CAPÍTULO III: DESARROLLO DEL TRABAJO DE SUFICIENCIA
3.1. Modelo de solución propuesto
1. Ans. (2)
Velocity of sound in rod v = Y
r Þ Y = rv2
Þ Stress = (Y) (Strain) = (rv2) (strain) = 4 × 103 × (5000)2 × 1
100 = 109 N/m2 2. Ans. (1)
For tuning fork 'A' l1
4 = 37.5 so n1 = v
l1 = v 4 37 5´ .
For tuning fork 'B' l2
4 = 38.5 \ n2 = v
l2 = v 4 38 5´ .
\ n1 – n2 = 8 Þ v
4 375´ . – v
4 38 5´ . = 8 \ v = (8 × 4 × 37.5 × 38.5) n1 = 8 4 37 5 38 5
4 37 5
´ ´ ´
´
. .
. = 308 Hz and n2 = 308 – 8 = 300 Hz
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XI
3. Ans. (2)
Let n1 be the frequency of the wire having tension T + DT and n2 be the frequency of the wire having tension T, then n
Difference between two consecutive resonating frequency n2 – n1 = 1 2l
When driver approaches to the policeman then observed frequency n' = næ + ö
ç ÷
7. Ans. (3) The sign of x & t are –ve and +ve respectively. Hence wave is travelling in +ve x-direction.
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XI
8. Ans. (3)
( )
( )y x,t A sin kx t B sin kx t 2
æ pö
= çè - w + ÷ø+ - w
It is combination of two wave of phase difference 2
p. \ Resultant amplitude = A2+B2 Compare with the wave y (x, t) = D sin(kx–wt +f) \ D2 = A2 + B2
9. Ans. (1)
vParticle = – (slope) (vwave)
At A : slope ® +ve so vP is –ve. OR At B : slope ® –ve so vP is +ve.
10. Ans. (1)
2 2 2
k a
a
p p p
= Þ = Þ l =
l l , v v b
k a
=w Þ =
11. Ans. (2)
yr = – Ar cos (bt–ax) where Ar = 0.64A= 0.8AÞyr = -0.8A cos bt ax( - ) 12. Ans. (3)
( )
t t
y =A cos ax bt+ where At = 0.36A 0.6A= 13. Ans. (3)
For (i) : 2A = 0.06 Þ Amplitude of constituent wave A = 0.03 m
For (ii) : Position of node when y=0 Þ sin (2px)=0 Þ 2px = np where n=1,2,3... Þ x =n 1 0.5m 2= 2= For (iii) : Position of antinode 0.06 = 0.06 sin (2px) Þ2px = (2n–1)2pwhere n=1,2,3... Þ x=1
4=0.25 m
For (iv) : Amplitude at x=121 mÞ A =0.06 sin 1
2 12
æ p ´ ö
ç ÷
è ø = 0.06 sin6p= 0.06 × 1
2 = 0.03 m 14. Ans. (2)
General equation y = f x t v æ + ö
ç ÷
è ø
15. Ans. (1)
Particle velocity = – (wave velocity) × slope
Q Slope = –ve so Particle velocity = vertical upward 16. Ans. (1)
S1 S2
4l Dx=0
Dx= Dl x=2l Dx=3l
Dx=4l
Dx=0 Dx=4l
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XI
17. Ans. (2)
Refer physics gutka 18. Ans. (1) Do yourself 19. Ans. (2) Do yourself 20. Ans. (3)
Frequency observed by hill C
C
Beat frequency observed by observer = 0 0
C C
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XI
SHORT ANSWER TYPE QUESTIONS
1.
3. (i) When one end of a rod is heated, the temperature of various points of the rod changes continuously but after some time a state is reached, when the temperature of each cross–section becomes steady which is called steady state. In this state the heat received by any section will be totally transfered to the next section so no heat is absorbed by any cross section.
(ii) Temperature gradient is defined as the rate of change of temperature with distance in the direction of flow of heat.
4.
