Physics

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New answer posted

a year ago

0 Follower 7 Views

V
Vishal Baghel

Contributor-Level 10

 R=82Ω=4Ω  (wheat stone bridge)

I=VR=404=10A

New answer posted

a year ago

0 Follower 5 Views

P
Payal Gupta

Contributor-Level 10

All the charge given to a conducting sphere resides on outer surface.

New answer posted

a year ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

 B=Nμ0l2μ

BαNμ

βxβy=Nxrx*ryNy

BxBy=20020*20400=12

New answer posted

a year ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

 VA=VB

⇒kQARA=kQBRB

⇒QAQB=RARB* (RBRA)2=12* (21)2=21

New answer posted

a year ago

0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

ΔE=13.6 (112−152)=13.6*2425eV

⇒hcλ=13.6*2425eV.......... (1)

With the help of conservation of linear momentum, we can write

hλ=mHvH⇒hcλ=cmHvH⇒vH=hcλcmH=13.6*2425*1.6*10−193*108*1.67*10−27=4.17m/s

New answer posted

a year ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

New answer posted

a year ago

0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

Time period of second pendulum is 2 seconds.

New answer posted

a year ago

0 Follower 6 Views

V
Vishal Baghel

Contributor-Level 10

 P1=cos? =Rz=RXL2+R2

P1=cos? =RR2=12

So,  P1P2=12

New answer posted

a year ago

0 Follower 6 Views

V
Vishal Baghel

Contributor-Level 10

 Speed  of  light  C=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

a year ago

0 Follower 7 Views

P
Payal Gupta

Contributor-Level 10

Consider the following image dddddddddddddddddddddddddddddddddddddd

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