physics ncert solutions class 11th

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a month ago

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R
Raj Pandey

Contributor-Level 9

ω = ω 0 + α t α = ω - ω 0 t = ( 3120 - 1200 ) 16 s r p m = 1920 16 * 2 π 60 r a d / s 2 = 4 π r a d / s 2

New answer posted

a month ago

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R
Raj Pandey

Contributor-Level 9

k = 1 m

k 1 k 2 = I 1 I 2 = m R 2 / 2 m R 2 / 4 = 2 : 1

New answer posted

a month ago

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R
Raj Pandey

Contributor-Level 9

X C M = 20 * 10 20 + 10 = 20 3 m

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a month ago

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A
alok kumar singh

Contributor-Level 10

τ ? = P ? * E ?

| τ ? | = P E s i n ? θ 4 = q * 2 a * E s i n ? 30 ?

q = 4 2 * 10 - 2 * 2 * 10 5 * 1 2

= 2 * 10 - 3 C

= 2 m C

New answer posted

a month ago

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A
alok kumar singh

Contributor-Level 10

Average speed = 4 v 2 3 v

= 4 v 3

(d) Initial velocity  = - v j ˆ

Final velocity = v i ˆ

Change in velocity  = v i ˆ - ( - v j ˆ )

= v ( i ˆ + j ˆ ) Momentum gain is along i ˆ + j ˆ

 Force experienced is along i ˆ + j ˆ

  Force experienced is in North-East direction.

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a month ago

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V
Vishal Baghel

Contributor-Level 10

|Δp| = 2mu sinθ = 2 * 5 * 10? ³ * 5√2 * (1/√2) = 5 * 10? ² kg m s? ¹

⇒ x = 5

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a month ago

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V
Vishal Baghel

Contributor-Level 10

v_rms = √ (3RT/M) and v_av = √ (8RT/πM)

⇒ v_rms / v_av = √ [ (3RT/M) / (8RT/πM)] = √ (3π/8)

New answer posted

a month ago

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V
Vishal Baghel

Contributor-Level 10

μ_min = (I tanθ)/ (I + mR²)
= [ (mR²/2)tanθ] / [mR²/2 + mR²] = (tanθ)/3 = (tan 60°)/3 = √3/3 = 1.732/3 = 0.5773

Since given coefficient of static friction is less than μ_min, so body will perform rolling with slipping and kinetic friction will act
F? = μN = μmg cosθ = (0.4) * mg cos 60° = mg/5

New answer posted

a month ago

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A
alok kumar singh

Contributor-Level 10

4t = 100 * (4/3)π * (40π * 10? * 0.01)³
t = 2.5 * 10? ¹¹ cm = 2.5 * 10? ¹³ m

New answer posted

a month ago

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R
Raj Pandey

Contributor-Level 9

According to principle of equi-partition of Energy, the average energy per molecules associated with each degree of freedom is (1/2)kT.

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