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The GATE Energy Science syllabus can be checked below -

• Energy resources and conversion technologies - fossil energy resources, nuclear energy resources, biomass, hydropower, etc.
• Energy storage, economics, environment, and efficiency - energy storage systems, energy management, environmental impacts of energy use, etc. 

The GATE paper code of Energy Science is GATE XE-I. 

Students can download GATE syllabus, following process below-

• Go to official website
• Select desired GATE paper (applying for)
• Download syllabus and start preparing for exam.

No, there is no change in GATE 2026 syllabus. However, a new sectional paper has been added under Engineering Sciences, XE, which is Energy Science and GATE paper code is XE-I. Candidates can download GATE syllabus 2026 PDF from official website- gate2026.iitg.ac.in.

LPU NEST is conducted multiple times in a year. There is no requirement to appear for the previous phase of the exam to appear for the next phase. The application form will be available for each phase separately. The candidates can log in to the official website and submit the application form, and appear for the exam. The exam is conducted from January to August. 

$forA,t_{\frac{1}{2}}=4sec$  

$=m{A}_0{e}^{-}\frac{0.693}{4}\times 16$

$m_A=m_{A0}{e}^{-4\times 0.693}$    -(1)

for B,

for B,  $t_{\frac{1}{2}}=8sec$  

$m_B=m_{B_0}{e}^{-2\times 0.693}$               -(2)

$\frac{m_A}{m_B}=\frac{m_{AO}}{m_{BO}}\frac{e^{-4}\times 0.693}{e^{-2}\times 0.693}$

$\frac{m_A}{m_B}=\frac{25}{100}=\frac{x}{100}x=25$

Let we take $1\mathcal{l}$  of solution

 Mass of solute = Volume × Density

= 0.5 $m\mathcal{l}\times 1.05gm/m\mathcal{l}$  

= 0.525 gram

Mass of solution = 1 kg. [considering very dilute solution]

Mass of solvent = 1000 – 0.525 = 999.475 gram

$\Rightarrow i=1.9$  

For maxima = y = (2n + 1) $\frac{\lambda }{2}\frac{D}{a}$

For 1^st maxima for l₁ wavelength (n = 1)

$y_1=\frac{3{\lambda }_1}{2}\frac{D}{a}$          - (1)

First maxima for l₂ wavelength

$y_2=\frac{3}{2}{\lambda }_2\frac{D}{a}$          - (2)

$\Rightarrow y_2-y_1=\frac{3}{2}\frac{D}{a}\left[{\lambda }_2-{\lambda }_1\right]$

$=\frac{3}{2}\times \frac{2\times 5\times 10^{-9}}{0.5\times 10^{-3}}$

$=\frac{3}{10^{-3}}\times 10^{-8}$

Value of n & $\mathcal{l}$ can't be same

Let l = l₀ cos wt

Then v = v₀ sinwt

at t = 0, v = 0

but l = l₀

$l_{rms}=\frac{l_0}{\sqrt[]{2}}$  

$V_{rms}=l_{rms}z$

$\Rightarrow 220=\frac{l_0}{\sqrt[]{2}}\left(X_L\right)$

$\Rightarrow 220=\frac{l_0}{\sqrt[]{2}}\left(2\pi \times 50\times 200\times 10^{-3}\right)$

$\Rightarrow 220=\frac{l_0}{\sqrt[]{2}}\left(20\pi \right)$

$l_0=\frac{220\sqrt[]{2}}{20\pi }$

$l_0=\frac{11\sqrt[]{2}}{\pi }=\frac{\sqrt[]{a}}{\pi }$

a = 121 × 2

a = 242

$S{O}_{2}C{l}_2+2H_{2}O\to H_{2}S{O}_4+2HCl$  

Let a moles of SO₂Cl₂ is taken

Then no. of moles of H₂SO₄ = a moles

No. of moles of HCl = 2a moles

No. of moles of NaOH required = 2a + 2a = 4a = 16

$\Rightarrow a=4moles$  

I_b = 10 µA

I_C = 1.5 mA

R_L = 50 kW or (Rc)

Base – emitter voltage = 10 mv

$R_B=\frac{V_B}{I_B}=\frac{10\times 10^{-3}}{10\times 10^{-6}}=10^3\Omega$  

$A_v=\left(\frac{\Delta l_C}{\Delta l_B}\right)\times \left(\frac{R_C}{R_B}\right)$

=  $\frac{1.5\times 10^{-3}}{10\times 10^{-6}}\times \frac{5\times 10^3}{10^3}$

A_v = 750

B₂H₆ has 4 2 c -2e bonds and 2 3c-2e bonds.

Bridging (B-H) bonds have more value of bond- length then terminal (B – H) bonds

Bridging bonds are in one plane, but terminal bonds are in perpendicular plane.

Due to presence of (3c-2e) bonds, it behaves as electrons deficient and prone to get attached by  lewis  base.

$3NaB{H}_4+4B{F}_3\underset{}{\overset{\Delta }{\to }}3NaB{F}_4+2B_2{H}_6$

f₁ = 15 cm

$P_1=\frac{1}{15}\times 100=\frac{100}{15}D$

$P_3=\frac{100}{15}D$

$\frac{1}{f^2}=\left({\mu }_2-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

$=\left(1.25-1\right)\times \frac{-2}{R}$

$\frac{1}{f_2}=\frac{0.25\times -2}{15}$

$\frac{1}{f_2}=\frac{-1}{30}c{m}^{-1}$

$\begin{array}{l}{p}_2=\frac{1}{f_2}=-\frac{100}{30}\\ P_R=P_1+P_2+P_3\\ P_R=\frac{100}{15}-\frac{100}{30}+\frac{100}{15}\\ P_R=10D\end{array}$

$P_R=\frac{1}{f_R}$

$f_R=\frac{1}{10}\times 100cm$

$f_R=10cm$

$V_{rms}\alpha \sqrt[]{T}|T_i=127°C=400k.$

$V_{rm{s}^1}=2V_{rms},m=0.056kg=56g$

T_f = 4T_i = 4 × 400 = 1600 k

$Q=n{c}_v\Delta T=\left(\frac{m}{M}\right)C_v\Delta T$

= $\left(\frac{56}{28}\right)\frac{5}{2}R\Delta T$

= $2\times \frac{5}{2}\times 2\times 1200$

Q = 12 × 10³ cal

= 12 k cal

Answer (3) 

Particles having phase difference of $?$ will move with same speed.

Due to higher extent of polarization by Li⁺ and Mg²⁺, LiCl and Mgcl₂ have covalent character. Therefore they are soluble in ethanol.

 Due to very high value of lattice energy, LiF is having very less solubility in water.

Kindly go through the solution

Answer (12.00)

The highest industrial consumption of hydrogen gas is in the synthesis of ammonia gas (Having manufacturing of N-based fertizers)