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2007 AIEEE PAPER - (Physics)

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x = 2 X 10^-2 cos πt
Differentiating both side
=> dx/dt = -0.02πsinπt
=> v = -0.02π sinπt
=> v is maximum when, πt = π/2
=> t = 1/2 = 0.5 sec

1 )

0.5 s

2 )

0.75 s

3 )

0.125 s

4 )

0.25 s

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In AC circuit average power is :
P_avg = V_rmsI_rms cosφ
In AC circuit voltage and current is represented as:
V = V₀sin(ωt+φ)
I = I₀sin(ωt)

So, φ = -π/2 = -90°
So, P_avg = V_rmsI_rms cos(-90°)
So, P_avg = 0

1 )

	P =	E₀I₀

2 )

P = zero

3 )

	P =	E₀I₀
		2

4 )

P =

E₀I₀

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Distance between point A and origin(0,0) = 2
Distance between point B and origin(0,0) = 2
As the distance from the electric charge of A and B are same
so both A and B will have same potential.
So potential difference between point A an B will be = 0

1 )

9 volt

2 )

zero

3 )

2 volt

4 )

4.5 volt

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Q = charge on the capacitor = CV
Work done by battery = Q.V = CV²
Energy stored in capacitor = (1/2)CV²
So, (Energy stored in capacitor)/(Work done by battery) = 1/2

1 )

2 )

3 )

1/4

4 )

1/2

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I=I₀[1-e^-Rt/L]
I₀ = 5/5 = 1
So, I=1-e^-5t/10
=> I=1-e^-t/2
As t=2, so
I=1-e^-1

1 )

( 1 - e )

2 )

3 )

e^-1

4 )	( 1 - e^-1 )

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Magnetic Field around a wire (B₁) when r is greater than the radius of the wire.

B₁ = μ₀ I

2 π r

where
I = current
r = distance from wire
and r ≥ Radius of the wire

Magnetic Field around a wire (B₂) when r is less than the radius of the wire.

B₂ = μ₀ I r

2 π R²

where
I = current
R = radius of wire
r = distance from wire
and r ≤ Radius of the wire (R)

B₂ = μ₀ I a

4 π a²

B₁ = μ₀ I

4 π a

So, B₂/B₁ = 1

1 )

1/4

2 )

3 )

4 )

1/2

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If electric current flowing through a hollow pipe, it will induce a magnetic field outside the pipe.
The magnetic field inside the pipe will be zero because a closed loop just inside the pipe will not have any current flowing through it.
This situation is similar to a Faraday cage where the electric field inside a hollow conducting shell is zero.

1 )	the magnetic field is zero only on the axis of the pipe

2 )	the magnetic field is different at different points inside the pipe

3 )	the magnetic field at any point inside the pipe is zero

4 )	the magnetic field at all points inside the pipe is the same, but not zero

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Nuclear binding energy = [mass of nucleus - mass of nucleons].C²
Nuclear binding energy = (M₀ - 8 M_P - 9 M_N) C²

1 )	(M₀ - 8 M_P) C²

2 )	(M₀ - 8 M_P - 9 M_N) C²

3 )

M₀ C²

4 )	(M₀ - 17 M_N) C²

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Gamma ray is aform electromagnetic radiation. It is produced by sub-atomic particle interactions, such as electron-positron annihilation or radioactive decay. It does does not involve any change in proton number or neutron number

1 )	both the neutron number and the proton number change

2 )	there is no change in the proton number and the neutron number

3 )	only the neutron number changes

4 )	only the proton number changes

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In the first half cycle, the diode is in forward biased.
In the next half-cycle, the diode is in reverse biased
Diode is forward biased in first half-cycle.
Voltage applied: 10 V (it difference the peak ie. +5V and -5V)
So the amplitude of signal: 5V.

1 )

2 )

3 )

4 )

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Relation between energy and frequency
E = hν
where
E = Energy
h = Planck's constant
ν = frequency As per de Broglie equation

λ = h = h

p mv

where
p = momentum
λ = wavelength
h = Planck's constant
v = velocity

By solving above two equation
p= hν/c

1 )

ν/c

2 )

h ν c

3 )

h ν /c²

4 )

h ν /c

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1 )	v₀ + 2g + 3f

2 )	v₀ + g/2 + f/3

3 )	v₀ + g + f

4 )	v₀ + g/2 + f

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Using perpendicular Axis Theorem:
I_AC = I_EF

1 )	I_AC = I_EF

2 )	I_AD = 3 I_EF

3 )	I_AC = I_EF

4 )	I_AC = I_EF

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x = x₀ cos (ω t - π/4)
dx/dt = v = -x₀ωsin(ωt-π/4)
dv/dt = a = -x₀ω²cos(ωt-π/4)
a = x₀ω²cos(ωt+3π/4)
By comparing it with , a = A cos (ω t + δ )
A = x₀ω²
δ = 3π/4

1 )	A = x₀ , δ = - π/4

2 )	A = x₀ ω² , δ = π/4

3 )	A = x₀ ω² , δ = - π/4

4 )	A = x₀ ω² , δ = 3π/4

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Electric field (E) is vector quantity.
Electric Potential (V) is scalar quantity.
By changing the charge Electric Potential (V) do not change, but the direction of Electric field (E) changes.

