keam 2014 physics solutions for all codes a1, a2, a3 & a4

1

SCIENCE INSTITUTE KERAL ENGG. 2014 PHYSICS

EXPLANATION FOR A1 VERSION

1. 2

lg

T∝

⇒% error in g = % error in l + 2x % error in T

⇒1 × 2 + 2 × 3

⇒8%

3. Displacement of the dropped body +

Displacement of the thrown body = Total

height

2 21 1

2 2gt ut gt h+ − =

⇒ h

tu

= ⇒100

425

t s= =

Distance to the meeting point from top

= 21

2gt

= 4.9 × 42

= 78.4

5. In first 3 seconds, 21

2S ut at= + ⇒

12 = u × 3 + 1

2× a × 32

12 = 3u + 9

2a ⇒3 = u + 1.5 a -------(1)

In first 6 seconds,

42 = u × 6 + 216

2a× ×

7 = u + 3a ----------(2)

Solve (1) & (2) to get a

6. Impulse = change in P

= 2mv

Impulse

2mv =

7. L r P= ×

1 2 1

2 1 1

i j k

= −−

9. X component of velocity xdx

vdt

=

8xv t=

Y component of velocity 6yv t=

Resultant velocity 2 2x yv v v= +

10t=

11. Acceleration 2120 6010 /

6

fa m s

m

−= = =

Tension = ( )m g a+

= 2 (10 + 10) = 20 N

12. 27

10TE mv=

13. By work-energy theorem

2

21. .........(1)

2 4

vF x m v

= −

21.( ) .........(2)

2F x y mv+ =

(2) 4

(1) 3

x y

x

+⇒ =

4

3

xy x= −

2

( x = the initial thickness penetrated and y the

additional thickness penetrated)

15. Applying law of conservation of angular

momentum,

1 1 2 2 1 2( )I I I Iω ω ω+ = +

3 31 12 10 5 ( 10 )4I I− −× + × = +

31 0.5 10I −⇒ = ×

16. 0t

t

ω ωα −= = t

t

ω

= 02

t

πυ

=2

π

( 0υ = 1800 rpm = 1800

30,60

= t = 2 × 60 s)

17. Centripetal force = 2mv

r-------(1)

Angular momentum L = mvr

L

vmr

⇒ =

(1) ⇒2

.m L

CFr mr =

19. Energy required = 1

nMgR

n +

(here n = 1)

20. Total energy = − KE

21. Area of the big drop = Area of the droplets

2 2 21 24 4 4R r rπ π π= +

⇒ 2 21 2R r r= +

23. Pressure difference = Pressure due to 10 mm

of Hg = 1 cm of Hg

h gω ωρ = Pressure due to 10mm of Hg

Hg Hghhω

ω

ρρ

= = 1 13.6

13.61

× = cm

24. 21

2W kx=

21' (2 )

2W k x=

21' 3

2W W k x− = ×

3W=

3J=

25. 3

rms

RTV

M=

3

A

RT

mN=

26. 2

1

1T

Tη = − 2

1

1T

Tη⇒ = −

2

1

0.8T

T⇒ = 2 10.8T T⇒ =

In the second case

2

1

500.6

T

T

− =

T2 – 50 = 0.6 T1

0.8 T1 – 50 = 0.6 T1

0.2 T1 = 50

T1 = 250 K

T2 = 200 K

27. 2V

nC R=

n = no of degrees of freedom

28. Tree translational and two rotational degrees

of freedom

29. Comparing with standard equation of SHM,

we get 2ω =

2

Tπ

ω=

31. Here, given y = 0.707 A

2

Ay =

3

KE = 2 2 21( )

2m A yω −

Put ,2

Ay =

KE = 2

2 21( )

2 2

Am Aω −

=2

TE

33.

