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Securing an admission to top MBA universities in Europe is a competitive affair, but not impossible. International students must fulfill the admission requirements listed below order to boost their chances of pursuing the course:

London Business School

• An undergraduate degree
• Excellent GMAT/GRE scores (GMAT: Over 600 | GRE: At least 160 in both verbal and quantitative sections)
• At least 2 years of work experience
• Impressive letters of recommendation
• GPA of 3.3 or higher (on a scale of 4.0)

HEC Paris

• A bachelor's degree
• 3-4 years of work experience
• GMAT: 690 | GRE: 320
• Impressive application and profile

Cambridge Judge Business School

• A 2.1 UK honours degree or international equivalent
• Over 2 years of work experience
• Competitive GMAT/GRE scores (GMAT: 680 - 690)
• Impressive English proficiency exam scores

IE Business School

• An undergraduate degree from a recognised university
• At least 3 years of work experience
• Competitive GMAT/GRE scores

IESE Business School

• A bachelor's degree
• Strong English proficiency skills
• Excellent GMAT/GRE scores
• Impactful motivation letter

To get admission into Pratibha Institute of Business Management (Kamala Education Society's), one must complete a set of steps. The process involves appear for the accepted entrance exam, getting shortlisted, going through document verification, and admission confirmation. Find below the general steps to get a seat at the institute:

Step 1: Appear for the accepted entrance test. The institute accepts students based on state-level tests for all of the courses, while national-level exams are also considered for MBA admission.

Step 2: Go through document verfication by providing the set of credentials specified by the college.

Step 3: Pay the course fees to secure your seat at the institute.

11.29 Wavelength of the radiation, $\lambda$ = 3300 Å = 3300 $\times {10}^{-10}$ m

Speed of light, c = 3 $\times {10}^8$ m/s

Planck’s constant, h = 6.626 $\times {10}^{-34}$ Js

The energy of the incident radiation is given as:

$E$ = $\frac{hc}{\lambda }$

= $\frac{6.626\times {10}^{-34}\times 3\times {10}^8\mathrm{}}{3300\times {10}^{-10}}$ = 6.024 $\times {10}^{-19}$ J = $\frac{6.024\times {10}^{-19}}{1.6\times {10}^{-19}}$ eV = 3.765 eV

It can be observed that the energy of incident radiation is greater than the work function of Na and K only. Mo and Ni having more work function will not show the photoelectric emission.

If the source of light is brought near the photocells and placed 50 cm away from them, then the intensity of radiation will increase, but it will not affect the energy of the radiation. Hence, the result will be the same as before. However, the photoelectrons emitted from Na and K will increase in proportion to intensity.

Northeastern University is one of the best universities for higher education in the United States. Northeastern University placement rate attracts international students the most. Northeastern University placement rate stands at 97%, and the graduates are employed within 9 months of employment. This figure shows that out of every 100 students, around 97 students secured employment. The university graduates are in high demand worldwide. Northeastern University graduates work in top companies such as Google, Apple, Microsoft, Meta, Tesla, etc. Furthermore, the Northeastern University graduates work in esteemed sectors that are listed below:

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Length of the mild steel wire, l = 1.0 m

Area of cross-section, A = 0.5 $\times {10}^{-2}$ ${cm}^2$ = 0.5 $\times {10}^{-6}{m}^2$

A mass of 100 gm is suspended at the midpoint.

m = 100 gm= 0.1 kg

Due to the weight, the wire dips, as shown in the figure.

Original length = XZ, depression = l

The final length of the wire after it dips = XO + OZ

Increase in length of the wire, Δl = (XO + OZ) – XZ ……(i)

From Pythagoras theorem

XO = OZ = $\sqrt{{0.5}^2+l^2}$

From equation (i)

Δl = 2 $\times \sqrt{{0.5}^2+l^2}$ - 1.0 = 2 $\times 0.5\times \sqrt{1+{\left(\frac{l}{0.5}\right)}^2}$ - 1.0 = $\sqrt{1+{\left(\frac{l}{0.5}\right)}^2}$ - 1.0

Neglecting the smaller terms, we can write, Δl = $\frac{l^2}{0.5}$

We know, Strain = $\frac{Increaseinlength}{Originallength}$

Let T be the tension in the wire, then

mg = 2T $\cos ?\theta$

From the figure

$\cos ?\theta$ = $\frac{OY}{OX}$ = $\frac{0.5}{\sqrt{{0.5}^2+l^2}}$ = $\frac{0.5}{0.5\times \sqrt{1+{\left(\frac{l}{0.5}\right)}^2}}$

