1. Distance of the centre of mass of a solid

Class : JEE Main 2015 Subject : Physics Paper Code : 1. 2 é
æ T M ö ù A ê
1 ç
÷ ú
1. è T ø ûú Mg
ëê
Distance of the centre of mass of a solid uniform cone from its vertex is z 0 . If the radius of its base is R and its height is h then z 0 is equal to : 1. 5 th 8 2. 3 h 2 8 R D é æ T ö2 ù A M 3. êç T ÷ - 1 ú Mg êë è
úû
ø
2 é
æ T ö ù A ê
ú
1 ç
÷
2. ê
TM ø ú Mg
è
ë
û
4. é æ T ö2 ù Mg ê ç M ÷ - 1 ú
êë è T ø
úû A Solution: (3) 1
F l
F/A
1 é l - l ù A = A ( l1 - l ) Þ
= ê l ú
Dl / l
Y ë
û F Extra Force F = mg Y = 3. h 2 4 R 4. 3 h 4 T = 2 p
Solution: (4) l / g and T m = 2 p
l1 / g
2 y c m =
ò
æ T m ö
l1 - l
By solving = ç
÷ ­1
l
è T ø
y d m m
3 h 4 4 2. A red LED emits light at 0.1 watt uniformly around it The amplitude of the electric field of the light at a distance of 1m from the diode is 1. 5.48 V/m 2. 7.75 V/m 3. 1.73 V/m 4. 2.45 V/m Solution: (4) Intensity of EM wave (I) = UC = eo E 2 C = eoEp 2 C 2
é æ T ö2 ù A M 1 ê
ú
1 ç
÷
= T
\
êë è
úû mg ø
Y For a simple pendulum, a graph is plotted between its kinetic energy (KE) and potential energy (PE) against its displacement d. Which one of the following represents these correctly? (graphs are schematic and not drawn to scale) 1.
... (1) Power P = ... (2) Area A From (1) and (2) I = 2.
2P E p 2 = 4pr 2 e C = 6 Þ E p = 6 = 4.245 v/m o 3. A pendulum made of a uniform wire of cross sectional area A has time period T. When an additional mass M is added to its bob, thetime period changes to T M . Young’s modulus of the 1 material of the wire is Y then is equal to: Y 3.
(g = gravitational acceleration) 1 E­92, South Ex. Part 1, New Delhi 49. Ph: 011­41070321, 8510070321 8 www.niveditaclasses.com 1. 2
p gL
tan q
m0 pl gL
tan q
m0 2. 4. pl gL
3. sin q m cos q
0 Solution: (4) KE = 1 mw w 2 (A 2 ­x 2 ) — Inverted Parabola 2 Solution: (4) Fe tan q = Fg PE = 1 2 mw w 2 x 2 — Parabola 5. A train is moving on a straight track with speed 20 ms­1. It is blowing its whistle at the frequency of 100 Hz. The precentage change in the frequency heard by a person standing near the track as the train passes him is (speed of sound = 320 ms ­1 ) close to: 1. 18% 2. 24% 3. 6% 4. 12% U0 I 2 Fe per unit length = 2p 2L Sinq
Fe Vi x 100 = 2LSinq
Fg per unit length = l g
m 0 I 2 tan
= Þ
q
4pL sin ql g
2V S x 100
C + VS Þ I = 2 sin q
» 12% 6. when 5 v potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is 2.5x10­4 ms ­1 . If the electron density in the wire is 8x10 28 m ­3 , the resistivity of the material is close to : 1. 1.6x10 ­6 Wm 2. 1.6x10 ­5 Wm 2. 1.6x10 ­8 Wm 3. 1.6x10 ­7 Wm Solution: (2) Ohm’s law in microscopic form E = J r
Þ r = L Fg c c V i = c - V ; V f = c + V
s s Vi - V f q
T Solution: (4) Observer at rest and source is moving. c = 320 m/s; V s = 20m/s; V 0 = 1000 H 3 % change = pl gL
4. 2 sin q m cos q
0 V/ l
E = neVd J Putting Values r » 1.6 x 10 ­5 Wm
pl Lg m0 cos q
8. In the circuit shown, the current in the 1W
resistor is: 1. 0.13 A, from Q to P 2. 0.13 A, from P to Q 3. 1.3 A, from P to Q 4. 0A Solution: (1) Using KVL 6 = 4 I 2 ­ I 1 — (1) 9 = 6I 1 ­ I 2 — (2) 7. 6v Two long current carrying thin wires, both with current I, are held by insulating threads of length L and are in equilibrium as shown in the figure, with threads making an angle ‘q’ with the vertical. If wires have mass l per unit length then the value of is : I 1 2W
I2­I 1 1W
I 2 (g = gravitational acceleration) 2 P E­92, South Ex. Part 1, New Delhi 49. Ph: 011­41070321, 8510070321 3W
9v
Q I 1 3W
8 www.niveditaclasses.com Solving I 1 = If student plots graphs of the square of 42 45 , I 2 = 23 23
3
» 0.13 A from Q to P 23
Assuming human pupil to have a radius of 0.25 cm and a comfortable viewing distance of 25cm, the minimum sepration between two objects that human eye can resolve at 500 nm wavelength is: \ I 2 ­ I 1 = 9. 1. 100mm 2. 300mm 3. 1mm 4. 30mm maximum charge ( Q 2 ) on the capacitor with Max time (t) for two different values L 1 and L 2 (L 2 >L 2 ) of L then which of the following represents this graph correctly? (plots are schematic and not drawn toscale) 1.
