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2008 AIEEE PAPER - (Physics)

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Momemntum = mass x velocity
Momemntum = 3.513 x 5.00 = 17.565 kg m sec^-1
Momemntum in three significant digit = 17.6 kg m sec^-1

1 )	17.565 kg ms^-1

2 )	17.56 kg ms^-1

3 )	17.57 kg ms^-1

4 )	17.6 kg ms^-1

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Moment of Intertia of a body is the intertia of a rotating body with respect to the axis of rotation.

Moment of interia of the square plate at the center of the axis (virtical axis) = ma²

6

Moment of interia at th axis passing thru corner = I_center + m d² = ma² + m( a )² = 2 ma²

6 3

1 )

1	m a²
12

2 )

7	m a²
12

3 )

2	m a²
3

4 )

5	m a²
6

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Speed of the sound in gas

R = gas constant(8.314 J/mol K)
T = the absolute temperature
M = the molecular weight of the gas (kg/mol)
γ = adiabatic constant = c_p/c_v
γ_O₂ = 7/5
γ_He = 5/3
M_O₂ = 32
M_He = 4
......(i)
......(ii)
By solving (i) and (ii) V_He = 1419 m sec^-1

1 )

500 ms^-1

2 )

650 ms^-1

3 )

330 ms^-1

4 )

460 ms^-1

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1 )	Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1

2 )	Statement-1 is true, Statement-2 is false

3 )	Statement-1 is false, Statement-2 is true

4 )	Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1

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5) This question contains Statement-1 and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements.

Statement-1:
For a mass M kept at the centre of a cube of side 'a', the flux of gravitational field passing through its sides is 4π GM. and
Statement-2:
If the direction of a field due to a point source is radial and its dependence on the distance 'r' from the source is given as 1/r² , its flux through a closed surface depends only on the strength of the source enclosed by the surface and not on the size or shape of the surface.

1 )	Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1

2 )	Statement-1 is true, Statement-2 is false

3 )	Statement-1 is false, Statement-2 is true

4 )	Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1

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Two full turns of the circular scale of a screw gauge cover a distance of 1 mm .
Therrefore one full turns of the circular scale of a screw gauge cover a distance of 0.5 mm .
Reading = MSR + CSR x LC - error L.C (least count) = 0.5/50 = 0.01
So the diameter of wire = 3mm + 35 x L.C - (-0.03)
the diameter of wire = 3mm + 35 x 0.01 mm + 0.03mm
the diameter of wire = 3.38

1 )

3.67 mm

2 )

3.38 mm

3 )

3.32 mm

4 )

3.73 mm

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Internal entergy of ideal gas (U) = c_vnRT
Assume the final temperature = T
Internal energy of first gas before removing the partition = c_vn₁RT₁
Internal energy of second gas before removing the partition = c_vn₂RT₂
Internal energy of first gas after removing the partition = c_vn₁RT
Internal energy of second gas after removing the partition = c_vn₂RT
Change in energy of first gas = c_vn₁RT - c_vn₁RT₁
Change in energy of second gas = c_vn₂RT - c_vn₂RT₂
As the container is insulated, so total change of energy = 0
(c_vn₁RT - c_vn₁RT₁ ) + (c_vn₂RT - c_vn₂RT₂ ) = 0
n₁R(T - T₁) + n₂R(T - T₂ ) = 0 .........(i)

n₁R= P₁V₁/T₁ .........(ii)
n₂R= P₂V₂/T₂ .........(iii)
By solving (i),(ii) and (iii) you get the answer as

T₁T₂ (P₁V₁ + P₂V₂)

P₁V₁T₂ + P₂V₂T₁

1 )

T₁T₂ (P₁V₁ + P₂V₂)

P₁V₁T₁ + P₂V₂T₂

2 )

T₁T₂ (P₁V₁ + P₂V₂)

P₁V₁T₂ + P₂V₂T₁

3 )

P₁V₁T₁ + P₂V₂T₂

P₁V₁ + P₂V₂

4 )

P₁V₁T₂ + P₂V₂T₁

P₁V₁ + P₂V₂

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Mutual inductance of two solenoid two long thin solenoids, one wound on top of the other
M = μ₀N₁N₂LA
N₁ = total number of turns per unit length for first solenoid
N₂ = number of turns per unit length for second solenoid
A = cross-sectional area
L = length of the solenoid.