A1

N4N3N2N1

A2 A3

A1, A2 & A 3 = Antinodes

N1, N2, N3 & N4 = Nodes

34. In stretched string all harmonics are present.

35. Electric field near the sheet F

Eq

=

12

19

1.6 10

1.6 10E

−

−

×=×

= 107

But 02

Eσε

=

02 Eσ ε⇒ =

Total charge = Areaσ ×

02 E Areaε= ×

(The charge will be negative)

38. Centripetal force = Coulomb’s force

2

1 22

0

1.

4

q qmv

r rπε=

2 1 2

04

q qv

rmπε⇒ =

Period 2 r

Tv

π=

2 2

22

4 rT

v

π=

2 2

2 0

1 2

4 4r rmT

q q

π πε×=

39. Electric field F

Eq

=

1000

2= = 500

Potential difference = E × distance

= 500 × 2 × 10 – 2 = 10 V

40. Current through 4 Ω resistor near xy =

4

14

A=

Current through the 4 Ω resistor parallel to

4 Ω = 1 A

Total current in the circuit = 2A

Total potential difference = ( .)effI R×

= 2 × 8

= 16 V

41. PD across R = 2V

IR = 2

⇒ ( )10

2500

RR

=+

⇒ 125R = Ω

42. s pI I=

5

55

E Err R R

=+ +

r R⇒ =

43. Power loss = I2R P

IV

=

2P

RV =

⇒ Power loss 2

1

v∝

44. Charge in internal energy =I2Rt

= 12 × 50 × 2 × 60

= 6 kJ

4

46. 2I

TmB

π=

2 24I

TmB

π=

22

4I

mT

π=

2 5

2 5

4 9 109

16 1016

π

π

−

−

× ×=× ×

= 4 Am2

2 2 2

2

q B rKE

m= -----(1)

22

qBqB m

mυ π υ

π= ⇒ =

2 2 2 24

(1)2

m rKE

m

π υ⇒ =

2 2 2 22 m rπ υ=

49. (1 cos )W mB θ= − 0( 180 )θ =

52. V(max) = BANω

53. ps s

p p s

IN V

N V I= =

330 4

5524

p ps

s

N IN

I

× ×⇒ = = =

54. Average energy density of magnetic field =

= 20

1

2Eε

2

121 18.85 10

2 2− = × × ×

= 1

2× 4.43 × 10 – 12 = 2.2 × 10 – 12 J

57. tanµ ρ=

tan 60=

3=

1

sinCµ

= 1 1sin

3C −

⇒ =

60. sind nθ λ= here n = 2

2

sind

λθ = ⇒ 7

6

2 5 10sin

2.5 10θ

−

−

× ×=×

2

sin5

θ⇒ =

61. 1

2

h

mmKEλ λ= ⇒ ∝

62. Time for 50% decay = One half life

Time for 87.5% decay = Three half lives

Time interval between 50% decay and 87.5%

decay = Two half lives

63. 1 2

23 30R R A R A= = ∝

Area ∝ R2

2/3

(1) 125 25

(2) 27 9

Area

Area = =

64. KE hυ φ= −

' 2KE h υ φ= × −

2hυ φ= − ⇒more than twice the initial value

67. The out put .y A B C= +

68. Voltage gain = Current gain × resistance gain

= 49 × 10

2

= 245

71. LOS distance = 2 2Rh Rh+

8Rh=

5

KERAL ENGG. 2014 PHYSICS


55. 2

lg

T∝


⇒1 × 2 + 2 × 3

⇒8%

57 Displacement of the dropped body +


height

2 21 1

2 2gt ut gt h+ − =

⇒ h

tu

= ⇒100

425

t s= =


= 21

2gt

= 4.9 × 42

= 78.4

59 In first 3 seconds, 21

2S ut at= + ⇒

12 = u × 3 + 1

2× a × 32

12 = 3u + 9

2a ⇒3 = u + 1.5 a -------(1)

In first 6 seconds,

42 = u × 6 + 216

2a× ×

7 = u + 3a ----------(2)



