Class 11th

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New answer posted

3 months ago

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V
Vishal Baghel

Contributor-Level 10

P = (RT/M? )ρ ⇒ P/ρT = constant
⇒ (12)/ (4 * 10? ³ * 10? ) * 280 = 6 * 10? * T ⇒ T = 1400K

New question posted

3 months ago

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New answer posted

3 months ago

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A
alok kumar singh

Contributor-Level 10

 

θ = 3 0 °

A B = R s i n θ = 2 R

New answer posted

3 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

f? = f? (330/ (330-V) f? = f? (330/ (330+V)
(f? - f? )/f? * 100 = 2% ⇒ (330/ (330-V) - (330/ (330+V) = 0.02
⇒ 330 [ (2V)/ (330)²-V²)] = 0.02 ⇒ 330V = 0.01 * (330)² ⇒ V = 3.3 m/s (approx.)

New answer posted

3 months ago

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V
Vishal Baghel

Contributor-Level 10

(K (4A)/l) (100 – θ) = (KA/l) (θ – 50)
⇒ 400 – 4θ = θ – 50 ⇒ 5θ = 450 ⇒ θ = 90°C

New answer posted

3 months ago

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V
Vishal Baghel

Contributor-Level 10

v? = 3tî v? = 24cos 60°î + 24sin 60°? = 12î + 12√3?
v? = v? – v? = (12 – 3t)î + 12√3?
It is minimum when 12 - 3t = 0 ⇒ t = 4sec

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3 months ago

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V
Vishal Baghel

Contributor-Level 10

Zero error = 0.0 + 4 * 0.01 = 0.04 cm
Reading = 2.5a + 7 * 0.01 = 2.57 cm
Correct reading = 2.57 cm - 0.04 cm = 2.53 cm

New answer posted

3 months ago

0 Follower 113 Views

A
alok kumar singh

Contributor-Level 10

In this case,

F = η v h l ( l x )            

v = F h η l ( l x ) = d x d t            

So,     t = 3 η l 3 8 F h = 5 s e c

New answer posted

3 months ago

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V
Vishal Baghel

Contributor-Level 10

at t = t? , x? = 0 ⇒ 0 = asin (ωt? + π/6)
⇒ ωt? + π/6 = π
at t = t? , x? = 0 ⇒ 0 = asin (ωt? + π/4)
⇒ ωt? + π/4 = π
From (i) and (ii)
ω (t? - t? ) = π/4 - π/6 ⇒ t? - t? = π/ (12ω)

New answer posted

3 months ago

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V
Vishal Baghel

Contributor-Level 10

ΔW = area under P — t graph
= (1/2) (4 + 6) * 7 = 35 J
Work done = change in KE ⇒ 35 = (1/2) * 2 * v² - (1/2) * 2 * (1)² ⇒ v = 6 m/s

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