AIEEE 2005 PHYSICS PAPER

PHYSICS
1. A car, starting from rest, accelerates at the rate f through a distance S, then continues at constant speed for time
f
t and then decelerates at the rate to come to rest. If the total distance traversed is 15 S, then
2
12 12 12
1) S= ft 2) S = ft 3) S= ft 4) S= ft
6 4 2
f fl2
Sol : t1 t 2t1
m
15 S
o
.c
12 1f 2
f1 + f1t + x 2t1 = 15S
2 22
e
S + ft1t + 2 S = 15S
⇒
g
ft1t = 12 S (1)
⇒
a
12
ft1 = S (2)
p
2
Dividing (1) by (2), we get
e
1
e
t1 =
6
ie
2
ft 2
1 t
S= f  =
⇒
.a
2 6 72
None of the given options is correct
w
2. A particle is moving eastwards with a velocity of 5 ms-1. In 10 seconds the velocity charges to 5 ms-1 northwards.
The average acceleration in this time is
w 1
1
1) ms-2 towards north 2) ms-2 towards north-east
w
2
2
1
3) ms-2 towards north-west 4) zero
2
r
Change in velocity ∆v
Sol: (2) Average acceleration = =
Time interval t v2
r
∆v = v 2 + (− v1 ) N
r r r
∆v = ( v 2 − v1 ) = v1 + v 2 + 2v1v 2 cos90
2
2
[As | v1 | = | v2 | = 5m / s]
= 52 + 52 + 0
90°
E
W
= 5 2 m/s v1
− v1
r
∆v 5 2 1
m/s 2 towards North - West
Average acceleration = = =
t 10 2
S
3. Out of the following pair, which one does NOT have identical dimensions is
1) impulse and momentus 2) angular momentum and Planck’s constant
3) work and torque 4) moment of inertia and moment of force
Sol: (4)
The relation between time t and distance x is t = ax2 + bx where a and b are constants. The acceleration is
4.
1) 2 b v3 2) -2 a b v2 3) 2 a v2 4) - 2 a v3
Sol: (4)
t = ax2 + bx; diff. with respect to time (t)
d dx
d dx
t = a (x2 ) + b = a.2 x + b.v.
dt dt
dt dt
1 = 2axv + bv = v (2ax + b)
1
.. diff.
2ax + b =
v
m
1
o
f
2av = -
v2
.c
f = -2av3
5. A smooth block is released at rest on a 45 incline and then slides a distance ‘d’. The time taken to slide is ‘n’
0
e
times as much a slide on rough incline than on a smooth incline. The coefficient of friction is
g
1 1 1 1
μk = 1- μk = 1 - μ8 = 1 - μ8 = 1 -
1) 2) 3) 4)
2 2 2
n2
a
n n n
p
Sol: 2)
eg s in θ − μ g c o s θ
e
ie
d
.a
smooth rough
w
1
d= (g sin θ )t2
2
w
1
g sinθ . t 2
d=
2
w 2d
2d
t1 = t2 =
g sin θ g sin θ - μg cos θ
2d 2d
=
n
g sin θ g sin θ - μg cosθ
1
n= 1− μ
1
n2 =
1− μ
1
1− μ =
n2
1
μ = 1-
n2
6. A parachutist after bailing out falls 50 m without friction. When parachute opens, it decelerates at 2 m/s2. He
reaches the ground with a speed of 3 m/s. At what height, did he bail out?
a) 182 m 2) 91 m 3) 111 m 4) 293 m
Sol : (4)
50m
v
a = -2 m/s
3 m/s
v = 2gh
v = 2 × 9.8 × 50 = 14 5
m
v 2 - u 2 32 - 980
o
S= = ≈ 244 m
2× 2 4
.c
Initially he has fallen 50 m
∴ Total hight from where he bailed out = 244 + 50 = 294 m
e
7. A bullet fired into a fixed target loses half of its velocity after penetrating 3 cm. How much further it will penetrate
g
before coming to rest further it will penetrate before coming to rest assuming that it faces consistant resistance to
a
motion ?
p
1) 2.0 m 2) 3.0 m 3) 1.0 m 4) 1.5 m
Sol: (3)
e
e
x
3cm
u
ie
u v=0
2
.a
2
u 2
  - u = 2.a .3
2

