Class 12th

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New answer posted

10 months ago

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Payal Gupta

Contributor-Level 10

Energy corresponding to a particle is mc2

hcλ=mc2hcλ= (x3h)c2x=3λc=310*1010*3*108 = 1101=10

New answer posted

10 months ago

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Payal Gupta

Contributor-Level 10

i2 = 2i1

i1 + i2 = 6

i1 = 2A.

New answer posted

10 months ago

0 Follower 6 Views

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Payal Gupta

Contributor-Level 10

Tsinθ=kq2r2

Tcosθ=mg

tanθ=kq2mgr2

Using the above equation find the value of 'q'

New question posted

10 months ago

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New answer posted

10 months ago

0 Follower 8 Views

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Payal Gupta

Contributor-Level 10

Since force will act in horizontal direction so vertical component of speed will be same

v1 cos α = v2 cos β

K 1 K 2 = ( v 1 v 2 ) 2 = c o s 2 β c o s 2 α

New answer posted

10 months ago

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Payal Gupta

Contributor-Level 10

In ferromagnetic material, below Curie's temperature, a domain is defined as macroscopic region with zero magentisation.

New answer posted

10 months ago

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Payal Gupta

Contributor-Level 10

Using Bohr's theory :

λ=hcE2E1=1242eVnm (3.4) (13.6)=121.8nm

New answer posted

10 months ago

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Payal Gupta

Contributor-Level 10

Using de-Broglie equation for wave nature of particle :

λ=hcmvλα1m

λeλP=1me1mP=mPme=1836

New answer posted

10 months ago

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Payal Gupta

Contributor-Level 10

Using diffraction formula of circular hole :

rα1D

size will decease.

New answer posted

10 months ago

0 Follower 4 Views

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Payal Gupta

Contributor-Level 10

λ a i r = c f c

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