Aieee 2006 physics

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v = α.
=> dx/dt = α

=> = α/2. t
So, x ∝ t²

1 )

t³

2 )

t²

3 )

4 )

t^1/2

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F×BC = Mg×AC
Flsin45° = Mg(l-lcos45°)
=> Mg(-1)

1 )

Mg(

-1)

2 )

Mg(

+1)

3 )

4 )

Mg/

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Suppose partice moved up to a distance = 5
using, v² = u² + 2as
=> 0 = 25 - 2gy
=> 2gy = 25
Work done by the gravity = W = Gravitational force)×(displacement)×cos180°
(direction between gravitational force and displacement is 180°)
Work done by the gravity = W = -0.1gy = -1.25 J

1 )

-0.5 J

2 )

-1.25 J

3 )

1.25 J

4 )

0.5 J

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V_max = Aω
Period of oscillation = T = 2π/ω
=> T = 2πA/V_max = 0.01 second

1 )

0.01 s

2 )

10 s

3 )

0.1 s

4 )

100 s

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Total initial momentun = 0
Total final momentum = m₁v₁ + m₂v₂ = 0
=> 12 × 4 + 4 × v₂ = 0
=> v₂ = -12
Kinetic energy of 4kg mass = 1/2 × 4 × (-12)²
Kinetic energy of 4kg mass = 288 J

1 )

96 J

2 )

144 J

3 )

288 J

4 )

192 J

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f = f₀( v )

v + v_s

where,
v=velocity of wave
v_s=velocity of source. It is positive if source of wave is moving away from observer. It is negative if source of wave is moving towards observer.

10000 = 9500( 300 )

300 + v_s

300 + v_s = 285
v_s = -15ms^-1 [negative means that wave-source is approching the observer]
So approching velocity of wave-source = 15ms^-1

1 )

30 ms^-1

2 )

ms^-1

3 )

15/

ms^-1

4 )

15 ms^-1

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Simple harmonic is defined as, x = Asinωt
velocity = dx/dt = v = Aωcosωt
Maximum velocity = Aω
Kinetic energy = 1/2 m v² = 1/2 m A²ω²cos²ωt
Maximun Kinetic energy = 1/2 m A²ω²
=> (3/4)(1/2 m A²ω²) = 1/2 m A²ω²cos²ωt
=> (3/4) = cos²ωt
=> ωt = &pi/6
=> (2π/T)t = &pi/6
=> (2π/T)t = &pi/6
t = 1/6 s

1 )

1/6 s

2 )

1/4 s

3 )

1/3 s

4 )

1/12 s

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Resonance of a open tube of air(approximate)

Approximate frequency = f = nv

2L

where,
L: length of the cylinder
n = 1, 2, 3...
v = speed of sound
315 = nv/2L .......... (i)
400 = ((n+1)v)/(2L) ................(ii)
Solving (i) and (ii)
n = 3
Lowest frequency of string stretched between fixed points = v/2L = 315/3 = 105 Hz

1 )

10.5 Hz

2 )

105 Hz

3 )

1.05 Hz

4 )

1050 Hz

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Stefan-Boltzmann Law
The energy radiated by a blackbody radiator per second = P
P = AσT⁴
So energy released by sun = P = 4πR²σT⁴
Radiant power, incident on Earth, at a distance r from the Sun = P_e
P_e = (πr₀²/4πr²)(4πR²σT⁴)

P_e = πr₀²R²σT⁴

r²

1 )

R²σT⁴

r²

2 )

4πr₀²R²σT⁴

r²

3 )

πr₀²R²σT⁴

r²

4 )

r₀²R²σT⁴

r²

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Refer Terms and definition

1 )

Wb/m²

2 )

Henry (H)

3 )

H/m²

4 )	Weber (Wb)

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Fs = mgh
F × 0.2 = 0.2 × 10 × 2.2
F = 22 N

1 )

4 N

2 )

16 N

3 )

20 N

4 )

22 N

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m₁ d + m₂ x = 0

x = m₁ d

m₂

1 )

m₂	d
m₁

2 )

m₁	d
m₁ + m₂

3 )

m₁	d
m₂

3 )

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Force = change in momentum per unit time
F = (mv-0)/0.1 = 0.15 × 20/0.1 = 30 N

1 )

150 N

2 )

3 N

3 )

30 N

3 )

300 N

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I_b = I_e - I_c = 5.60 - 5.488 = 0.112
β = I_c/I_b = 5.488/0.11 = 49

1 )

2 )

3 )

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Antimony appears first in Seeback’s thermal electric series.

1 )	flow from Antimony to Bismuth at the hot junction

2 )	flow from Bismuth to Antimony at the cold junction

3 )	not flow through the thermocouple

3 )	flow from Antimony to Bismuth at the cold junction

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&lambda = (1242 eVnm)/(11.2) = 1100 A
Therefore, ultra-violet region

1 )	ultra-violet region

2 )	infra-red region

3 )	visible region

3 )	X-ray region

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Final result and score
Q No	Ans	Correct Ans	Score	Q No	Ans	Correct Ans	Score
1	-		0	2	-		0
3	-		0	4	-		0
5	-		0	6	-		0
7	-		0	8	-		0
9	-		0	10	-		0
11	-		0	12	-		0
13	-		0	14	-		0
15	-		0	16	-		0
Final score : 0

Answer sheet
Q No	Ans	For review	Q No	Ans	For review
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3			4
5			6
7			8
9			10
11			12
13			14
15			16

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