Engineering Entrance Exam

Explanation -Given threshold frequency of A is given by v_0A= 5 $\times {10}^{14}$ hz

V_OB= 10 $\times$ 10¹⁴hz

$?=h{v}_0$

$\frac{{?}_{OA}}{{?}_{OB}}=\frac{5\times {10}^{14}}{10\times {10}^{14}}1$

${?}_{OA}$ < ${?}_{OB}$

$\left(ii\right)$  for metal A slope=h/e= $\frac{2}{(10-5){10}^{14}}$

$h=\frac{2e}{5\times {10}^{14}}=\frac{2\times 1.6\times {10}^{-19}}{5\times {10}^{14}}$ = 6.4 $\times {10}^{-34}$ js

$\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{a}\mathrm{l}\mathrm{}\mathrm{B}$ slope=h/e= $\frac{2.5}{(15-10){10}^{14}}$ = 8 $\times {10}^{-34}$ js

Explanation- F_x= $\frac{1}{4}\frac{q^2}{4?{?}_0{x}^2}$

W= ${?}_d^{?}fdx$ = ${?}_d^{?}\frac{q^{2}dx}{4*4?{?}_0}\frac{1}{x^2}$

= $\frac{q^2}{4*4?{?}_0}\frac{1}{d}$

= $\frac{{(1.6*{10}^{-19})}^2*9*{10}^9}{4*{10}^{-10}}$ J= 3.6eV 

Explanation- A= 10^-4m²

So d= 10^-3 and i= 100^-4A

I= 100W/m²

$\mathrm{?}=600\mathrm{n}\mathrm{m}=600\times {10}^{-9}$

${\mathrm{?}}_{\mathrm{N}\mathrm{a}}=0.97\mathrm{k}\mathrm{g}/{\mathrm{m}}^3$

$\mathrm{v}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}=\mathrm{A}\times \mathrm{d}$ 10^-4(10^-3)=10^-7m³

$\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{}23\mathrm{k}\mathrm{g}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{s}\mathrm{o}\mathrm{i}\mathrm{d}\mathrm{u}\mathrm{m}$

So volume Na atoms=23/0.97m³

Volume occupied by one Na atom= $\frac{23}{0.97\times 6\times {10}^{26}}=3.95\times {10}^{-26}{\mathrm{m}}^3$

Number of Na atoms in target $\frac{{10}^{-7}}{3.95\times {10}^{-26}}=2.53\times {10}^{18}$

So energy falling per sec= $\frac{\mathrm{n}\mathrm{h}\mathrm{c}}{\mathrm{?}}=\mathrm{I}\mathrm{A}$

So n= $\frac{\mathrm{I}\mathrm{A}\mathrm{?}}{\mathrm{h}\mathrm{c}}$ = $\frac{100\times {10}^{-4}\times 660\times {10}^{-9}}{6.62\times {10}^{-34}\times 3\times {10}^8}=3.3\times {10}^{16}$

N=P $\times \mathrm{n}\times \mathrm{N}\mathrm{a}=\mathrm{P}\times 3.3\times {10}^{16}\times 2.53\times {10}^{18}$

I = 100 $\times {10}^{-6}={10}^{-4}$ A

I=Ne= $\mathrm{P}\times 3.3\times {10}^{16}\times 2.53\times {10}^{18}$ ( ${10}^{-4}$&nb

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Last year, the cut-off percentile for Electrical Engineering at NIT Durgapur was around 97-98 percentile in the JEE Main exam for general category students. For reserved categories, the cut-off was typically lower, around 90-92 percentile. The exact percentile varies slightly depending on the openin