Ask & Answer: India's Largest Education Community

Choosing the best books for GRE prep is not a tough task. You need to keep few things in mind:

1. Choose only official ETS materials for accurate question types. 
2. Look for books that explain GRE concepts clearly and offer plenty of practice questions with detailed explanations.
3. If the books cover all sections: Verbal, Quant, and AWA then its good or you can buy section-wise books of reputed publishers.
4. Read reviews to see what other students found helpful for their learning style.

To get selected in group discussion, candidates must:

• Focus on the topic given
• Listen actively
• Communicate the ideas clearly
• Demonstrate strong communication skills
• Maintain group dynamics and etiquettes

Kindly go through the solution

Lady Shri Ram College offers UG, PG, and other courses to students. These programmes are offered across BA,  Maths,  and various other streams. At the UG level, the college offers BSc, BA, BCom, and other courses to students. At the postgraduate level, the college offers MA,  MSc, and other courses to students. The college also provides Diploma and other courses to students

W₀ = 1.9 eV = 1.9 x 1.602 x 10^-19 J

Threshold frequency, v₀ = W₀ / h = (1.9 x 1.602 x 10^-19 J) / (6.626 x 10^-34Js)

= 0.459 x 10¹⁵ s^-1 = 4.59 x 10¹⁴ s^-1

Threshold wavelength? ₀ = c / v₀ = (3 x 10⁸ ms^-1) / (4.59 x 10¹⁴ s^-1)

= 0.6536 x 10^-6 m = 653.6 nm? 654 nm

Kinetic energy, E = E₀ + ½ mv²

(1/2 mv²) = E – E₀ = hc [ (1/? ) – (1/? ₀)]

= (6.626 x 10^-34Js) x (3 x 10⁸ ms^-1) / (10^-9) x [ (1/500) – (1/654)]

= 6.626 x 3 x 154 x 10-34+8+9) / (500 x 654)

= 9.36 x 10^-20 J

Velocity, v = [ (2 x 9.36 x 10^-20) / m]^1/2

= [ (2 x 9.36 x 10^-20) kg m² s^-2 / 9.1 x 10^-31 kg]^1/2

= (2.057 x 10¹¹ m²s^-2)^1/2= (20.57 x 10¹⁰ m²s^-2)^1/2

= 4.5356 x 10⁵ ms^-1

MBA colleges which have group discussion as admission criteria are:

• IIM Kozhikode
• XLRI
• MDI Gurgaon
• SIBM
• NMIMS
• IMT Ghaziabad
• TAPMI

Dronacharya Group of Institutions admission is entrance-based, followed by counselling. To get admitted to DGI Greater Noida, candidates must meet the eligibility criteria set by the college. Further, candidates seeking admission to the BTech programme must be Class 12 passed with a minimum of 50% aggregate in Physics,  Chemistry, and Mathematics. 

To prepare for Group Discussion (GD) for MBA admission, candidates must focus on understanding the topic, enhancing communication and analytical skills and staying updated on current affairs and important topics. To excel in group discussion, candidates must:

• Engage in mock GDs and public events to improve communication skills.
• Play close attention to important details told by speaker
• Watch online mocks of GD through YouTube or any other relevant source.

Kindly go through the solution

The Sasurie College of Engineering has released the placement details for 2023-2025. Check out the tabulated data given below to know more about Sasurie College of Engineering placement record during season 2023-2025:

λ₁ = 589 nm = 589 x 10^-9 m

ν₁ = c / λ₁ = (3 x 10⁸ ms^-1) / (589 x 10^-9 m) = 5.0934 x 10¹⁴ s^-1

λ₂ = 589.6 nm = 589.6 x 10^-9 m

ν₂ = c / λ₂ = (3 x 10⁸ ms^-1) / (589.6 x 10^-9 m) = 5.0882 x 10¹⁴ s^-1

ΔE = E₁ – E₂ = h [ν₁ – ν₂]