1 )	E remains unchanged, V changes

2 )	both Both E and V change

3 )	E and V remains unchanged

4 )	E changes, V remains unchanged

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Half life of radioactive element

t_1/2 = ln(2)

λ

Average life of radioactive element

τ = 1

λ

So,

ln(2) = 1

λ_x λ_y

So, 1.4λ_x = λ_y
λ_x < λ_y
So element Y will decay faster than element X

1 )	X will decay faster than Y

2 )	Y will decay faster than X

3 )	X and Y have same decay rate initially

4 )	X and Y decay at same rate always

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Efficiency of Carnot cycle

η = 1 - T_c

T_h

W = η.Q where Q=total energy put into system
=> 10=(1/10)Q
=> Q = 100
=> So energy absorbed by the system = (total energy put into system)-(Work done)
=> energy absorbed by the system = 100-10 = 90

1 )

99 J

2 )

90 J

3 )

1 J

4 )

100 J

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1 )	The number of free conduction electrons is significant in C but small in Si and Ge

2 )	The number of free conduction electrons is negligible small in all the three

3 )	The number of free electrons for conduction is significant in all the three

4 )	The number of free electrons for conduction is significant only in Si and Ge but small in C

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Magnetic force acting on charge = F_m = q v x B
Electric force acting on charge = F_e = q E
So, q v x B = q E
=> q v x B = q E
=> v =(E x B)/B²

1 )	v = E X B/B²

2 )	v = B X E/B²

3 )	v = E X B/E²

4 )	v = B X E/E²

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V(x) = 20

x²-4

E = - dV = 20 (2x-0) = 160 = 10 (+ve)

dX (x²-4)² 144 9

1 )	5/3 Volt/μ m and in the -ve x direction

2 )	5/3 Volt/μ m and in the +ve x direction

3 )	10/9 Volt /μ m and in the -ve x direction

4 )	10/9 Volt/μ m and in the +ve x direction

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Emitting Photons(Rydberg Formula)

E_photon = E₀( 1 - 1 )

n₁² n₂²

where
n₁ < n₂
E₀ = 13.6 eV

By using above formula E is maximum when n=2 to n=1
As, E=hν
So ν is maximum if E is maximum.

1 )	n = 2 to n = 6

2 )	n = 6 to n = 2

3 )	n = 2 to n = 1

4 )	n = 1 to n = 2

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Assuming acceleration of both blocks are: a

a = F

M+m

So force acting on m = F_m = Fm

M+m

1 )

2 )

(M + m)F

3 )

M+m

4 )

M+m

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P=P₁+P₂ = -15+5=-10
The focal length of the combination = 1/P = -1/10 m = -10cm

1 )

- 20 cm

2 )

- 10 cm

3 )

+ 20 cm

4 )

+ 10 cm

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Assume that the temperature at the interface = T

(T₁-T)k₁ = (T-T₂)k₂

l₁ l₂

By solving above equation, T = T₁k₁l₂+T₂k₂l₁

k₁l₂+k₂l₁

1 )	( k₂ l₂ T₁ + k₁ l₁ T₂ ) / ( k₁ l₁ + k₂ l₂ )

2 )	( k₂ l₁ T₁ + k₁ l₁ T₂ ) / ( k₂ l₁ + k₁ l₂ )

3 )	( k₁ l₂ T₁ + k₂ l₁ T₂ ) / ( k₁ l₂ + k₂ l₁ )

4 )	( k₁ l₁ T₁ + k₂ l₂ T₂ ) / ( k₁ l₁ + k₂ l₂ )

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Sound intensity is sound power Pac per unit area A
Sound intensity I = 10log₁₀(I/I₀)
B₁ = 10log₁₀(I₁/I₀)
B₂ = 10log₁₀(I₂/I₀)
B₁-B₁=20
20 = 10log₁₀(I₁/I₂)
=> 2 = log₁₀(I₁/I₂)
=> I₁/I₂ = 100
=> I₁ = 100I₂

1 )

1000

2 )

10000

3 )

4 )

100

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c_p - c_v = R [for one mole of gas]
=> c_p - c_v = R [for one 28 gm of nitrogen gas]
=> c_p - c_v = R/28 [for one 1 gm of nitrogen gas]

