Physics NCERT Exemplar Solutions Class 12th Chapter Eight

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New answer posted

3 months ago

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V
Vishal Baghel

Contributor-Level 10

 SpeedoflightC=wk=1.5*10110.5*103=3*108m/s

So, E0 = B0 C

= 2 * 10-8 * 3 * 108

= 6v/m

Direction will be along z – axis

New answer posted

3 months ago

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P
Payal Gupta

Contributor-Level 10

Frequency increases on filing. So, initial frequency of A is 335 Hz.

f = 340 – 5 = 335 Hz

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3 months ago

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V
Vishal Baghel

Contributor-Level 10

A = 36 cm2= 36 * 104 m2

? F? =7.2*109N, t=20min.

? F? =ΔPΔt=ECΔtEΔt=? F? cEAΔt=? F? cA

= 0.06 W/cm2

New answer posted

3 months ago

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alok kumar singh

Contributor-Level 10

Kindly go through the solution

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3 months ago

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alok kumar singh

Contributor-Level 10

 g=Axx2+a23/2

v0?dV=-x?gdx

O-V=-Axa2+x23/2

Let, a2+x2=t2

2xdx=2tdt

xdx=tdt

V=Atdtt3-At-Aa2+x2x

V = A a 2 + x 2

New answer posted

3 months ago

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A
alok kumar singh

Contributor-Level 10

 g=Axx2+a23/2

v0?dV=-x?gdx

O-V=-Axa2+x23/2

Let, a2+x2=t2

2xdx=2tdt

xdx=tdt

V=Atdtt3-At-Aa2+x2x

V = A a 2 + x 2

New answer posted

3 months ago

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P
Payal Gupta

Contributor-Level 10

E0=2.25vm Speed of declromagnetic signal,  v=E0B0=2.25v/m1.5*108T

B0=1.5*108Tv=225150*108=1.5*108m/s

Time to reach each to the same

radar,  t=2*3*103m1.5*108m/s=4*105S

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3 months ago

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Payal Gupta

Contributor-Level 10

According to kepler's third law of time period –

(TATB)2= (rArB)3rArB=41/3

r43=4rB3

New answer posted

4 months ago

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Vishal Baghel

Contributor-Level 10

 g (ath)=g (atdepthαh) h << R

g (12hR)=g (1αhR)

12hR=1αhR2hR=αhRα=2

New answer posted

4 months ago

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V
Vishal Baghel

Contributor-Level 10

Based an theoretical data.

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