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New answer posted

a year ago

0 Follower 11 Views

V
Vishal Baghel

Contributor-Level 10

For angle of incidence 'i' :-

cos I = 548+27+25=5100

i = 60°

Using snell's law :-

μ1sini=μ2sinr

sin r = 23sin60°=23*32=12

∴r=45°

So, difference, I – r = 60° - 452 = 15°

New answer posted

a year ago

0 Follower 8 Views

V
Vishal Baghel

Contributor-Level 10

l=2πr

⇒314cm=2*3.14r

⇒r=12m=0.5m

μ=iA

=14*πr2

=142*227*14=11

New answer posted

a year ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

IA=9l+4l+29l*4lcos0°

=13l+12l=25l

IB=9l+4l+29l*4lcosπ

=13l−12l=l

∴IA−IB=24l

New answer posted

a year ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

 Bcentre=Nμ0i2r

⇒100*4π*10−7*i2*5*10−2=37.68*10−4

∴i=37.684*3.14=3A

New answer posted

a year ago

0 Follower 7 Views

V
Vishal Baghel

Contributor-Level 10

i = v R N e t = 6 3 = 2 A

New answer posted

a year ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

 Cv=α2R4J/mol−k

As, Cv (mix) = 1*32R+3*52R4=9R4=α2R4

∴α=3

New answer posted

a year ago

0 Follower 7 Views

V
Vishal Baghel

Contributor-Level 10

For adiabatic process – PVY = const

T1V1Y−1=T2V2Y−1

T2T1= (v1v2)Y−1= (d2d1)Y−1= (32) (75−1)= (32)2/5

= (2)2=4

New question posted

a year ago

0 Follower 3 Views

New answer posted

a year ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

 g (ath)=g (at  depth  αh) h << R

⇒g (1−2hR)=g (1−αhR)

1−2hR=1−αhR⇒2hR=αhRα=2

New answer posted

a year ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

 R=u2sin (2*45°)g=u2g

R2=u22g=u2sin20g

sin2θ=12

⇒2θ=30°⇒θ=15°

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