= 2mv

Impulse

2mv =

61 L r P= ×

1 2 1

2 1 1

i j k

= −−


vdt

=

8xv t=



10t=


6

fa m s

m

−= = =

Tension = ( )m g a+

= 2 (10 + 10) = 20 N

66. 27

10TE mv=

67 By work-energy theorem

2

21. .........(1)

2 4

vF x m v

= −

21.( ) .........(2)

2F x y mv+ =

(2) 4

(1) 3

x y

x

+⇒ =

4

3

xy x= −




momentum,

1 1 2 2 1 2( )I I I Iω ω ω+ = +

6

3 31 12 10 5 ( 10 )4I I− −× + × = +

31 0.5 10I −⇒ = ×

70. 0t

t

ω ωα −= = t

t

ω

= 02

t

πυ

=2

π

( 0υ = 1800 rpm = 1800

30,60

= t = 2 × 60 s)


r-------(1)


L

vmr

⇒ =

(1) ⇒2

.m L

CFr mr =

1. Energy required = 1

nMgR

n +

(here n = 1)



2 2 21 24 4 4R r rπ π π= +

⇒ 2 21 2R r r= +

5 Pressure difference = Pressure due to 10 mm

of Hg = 1 cm of Hg


Hg Hghhω

ω

ρρ

= = 1 13.6

13.61

× = cm

6 21

2W kx=

21' (2 )

2W k x=

21' 3

2W W k x− = ×

3W=

3J=

7 3

rms

RTV

M=

3

A

RT

mN=

8. 2

1

1T

Tη = − 2

1

1T

Tη⇒ = −

2

1

0.8T

T⇒ = 2 10.8T T⇒ =

In the second case

2

1

500.6

T

T

− =

T2 – 50 = 0.6 T1

0.8 T1 – 50 = 0.6 T1

0.2 T1 = 50

T1 = 250 K

T2 = 200 K

9 2V

nC R=



of freedom

11 Comparing with standard equation of SHM,

we get 2ω =

2

Tπ

ω=


2

Ay =

KE = 2 2 21( )

2m A yω −

Put ,2

Ay =

KE = 2

2 21( )

2 2

Am Aω −

7

=2

TE

15

A1

N4N3N2N1

A2 A3


N1, N2, N3 & N4 = Nodes


17 Electric field near the sheet F

Eq

=

12

19

1.6 10

1.6 10E

−

−

×=×

= 107

But 02

Eσε

=

02 Eσ ε⇒ =


02 E Areaε= ×



2

1 22

0

1.

4

q qmv

r rπε=

2 1 2

04

q qv

rmπε⇒ =

Period 2 r

Tv

π=

2 2

22

4 rT

v

π=

2 2

2 0

1 2

4 4r rmT

q q

π πε×=

21 Electric field F

Eq

=

1000

2= = 500


= 500 × 2 × 10 – 2 = 10 V

22. Current through 4 Ω resistor near xy =

4

14

A=


4 Ω = 1 A



= 2 × 8

= 16 V

23 PD across R = 2V

IR = 2

⇒ ( )10

2500

RR

=+

⇒ 125R = Ω

24 s pI I=

5

55

E Err R R

=+ +

r R⇒ =

25 Power loss = I2R P

IV

=

2P

RV =

⇒ Power loss 2

1

v∝


= 12 × 50 × 2 × 60

= 6 kJ

28 2I

TmB

π=

2 24I

TmB

π=

22

4I

mT

π=

8

2 5

2 5

4 9 109

16 1016

π

π

−

−

× ×=× ×

= 4 Am2

2 2 2

2

q B rKE

m= -----(1)