w
3u2
= 2.a.3
w
u
u2
w
a=
8
2
u
0 -  = 2.a .x
2
u2 u2
= 2. ×x
8
4
x = 1 cm
An annular ring with inner and outer radii R1 and R2 is rolling without slipping with a uniform angular speed. The
8.
F1
ration of the forces experienced by the two particles situated on the inner and outer parts of the ring, F is
2
2
 R1  R2 R1
  2) 3) 4) 1
1) R  R1 R2
2 
Sol: 3)
a2
R2
V2 = ω R 2
a1 R1
V1 = ω R 1
v12 ω 2 R12
= ω 2 R1
a1 = =
R1 R1
v22
= ω 2 R2
a2 =
R2
F1 ma 1 a1 R 1
m
= = =
taking particle masses equal F2 ma 2 a 2 R 2
o
9. A projectile can have the same range ‘R’ for two angles of projection. If ‘t1’ and ‘t2’ be the times of flights in the
.c
two cases, then the product of the two time of flights is proportional to
1 1
e
2) 3) R 4)
1) R2
R2 R
g
Sol: 3)
a
2R
T1 T2 = g (It is a formula)
p
e
T, T2 ∞ R
e
10. The upper half of an inclined plane with inclination φ is perfectly smooth while the lower half is rough. A body
starting from rest at the top will again come to rest at the bottom if the coefficient of friction for the lower half is
ie
given by
2 cos φ 2 sin φ tan φ 2 tan φ
1) 2) 3) 4)
.a
Sol: 4)
Acclaration of block while sliding down upper half = g sin θ ;
w
retardation of block while sliding down lower half = - ( g sin θ - μg cosθ )
w
for the block to come to rest at the bottom, acclaration in I half = retardation in II half.
g sin θ = - ( g sin θ - μg cosθ )
w
⇒ μ = 2 tan φ
11. A mass ‘m’ moves with a velocity ‘v’ and collides inelastically with another identical mass. After collision the Ist
v
mass moves with velocity in a direction perpendicular to the initial direction of motion. Find the speed of the
3
2nd mass after collision.
v
3 after
collision
m m
before
collision
v 2
1) 2) v 3) 4) v
3v 3 3
Sol: 4)
v
=v
3
u2 = 0
u1 = V
m m
In x direction : mv + O = m (O) + m(v2) x
v
In y direction : O + O = m   + m (v 2 ) y is
 
 3
V
(V2 ) y = (V2 ) x = V
3
m
2
V V2
o
4
∴V2 =   + V2 = + V2 = v
  3
3
 3
.c
2V
=
e
3
g
12. A particle of mass 0.3 kg is subjected to a force F = - kx with k = 15 N / m. What will be its initial acceleration if
it is released from a point 20 cm away from the origin ?
a
1) 15 m/s2 2) 3 m/s2 3) 10 m/s2 4) 5 m/s2
p
Sol: 3)
e
m = 0.3 ⇒ F = m.a. = -15 x
e
15 - 150
a=− x= x = - 50x
0.3 3
ie
m
a = - 50 x 0.2 = 10
.a
s2
13. The block of mass M moving on the frictionless horizontal surface collides with the spring of spring constant K
and compresses it by length L. The maximum momentum of the block after collision is
w
w M
w
KL2 ML2
1) 2) 3) 4) Zero
MK L
2M K
Sol: 2)
M
K
1
1
mv 2 = kl 2 : V = .L
m
2
2
K
.L
Momentam = m x v = m x
m
= km . L
14. A block is kept on a frictionless inclined surface with angle of inclination ‘α ’. The incline is given an acceleration
‘α ’ to keep the block stationary. Then α is equal to
1) g cosec α 2) g / tan α 3) g tan α 4) g
Sol: 3)
a cos
a
m
a
g sin α
o
.c
g sin α = a cos α
a = g tan α
e
15. A spherical ball of mass 20 kg is stationary at the top of a hill of height 100 m. It rolls down a smooth surface to
g
the ground, then climbs up another hill of height 30 m and finally rolls down to a horizontal base at a height of 20
a
m above the ground. The velocity attained by the ball is
p
10 30 m/s 4) 10 m/s
1) 20 m/s 2) 40 m/s 3)
e
Sol: 2)
e
ie
.a
30
100
20
w
mgh 1
mv 2 + mgh
w
2
w
12
10 x 100 = v + 10 x 20
2
12 m
v = 800, v = 1600 = 40
2 s

