Engineering Entrance Exam

Explanation- according to law of conservation of momentum

S0 m_Av+m_b0=m_Av₁+m_Bv₂

So m_A(v-v₁)= m_Bv₂

according to law of conservation of kinetic energy

1/2m_Av²=1/2m_Av₁²+1/2m_Bv₂²

So m_A(v²-v₁²)= m_Bv₂²

From above eqn we can say that v+v₁=v₂ or v=v₂-v₁

So v₁= $\frac{m_A{-m}_B}{m_A{+m}_B}$ v  and v₂= $\frac{2m_A}{m_A{+m}_B}$ v

_$?$ initial=h/m_Av

_$?$ final=h/m_Av₁= $\frac{h\left(m_A{+m}_B\right)}{m_{a\left(m_A{-m}_B\right)v}}$

$d?$ = $?$ _final- $?$ _initial= $\frac{h}{m_{A}v}\{\frac{m_A{+m}_B}{m_A{-m}_B}-1\}$

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

V_OB= 10 $*$ 10¹⁴hz

$?=h{v}_0$

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

${?}_{OA}$ < ${?}_{OB}$

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

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

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

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

$\mathrm{v}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}=\mathrm{A}*\mathrm{d}$ 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= $\frac{23}{0.97*6*{10}^{26}}=3.95*{10}^{-26}{\mathrm{m}}^3$

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

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

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

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

I = 100 $*{10}^{-6}={10}^{-4}$ A

I=Ne= $\mathrm{P}*3.3*{10}^{16}*2.53*{10}^{18}$ ( ${10}^{-4}$ A)

P= 7.48 $*{10}^{-21}$ it is less than 1.

BSEB 12th books are based on NCERT patterns, especially for Science subjects. These books help in building strong basics which are useful for exams like NEET, JEE, and other entrance tests. Subjects like Physics, Chemistry, and Biology in BSEB books explain core concepts clearly. After reading the textbooks, students can move to advanced-level reference books. Regular revision of BSEB books can give students a solid foundation for competitive exams.

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 opening and closing ranks each year.