A = 10cm² = 10/10000 = 0.001m²
L = 20cm = 0.2 m
N₁ (turns per unit length) = 300/0.2 = 1500
N₂ (turns per unit length) = 400/0.2 = 2000
M = 4π x 10^-7 x 1500 x 2000 x 0.001 x 0.2
M = 2.4π x 10^-4 H

1 )	2.4π X 10^-4 H

2 )	2.4π X 10^-5 H

3 )	4.8π X 10^-4 H

4 )	4.8π X 10^-5 H

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Soap solution has less surface tension than water.
Soap solution and water have almost same density.
As water has more surface tension so it has more height.

Refer following formula
The height to which the liquid can be lifted is given by
h=height of the liquid lifted
T=surface tension
r=radius of capillary tube

h= 2T

ρrg

1 )

2 )

3 )

4 )

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As per Bragg's law
nλ = 2d sinθ

where
θ = angle between the surface and the ray = 90-30=60°
nλ = 2 . 10^-10 . sin30°
(nλ)² = 3 x 10^-20
As per Davisson and Germer experiment

λ = h

2meV = (h/λ)²
29.12 x 10^-50 V = ((6.6 x 10^-34)/λ)²
Vλ² = 1.496 x 10^-18
V x 3 x 10^-20 = 1.496 x 10^-18 n²
V = 50 n²
n → it is integer value (it can 1,2,3,4,...)
If we replace n=1 then we get V = 50.

1 )

1000 V

2 )

2000 V

3 )

50 V

4 )

500 V

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As per Bragg's law
nλ = 2d sinθ

where
n = integer (based upon order)
λ = wavelength
d = distance between the planes
θ = angle between the surface and the ray

So,
θ = 90° - i
nλ = 2d sin(90°-i)
nλ = 2d cosi

11) Wave property of electrons implies that they will show diffraction effects. Davisson and Germer demonstrated this by diffracting electrons from crystals. The law governing the diffraction from a crystal is obtained by requiring that electron waves reflected from the planes of atoms in a crystal interfere constructively (see figure)

If a strong diffraction peak is observed when electrons are incident at an angle 'i' from the normal to the crystal planes with distance 'd' between them (see figure), de Broglie wavelength λ_dB of electrons can be calculated by the relationship (n is an integer)

1 )	dcosi = nλ_dB

2 )	dsini = nλ_dB

3 )	2dcosi = nλ_dB

4 )	2dsini = nλ_dB

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12) Wave property of electrons implies that they will show diffraction effects. Davisson and Germer demonstrated this by diffracting electrons from crystals. The law governing the diffraction from a crystal is obtained by requiring that electron waves reflected from the planes of atoms in a crystal interfere constructively (see figure)

In an experiment, electrons are made to pass through a narrow slit of width 'd' comparable to their de Broglie wavelength. They are detected on a screen at a distance 'D' from the slit (see figure)

Which of the following graphs can be expected to represent the number of electrons 'N' detected as a function of the detector position 'y' (y = 0 corresponds to the middle of the slit)?

1 )

2 )

3 )

4 )

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Escape velocity
Escape velocity from a body of mass M and radius r is

So escape velocity is directly proportional to root of mass and inversely proportional to root of radius
So the escape velocity from the surface of the planet would be
V_e = 11 x x
V_e = 110 km s^-1

1 )	0.11 km s^-1

2 )	1.1 km s^-1

3 )	11 km s^-1

4 )	110 km s^-1

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1 )

2 )

3 )

4 )

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ρ₂ has the maximum density as it is at the bottom
ρ₁ has the least density as it is at the top
Therefore , ρ₁ < ρ₃ < ρ₂

1 )	ρ₁ < ρ₂ < ρ₃

2 )	ρ₁ < ρ₃ < ρ₂

3 )	ρ₃ < ρ₂ < ρ₁

4 )	ρ₁ > ρ₃ > ρ₂

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1 )	It is a pnp transistor with R as emitter

