Physics NCERT Exemplar Solutions Class 12th Chapter Six

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2 months ago

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A
alok kumar singh

Contributor-Level 10

Kindly go through the solution 

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

 VP=8kV, VS=160vP=80kW

(Purely resistive load)

P=VI=VP2RP=VS2RS

RS=VS2P=160*16080*103=320*103=0.32Ω

New answer posted

2 months ago

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A
alok kumar singh

Contributor-Level 10

ρvg-mg=ma

ρvgm=g+a

m=ρvgg+a

10343π*10-6 (9.8)9.898

4.15gm

New answer posted

2 months ago

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A
alok kumar singh

Contributor-Level 10

V?1=2Ucm?-U1?2m/2Vom2+m3-Vo

65Vo-VoV05

λo=hcm2Voλf=hcM2Vo5

Δλ=8hcmVo

 

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2 months ago

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A
alok kumar singh

Contributor-Level 10

BA=μ0Iθ4πR

BABB=IAθARBIBθBRA

23π2 (4)35π3 [2]

65 .

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2 months ago

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P
Payal Gupta

Contributor-Level 10

Using conservation of momentum:

60 * v = (60 + 120) * 2

⇒  v = 6m/s

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2 months ago

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P
Payal Gupta

Contributor-Level 10

From conservation of Energy:

mgl (1cos60°)=12mv2

v2=2gl (12)=gl

=10*2510=5m/s

New answer posted

3 months ago

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V
Vishal Baghel

Contributor-Level 10

L = 1H, R = 100 Ω

As,i=i0et/τ

Fori=i02i02=i0et/τ

ln2=t/τ

i=6100e151/100=.06e1500*103=0.06e1.5=0.06*0.25=.0.015A

So, U=12Li2=12*1*(15*103)2=12(225)*106=112.5*106J=0.1125mJ

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3 months ago

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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

3 months ago

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A
alok kumar singh

Contributor-Level 10

Sol. ? v

mv=16mv1

V1=V16

Δk loss =12mv2-12 (16m)V162

12mv21516

% loss =1516*100=93.75%

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