5. Yes, torque and energy have same dimention
6. For a monoatomic gas there are 3 degrees of freedom so Cv = dU
7. Yes, by reducing pressure of water, boiling point of water can be brought down to room temperature
8. Given that : A B Cr = +r r
make a triangle with angle between Br
and Cr
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XII
1. Ans. (1)
Given circuit can be reduced to
Q Heat loss in R = Heat loss in 4W
where Acu & AAg denotes atomic weights and V1 & V2 denotes valencies of Cu and Ag respectively
\ MMCu
voltage on capacitor is more than that of supply voltage because the phase difference between VL and VC is 180° (i.e. out of phase)
4. Ans. (4)
Let q be the charge on inner sphere then potential at the surface of inner surface
Vinner = KQ
By using formula for finite length wire B = m p0 The direction of r
B at P is perpendicular to the plane of the page and coming out of the page.
OR
From Biot-Savart's law, B at P due to horizontal part of the current is zero.
Due to vertical part, B is half of that due to an infinitely long current carrying wire.
Therefore B = 1
The direction of Br at P perpendicular to the plane of the page and coming out of the page 6. Ans. (2)
The phase difference between VC and VR2 is p
2 rad or 90°
ELECTRODYNAMICS SOLUTIONS
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XII
Power dissipation in series connection 1 P = 1
Power dissipated in circuit P = V Req
No silver will be liberated because with AC, anode and cathode are interchanged in each half cycle.
13. Ans. (2)
The lines of electric field are outwards from the faces, hence the charge should the positive
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XII
14. Ans. (3)
The system is equivalent to two condensers in parallel
\ C = C1 + C2 = K A Magnetic field (2) at D due to straight current KA & CL is zero
and magnetic field due to a semi circular current at the centre is B =m a p
At resonance the net potential drop across L and C is zero. Hence whole of the potential drop of 220 V is across the resistance.
Applying Gauss's law to the surface S. Charge –q and +q are in duced on the inner & outer surfaces of the shell.
Vp =
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XII
Force on small element at a distance r of wire
of length L is dF = i(dr) m
Total energy = Indial energy on capacitor = 1
2 CV2, Magnetic field energy = 1
Magnetic moment of current carrying triangular loop M = IA
M = 0.2 For curved surface r
E and r
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XII
Root mean square value <V> =
T / 4
% Efficiency = Output power
Input power × 100 Þ 0.8 = s.. s
here induced emf is positive so direction of induced current is in the direction of main current 35. Ans. (2)
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XII
36. Ans. (4)
Current through ab = 10 10 1A
As electric and magnetic energy densities are equal in an em wave so 0 2 2
0 0 0
Initial charging rate = initial current in the line of capacitor =2E 5R Steady state p.d. across capacitor V0 =2
3E Þ q0 =CV0=2
Steady-state current in 0
1
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XII
Restoring force is zero on z–axis so the equilibrium of proton is neutral w.r.t. its displacement along z–axis.
51. Ans. (2) Assume capacitor as dipole and use F = q E, E =2kP3
We can break the motion along x & y–direction. Consider motion in y–direction, we have h 1gt2
= 2 Þ 2h
Distribution of charge on different surfaces of the plates Take a point P on right most plate P
Note : Charge on outer surface =Net charge on plates
2 = Q 2Q Q
2 2
- =
-Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XII
63. Ans. (4) See solution of Q. 62 64. Ans. (2) See solution of Q. 62 65. Ans. (1)
These five plates constitute four identical capacitor in parallel, each of capacity 0 A d
Î Þ q1 = 0 AV d Î
66. Ans. (4) 4 0
2 AV
q d
= - Î
67. Ans. (2) Total energy stored =
2
2 2A 0 V
4 1CV
2 d
´ = Î
68. Ans. (3) I = constant
69. Ans. (2) d d
I neAv constant v 1
= = Þ µA
70. Ans. (1)
d
Time taken= A vl µ
71. Ans. (1) e = eeq 5 & req =5r
72. Ans. (2)
73. Ans. (2) ,
eeq = 3e, req = r + r + r 7r 3= 3
74. Ans. (2) Q V2 R t and R=
t R A
= Þ µ rl
75. Ans. (3) In parallel combination
i
1 1
P = SP
76. Ans. (2)
Heat generated in 5 W=V2 10cal / sec
5 = , Heat generated 4W = 4 2
10V 2 cal/sec 4
æ ö
ç ÷
è ø
=
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XII
Current through branch BG : 3
x 6
Final potential difference across 3mF capacitor = 6 C 3 F 2V
m =
85. Ans. (3) m
Fraction of energy lost =
1 2 2
87. Ans. (4) Magnetic force in a uniform magnetic field on a loop is always zero.
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XII
96. Ans. (1) Do yourself 97. Ans. (2)
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XII
104. Ans. (2) At low frequencies
2
As current is leading the source voltage, so circuit should be capacitive in nature and as phase difference is not 2
p, it must contain resistor also.