1
16. A body A of mass M while falling vertically downwards under gravity breaks into two parts; a body B of mas
3
2
M and a body C of mass M. The centre of mass of bodies B and C taken together shifts compared to that of
3
body A towards
1) does not shift 2) depends on height of breaking
3) body B 4) body C
Sol: 1)
does not shift as no external face acts.
17. The moment of inertia of a uniform semicircular disc of mass M and radius r about a line perpendicular to the
plane of the disc through the centre is
2 1 1
Mr 2 Mr 2 Mr 2
1) 2) 3) 4) Mr 2
m
5 4 2
Sol: 3)
o
1
.c
Mr 2
2
18. The change in the value of ‘g’ at a height ‘h’ above the surface of the earth is the same as at a depth ‘d’ below
e
the surface of earth. When both ‘d’ and ‘h’ are much smaler than the radius of earth, then which one of the
g
following is correct?
a
3h h
1) d= 2) d= 3) d=h 4) d=2h
p
2 2
Sol. 4)
e
 2h   d
e
g h1 = g 1 -  ; g d1 = g 1 - 
 R  R
ie
∴ d = 2h
.a
19. A particle of mass 10 g is kept on the surface of a uniform sphere of mass 100 kg and radius 10 cm. Find the work
to be done against the gravitational force between them to take the particle far away from the sphere
(you may take G = 6.67 x 10-11 Nm2 / Kg2)
w 13.34 x 10-10 J 3) 6.67 x 10-10 J 4) 6.67 x 10-9 J
3.33 x 10-10 J
1) 2)
w
Sol. 3)
w
GM
V=−
R
GM 6.67 x 10-11 x 100
V =- =-
R 0.1
= - 6.67 x 10 -8
‘V’ at infinity (at large distance) = 0
10
x 6.67 x 10 -8
ω= m x v =
1000
= 6.67 x 10-10 J
20. A ‘T’ shaped object with dimensions shown in the figure, is lying on a smooth floor. A force ‘ F ’ is applied at the
point P parallel to AB, such that the object has only the translational motion without rotation. Find the location of
P with respect to C.
l
B
A
P
2l
F
C
2 4
3
l l
l
1) 2) 3) l 4)
3 3
2
Sol. 4)
m
o
l
B
A
.c
P
2l
(0, l )
e
F
g
C
a
To have linear motion. The force F has 0.2l to be applied at centre of mass.
p
i.e. the paint ‘p’ has to centre of mass
e
4l 2 4l
2l x l + l x 2l
e
=
= Y= =
l + 2l 3
3l
ie
21. Which of the following is incorrect regarding the first law of thermodynamics?
.a
1) It is a restatement of the principle of conservation of energy
2) It is not applicable to any cyclic process
3) It introduces the concept of the entropy
w
4) It introduces the concept of the internal energy
w
Sol. 2)
w
22. Consider a car moving on a straight road with a speed of 100 m/s. The distance at which car can be stopped is
[μ k = 0.5]
1) 1000 m 2) 800 m 3) 400 m 4) 100 m
Sol. 1)
v 2 − u 2 = 2as
0 − u 2 = 2 (-μg) s
1
− 100 2 = 2 x - x 10 x s
2
s = 1000 m
23. A body of mass m is accelerated uniformly from rest to a speed v is a time T. The instantaneous power delivered
to the body as a function of time is given by
mv 2 1 mv 2 2
mv 2 1 mv 2
.t 2 .t .t .t
2) 3) 4)
1)
T2 2 T2
T2 2 T2
Sol: 2)
u = 0; v = u + aT; v = a.T.
Instantaneous power = F x v = m.a.v = m.a.at = m.a2.t
v2
∴ instantaneous power = m t
T2
24. Average density of the earth
1) is a complex function of g 2) does not depend on g
3) is inversely proportional to g 4) is directly proportional to g
Sol: 4)
Gm G ρ x v
g= =
R2 R2
m
4
o
G × P × πR 3
3
.c
g=
R2
e
4
g= ρπ G.R.
g
3
a
ρ → average density
p
 3g 
ρ= 
e
 4π G R 
e
‘ ρ ’ is directly profactional to (g)
ie
25. If ‘S’ is stress and ‘Y’ is Young’s modulus of material of a wire, the enrgy stored in the wire per unit volume is
.a
S2 2Y
S
2S2Y
2) 3) 4)
1)
S2
2Y
2Y
w
Sol. 1)
Energy stored per unit volume
w
Stress 