2 )	It is an npn transistor with R as collector

3 )	It is an npn transistor with R as base

4 )	It is a pnp transistor with R as collector

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1 )

0.67 J

2 )

0.34 J

3 )

0.16 J

4 )

1.00 J

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1 )

α =	0.04	, β =	1.0
	π		π

2 )

α = 12.50 π , β =	π
	2.0

3 )	α = 25.00 π , β = π

4 )

α =	0.08	, β =	2.0
	π		π

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1 )

2 )

3 )

4 )

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Surface area of hemisphere = 2πr²

E = ρ j = ρ(I/2πr²)

By solving above equation
V_B-V_C due to current I at point A will be

ρ I - ρ I

2 π a 2 π a (a + b) ..................... (i)

In the sanme way, V_C-V_B due to current I at point D will be

- ρ I + ρ I ..................... (ii)

2 π a 2 π a (a + b)

By adding (i) and (ii)

V_B-V_C = ρ I - ρ I

π a π a (a + b)

20) Consider a block of conducting material of resistivity 'ρ' shown in the figure. Current 'I' enters at 'A' and leaves from 'D'. We apply superposition principal to find voltage 'ΔV' developed between 'B' and 'C' . The calculation is done in the following steps:
(i) Take current 'I' entering from 'A' and assume it to spread over a hemispherical surface in the block.
(ii) Calculate field E(r) at distance 'r' from A by using Ohm's law E = ρ j , where 'j' is the current per unit area at 'r'.
(iii) From the 'r' dependence of E(r), obtain the potential V(r) at 'r'
(iv) Repeat (i), (ii) and (iii) for current 'I' leaving 'D' and superpose results for 'A' and 'D'.

Δ V measured between B and C is

1 )

ρ I	-	ρ I
2 π a	-	2 π a (a + b)

2 )

ρ I

2 π a (a - b)

3 )

ρ I	-	ρ I
π a	-	π a (a + b)

4 )

ρ I	-	ρ I
a	-	(a + b)

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Surface area of hemisphere = 2πr²

ρ(I/2πr²) V_B-V_C due to current I at point A will be

j= I

2πr²

Therefore, E = ρj = ρI

2πr²

21) Consider a block of conducting material of resistivity 'ρ' shown in the figure. Current 'I' enters at 'A' and leaves from 'D'. We apply superposition principal to find voltage 'ΔV' developed between 'B' and 'C' . The calculation is done in the following steps:
(i) Take current 'I' entering from 'A' and assume it to spread over a hemispherical surface in the block.
(ii) Calculate field E(r) at distance 'r' from A by using Ohm's law E = ρ j , where 'j' is the current per unit area at 'r'.
(iii) From the 'r' dependence of E(r), obtain the potential V(r) at 'r'
(iv) Repeat (i), (ii) and (iii) for current 'I' leaving 'D' and superpose results for 'A' and 'D'.

For current entering at A, the electric field at a distance 'r' from A is

1 )

ρ I

2 π r²

2 )

ρ I

4 π r²

3 )

ρ I

8 π r²

4 )

ρ I

r²

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a vernier scale provided on the microscope

1 )	a meter scale provided on the microscope

2 )	a screw gauge provided on the microscope

3 )	a vernier scale provided on the microscope

4 )	a standard laboratory scale.

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Magnetic Field around a wire (B)

B = μ₀ I

2 π r

where
I = current
r = distance from wire

B = 4 π X 10^-7 x 100 = 5 x 10^-6 southward

2 x π x 4

Direction is southward: Try to do cross product between the direction of current and radius. You will get the direction as southward.

