Physics NCERT Exemplar Solutions Class 12th Chapter Four

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New answer posted

4 months ago

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V
Vishal Baghel

Contributor-Level 10

 B1=μ0i2R

B2=μ0iR22 (R2+ (3R)2)3/2=μ0iR22*8R3=μ0i8*2R=B18

B1B2=8:1

New answer posted

4 months ago

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V
Vishal Baghel

Contributor-Level 10

 B=0.5T

FCD=il*B

ilBsin60

=10*5100*0.5*32=38=1.7328

= 0.216 N

New answer posted

4 months ago

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A
alok kumar singh

Contributor-Level 10

 U? =5jˆ

a? =10iˆ+4jˆ

S? =U? t+12 (a? )t2

20iˆ+y0jˆ=5t2iˆ+5t+2t2jˆ

20=5t2y0=5t+2t2t=218m. 

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4 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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4 months ago

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A
alok kumar singh

Contributor-Level 10

m (L)=m1S1 (ΔT)

m3.4*105= (200) (4200) (25)

m=61.7

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4 months ago

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P
Payal Gupta

Contributor-Level 10

 r=mvqB=2kmqB

2*5*103*1.6*1019*24*1.66*10271.6*1019*12

= 9.975 * 102 cm

= 9.975 cm

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4 months ago

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P
Payal Gupta

Contributor-Level 10

 x=4sin (π2ωt) - (i)

y=4sin (ωt) - (ii)

From (i) and (ii) cos2 ωt+sin2ωt= (x4)2+ (y4)2=1

x2+y2= (4)2

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4 months ago

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V
Vishal Baghel

Contributor-Level 10

 E=12mu2

EHighestpoint=12m (u2)2=18mu2=E4

New answer posted

4 months ago

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V
Vishal Baghel

Contributor-Level 10

N = m v 2 R

 

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4 months ago

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P
Payal Gupta

Contributor-Level 10

rd = (Radius of deuteron)

= m d v d q d B                

= 2 m d k . E q d B                

= 2 m P m P * q P q d [ q P = q d = 1 . 6 * 1 0 C 1 9 ]                     

= 2 : 1              

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