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I ITJ EE PHYSICS SAMPLE PAPER - ISOLUTIONS
SECTIONI
Straight Objective Type
1. uM
vM
Tdt22
C
Tdt
v
..... (i)
gll
gu2
2
MvTdt v
Tdt
A
T1
M ..... (ii)
MaMgT1
Mg
T1
B
a..... (iii)
al
v2
From (i) and (ii) uM
vM
22
3
3
uv
999
2 g
l
gl
l
ua .
(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. 1 2
V3V2 V1
(b)
3. Torque about O is zero as well angular momentum hence = 0.
(d)
4. )(2
1)(
2
1
2vvMyHMg 21
2
1Mv
wIvM cc2
1)(
2
1 2 .....(i)
cMvMv12 ..... (ii)
vvc2 ..... (iii)
yvv1
Solving (i), (ii) and (iii) )(4
3 2
yHgv g
H
tyH
g
dt
dy
v
3
)(3
4
(d)
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5. Both upper half and lower half will have same effective area of
2
2
R so charge in flux will be same and induced emf will have
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.
010riRE ..... (i)
0riRE ..... (ii)
092 irRE
R
ir
E 2
9
r
bR
r
E
i 1111
2
r
bR
R
rE
112
9 2= Rb
22
9
10 r
i10r
E
+
E
+
(b)
6. twwtatwaE 00 coscoscos )]cos()[cos(2
cos 000 wwwwa
twa
Highest possible energy for photon corresponds to frequency 0ww hence.
Qww
hKE
2
)( 0max
(c)
7. ghhrWST2 ,
2
2 ghhrWg
Heat = Q = gST WW =2
22 ghrand
gr
Sh
2
g
SQ
22
(a)
8.svhat
22
1 sp vh
vat2
21 tt
1
2
2
s
p
s
p
v
v
hv
v
a
(a)
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SECTION II
Reasoning Type
9. PV= nRT
V
nR
T
P
)(
1
slopeV
P
T
12
v2 = constant
v1 = constant
(b)
10. Potential atEandKare different due to which current flows betweenEandKmake current
flow betweenAB 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-IWhen the temperature of rods in increased there will be increase in their lengths and thereby
the springs are compressed, let21 , xx and 3x be the compression in the three springs
respectively. ThenL
Kx1 2Kx2 2Kx2 3Kx3
2212
xxxTL
TL
21 2KxKx and 32 32 KxKx
321 32 xxx
TLxx
x2
3
32
111
TLx
11
91
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2
2
1111
9
2
1
2
1TLKKxE 222 )(
242
81TKL
Similarly
222
2484
81TKLE
222
3242
27TKLE
13. (b)
14. (a)
15. (a)
Passage-II16. The charge q on the small disc can he calculated by applying Gauss law
qdsE 0
Since one side of the small disc is in contact with plate
xvVd
rrEq
2
0
2
0 d
rx 0
2
.
(b)
17. Let sv be the steady velocity of the small disc just after its collision with the bottom plate
then the steady state kinetic energy sk of the disc just above the bottom plate is given by
2
2
1ss mvk
for each round trip disc gains electrostatic energy by
U= 2qV
afterbeforeloss kkk beforeke )1(2
afterk112
.
Since sk is the energy after collision at the bottom plate and )( mgdqVks is the
energy before the collision at the top plates total energy loss can be written as
)()1(1
1 22 mgdqVkkk sstot
In its steady state Ushould be compensated by totalk
)()1(11
2 22
mgdqVkkqV ss
])1()1[(1
22
4
2
mgdqVks
mgdqVmvs 2
2
2
22
112
1
)2(1
2
1 22
2
2gd
m
xVvs
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2Vvs .
(c)18. The disc will lose kinetic energy eventually cease to move when the disc cannot reach the
top plate. In other words, the threshold voltage cV 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
0mgdqVkK css
sk is steady state kinetic energy at the bottom plate
011 2
2
2
2
mgdqVmgde
eqV
e
ecc
xmgd
eeVc 2
2
11
(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.
Convexobject 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
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2.
q, v, m
y0O
F
B
2B
x
R/2
MA
Ry0B
R
C
D
qB
mvR ; 1cos 0
R
yR
= 2 ;)2( Bq
mtAB ;
)(Bq
mtBC
))1((cos3)3( 1
qB
m
qB
mttT BCAB
sinRXA ; sin2RABXX AB
sin4RBCXX BC
sin5RCDXX CD
When velocity be come parallel
qB
m
qB
m
Bq
m
qB
mtt AMFA
2
33
22
(A)2; (B)1; (C)3, 4; (D)1, 2
SECTION V
Subjective or Numerical Problems
1. 2Tcos = W+ 2w
2Ncos (90
) = W2Nsin = W
Taking torque aboutB
T AB sin 2 = cossin2
1OBNABw
T 4 sin 2 = w 2 sin +N rcot
(AB = 4 m ; OB = rcosec )
Solving from above r= 3m
D
2
90
CA
90
O
NN
w w
W
B
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2.a
xay seclog ; w
td
d(Constant)
a
x
dx
dytan ;
a
x
adx
yd 22
2
sec1
Radius of curvature =
2
2
2 23
1
dx
yd
dx
dy
=a
xa sec
a
x
dx
dytantan
a
x; x = a
awdt
da
dt
dx; 0
2
2
dt
xd; aw
a
x
dt
dx
dx
dy
dt
dytan
dxaa
xaw
dt
yd 1sec
2
2
2
a
xaw 22 sec
Now resultant acceleration
2
2
22
2
2
dt
yd
dt
xda
a
xwa 442 sec0
22
22 sec Ra
w
a
xaw
a
a
4sec
2
1
2
1
22 22
= 2m/sec424
18