Q Y = Strain 
w


 Stress 
Q Strain = Y 
 
1
x stress x strain
=
2
1 strain
x stress x
=
2 Y
1 S2
= .
2Y
26. A 20 cm long capillary tube is dipped in water. The water rises up to 8 cm. If the entire arrangement is put in a
freely falling elevator the length of water column in the capillary tube will be
1) 10 cm 2) 8 cm 3) 20 cm 4) 4 cm
Sol. 3)
Water fills the tube entirely in gravity less condition.
∴ 20 cm.
27. The figure shows a system of two concentric spheres of radii r1 and r2 kept at temperatures T1 and T2, respectively.
The radial rate of flow of heat in a substance between the two concentric spheres is proportional is
r1
T1
T2
r2
r  r1 r2
In  r 
2
1) 2) (r2 - r1) / (r1r2) 3) (r2 - r1) 4)
 (r2 - r1 )
 1
m
Sol: 4)
o
.c
T − ∆T
dr
e
T1
g
r1
a
T2
p
r2
e
Consider a shall of thickness (dr) and of radii (r).
e
It the temperature of inner and outer surfaces of this shell be
T, (T -dT)
ie
dθ
= rate of flow of heat though it
.a
dt
kA((T − dT ) − T )
=
w
dr
−kA - dt dT
w
= - 4π kr 2
=
dr dr
w
- (dr/dT) dr
dt =
4πk r 2
integrating between the proper limits
T2 r2
 dQ  1 dr
∫ ∫r
dt = -  
 dt  4πK 2
T r1
1
dQ  1 1 

