Engineering Entrance Exam

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New question posted•a year ago
New question posted•a year ago
New question posted•a year ago
New answer posted•a year ago
S
a year ago

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.

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New answer posted•a year ago
S
a year ago

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

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New answer posted•a year ago

What is the JEE Main Counselling 2025 fees?

5 Views|Posted a year ago
Asked by Parul
S
a year ago

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 candida

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New question posted•a year ago
New answer posted•a year ago
S
a year ago

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

And ? =hmv=hm2gh

? =h -1

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New answer posted•a year ago
S
a year ago

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

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New answer posted•a year ago
S
a year ago

Explanation- according to law of conservation of momentum

S0 mAv+mb0=mAv1+mBv2

So mA(v-v1)= mBv2

according to law of conservation of kinetic energy

1/2mAv2=1/2mAv12+1/2mBv22

So mA(v2-v12)= mBv22

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

So v1= mA-mBmA+mB v  and v2= 2mAmA+mB v

? 

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