Document 6529736

IITJEE PHYSICS SAMPLE PAPER - I
SOLUTIONS
SECTION – I
Straight Objective Type
Tdt
1.
M
v
2
Tdt
u
2g
Tdt
l
2
M
u
2
v
gl
Tdt

v
T1
Mv
T1 Mg
..... (i)
C
Ma
T1
M
..... (ii)
A
a
..... (iii)
B
Mg
v2
l
a
From (i) and (ii)
3M
v
2
M
u
2
v
u
3
a
u2
9l
gl
9l
g
.
9
(b)
2.
All the equipotential surfaces of the field between the
sphere and the plate are convex down ward. Hence on
any straight line parallel to plate the points farther from
the sphere will have potential lower than those closer to
sphere.
(b)
3.
Torque about O is zero as well angular momentum hence
(d)
4.
Mg ( H
y)
1
M (vc )2
2
2Mv1 Mv c
2v c v
1
M (v 2
2
1
Ic w
2
v12 )
3
Solving (i), (ii) and (iii) v 2
4
(d)
V
V3 V2 1
1
2
= 0.
1
Mv12
2
.....(i)
v y
v1
..... (ii)
..... (iii)
g (H
y)
v
dy
dt
4g
(H
3
y)
t
3H
g
5.
6.
Both upper half and lower half will have same effective area of
R2
so charge in flux will be same and induced emf will have
2
some value. But since the resistance is different due to which
current must be different but ring is as a whole is closed circuit
so electric field will be generated to make the current flow in
both parts to be same.
E R i10r 0
..... (i)
E R ir 0
..... (ii)
2E R 9ir 0
E
R 2b
9ir
i
E
11r
11r
2 R
2
9r
Rb
9
=
E
Rb
2 R 11r
22
(b)
a cos w0t a cos wt cos w0t
h
( w w0 )
2
a cos w0t
Q
(c)
8.
i
10r
E
–
+
–
+
a
[cos(w w0 ) cos( w w0 )]
2
Highest possible energy for photon corresponds to frequency w w0 hence.
E
KEmax
7.
10 r
r 2h gh
WST
r h gh , Wg
2
r 2 gh 2
Heat = Q = WST Wg =
and h
2
(a)
2
a 2 h2
vs
t1
t1
t2
t2
2
a
vp
vs
(a)
vp
vs
h
2
1
a
vp
h
vs
2S
r g
Q
2 S2
g
E
SECTION II
Reasoning Type
9.
PV = nRT
P
nR
T
V
V
1
( slope)
v1 = constant
P
1
2
v2 = constant
T
(b)
10.
Potential at E and K are different due to which current flows between E and K make current
flow between AB and CD also possible.
(d)
11.
Assertion and Reason correct and correct explanation.
(a)
12.
Acceleration relative to cart parallel to incline is always zero only the acceleration
perpendicular to incline will change in different situation due to which change in tension but
angle will remain same and string always remain perpendicular to the incline.
(d)
SECTION III
Linked Comprehension Type
Passage-I
When the temperature of rods in increased there will be increase in their lengths and thereby
the springs are compressed, let x1 , x 2 and x 3 be the compression in the three springs
respectively. Then
L
Kx1
L
Kx1
T
L
T x1 x2 x2
2
2Kx2 and 2 Kx2 3Kx3
x1 2 x 2 3x 3
x x1 3
x1 1
L T
2 3 2
9
x1
L T
11
2Kx2
2Kx2
3Kx3
E1
1 2
Kx1
2
1
9
K
L
2 11
2
81
KL2
242
T
2
( T )2
Similarly
E2
E3
13.
14.
15.
16.
81
KL2
484
27
KL2
242
2
T2
2
T2
(b)
(a)
(a)
Passage-II
The charge q on the small disc can he calculated by applying Gauss law
E · ds
0
q
Since one side of the small disc is in contact with plate
r2
r2 0
2
.
q
E
·
r
V
xv
x
0
0
d
d
(b)
17.