22

qBqB m

mυ π υ

π= ⇒ =

2 2 2 24

(1)2

m rKE

m

π υ⇒ =

2 2 2 22 m rπ υ=

31. (1 cos )W mB θ= − 0( 180 )θ =

34 V(max) = BANω

35 ps s

p p s

IN V

N V I= =

330 4

5524

p ps

s

N IN

I

× ×⇒ = = =


= 20

1

2Eε

2

121 18.85 10

2 2− = × × ×

= 1

2× 4.43 × 10 – 12 = 2.2 × 10 – 12 J

39 tanµ ρ=

tan 60=

3=

1

sinCµ

= 1 1sin

3C −

⇒ =

42 sind nθ λ= here n = 2

2

sind

λθ = ⇒ 7

6

2 5 10sin

2.5 10θ

−

−

× ×=×

2

sin5

θ⇒ =

43. 1

2

h

mmKEλ λ= ⇒ ∝

44 Time for 50% decay = One half life




45 1 2

23 30R R A R A= = ∝

Area ∝ R2

2/3

(1) 125 25

(2) 27 9

Area

Area = =

46 KE hυ φ= −

' 2KE h υ φ= × −



50 Voltage gain = Current gain × resistance gain

= 49 × 10

2

= 245

53. LOS distance = 2 2Rh Rh+

8Rh=

9



32. 2

lg

T∝


⇒1 × 2 + 2 × 3

⇒8%

34. Displacement of the dropped body +


height

2 21 1

2 2gt ut gt h+ − =

⇒ h

tu

= ⇒100

425

t s= =


= 21

2gt

= 4.9 × 42

= 78.4


2S ut at= + ⇒

12 = u × 3 + 1

2× a × 32

12 = 3u + 9

2a ⇒3 = u + 1.5 a -------(1)

In first 6 seconds,

42 = u × 6 + 216

2a× ×

7 = u + 3a ----------(2)


37 Impulse = change in P

= 2mv

Impulse

2mv =

38. L r P= ×

1 2 1

2 1 1

i j k

= −−


vdt

=

8xv t=



10t=


6

fa m s

m

−= = =

Tension = ( )m g a+

= 2 (10 + 10) = 20 N

43. 27

10TE mv=


2

21. .........(1)

2 4

vF x m v

= −

21.( ) .........(2)

2F x y mv+ =

(2) 4

(1) 3

x y

x

+⇒ =

4

3

xy x= −




momentum,

1 1 2 2 1 2( )I I I Iω ω ω+ = +

10

3 31 12 10 5 ( 10 )4I I− −× + × = +

31 0.5 10I −⇒ = ×

47. 0t

t

ω ωα −= = t

t

ω

= 02

t

πυ

=2

π

( 0υ = 1800 rpm = 1800

30,60

= t = 2 × 60 s)


r-------(1)


L

vmr

⇒ =

(1) ⇒2

.m L

CFr mr =

50 Energy required = 1

nMgR

n +

(here n = 1)



2 2 21 24 4 4R r rπ π π= +

⇒ 2 21 2R r r= +


of Hg = 1 cm of Hg


Hg Hghhω

ω

ρρ

= = 1 13.6

13.61

× = cm

55. 21

2W kx=

21' (2 )

2W k x=

21' 3

2W W k x− = ×

3W=

3J=

56. 3

rms

RTV

M=

3

A

RT

mN=

57 2

1

1T

Tη = − 2

1

1T

Tη⇒ = −

2

1

0.8T

T⇒ = 2 10.8T T⇒ =

In the second case

2

1

500.6

T

T

− =

T2 – 50 = 0.6 T1

0.8 T1 – 50 = 0.6 T1

0.2 T1 = 50

T1 = 250 K

T2 = 200 K

58. 2V

nC R=



of freedom


we get 2ω =

2

Tπ

ω=


2

Ay =

KE = 2 2 21( )

2m A yω −

Put ,2

Ay =

KE = 2

2 21( )

2 2

Am Aω −

11

=2

TE

64

A1

N4N3N2N1

A2 A3


N1, N2, N3 & N4 = Nodes

65 In stretched string all harmonics are present.


Eq

=

12

19

1.6 10

1.6 10E

−

−

×=×

= 107

But 02

Eσε

=

02 Eσ ε⇒ =


02 E Areaε= ×



2

1 22

0

1.

