01_Sol_Phy
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I I TJEE PHYSICS SAMPLE PAPER - I
SOLUTIONS
SECTION – I
Straight Objective Type
1. u M
v M
Tdt 22 C
Tdt
v
..... (i)
gl l
g u2
2
MvTdt v
Tdt A
T 1
M ..... (ii)
Ma Mg T 1
Mg
T 1
B
a..... (iii)
al
v2
From (i) and (ii) u M
v M
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
V3 V2 V1
(b)
3. Torque about O is zero as well angular momentum hence = 0.
(d)
4. )(2
1)(
2
1
2vv M y H g 21
2
1 Mv
w I v M cc2
1)(
2
1 2 .....(i)
c Mv Mv12 ..... (ii)
vvc2 ..... (iii)
yvv 1
Solving (i), (ii) and (iii) )(4
3 2
y H g v g
H
t y H
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.
010r i R E ..... (i)
0r i R E ..... (ii)
092 ir R E
R
ir
E 2
9
r
b R
r
E
i 1111
2
r
b R
R
r E
112
9 2 = Rb
22
9
10 r
i 10r
E
– +
E
– +
(b)
6. t wwt at wa E 00 coscoscos )]cos()[cos(2
cos 000 wwwwa
t wa
Highest possible energy for photon corresponds to frequency 0ww hence.
Qww
h KE
2
)( 0max
(c)
7. ghhr W ST 2 ,
2
2 ghhr W g
Heat = Q = g ST W W =2
22 ghr and
g r
S h
2
g
S Q
22
(a)
8. svhat
22
1 s p vv
at 2
21 t t
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 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-IWhen the temperature of rods in increased there will be increase in their lengths and thereby
the springs are compressed, let21 , x and 3 be the compression in the three springs
respectively. ThenL
Kx 1 2Kx 2 2Kx 2 3Kx 3
2212
x x xT L
T L
21 2 Kx Kx and 32 32 Kx Kx
321 32 x x
T L x x
2
3
32
111
T L
11
91
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2.
q, v , m
y 0 O
F
B
2B
x
R /2
M A
R –y 0 B
R
C
D
qB
mv R ; 1cos 0
R
y R
= 2 ;)2( Bq
mt AB ;
)( Bq
mt BC
))1((cos3)3( 1
qB
m
qB
mt t T BC AB
sin R A ; sin2 R AB X A B
sin4 R BC X BC
sin5 RCD X C D
When velocity be come parallel
qB
m
qB
m
Bq
m
qB
mt t AM FA
2
33
22
(A) –2; (B) –1; (C) –3, 4; (D) –1, 2
SECTION V
Subjective or Numerical Problems
1. 2T cos = W + 2w
2 N cos (90 – ) = W2 N sin = W
Taking torque about B
T × AB sin 2 = cossin2
1OB N ABw
T × 4 × sin 2 = w × 2 sin + N × r cot
( AB = 4 m ; OB = r cosec )
Solving from above r = 3m
D
2
90º
C A
9 0 º –
O
NN
w w
W
B
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2.a
xa seclog ; w
t d
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
d a
dt
dx; 0
2
2
dt
xd ; aw
a
x
dt
dx
dx
dy
dt
dytan
dxaa
xaw
dt
yd 1·sec·
2
2
2
a
xaw 22 sec
Now resultant acceleration
2
2
22
2
2
dt
yd
dt
xd a
a
xwa 442 sec0
22
22 sec Ra
w
a
xaw
a
a
4sec
2
1
2
1
22 22
= 2m/sec424
18
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