Physics NCERT Exemplar Solutions Class 12th Chapter Nine

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New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

For circular motion:-

Tmg=mv2r

T=m (g+v2r)

=2 (10+251/2)=120N

Strain=stresY

= 30 * 10-5

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

Power of a lens, P = 1.25 m-1

μlens=1.5, r1=20cm, r2=40cm

As, P = 1f=1.25

1.5μ1=16+1=76μ1=67*1.5=97

 

New answer posted

2 months ago

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A
alok kumar singh

Contributor-Level 10

M=1fo1+DFe

100=2011+25Fe

1+25Fe=5

Fe=254=6.25cm

New answer posted

2 months ago

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P
Payal Gupta

Contributor-Level 10

Resolving power, = d1.22λ=24.4*1021.22*2440*1010

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P
Payal Gupta

Contributor-Level 10

At minimum deviation –

δ=2iAi=δ+A2, r=A2

Using snell's low : - sini=μ=sinr

δ+A2=πA2δ=π2A

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2 months ago

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P
Payal Gupta

Contributor-Level 10

Pressure * time =FtA= [MLT2] [T] [L2]= [ML1T1]

Coefficient of viscosity,  η= [ML1T1]

Fviscous=ηAΔvΔy

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

As, Ed = VB

E=VBd=0.66*106=1*105N/C

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2 months ago

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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

2 months ago

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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

2 months ago

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V
Vishal Baghel

Contributor-Level 10

IA=9l+4l+29l*4lcos0°

=13l+12l=25l

IB=9l+4l+29l*4lcosπ

=13l12l=l

IAIB=24l

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