iit jee 2005 physics sol screening)
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8/8/2019 Iit Jee 2005 Physics Sol Screening)
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Tvm Branch: T.C.No: 5/1703/30, Golf Links Road, H.B. Clony, Kowdiar Gardens, Trivandrum,(: 0471-2438271
Kochi Branch: Bldg.No.41/352, Mulloth Ambady Lane, Chittoor Road, Kochi - 011,(: 0484 - 2370094,93884659441
Note: Based on the memory
IIT - JEE 2005 Physics Solution (Screening)
1 Sol: Ans. C
mv
h
=λ
v is increased, λ is decreased.
.decreasesd / D β⇒λ=β .
2 Sol: Ans. A
2
2
0
2 )s / m(50
A
A1
gh2u =
−
= h
52 .5 cm
3 m
3 Sol: Ans. A
3
2
P
P
convex
concave =
convexconcave f
1
f
1
F
1+=
,cm10f f 3
1
f
1
f 3
2
30
1=⇒=+
−= where f is focal length of convex lens.
4 Sol: Ans. D
ρ=
B1f 1
l
ρ=
B
4
nf 2
l
12
2
1 f 4
nf
n
4
f
f =⇒=
For the first resonance n = 5, 12 f 4
5f = (as frequency increases)
5 Sol: Ans. C
( ) ( )( ) ( ) ( )k ˆ2
k ˆ2
2k ˆ
2E
000
P ∈σ−
+∈
σ−+−
∈σ
=r
k ˆ2
0∈σ−
=
6 Sol: Ans. C
Rate of heat gain = 100 - 160 = 840 J/s
∴ Required time = sec20min8sec500840
)2777(102.42 3==−××× .
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Tvm Branch: T.C.No: 5/1703/30, Golf Links Road, H.B. Clony, Kowdiar Gardens, Trivandrum,(: 0471-2438271
Kochi Branch: Bldg.No.41/352, Mulloth Ambady Lane, Chittoor Road, Kochi - 011,(: 0484 - 2370094,93884659442
Note: Based on the memory
7 Sol: Ans. B
8 Sol: Ans. A
10.2 eV photon on collision will excite H-atom to first excited state but Hydrogen atom will return to ground
state before next collision. Second photon will provide ionization energy to Hydrogen atom, i.e., electron will be
ejected with energy = 1.4 eV.
9 Sol: Ans. A
ttancons)1Z(2 =λ−
6Z)1Z(4)10(22 =⇒−λ=λ∴ .
10 Sol: Ans. A
Temperature of sun would be maximum out of the given three
as =λ Tm constant
mλ for Sun is minimum
11 Sol: Ans. A
According to Kirchoff’s junction rule no current passes through 2 Ω resistor.
∴ i = 0
12 Sol: Ans. D
For equilibrium, f = Mg.
F = N
For rotational equilibrium normal will shift downward.
x
N
F
f
M g
Hence torque due to friction about centre of mass = Torque due to Normal reaction about centre of mass.
13 Sol: Ans. A
RC
t
RC
t
eR
ie1Cq−− ε
=⇒
−ε=
C=4µF
ε=12V
2. 5 Ωm
CR VV3 =
4 / 1ee3e1RC / tRC
t
RC
t
=⇒ε=
−ε⇒ −−−
.
t/RC = 2ln2
∴ t = 20× (0.693) = 13.86 sec.
14 Sol: Ans. C
MeV24.105.931]9994.150026.44[mcE2
=×−×=∆=
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Tvm Branch: T.C.No: 5/1703/30, Golf Links Road, H.B. Clony, Kowdiar Gardens, Trivandrum,(: 0471-2438271
Kochi Branch: Bldg.No.41/352, Mulloth Ambady Lane, Chittoor Road, Kochi - 011,(: 0484 - 2370094,93884659443
Note: Based on the memory
15 Sol: Ans. C
f 4
v3eand
f 4
ve 21 =+=+ ll
f 4
v212 =− ll
v
v
)(
)(
12
12 ∆=
−−∆ll
ll
)(f 2)(f 2v 2112 llll ∆+∆=−=∆=∆ (For maximum error)
s / cm8.2042.05122 =××= .
16 Sol: Ans. A
2
222
0 MR43
R2
M2
)3 / R(M
2
MR9
l =
+−=O
O '
17 Sol: Ans. D
Heating of glass bulb is by radiation.
18 Sol: Ans. A
19 Sol: Ans. C
By definition.
20 Sol: Ans. C
After first refraction, position of the image = cm2533.125.33 =
From reflection,40
1
v
1
1525
1
v
1
f
1−=
+−=
From the second refraction position of the object =33.1
25
.cm31.18f 40
1
33.1
2515
1
f
1−=⇒−
+−=
Hence magnitude of focal length of convex lens is 18.31 cm.
The nearest possible matching answer is 20 cm.
21 Sol: Ans. D
When a non black body is placed inside a hollow enclosure the total radiation from the body is the sum of what
it would emit in the open (with e < 1) and the part (1 - a) of the incident radiation from the walls reflected by
it. The two add up to a black body radiation. Hence the total radiation emitted by the body is 1.0 σ AT4.
22 Sol: Ans. A
( ) 55
1055.11.0
10)01.1165.1(V / VPB ×=
×−−=∆∆−= .
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Tvm Branch: T.C.No: 5/1703/30, Golf Links Road, H.B. Clony, Kowdiar Gardens, Trivandrum,(: 0471-2438271
Kochi Branch: Bldg.No.41/352, Mulloth Ambady Lane, Chittoor Road, Kochi - 011,(: 0484 - 2370094,93884659444
Note: Based on the memory
23 Sol: Ans. B
Acceleration of the point of suspension
2
2
2
s / m2k 2dt
yda ===
12
L2Tand
10
L2T
g
L2T 21
eff
π=π=⇒π=
5
6
T
T22
21 =∴
24 Sol: Ans. A
1001
100R
R100
R1001.0 =′⇒
′+′
=0 .1Ω
1 0 0Ω
)101001(R)10100()100(66 −− ×−′=×
mA1.100I =∴
25 Sol: Ans. B
UQ0W ∆=∆∴=∆
kJ30)605()100()1(tRIUQ 22 =×=∆=∆=∆ .
26 Sol: Ans. C
1 = 1max
cos2( φ /2)
3
2sind
2and3 / 2
π=θ
λπ
π=φ⇒θ
λ
=θ∴ −
d3sin
1
27 Sol: Ans. A
Equation of curve is
1x
x
v
v
00
=+
a
x
0
0
v
x
x1v
−=∴
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Tvm Branch: T.C.No: 5/1703/30, Golf Links Road, H.B. Clony, Kowdiar Gardens, Trivandrum,(: 0471-2438271
Kochi Branch: Bldg.No.41/352, Mulloth Ambady Lane, Chittoor Road, Kochi - 011,(: 0484 - 2370094,93884659445
Note: Based on the memory
−−=−==∴
00
20
0
0
x
x1
x
v)v(
x
v
dt
dva
Alternative: ;dx
dvva
−= but dv/dx is negative and v is decreasing with the increase in x.
Hence ‘a’ should increase with increase of ‘x’.
28 Sol: Ans. A
Since B is constant
0dt
d=
φ∴
0I =∴