Let v s be the steady velocity of the small disc just after its collision with the bottom plate
then the steady state kinetic energy k s of the disc just above the bottom plate is given by
1
ks
mv s2
2
for each round trip disc gains electrostatic energy by
U = 2qV
1
1 k after .
k loss k before k after (1 e 2 ) k before
2
Since k s is the energy after collision at the bottom plate and (ks qV mgd ) is the
energy before the collision at the top plates total energy loss can be written as
1
ktot
1 ks (1 2 ) (ks qV mgd )
2
In its steady state U should be compensated by k total
1
2qV
2
1 ks
(1
2
) qV
2
) (ks
qV
2
ks
4
1
2
1 2
mvs
2
1
2
vs
1
[(1
2
2
(1
) mgd ]
2
2
qV
2 xV 2
m
1
1
2
mgd
2
(2 gd )
mgd )
vs
18.
V2
.
(c)
The disc will lose kinetic energy eventually cease to move when the disc cannot reach the
top plate. In other words, the threshold voltage V c can be determined from the condition
that velocity of the disc at the top plate is zero. In order to have velocity zero at top plate
kinetic energy at top plate must satisfy the relation
Ks ks qVc mgd 0
k s is steady state kinetic energy at the bottom plate
e2
qVc
1 e2
Vc
1 e2
1 e2
e2
mgd
1 e2
qVc
mgd
0
mgd
x
(d)
SECTION IV
Matrix Match Type
1.
Concave
Convex
(A)
m < 0 |m| < 1
Object as well image is real in case of concave mirror object and image is virtual
incase of convex.
(B)
m < 0 |m| > 1
Both real in case of concave both virtual incase of convex.
(C)
|m| < 1 m > 0
Concave object virtual image real.
Convex object real image virtual.
(D)
|m| > 1 m > 0
Concave object real image virtual.
Convex object virtual, image real.
(A) –2, 4; (B) –1, 3; (C) –2, 3; (D) –2, 3
2.
q, v, m
B
F
R/2
y0
O
D
M
A
R–y0
x
B
C
R
2B
mv
; cos
qB
R
y0
R
m
q (2 B)
= 2 ; t AB
T
R
; t BC
m
(3 )
qB
t AB t BC
; XB
1
XA
R sin
XC
XB
BC
4R sin
XD
XC
CD
5R sin
m
q (B )
m
3 (cos 1 (1
qB
XA
AB
))
2R sin
When velocity be come parallel
t FA t AM
m
qB
m
q2B
m
3
2qB
3m
2qB
(A) –2; (B) –1; (C) –3, 4; (D) –1, 2
SECTION V
Subjective or Numerical Problems
2T cos
D
= W + 2w
2 N cos (90 – ) = W
2 N sin
=W
Taking torque about B
T × AB sin 2
1
= w AB sin
2
T × 4 × sin 2 = w × 2 sin
A
N OB cos
+ N × r cot
(AB = 4 m ; OB = r cosec )
Solving from above r = 3m
2
C
N
N
O
90º
90º–
1.
w
B
W
w
2.
y
d
x
;
dt
a
a log sec
dy
dx
tan
w (Constant)
d2y
dx 2
x
;
a
1
x
sec2
a
a
dy
dx
1
Radius of curvature =
dy
dx
tan
dx
dt
a
d2y
dt 2
tan
d
dt
aw · sec2
2
d y
dx 2
x
a
0;
x 1
· dx
a a
Now resultant acceleration
a
d 2x
dt 2
2
d2y
dt 2
0 a 2 w 4 sec 4
aw 2 sec 2
x
a
2
3
2
= a sec
x
a
x
; x=a
a
d 2x
dt 2
aw ;
2
2
x
a
w2
R2
a
2 2 1
a
sec2
1
2
4a
2
1
=8
2 4 m/sec 2
4
dy
dt
dy
dx
aw 2 sec 2
dx
dt
x
a
tan
x
a
aw