physics ncert exemplar solutions class 12th chapter two

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New answer posted

2 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

 C0=ε0Ad

C1=4ε0Ad=4C0

CZ=kε0A34d=4kc03

Series

Ca=C1C2C1+C2

16k3dC024C0 (1+k3)

=4kC0k+3

New answer posted

2 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

K=QrΔxΔAT

ML2T-2 (L)L2 (θ) (T)M1L1-T-3θ-1

New answer posted

2 months ago

0 Follower 5 Views

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

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

Sol. V?1=2Ucm?-U1?2m/2Vom2+m3-Vo

65Vo-VoV05

λo=hcm2Voλf=hcM2Vo5

Δλ=8hcmVo

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

MS – Reading = 2.5 mm.

50 division on CS = 0.5 mm on MS.

45thdivisiononCS=0.550*45mmM.S.

= .45 mm on M.S.

So, diameter = 2.5 mm

+0.45 mm

+0.03 mm

= 2.98 mm

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

MSD = 20 divisions per cm 1 MSD = 120cm.

VSD = 50 divisions

As, 25 VSD = 24 MSD

1VSD=2425MSD

L.C. = 1 MSD – 1 VSD

=1500cm=0.002cm

New answer posted

2 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

 T=kρr3S3/2

[T]  [ML3]1/2 [L3]1/2 [MLT2L1]34

[T] = [M1/2L32L32M34T32]

= [M14T3/2]

New answer posted

2 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

T=kρr3S3/2

[T]  [ML3]1/2 [L3]1/2 [MLT2L1]34

[T] = [M1/2L32L32M34T32]

= [M14T3/2]

New answer posted

2 months ago

0 Follower 4 Views

P
Payal Gupta

Contributor-Level 10

C 1 = 0 A 5 b

C 2 = 0 A 3 b

C 3 = 0 A b

C e q = C 1 + C 2 + C 3

= 0 A b [ 1 5 + 1 3 + 1 ]

C e q = 2 3 1 5 0 A b

New answer posted

2 months ago

0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

Sol.

BA=μ0Iθ4πR

BABB=IAθARBIBθBRA

23π2 (4)35π3 [2]

65 .

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