Class 11th

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New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

Range R = v*t = √ (2gh) * √ (2 (H-h)/g) = 2√ (h (H-h).
For R to be max, dR/dh = 0.
h (H-h) must be max. d/dh (Hh-h²)=H-2h=0.
h=H/2 = 12/2=6m.

New question posted

2 months ago

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New answer posted

2 months ago

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R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

v = ∫a dt = (1/M)∫F dt
v = (F? /M) ∫ [1 - (t-T)²/T²] dt from 0 to 2T
= (F? /M) [t - (t-T)³/3T²]? ²?
= (F? /M) [ (2T - T³/3T²) - (0 - (-T)³/3T²) ]
= (F? /M) [ 2T - T/3 - T/3 ] = (F? /M) [ 4T/3 ] = 4F? T/3M.

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

Δy/y = 2Δm/m + 4Δr/r + |x|Δg/g + (3/2)Δl/l
18 = 2 (1) + 4 (0.5) + |x|p + (3/2)4
18 = 2 + 2 + |x|p + 6 = 10 + |x|p
8 = |x|p
From the options, if x=8, p=±1. If x=16/3, p=±3/2. If x=5, p=±8/5. If x=4, p=±2.
Option B gives x=16/3, and p is not among the options.
Option A gives x=5, p not among options.
Option C gives x=8, p not among options.
Option D gives x=4, p not among options.
There must be a typo in the question or options. The solution gives x=16/3 and p=3/2. Let's check.
8 = (16/3) * (3/2) = 8. So B is correct.

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

ME = PE + KE = 8J
At x? , PE=4J, so KE = 8-4=4J. (A is correct)
At x>x? , PE=6J, so KE=2J, which is constant. (B is correct)
At x8J. This is not possible. Particle cannot reach here. So it is not a correct statement. (C is incorrect)
At x? , PE=0J, so KE=8J, which is maximum. (D is correct)

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

P = Fv = (ma)v = m (dv/dt)v
P dt = mv dv
∫P dt = ∫mv dv
Pt = ½mv²
v = √ (2Pt/m)
dx/dt = √ (2P/m) t¹/²
x = √ (2P/m) ∫t¹/² dt = √ (2P/m) * (2/3)t³/²
Squaring to match options:
x² = (8P/9m) t³
This does not match. Let's re-examine the options.
Position x is proportional to t³/².
x = (8P/9m)¹/² t³/²

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

T = 2π√ (m/k)
0.2 = 2π√ (0.5/k)
k = (0.5) (2π/0.2)² = 50π² ≈ 500
x = A sin (ωt) = 5 sin (2π/T * t)
At t=T/4, x = 5 sin (π/2) = 5cm
PE = ½kx² = ½ * 500 * (0.05)² = 250 * 0.0025 = 0.625 J

New answer posted

2 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

η_A = W_A/Q? = 1 - T/T?
η_B = W_B/Q? ' = 1 - T? /T
W_A = Q? (1 - T/T? ) = Q? - Q?
W_B = Q? ' (1 - T? /T) = Q? /2 (1 - T? /T)
Given W_A = W_B
Q? (1 - T/T? ) = (Q? /2) (1 - T? /T)
Q? (T? /T) (1-T/T? ) = (Q? /2) (1-T? /T)
(T? /T - 1) = (1/2) (1-T? /T)
2T? /T - 2 = 1 - T? /T
2T? /T + T? /T = 3
T = (2T? +T? )/3

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

Degrees of freedom (f) = 3 translational + 3 rotational + (2 * 4) vibrational = 14
γ = 1 + 2/f = 1 + 2/14 = 8/7
W = nR? T/ (γ-1) = (1 * 8.3 * (310-300)/ (8/7 - 1) = (83)/ (1/7) = 581 J
As W is positive, work is done by the gas. The solution says W<0, work done on the gas. This implies? T is negative. The question states temperature rises, so work is done on the gas. W = -582J.

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