108. Ans. (1)
= = , Anticlockwise 112. Ans. (1)
We add a resistance R in series to achieve a voltmeter of range 50V Þ 50 × 10–6 = 50
100 R+ Þ R = 106 W Now, when a current of 10 mA is passed through the ammeter, 50 mA should go through the coil. We add a resistance S in parallel to the ammeter, 50 mA should go through the galvanometer is :
50 × 10–6 = (10 × 10–3) S
S 100+ Þ S = 0.5 W 113. Ans. (4)
For (1) : After a sufficiently long time VC = V0 Þ VR = 0 For (2) : As time passes VR decreases exponentially.
For (3) : Time constant = RC = (1×103) × ( 1× 10–6) = 10–3 s = 1 ms
-Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XII
Potential difference across 2mF capacitor = 1V Charge on 2mF capacitor = 2mC
Charge on mF = Charge on 3mC= 0
Potential difference across C1 : q 11=55
For the above condition to hold, (A, B or C) may be correct.
118. Ans. (1)
Capacitance of 1st row is maximum, hence charge stored will be also maximum.
GEOMETRICAL OPTICS
Multiplying all the three equations sin sin
i
r3 = m since i = r3 therefore m = 1
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XII
When light is incident at Brewster's angle, the reflected and refracted rays are perpendicular to each other. In addition, the reflected light is completely polarized.
7. Ans. (2)
Since along the surface, conservation principle of momentum is applicable.
glass
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XII
11. Ans. (2)
r1=0, A = 60°, e= 90° Þ r2 = 60° Þ msin60° = sin90° Þ m= 2 3 12. Ans. (2)
In quadilateral ECDF
15° + +r 135° +60° +90° + =r 360° Þ =r 30° For TIR
1 C
75 r 45 sin 1
n
- æ ö
° - ³ q Þ ° ³ ç ÷è ø Þsin45° ³1n Þ ³n sin 451 °Þ ³n 2
13. Ans. (4) Do yourself 14. Ans. (1) Do yourself 15. Ans. (4) Do yourself 16. Ans. (3)
As for telescope lenses having highest focal length & lowest focal length should be used to get highest magnification
as 0
e
M f
= f 17. Ans. (2)
As for microscope lenses having shortest focal length i.e. high powers should be used to get highest magnification,
as 0 e
L D
M 1
f f
æ ö
= - çè + ÷ø 18. Ans. (4)
For blue, wavelength is less than red so resolution increases.
19. Ans. (4)
Lens equation 1 1 1 1 1 1 v u f- = Þ = +v u f For (i) and (iii) :
Converging lens Þ f= Åve
For real object u is negative Þ v is either positive or negative Þ image is either real or virtual For virtual object u is positive Þ v is positive Þ image is real.
For (ii) :
Diverging lens Þ f = – ve for real object u is negative Þ v is negative Þ Image is virtual For virtual object u is positive Þ v is either positive or negative Þ image is either real or virtual.
For (iv) :
Mirror equation 1 1 1 1 1 1 v u f+ = Þ = - +v u f System behave as a convex mirror Þ f= Åve
For real object u is negative Þ v is positive Þ image is virtual.
For virtual object u is positive Þ v is either positive or negative Þ image is either real or virtual 20. Ans. (4)
For a magnifying lens D D
f £m 1£ + f 25 25
m 1 2.5 m 3.5
10 10
Þ £ £ + Þ £ £
21. Ans. (1) Do yourself 22. Ans. (1)
xy = C Þ P2V= C Þ P·PV = C2 ÞPnRT = C2 ÞPT = constant
Data2\2011\AIPMT Mains Que\Phy\B.M. Sir\Solution\Class XII
Kinetic energy of electron = 500 keV & Rest mass energy of electron = 511 keV Total energy = mc2 = m0c2 + KE = (511 + 500) keV
Energy of emitted photon = 13.6 – 0.85 = 12.75 eV 5. Ans. (1)
The expression of decay n ® p + e– + n
The number of undecayed neutron would be 150 by using N = N0e–lt 150 = 600e–lt Þ t = 2T1/2= 1200 sec