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New answer posted

a year ago

0 Follower 14 Views

A
alok kumar singh

Contributor-Level 10

20

Sol. Angular momentum conservation:

I1ω1+I2ω2=I1+I2ωfMR22ωo=MR22+MR28ωfωf=45ωoKEfinal =12I1+I2ωf2=MR2ω25KEinitial =12I,ω02=MR2ω24% loss 20%.

New answer posted

a year ago

0 Follower 15 Views

A
alok kumar singh

Contributor-Level 10

10553.33

Sol. 1λmin =R11-1=R[n=n=1]

1λmax=R11-14=3R4[n=2n=1]

Δλ43R-1R13R=340

For Paschan 1λmin =R19[n=h=3]

1λmax=R19-116=7R144

Δλ=817R

New answer posted

a year ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

E = 440 sin 100 t ω=100π

L=2πH. XL=ωL=100π.2π=1002Ω.

=220100=2.2A

New answer posted

a year ago

0 Follower 6 Views

V
Vishal Baghel

Contributor-Level 10

R = R1 + R2

2 l σ A = l σ 1 A + l σ 2 A 2 σ = 1 σ 1 + 1 σ 2 σ = 2 σ 1 σ 2 σ 1 + σ 2

New answer posted

a year ago

0 Follower 12 Views

A
alok kumar singh

Contributor-Level 10

Sol.

AHH2H=3a2AGH3Al=H23MABC=MMADE=M4

I G = M a 2 12 - M 4 a 2 2 12 + M 4 a 2 3 2 = 11 16 M a 2 12

New answer posted

a year ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

'A' at static equilibrium, E (inside conductor) = 0

New answer posted

a year ago

0 Follower 10 Views

V
Vishal Baghel

Contributor-Level 10

 ρr= {ρ0 (34rR)forrR0forr>R

Pr=E4πr2=qencε0

As,  E4πr2=πρ0r3ε0 (1rR)

E=ρ0r4ε0 (1rR)

New answer posted

a year ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

Sol. M=1fo1+DFe

100=2011+25Fe

1+25Fe=5

Fe=254=6.25cm

New answer posted

a year ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

goff = g cos

T=2πLgcosα

New answer posted

a year ago

0 Follower 23 Views

A
alok kumar singh

Contributor-Level 10

Sol. 

n1T1+n2T2=nT

(0.1) (200)+ (0.05) (400)= (0.15)T

T=266.67

 

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