 – k) = 6

K = 21

Let f (x) = x⁶ + ax⁵ + bx⁴ + ax³ + dx² + ex + f

$\underset{x\to 0}{lim}\frac{f\left(x\right)}{x^3}=1$    Non zero finite

So, d = e = f = 0

f (x)  = x⁶ + ax⁵ + bx⁴ + cx³

$\underset{x\to 0}{lim}\frac{f\left(x\right)}{x^3}=1$ Non zero finite

f’ (x) = $6x^5+5a{x}^4+4b{x}^3+3x^2$

f’ (1) = 0

6 + 5a + 4b + 3 = 0

5a + 4b = - 9 . (i)

f’ (-1) = 0

-6 + 5a – 4b + 3 = 0 . (ii)

Solving (i) and (ii)

a  -3/5, b = -3/2

$\therefore f\left(x\right)=x^6+\left(\frac{-3}{5}\right)x^5+\left(\frac{-3}{2}\right)x^4+x^3$

$\therefore 5.f\left(2\right)=144$

Georgetown Law is a very reputable university in terms of post-study employment opportunities for its students. 
 
The institution has strong connections within the industry to place its students at some of the best law firms in the US. 
 
As per the latest employment data of 2023 that is available on the official website of the university, 685 members of the Class of 2023 applied for placements, 655 were able to successfully secure full-time positions at some top law firms, at a rate of 95.62%. 

  $\frac{GM}{{\left(3R/2\right)}^2}=\frac{GM}{R^3}\times r$

$\Rightarrow OA=\frac{4R}{9}=r$

$AB=R-\frac{4R}{9}=\frac{5R}{9}\Rightarrow OA:AB=4:5=x:y\Rightarrow x=4$

K $=\frac{4\sqrt[]{3}}{\sqrt[]{3}x+y}$ $\frac{\text{exception 25:}}{\text{exception 25:}}$

K = $\frac{\sqrt[]{3}x-y}{4\sqrt[]{3}}$ $\frac{\text{exception 25:}}{\text{exception 25:}}$

$\therefore \frac{4\sqrt[]{3}}{\sqrt[]{3}x+y}=\frac{\sqrt[]{3}x-y}{4\sqrt[]{3}}$

$\frac{x^2}{16}-\frac{y^2}{48}=1$

48 = 16 (e2 – 1)

$e=\sqrt[]{4}=2$

With the help of conservation of volume, we can write

  $27\times \frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3\Rightarrow R=3r.......\left(1\right)$                            

With the help of conservation of charge, we can write

Q = 27 q.(2)

Potential energy of single drop = U₁ =   $\frac{q^2}{8\pi {\epsilon }_{0}r}$

Potential energy of bigger drop =   $U_2=\frac{Q^2}{8\pi {\epsilon }_{0}R}=\frac{27\times 27\times q^2}{8\pi {\epsilon }_0\left(3r\right)}=243\left(\frac{q^2}{8\pi {\epsilon }_{0}r}\right)=243U_1$

$\Rightarrow \frac{U_2}{U_1}=243$           

Let a_n be the side of square A_n

$a_n=\sqrt[]{2}{a}_{n+1}$

a₁ = 12

a_n = 12 $\times {\left(\frac{1}{\sqrt[]{2}}\right)}^{n-1}$

${\left(a_n\right)}^2<1$

$\frac{144}{2^{\left(n-1\right)}}<1$

$\Rightarrow 2^{\left(n-1\right)}>144$

$\Rightarrow n-1\ge 8$

$\Rightarrow n\ge 9$

$l=\frac{90?30}{4000}=15mA$

$I_1=\frac{30}{5000}=6mA\&l_2=9mA$