1 )	C_p - C_v = R/28

2 )	C_p - C_v = R/14

3 )	C_p - C_v = R

4 )	C_p - C_v = 28 R

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Force acting on the particle = F = q v x B
The force acts perpendicular to the velocity so there is no work done on particle. So kinetic energy will not change.
As force acts on the particle so there is change in its momentum

1 )	the momentum changes but the kinetic energy is constant

2 )	both momentum and kinetic energy of the particle are not constant

3 )	both, momentum and kinetic energy of the particle are constant

4 )	kinetic energy changes but the momentum is constant

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Force acting at a distance d from O, due to wire AOB = F₁ = μ₀I₁

2 π d

Force acting at a distance d from O, due to wire COD = F₂ = μ₀I₂

2 π d

F₁ and F₂ are perpendicular to each other, So net force will be

μ₀ ( I₁² + I₂² )^½

2 π d

1 )

		μ₀	(	I₁ + I₂	)^½
		2 π		d

2 )

		μ₀	( I₁² + I₂² )^½
		2 π d

3 )

		μ₀	( I₁ + I₂ )
		2 π d

4 )

		μ₀	( I₁² + I₂² )
		2 π d

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5=R₀(1+50α) ..................... (i)
6=R₀(1+100α) ..................... (ii)
By solving (i) and (ii)
α = 1/200
R₀ = 4Ω

1 )

2 ohm

2 )

1 ohm

3 )

4 ohm

4 )

3 ohm

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Net work done by the system is zero because there is no change in energy of the system.

1 )	1/2 (K - 1) C V²

2 )	C V²(K - 1)K

3 )	(K - 1)C V²

4 )

zero

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Net work done by the system is zero because there is no change in energy of the system.

1 )

2 )

3 )

g_E/g_m

4 )

g_m/g_E

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Let mass of the circular disc = M
So mass of the removed disc = M/4
So mass of the remaining desc = 3M/4
Assume that centre of mass of remaining disc is at a distance x from centre
So, (3M/4).x = (M/4).R
=> x = R/3 = αR
=> α = 1/3

1 )

1/3

2 )

1/2

3 )

1/6

4 )

1/4

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Assume that acceleration = a
So angular acceleration = a/R

mgsinθ - f = ma .............. (i)
fR = I(a/R) ....................... (ii)
By solving (i) and (ii)

g sin θ

1 + (I/(MR²))

1 )

		g sin θ
		1 + (I/(MR²))

2 )

		g sin θ
		1 + ((MR²)/I)

3 )

		g sin θ
		1 - (I/(MR²))

4 )

		g sin θ
		1 - ((MR²)/I)

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The direction of the force that acts on the rotating particle passes thru the center. So there is no torque acting on the particle. As no torque acts on the particle so angular momentum is constant.

1 )	Constant Force

2 )	Constant linear momentum

3 )	Zero Torque

4 )	Constant Torque

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Assume that spring is compressed by distance=x
Kinetic energy of the block = (1/2)×2×4² = 16
While compressing the spring, energy loss due to friction = 10x
Energy transferred to spring due to block = (1/2)×10000×x²
(1/2)×10000×x² + 10x = 16 ...................(i)
By solving equation (i), x = 5.5 (approx.)

1 )

5.5 cm

2 )

2.5 cm

3 )

11.0 cm

4 )

8.5 cm

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Kinetic energy of the particle at the beginning : K = (1/2)mv²
Horizontal velocity of the particle = vcos60° Kinetic energy of the particle at the beginning : K = (1/2)mv²
Kinetic energy of the particle at the hightest point = (1/2)m(vcos60°)²
Kinetic energy of the particle at the hightest point = (1/2)mv²cos²60° = K/4

1 )

2 )

Zero

3 )

K/2

4 )

K/4

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Now, I/I₀ = cos²&phi/2 = cos²(2π/λ)(λ/12) = cos²30°
I/I₀ = 3/4

1 )

2 )

3 )

1/2

4 )

3/4

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1 )

f/2

2 )

f/4

3 )

4 )

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As per the first law of thermodynamics :
ΔQ = ΔU + ΔW
For the route iaf, 50 = 20 + ΔU
ΔU = 30
For the route ibf, 36 = ΔW + 30
ΔW = 6

1 )

6 cal

2 )

16 cal

3 )

66 cal

4 )

14 cal

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Kinetic Energy = (1/2)mω²a²sin²ωt
and ω = 2πν

Kinetic Energy = π²ma²ν²

1 )	&pi² m a²ν²

2 )	(1/4) &pi² m a²ν²

3 )	4 &pi² m a²ν²

4 )	2 &pi² m a²ν²

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1

2

2

4

4

3

3

2

2

4

4

2

3

4

4

2

2

4

1

4

3

3

3

3

4

4

1

2

3

4

1

1

1

3

1

4

4

4

1

1

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	36	-		0
37	-		0	38	-		0
39	-		0	40	-		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			36
37			38
39			40