Ask & Answer: India's Largest Education Community

The admission process designed by the Magadh University for the BA (H) course consists of four basic steps. To start, students who are eligible must visit the official website of the university and fill out an application form. Upon filling out the form, they would be able to download the offer letter.

Further, students must take this application form to the preferred institute/college. The college will further verify the student details. After verification, students are required to log in to the student account, fill out the required details, and pay the registration fees. After that, students who are offered admissions must pay the fees prescribed by the university.

2.25 Let C' be the equivalent capacitance for capacitors $C_2$ and $C_3$ connected in series.

Hence, $\frac{1}{C\text{'}}$ = $\frac{1}{200}$ + $\frac{1}{200}$ . So C' = 100 pF

Capacitors C' and $C_1$ are in parallel, if the equivalent capacitance be C”, then

C” = C' + $C_1$ = 100 + 100 = 200 pF

Now C” and $C_4$ are connected in series. If the total equivalent capacitance of the circuit be $C_{Total}$ , then $\frac{1}{C_{Total}}$ = $\frac{1}{C"}$ + $\frac{1}{C_4}$ = $\frac{1}{200}$ + $\frac{1}{100}$ , $C_{Total}$ = $\frac{200}{3}$ pF = $\frac{200}{3}*{10}^{-12}$ F

Let V” be the potential difference across C” and $V_4$ be the potential difference across $C_4$

Then, V” + $V_4$ = 300 V

Now, charge $Q_4$ on $C_4$ is given by $Q_4$ = $C_{Total}*V$ = $\frac{200}{3}*{10}^{-12}$ $*300$ = 2 $*{10}^{-8}$ C

$V_4=\frac{Q_4}{C_4}$ = $\frac{2\mathrm{}*{10}^{-8}}{100*{10}^{-12}}$ = 200. So V” = 100 V

So potential difference across C' and $C_1$ is V” = 100 V

Charge on $C_1$ is given by $Q_1$ = 100 $*{10}^{-12}*100$ C = 1 $*{10}^{-8}$ C

$C_2$ and $C_3$ are having same capacitance and have a potential difference of 100V together. Since $C_2$ and $C_3$ are in series, the potential difference is given by $V_2$ = $V_3$ = 50V

Hence the charge on $C_2$ is given by $Q_2$ = 200 $*{10}^{-12}*50$ = ${10}^{-8}$ C and the charge on $C_3$ is given by $Q_3$ = 200 $*{10}^{-12}*50$ = ${10}^{-8}$ C

Therefore, the equivalent capacitance of the circuit is $\frac{200}{3}$ pF and charge and voltage at all capacitance is given as

For $C_{1:}{Q}_1$ = ${10}^{-8}$ C, $V_1$ = 100 V

For $C_{2:}{Q}_2$ = ${10}^{-8}$ C, $V_2$ = 50 V

For $C_{3:}{Q}_3$ = ${10}^{-8}$ C, $V_3$ = 50 V

For $C_{4:}{Q}_4$ = 2 $*{10}^{-8}$ C, $V_4$ = 200 V

Chennai offers a Master of Arts (M.A.) in Public Administration as part of its postgraduate programs. The eligibility criteria for this course require candidates to have completed a Bachelor's degree from any recognized university with a minimum aggregate of 50%. Admission is based on merit, and there is no entrance examination for this program.  

Regarding the cutoff for the M.A. in Public Administration at MCC, the college does not publish specific cutoff marks for this program. Admission is determined by the merit of the applicants, which includes their undergraduate academic performance. The college follows the reservation norms of the Government of Tamil Nadu during the selection process

2.23 Requirements:

Capacitance, C = 2 $\mu F$

Potential difference, V = 1 kV = 1000 V

Available:

Each capacitor, capacitance, $C_1$ = 1 $\mu F$

Potential difference, $V_1$ = 400 V

Assumption:

A number of capacitors are connected in series and these series circuits are connected in parallel to each other. The potential difference across each row in parallel connection must be 1000 V and potential difference across each capacitor must be 400 V.

Hence, number of capacitors in each row is given as $\frac{1000}{250}$ = 2.5

So the number of capacitor in each row = 3

The equivalent capacitance in each row is given as, $\frac{1}{C_R}$ = $\frac{1}{1}+\frac{1}{1}$ + $\frac{1}{1}$ = 3, so $C_R=\frac{1}{3}$ $\mu$ F

Let there be ‘n’ rows, each having 3 capacitors, which are connected in parallel

The total equivalent capacitance of the circuit is given as

$C_{eff}$ = $\frac{1}{3}$ + $\frac{1}{3}$ + $\frac{1}{3}$ + $\frac{1}{3}$ + ……. upto n = $\frac{n}{3}$

It is given $C_{eff}$ = 2. So n = 6

So the final circuit will be 3 capacitors in each row and 6 rows connected in parallel, Total capacitors required = 3 $\times 6=18$

Marri Laxman Reddy Institute of Technology and Management (MLRITM) provide an online mode of application. Applicants searching for admission to any courses in the field of Pharmacy are required to visit the official website of the college to download the application form for their desired courses. Below is the detailed process to fill out Marri Laxman Reddy Institute of Technology and Management application form:

Step 1: Go through the official website of the college.

Step 2: Click on the “Admission Tab”.

Step 3: Download the application form for the desired courses.

Step 4: Print the form and fill out all the required details.

Step 5: Take the print of all the required documents.

Step 6: Submit the form to the college along with documents.

2.22 Four charges of same magnitudes are placed at points X, Y, Y and Z respectively, as shown in the figure.

It can be considered that the system of the electric quadrupole has three charges.

Charge +q placed at point X

Charge -2q placed at Y

Charge +q placed at point Z.

XY = YZ = a

YP = r

PX = r + a

PZ = r – a

Electrostatic potential caused by the system of three charges at point P is given by,

V = $\frac{1}{4\pi {\epsilon }_0}[\frac{q}{XP}$ - $\frac{2q}{YP}$ + $\frac{q}{ZP}$ ]

= $\frac{1}{4\pi {\epsilon }_0}[\frac{q}{r+a}$ - $\frac{2q}{r}$ + $\frac{q}{r-a}$ ]

= $\frac{q}{4\pi {\epsilon }_0}[\frac{r\left(r-a\right)-2\left(r-a\right)\left(r+a\right)+r(r+a)}{r\left(r-a\right)(r+a)}$ ]

= $\frac{q}{4\pi {\epsilon }_0}[\frac{r^2-ra-2r^2+2a^2+r^2+ra}{r\left(r-a\right)(r+a)}$ ]

= $\frac{q}{4\pi {\epsilon }_0}[\frac{2a^2}{r(r^2-a^2)}$ ]

= $\frac{2q{a}^2}{4\pi {\epsilon }_0{r}^3(1-\frac{a^2}{r^2})}$

Since $\frac{r}{a}?1,$ therefore $\frac{a}{r}?1$ . So $\frac{a^2}{r^2}$ is taken as negligible.

Therefore V = $\frac{2q{a}^2}{4\pi {\epsilon }_0{r}^3}$

It can be inferred that V $\propto \frac{1}{r^3}$ . It is known that for a dipole V $\propto \frac{1}{r^2}$ and for monopole V $\propto \frac{1}{r}$