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Weber ampere per meter is equal to :

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Weber ampere per meter is equal to :

Name: Video solution 1: Weber ampere per meter is equal to :
Uploaded: 2022-10-10T08:48:00.000Z
Duration: 24 min
Description: Video solution 1: Weber ampere per meter is equal to :

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Question 2

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Sura'in a X1 Sed - 5hyic a Hblume 1 \& Uelr 05 (iii) It shows that if the moment of incria I increases, the angular velocity

e

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\sum m = M

, mass of the wbole bos Thus (is) Therefore, The moment of inertia of the expressed as, 8 State and prove parallel axis theorem. about any axis is moment of inertia of a body Ans. (i) of inertia about a parallel axis through its center of mass a parallel axis through its the body and the square of the perpendicular distance between the two axes. To Prove : Ic

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I_{C}

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d

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d d = m (x + d)^{2}

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I = \sum m (x + d)^{2}

(v) The moment of inertia of the whole about the Z-axis is,

I_{2} = \sum m r^{2}

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51. When the energy of the incident radiation is increased by

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0.65 eV

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1.0 eV

(c)

1.3 eV

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1.5 eV

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Question 3

zoro to superhoro sorios quantum numbers of the two states. Assume the Bohr model to be valid. The time period of the Modern electron in the initial state is eight times that in the final state. The possible values of

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n_{1} = 4, n_{2} = 2

n_{1} = 8, n_{2} = 1

n_{1} = 4, n_{2} = 1

captain_shreyas

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Question 4

\Rightarrow n = \frac{E ^{'}}{E} = \frac{16 \times 1 0 ^{- 8}}{10 \times 1.6 \times 1 0 ^{- 19}} ∴ n = 1 0^{11} Maximum energy = h v - h v_{0} = 10 eV - 5 eV = 5 eV

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6.4 \times 1 0^{- 5} W / cm^{2}

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λ = 310 nm

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1 0^{3}

photons ejects an electron, total number of electrons ejected in

1 s

1 0^{x}

[h c = 1240 eV - nm, 1 eV = 1.6 \times 1 0^{- 19} J

, the the value of

x

is [JEE Jan. - 2020]

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Question Text	Weber ampere per meter is equal to :
Updated On	Oct 10, 2022
Topic	Modern Physics
Subject	Physics
Class	Class 12
Answer Type	Video solution: 1
Upvotes	105
Avg. Video Duration	24 min