-
(T2 - T1 ) = -
dt. 4πk  r1 r2 
 
dQ 4πK r1r2 (T1 − T2 )
=
dt r2 - r1
dQ r1 r2
∴ x
dt r2 - r1
28. A system goes from A to B via two processes I and II as shown in figure. If ∆ U1 and ∆ U2 are the changes in
internal energies in the processes I and II respectively, then
p
II
B
A
I
V
1) relation between ∆ U1 and ∆ U2 can not be determined 2) ∆ U1 = ∆ U2
3) ∆ U2 < ∆ U1 3) ∆ U2 > ∆ U1
Sol: 2)
As in a cyclic process, net internal energy is zero
∆U1 = ∆U 2
m
29. The temperature-entropy diagram of a reversible engine cycle is given in the figure. Its efficiency is
o
.c
T
e
< q1 =" To" toso =" To" 2 =" To" q3 =" 0" 21 =" 1−" 2 =" 1−" r2 =" x" 6 =" 6" undecayed =" N" 3=" n="" t =" half" period =" ="" e ="" 1 ="" e =" " i =" I" d =" log" 36 =" log" d =" log" 3 =" 1" d=" ="" 5 =" .a" w =" 1.63" n="4" n="3" n="2" n="1" 11 =" hc" e =" hc" h =" 4" 5 =" hc" h =" a" e =" hc" h =" L" e =" hc" h ="1→3" a =" 10" z =" 5" t=" Input" f =" 50" t1 =" ="" f1 =" 100" y1 =" 0.1" y2 =" cos" v1 =" 0.1×100π" v2 =" −0.1" t =" 0.1π" diff =" φ1" c =" ="" c =" ="" r=" cm" light =" 500" max =" 5" m =" 160" pm =" ="" f =" sin" 2 =" 2" i =" Io" y =" 0" n2 =" 6" n2 =" 6" 2 =" 6" 2 =" 1944" x =" 0" 2x =" dx" t=" ="" n ="n" 5 =";" 100 =" 20%" x =" 0" x =" L" x="0" x="L" 2q =" 2" 1 =" 2x" x =" L" x =" 0," distance =" L" l =" 2L" i =" I" light =" 2" light =" I" q =" −" v =" v1" 2 =" .+" 2q =" .−" f =" Eq" velocity =" V" v =" 2V" 2 =" m.s" v =" Sol:" p =" 0" i2 =" (attractive)" h=" (as" 2 =" (5" i=" 50" i=" ="" r =" 6" r =" 100" y =" ⇒G" 150 =" 15mA" g =" Full" current =" 10" 5 =" 9995Ω" v =" voltage" measured =" 15"> R1). If the potential difference across the source having internal resistance R2 is zero, then
R = R2 - R1 2) R = R2 x (R1 + R2) / (R2 - R1)
1)
R = R1 R2 / (R2 - R1) 4) R = R1 R2 / (R1 + R2)
3)
Sol: 1)
ε ε
R1 R2 I
2v
R
2ε
I=
R + R 1 + R2
Potential difference across L cell
= V = ε − i R2 = 0
m
2ε
o
ε− .R2 = 0
R + R1 + R2
.c
R + R1 + R2 − 2 R2 = 0
e
R + R1 − R2
g
∴ R = R2 − R1
a
64. Two voltameters, one of copper and another of silver, are joined in parallel. When a total charge q flows through
p
the voltameters, equal amount of metals are deposited. If the electrochemical equivalents of copper and silver are
z1 and z2 respectively the charge which flows through the silver voltameter is
e
e
q q
z2 z1
z z1
ie
q q
2) 3) 4)
1) 1+ 2 1+ z1 z2
z1 z2
.a
Sol: 1)
z q
1
⇒ 1 = 2 .......(i )
m = zq ⇒ z ∝
w
q z 2 q1
w
q
q q
= 1 +1⇒ q2 =
also q = q1 + q 2 ⇒ ...(ii )
q
q 2 q2
1+ 1
q2
w q
q2 =
z
From equation (1) and (ii) 1+ 2
z1
65. Two concentric coils each of radius equal to 2π cm are placed at right angles to each other. 3 ampere and 4
ampere are the currents flowing in each coil respectively. The magnetic induction in Weber/m2 at the centre of the
coils will be
( μ 0 = 4π x 10-7 Wb / A..m.)
10-5 12 x 10-5 7 x 10-5 5 x 10-5
1) 2) 3) 4)
Sol: 4)
(1)
(2)
μ 0 i1 μ 0 × 3
B1 = =
2(2π ) 4π
μ0
2 2
B2 = B1 + B2 = .5
4π
= 10 −7 × 5 × 10 2
= 5 ×10−5 ωb /n 2
m
66. In a potentiometer experiment the balancing with a cell is at length 240 cm. On shunting the cell with a resistance
o
of 2Ω , the balancing length becomes 120 cm. The internal resistance of the cell is
.c
2) 3) 4)
1) 0 .5 Ω 1Ω 2Ω 4Ω
e
Sol: 3)
g
l −l  240 − 120
r =  1 2 × R = ×2
l  120
a
2
p
= 2Ω
e
67. A charged particle of mass m and charge q travels on a circular path of radius r that is perpendicular to a magnetic
field B. The time taken by the particle to complete one revolution is
e 2π m
ie
2π m q 2πq B
2π q 2 B
1) 2) 3) 4)
qB
B m
m
.a
Sol: 3)
mv 2
= qvB
w
r
2π r 2π m
w
T= =
v qB
w
68. A magnetic needle is kept in a non-uniform magnetic field. It experiences
1) neither a force for a torque 2) a torque but not a force
3) a force but not a torque 4) a force and a torque
Sol: 4)
A magnetic needle kept in non-uniform magnetic field experience face and torque due to unequal forces
acting on poles
69. The resistance of hot tungsten filament is about 10 times the cold resistance. What will be the resistance of 100
W and 200 V lamp when not in use ?
1) 2) 3) 4)
20 Ω 40 Ω 200 Ω 400 Ω
Sol: 2)
V2
P = Vi =
R
V 2 200 × 200
R= = = 400 Ω
100
P
400
R cold = = 40Ω
10
70. The self inductance of the motor of an electric fan is 10 H. In order to impart maximum power at 50 Hz, it should
be connected to a capacitance of
8μ F 4μ F 2μ F 1μ F
1) 2) 3) 4)
Sol: 4)
m
1
P=n=
o
2π 2c
.c
1
50 =
2π 50 × c
e
C = 1 uf.
g
71. An energy source will supply a constant current into the load if its internal resistance is
a
1) very large as compared to the load resistance 2) equal to the resistance of the load
p
3) non-zero but less than the resistance of the load 4) zero
e
E
e
Sol: 4) I = , Internal resistance (r) is
R+r
ie
E
zero, I = = cons tan t
R
.a
π
72. The phase difference between the alternating current and emf is . Which of the following cannot be the
2
w
constituent of the circuit?
w
1) R, L 2) C alone 3) L alone 4) L, C
Sol: 4)
w
73. A circuit has a resistance of 12 ohm and an impedance of 15 ohm. The power factor of the circuit will be -
1) 0.4 2) 0.8 3) 0.125 4) 1.25
Sol:.2)
R 12 4
Power factor = cos φ = = = = 0 .8
Z 15 5
74. A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If an
electron is projected along the direction of the fields with a certain velocity then -
1) its velocity will increase 2) its velocity will decrease
3) it will turn towards left of direction of motion 4) it will turn towards right of direction of motion
Sol:.2)
due to electricfield, it experiences force and accelerates i.e, its velocity decreases.
75. A coil of inductance 300 mH and resistance 2Ω is connected to a source of voltage 2 V. The current reaches half
of its steady state value in
1) 0.1 s 2) 0.05 s 3) 0.3 s 4) 0.15 s
Sol: 1)
 − −
It
i = i0  1 − e L  

 
i0
= i0 (1 − e)
2
L
× 0.693 = t
R
300 × 10 −3
= 0.693 ×
m