= (6.626 x 10^-34Js) x [ (5.0934 x 10¹⁴ s^-1) – (5.0882 x 10¹⁴ s^-1)

= 3.31 x 10^-22 J

Time duration, t = 2 ns = 2 x 10^-9 s

Frequency, ν = 1 / t = 1 / 2 x 10^-9 s = 10⁹ / 2 s^-1

Energy of one photon, E = hν = 6.626 x 10^-34Js) x (10⁹ / 2 s^-1) = 3.25 x 10^-25 J

No. of photons = 2.5 x 10⁵

Energy of source = 3.3125 x 10^-25 J x 2.5 x 10¹⁵ = 8.28 x 10^-10 J

Students who belong to the General AI category need a JEE Main rank of at least 22339 or below to be eligible for admission at IIITM Gwalior for any BTech branch. Students can refer to the table below to know the last round cutoff for all courses. 

Please remember that these are the ranks for the General AI category; other categories will have different cutoffs accordingly.

The institution has released the BE/BTech branch-wise packages recorded during the placements held during 2023 till 2025. Refer to the table below for more clarity:

Energy of one photon, E = hν = 6.626 x 10^-34Js) x (3 x 10⁸ / 2 ms^-1) / 600 x 10^-9

= 3.31 x 10^-19 J

No. of photons = (3.15 x 10^-18) / 3.31 x 10^-19 = 9.52 ≈ 10.

The highest package offered at Sasurie College of Engineering placements 2025 for BE CSE stood at INR 4.7 LPA. Refer to the table to know the BE CSE the highest package provided in placement 2023-2025:

$={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{co{s}^{2}x}{co{s}^{2}x+4si{n}^{2}x}}dx$

= ${\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{co{s}^{2}x}{co{s}^{2}x+4\left(1-co{s}^{2}x\right)}}dx\{?1=co{s}^{2}x+si{n}^{2}x.$

= ${\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{co{s}^{2}x}{co{s}^{2}x+4-4co{s}^{2}x}}dx$

= ${\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{co{s}^{2}x}{4-3co{s}^{2}x}}dx$

= $-\frac{1}{3}{\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{-3co{s}^{2}x}{4-3co{s}^{2}x}}dx$

= $-\frac{1}{3}{\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{\left(4-3co{s}^{2}x\right)-4}{4-3co{s}^{2}x}}dx$

= $-\frac{1}{3}\left[{\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}dx}-{\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{4}{4-3co{s}^{2}x}}dx\right]$

= $-\frac{1}{3}\left\{{[x]}_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-{\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{4}{4-3\times \frac{1}{se{c}^{2}x}}}dx\right\}$

= $-\frac{1}{3}\left\{\frac{\pi }{2}-{\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{4se{c}^{2}x}{4se{c}^{2}x-3}}dx\right\}$

= $-\frac{1}{3}\left\{\frac{\pi }{2}-{\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{4se{c}^{2}x}{4\left(1+ta{n}^{2}x\right)-3}}dx\right\}\left\{se{n}^{2}x=1+ta{n}^{2}x\right\}$

= $\frac{-\pi }{6}+\frac{2}{3}{}_1$

where I₁ = ${\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{2ta{n}^{2}x}{1+4ta{n}^{2}x}}dx.$

Putting 2tan x = t =>2 sin²xdx = dt& when x = 0, t = 2 tan (0) = 0

x = $\frac{\pi }{2}$ , t = 2 tan $\frac{\pi }{2}$ = ∞

I1 = ${\displaystyle {\int }_0^{\infty }\frac{dt}{1+t^2}}.$

= $[ta{n}^{-}t]{}_0^{\infty }$

= tan^–1 (∞) – tan^–10

= $\frac{\pi }{2}-0$ = $\frac{\pi }{2}.$

? I = $\frac{-\pi }{6}+\frac{2}{3}\times \frac{\pi }{2}=-\frac{\pi }{6}+\frac{\pi }{3}=\frac{\pi }{6}.$