Engineering Entrance Exam

Get insights from 45.7k questions on Engineering Entrance Exam, answered by students, alumni, and experts. You may also ask and answer any question you like about Engineering Entrance Exam

Follow Ask Question
45.7k

Questions

23

Discussions

484

Active Users

6.2k

Followers

New question posted

6 months ago

0 Follower 4 Views

New question posted

6 months ago

0 Follower 3 Views

New answer posted

6 months ago

0 Follower 15 Views

K
Kamini Padyal

Contributor-Level 9

Yea. You can easily get the ECE with your rank in the upcoming round. As a rank with 6056 is already very close and the Amrita CBE have always a slight range distance for ECE every year.

New answer posted

6 months ago

0 Follower 12 Views

S
Shikha Dixit

Contributor-Level 7

The admission to BTech and BTech courses has been started through JEE Main. Additionally, the institute has released the Round 1 seat allotment result wherein B.Tech. in Computer Science and Engineering (Internet of Things, Cyber Security including Block Chain Technology) closing rank stood at 47166 for the General AI quota.

Moreover, the JEE Advanced Counselling 2024 was conducted in six rounds for admission to the BTech and other UG courses. The JEE Advanced Counselling Process includes steps like registration, choice filling, and paying the counselling fee. Candidates will then be called for document verification, where their el

...more

New answer posted

6 months ago

0 Follower 5 Views

C
Chandra Jain

Contributor-Level 7

Recently, NIELIT Patna cutoff 2025 for the Round 1 seat allotment has been released by the institute for the admission to BTech and BTech Integrated courses. As per the JEE Main 2025 cut off data, the B.Tech. in Computer Science and Engineering rank stood at 50162 for the General AI category candidates. Similarly, for the OBC AI quota, the rank stood at 18790.

Candidates seeking admission to NIELIT Patna via JEE Main 2025 must need to sit for the counselling rounds to get selected for the admission to any BTech or BTech Integrated course of thier choice. Although, he/she must have to confirm their seat if someone is willing to opt

...more

New question posted

6 months ago

0 Follower 4 Views

New answer posted

6 months ago

0 Follower 1 View

P
Pallavi Pathak

Contributor-Level 10

Explanation- velocity of a freely falling body is v= 2gh

And ? =hmv=hm2gh

? =h -1

New answer posted

6 months ago

0 Follower 5 Views

P
Pallavi Pathak

Contributor-Level 10

Explanation-number of photon emitted per second n= phc?=p?hc=20*5000*10-106.62*10-34*3*108

=5*1019s-1

(ii) E=hc /? = 6.62*10-34*3*1085000*10-10*1.6*10-19=2.48eV  this enegy is greater than 2 so emission is possible

(iii) work function ? = p4?d2*?r2?t = ?o

?t = 4?d2pr2 = 4*2*16*1.6*10-19*2220*(1.5*10-10)-2=28.4s

(iv) N= n?r24?d2*?t

 = 5*1019*(1.5*10-10)2*28.44*(2)2 =2

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

Get authentic answers from experts, students and alumni that you won't find anywhere else

Sign Up on Shiksha

On Shiksha, get access to

  • 66k Colleges
  • 1.2k Exams
  • 680k Reviews
  • 1800k Answers

Share Your College Life Experience

×
×

This website uses Cookies and related technologies for the site to function correctly and securely, improve & personalise your browsing experience, analyse traffic, and support our marketing efforts and serve the Core Purpose. By continuing to browse the site, you agree to Privacy Policy and Cookie Policy.