Iit Jee 2005 Physics Sol Screening)

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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==−××× .






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

=×−×=∆=






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 ×=

×−−=∆∆−= .






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

−=∴






−−=−==∴

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 =∴

Iit Jee 2005 Physics Sol Screening)

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