4

q qmv

r rπε=

2 1 2

04

q qv

rmπε⇒ =

Period 2 r

Tv

π=

2 2

22

4 rT

v

π=

2 2

2 0

1 2

4 4r rmT

q q

π πε×=


Eq

=

1000

2= = 500


= 500 × 2 × 10 – 2 = 10 V

71 Current through 4 Ω resistor near xy =

4

14

A=


4 Ω = 1 A



= 2 × 8

= 16 V

72 PD across R = 2V

IR = 2

⇒ ( )10

2500

RR

=+

⇒ 125R = Ω

31. s pI I=

5

55

E Err R R

=+ +

r R⇒ =


IV

=

2P

RV =

⇒ Power loss 2

1

v∝


= 12 × 50 × 2 × 60

= 6 kJ

35. 2I

TmB

π=

2 24I

TmB

π=

22

4I

mT

π=

12

2 5

2 5

4 9 109

16 1016

π

π

−

−

× ×=× ×

= 4 Am2

2 2 2

2

q B rKE

m= -----(1)

22

qBqB m

mυ π υ

π= ⇒ =

2 2 2 24

(1)2

m rKE

m

π υ⇒ =

2 2 2 22 m rπ υ=

38. (1 cos )W mB θ= − 0( 180 )θ =

41. V(max) = BANω

42. ps s

p p s

IN V

N V I= =

330 4

5524

p ps

s

N IN

I

× ×⇒ = = =

43 Average energy density of magnetic field =

= 20

1

2Eε

2

121 18.85 10

2 2− = × × ×

= 1

2× 4.43 × 10 – 12 = 2.2 × 10 – 12 J

46 tanµ ρ=

tan 60=

3=

1

sinCµ

= 1 1sin

3C −

⇒ =

49. sind nθ λ= here n = 2

2

sind

λθ = ⇒ 7

6

2 5 10sin

2.5 10θ

−

−

× ×=×

2

sin5

θ⇒ =

50 1

2

h

mmKEλ λ= ⇒ ∝





52. 1 2

23 30R R A R A= = ∝

Area ∝ R2

2/3

(1) 125 25

(2) 27 9

Area

Area = =

53. KE hυ φ= −

' 2KE h υ φ= × −


56 The out put .y A B C= +


= 49 × 10

2

= 245

60 LOS distance = 2 2Rh Rh+

8Rh=

13



13. 2

lg

T∝


⇒1 × 2 + 2 × 3

⇒8%

15 Displacement of the dropped body +


height

2 21 1

2 2gt ut gt h+ − =

⇒ h

tu

= ⇒100

425

t s= =


= 21

2gt

= 4.9 × 42

= 78.4


2S ut at= + ⇒

12 = u × 3 + 1

2× a × 32

12 = 3u + 9

2a ⇒3 = u + 1.5 a -------(1)

In first 6 seconds,

42 = u × 6 + 216

2a× ×

7 = u + 3a ----------(2)



= 2mv

Impulse

2mv =

19 L r P= ×

1 2 1

2 1 1

i j k

= −−

21 X component of velocity xdx

vdt

=

8xv t=



10t=

23 Acceleration 2120 6010 /

6

fa m s

m

−= = =

Tension = ( )m g a+

= 2 (10 + 10) = 20 N

24. 27

10TE mv=


2

21. .........(1)

2 4

vF x m v

= −

21.( ) .........(2)

2F x y mv+ =

(2) 4

(1) 3

x y

x

+⇒ =

4

3

xy x= −



27 Applying law of conservation of angular

momentum,

1 1 2 2 1 2( )I I I Iω ω ω+ = +

3 31 12 10 5 ( 10 )4I I− −× + × = +

14

31 0.5 10I −⇒ = ×

28. 0t

t

ω ωα −= = t

t

ω

= 02

t

πυ

=2

π

( 0υ = 1800 rpm = 1800

30,60

= t = 2 × 60 s)


r-------(1)


