JEE

The angle of projection is used to find the trajectory, horizontal range of a projectile, maximum height, and time of flight. For example, the maximum range on level ground is given by the 45-degree angle.

The vectors like velocity, displacement, and acceleration act along different directions in the two-dimensional motion. Resolving the vectors into the vertical and horizontal components allows the application of one-dimensional kinematic equations in each direction separately. It helps solve the problems more accurately and also simplifies the analysis.

PEC Chandigarh provide admissions to its Engineering BTech courses through the Junior Entrance Examination (JEE) cutoffs. Students can note that for the BTech, Integrated BTech, and BArch courses, the JoSAA cutoff 2024 was recorded between 10291 and 51699. Round 4 cutoff for BTech CSE was 10778 in 2023, while it was 10607 in 2022. The least competitive course for the 2024 cutoffs was BTech in Materials and Metallurgical Engineering with the cutoff of 51699.

Explanation- velocity of a freely falling body is v= $\sqrt[]{2gh}$

And $?=\frac{h}{mv}=\frac{h}{m\sqrt[]{2gh}}$

$?=h$ ^-1

Explanation-number of photon emitted per second n= $\frac{p}{\frac{hc}{?}}=\frac{p?}{hc}=\frac{20*5000*{10}^{-10}}{6.62*{10}^{-34}*3*{10}^8}$

$=5*{10}^{19}{s}^{-1}$

(ii) E=hc $/?$ = $\frac{6.62*{10}^{-34}*3*{10}^8}{5000*{10}^{-10}*1.6*{10}^{-19}}=2.48eV$  this enegy is greater than 2 so emission is possible

(iii) work function $?$ = $\frac{p}{4?d^2}*?r^2?t$ = ${?}_o$

$?t$ = $\frac{4?d^2}{p{r}^2}$ = $\frac{4*2*16*1.6*{10}^{-19}*2^2}{20*{(1.5*{10}^{-10})}^{-2}}=28.4s$

(iv) N= $\frac{n?r^2}{4?d^2}*?t$

 = $\frac{5*{10}^{19}*{(1.5*{10}^{-10})}^2*28.4}{4*({2)}^2}$ =2

(v) as the time of emission is 11.04s so photoelectric is not spontaneous.