Expanding the expression and eliminating the higher terms:

$\cos ?\theta$ = $\frac{l}{0.5\times \sqrt{1+{\left(\frac{1}{2}\right)\left(\frac{l}{0.5}\right)}^2}}$

Since ( 1+ $\frac{l^2}{0.5}$ ) = 1 for small l, $\cos ?\theta$ = $\frac{1}{0.5}$

Then T = $\frac{mg}{2\left(\frac{l}{0.5}\right)}$ = $\frac{mg\times 0.5}{2l}$ = $\frac{mg}{4l}$

Stress = $\frac{Tension}{Area}$ = $\frac{mg}{4l\times A}$

Young’s modulus = $\frac{Stress}{Strain}$

Y = $\frac{mg\times 0.5}{?l\times A\times l^2}$

l= $\sqrt{\frac{mg\times 0.5}{4YA}}$

For steel, Y = 2 $\times {10}^{11}$ Pa, then

l = $\sqrt{\frac{0.1\times 9.8\times 0.5}{4\times 2\times {10}^{11}\times 0.5\times {10}^{-6}}}$ = 0.0106 m

Hence, the depression at the midpoint is 0.0106 m

Cross-sectional area of wire A, $a_1$ = 1 ${mm}^2$ = 1 $*{10}^{-6}{m}^2$

Cross-sectional area of wire B, $a_2$ = 2 ${mm}^2$ = 2 $*{10}^{-6}{m}^2$

Young's modulus for steel, $Y_1$ = 2 $*{10}^{11}$ N/ $m^2$

Young's modulus for aluminium, $Y_2$ = 7 $*{10}^{10}$ N/ $m^2$

Stress in the wire = $\frac{Force}{area}$ = $\frac{F}{a}$

If the two wires have equal stresses, then

$\frac{F_1}{a_1}$ = $\frac{F_2}{a_2}$ or $\frac{F_1}{F_2}$ = $\frac{a_1}{a_2}$ = $\frac{1}{2}$ ………(i)

Where $F_1$ is the force exerted on steel wire and $F_2$ is the force exerted on aluminium wire

Taking a moment around the point of suspension, we get

$F_{1}y$ = $F_2(1.05-y)$

$\frac{F_1}{F_2}$ = $\frac{(1.05-y)\mathrm{}}{y}$ ……(ii)

Using equation (i) and (ii), we can write

$\frac{1}{2}=\frac{(1.05-y)\mathrm{}}{y}$ or y = 2.1 – 2y or y = 2.1/3 = 0.7 m

We know Young's modulus Y = Stress / Strain

Y = (F/A)/ Strain or Strain = (F/A)/Y

In case of Steel wire, ${Strain}_{steel}$ = $\frac{F_1}{a_1*Y_1}$

In case of Aluminium wire, ${Strain}_{alu}$ = $\frac{F_2}{a_2*Y_2}$

Since both the strains are equal, we get

$\frac{F_1}{F_2}$ = $\frac{a_1*Y_1}{a_2*Y_2}$ = $\frac{1*{10}^{-6}*2*{10}^{11}}{2*{10}^{-6}*7*{10}^{10}}$ = 1.429

From (ii) we get

$\frac{(1.05-y)\mathrm{}}{y}$ = 1.429

y = 0.432m

In order to produce an equal strain in the two wires. The mass should be suspended at a distance of 0.432m from the end A

11.28 Einstein’s photoelectric equation is given as:

$e{V}_0$ = $h\nu -{\varnothing }_0$

$V_0=\frac{h}{e}\nu -\frac{{\varnothing }_0}{e}$ ……….(1)

where

$V_0=$ Stopping potential

h = Planck’s constant

e = Charge of an electron

$\nu =\mathrm{F}\mathrm{r}\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{y}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{r}\mathrm{a}\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}$

${\varnothing }_0=\mathrm{W}\mathrm{o}\mathrm{r}\mathrm{k}\mathrm{}\mathrm{f}\mathrm{u}\mathrm{n}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{a}\mathrm{}\mathrm{m}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}$

Speed of light, c = 3 $\times {10}^8$ m/s

It can be concluded from equation (1) that $V_0$ is directly proportional to frequency $\nu$

Now frequency $\nu$ can be expressed as

$\nu =\frac{c}{\lambda }$