L

vmr

⇒ =

(1) ⇒2

.m L

CFr mr =

31 Energy required = 1

nMgR

n +

(here n = 1)



2 2 21 24 4 4R r rπ π π= +

⇒ 2 21 2R r r= +


of Hg = 1 cm of Hg


Hg Hghhω

ω

ρρ

= = 1 13.6

13.61

× = cm

36. 21

2W kx=

21' (2 )

2W k x=

21' 3

2W W k x− = ×

3W=

3J=

37 3

rms

RTV

M=

3

A

RT

mN=

38 2

1

1T

Tη = − 2

1

1T

Tη⇒ = −

2

1

0.8T

T⇒ = 2 10.8T T⇒ =

In the second case

2

1

500.6

T

T

− =

T2 – 50 = 0.6 T1

0.8 T1 – 50 = 0.6 T1

0.2 T1 = 50

T1 = 250 K

T2 = 200 K

39 2V

nC R=



of freedom


we get 2ω =

2

Tπ

ω=


2

Ay =

KE = 2 2 21( )

2m A yω −

Put ,2

Ay =

KE = 2

2 21( )

2 2

Am Aω −

15

=2

TE

45

.

A1

N4N3N2N1

A2 A3


N1, N2, N3 & N4 = Nodes



Eq

=

12

19

1.6 10

1.6 10E

−

−

×=×

= 107

But 02

Eσε

=

02 Eσ ε⇒ =


02 E Areaε= ×



2

1 22

0

1.

4

q qmv

r rπε=

2 1 2

04

q qv

rmπε⇒ =

Period 2 r

Tv

π=

2 2

22

4 rT

v

π=

2 2

2 0

1 2

4 4r rmT

q q

π πε×=


Eq

=

1000

2= = 500


= 500 × 2 × 10 – 2 = 10 V

52 Current through 4 Ω resistor near xy =

4

14

A=


4 Ω = 1 A



= 2 × 8

= 16 V

53. PD across R = 2V

IR = 2

⇒ ( )10

2500

RR

=+

⇒ 125R = Ω

54. s pI I=

5

55

E Err R R

=+ +

r R⇒ =


IV

=

2P

RV =

⇒ Power loss 2

1

v∝

56 Charge in internal energy =I2Rt

= 12 × 50 × 2 × 60

= 6 kJ

58. 2I

TmB

π=

2 24I

TmB

π=

22

4I

mT

π=

16

2 5

2 5

4 9 109

16 1016

π

π

−

−

× ×=× ×

= 4 Am2

2 2 2

2

q B rKE

m= -----(1)

22

qBqB m

mυ π υ

π= ⇒ =

2 2 2 24

(1)2

m rKE

m

π υ⇒ =

2 2 2 22 m rπ υ=

61 (1 cos )W mB θ= − 0( 180 )θ =

64. V(max) = BANω

65. ps s

p p s

IN V

N V I= =

330 4

5524

p ps

s

N IN

I

× ×⇒ = = =


= 20

1

2Eε

2

121 18.85 10

2 2− = × × ×

= 1

2× 4.43 × 10 – 12 = 2.2 × 10 – 12 J

69 tanµ ρ=

tan 60=

3=

1

sinCµ

= 1 1sin

3C −

⇒ =

72 sind nθ λ= here n = 2

2

sind

λθ = ⇒ 7

6

2 5 10sin

2.5 10θ

−

−

× ×=×

2

sin5

θ⇒ =

1. 1

2

h

mmKEλ λ= ⇒ ∝





3. 1 2

23 30R R A R A= = ∝

Area ∝ R2

2/3

(1) 125 25

(2) 27 9

Area

Area = =

4 KE hυ φ= −

' 2KE h υ φ= × −




= 49 × 10

2

= 245

11 LOS distance = 2 2Rh Rh+

8Rh=

keam 2014 physics solutions for all codes a1, a2, a3 & a4

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