1 )	5 X 10^-6 T southward

2 )	2.5 X 10^-7 T northward

3 )	2.5 X 10^-7 T southward

4 )	5 X 10^-6 T northward

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1 )	0.03 A P₂ to P₁

2 )	0.27 A P₁ to P₂

3 )	0.27 A P₂ to P₁

4 )	0.03 A P₁ to P₂

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Magnetic field is a vector field in a space that exert force on a moving electic charge. Init of magnetic field strength is Tesla.
One Tesla is a strength measured as force (Newton) on a wire of unit length (meter) with unit electric current(Ampere). Nm^-1A^-1
MT^-1C^-1

1 )	M T^-1C^-1

2 )	M T^-2C^-1

3 )	M L T^-1C^-1

4 )

M T²C^-2

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Parallel Plate Capacitor

C = 9pF = κ ε₀ A

d

where
C = [Farad (F)]
κ = dielectric constant
A = Area of plate
d = distance between the plate
ε₀ = permittivity of free space (8.85 X 10^-12 C²/N m²)

Initially there was no dielectric, so κ = 1, so

C = 9pF = ε₀ A ............................. (i)

d

After applying the dielectric, assume that the equivalent capacitance is = C

1 = 1 + 1 = d/3 + 2d/3 ............................. (ii)

C C₁ C₂ Aε₀κ₁ Aε₀κ₂

Replace κ₁=3 and κ₁=6 and solve (i) and (ii)
The equivalent capacitance C = 40.5 pF

1 )

40.5 pF

2 )

20.25 pF

3 )

1.8 pF

4 )

45 pF

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Kinetic Entergy (K) = 1 m v²

2

Mass of athlete is between 45 to 90 kg.
Assume mass of athlete is between 70 kg.
velocity of athlete 100/10 = 10 m/sec
K = 1/2 x 70 x 10 x 10 = 3500 J
So answer is 2,000 J - 5,000 J

1 )	20,000 J - 50,000 J

2 )	2,000 J - 5,000 J

3 )	200 J - 500 J

4 )	2 X 10⁵ J - 3 X 10⁵ J

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1 )

2 )

V g ( ρ₁ - ρ₂ )

3 )

4 )

V g ρ₁

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1 )

110 Ω

2 )

55 Ω

3 )

13.75 Ω

4 )

220 Ω

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1 )	54 > x > 36

2 )	36 > x > 18

3 )

18 > x

4 )

x > 54

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ε_r is greater than one for any type of material
Value of μ_r is between 1 and 0 for diamagnetic material.

1 )	ε_r =0.5 , μ_r = 0.5

2 )	ε_r =1.5 , μ_r = 1.5

3 )	ε_r =0.5 , μ_r = 1.5

4 )	ε_r =1.5 , μ_r = 0.5

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1 )

2 )

3 )

4 )

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1 )

2 )

3 )

4 )

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The current will flow into the resistor if any of the input (A or B) has the value as 1 (i.e. TRUE).
The current will NOT flow into the resistor if both the input (A or B) has the value as 0 (i.e. FALSE).
A(true) OR B(true) : current will flow into the resistor
A(true) OR B(false) : current will flow into the resistor
A(false) OR B(true) : current will flow into the resistor
A(false) OR B(false) : current will NOT flow into the resistor

So the circuit is OR.

1 )

NAND gate

2 )

OR gate

3 )

NOR gate

4 )

AND gate

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Electron is attracted towards the origin by a force k/r where 'k' is a constant and 'r' is the distance
Therefore,

k = mv²

r r

Kinetic Energy= 1 mv² = k = Constant, so T_n independent of n

2 2

As per Bohr's model

L = (angular momentum) nh

2 π

mvr = nh

2 π

= nh So r is proportional to n (h is constant (Planck's constant))

2 π

1 )

T_n ∝	1	, r_n ∝ n
	n

2 )

T_n ∝	1	, r_n ∝ n²
	n

3 )

T_n ∝	1	, r_n ∝ n²
	n²

4 )	T_n independent of n , r_n ∝ n

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4

3

none

2

4

2

2

1

1

3

3

1

4

2

2

3

1

3

1

3

1

3

1

1

1

1

2

3

4

4

4

3

4

2

4

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
17	-		0	18	-		0
19	-		0	20	-		0
21	-		0	22	-		0
23	-		0	24	-		0
25	-		0	26	-		0
27	-		0	28	-		0
29	-		0	30	-		0
31	-		0	32	-		0
33	-		0	34	-		0
35	-		0
Final score : 0

Answer sheet
Q No	Ans	For review	Q No	Ans	For review
1			2
3			4
5			6
7			8
9			10
11			12
13			14
15			16
17			18
19			20
21			22
23			24
25			26
27			28
29			30
31			32
33			34
35