Solution: (4) 0.25
2.
q
25 Resolving Power = 1.22 l
2m sin q
= 3.05 x 10 ­5 » 30 mm
10. An inductor (L=0.03H) and resistor (R=0.15 kW) are connected in series to a battery of 15 V EMF in a circuit shown below. The key K 1 has been kept closed for a long time. Then at t = 0 k 1 i s i s open ed an d key k 2 i s cl osed simultaneously. At t = 1 ms, the current in the circuit will be : (e5 @ 150)
1. 6.7 mA 2. 0.67 mA 3. 100 mA 4. 67 mA 3.
4. Solution: (3) Q max = Q 0 e ­kt K a 1 L Correct option can selected by looking the options. 12. In the given circuit, charge Q 2 on 2mF to 3mF. Q 2 as function of ‘C’ is given properly by : (figures are drawn schematically and are not to scale) Solution: (2) 1 -5 1 e = » 0.67 mA 10
1500
11. An LCR circuit is equivalent to a damped pendulum. In an LCR circuit the capacitor is charged to Q 0 and then connected to the L and R as shown below: I = I0 e­t/ t = 1.
3 E­92, South Ex. Part 1, New Delhi 49. Ph: 011­41070321, 8510070321 8 www.niveditaclasses.com Solution: (1) Side of cube (a) = 2R 3 2.
Mass of cube (M c ) = I CM (for cube) = 2M
M x vol.of cube = vol.of sphere 3p
Mc a2 = 4MR 2 9 3p
6 14. The period of oscillation of a simple pendulum 3.
is T = 2 p
L . Measured value of L is 20.0 cm g
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1s resoulation. The accuracy in thedetermination of g is: 4. Solution: (4) E =
2
Þ Q 1 = Q= Q 1 +Q 2 = 4. 3% T 2 4p2 e
Dg
2 x 1 0.1 2DT Dl
+
= + Þ g = » 0.03 90 20
T
l
= 3% Q2, 2m f 1
3. 2% Rearranging g = Q1, 1mf Q 2 2. 5% Solution: (4) C Q Q1
1. 1% 15. On a hot summer night, the refractive index of air is smallest near the ground and increases with height from the ground. When a light beam is directed horizontally, the Hugyens’ principle leads us to conclude that as it travels, the light beam: Q 2 2 3Q 2 2 Total Capacitance of ckt = 3c c+3
3CE 3Q 2 2CE =
Þ Q = \ Q = 2 C+3
2
C+3
1. bends downwards 2. bends upwards 3. becomes narrower 4. goes horizontally without any deflection C Þ Q 2 a
Solution: (2) C+3
13. From a solid sphere of mass M and radius R a cube of maximum possible volume is cut. Moment of inertia of cube about an axis passing through its center and prependicular to one of its faces is: 1. 3. 4 4 MR2 9 3 p
MR2 32 2 p
2. 4. 4 MR2 3 3 p
MR2 16 2 p
Velocity of rays away from the ground will be less as reprochive index is more. This will deform the wave front resulting into bending of the beam upwards. 16. A single of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz. The frequencies of the result signal is / are:
E­92, South Ex. Part 1, New Delhi 49. Ph: 011­41070321, 8510070321 8 www.niveditaclasses.com 1. 2005 kHz, 200 kHz and 1995 kHz 2. 2000 kHz and 1995 kHz 3. 2 MHz only 4. 2005 kHz and 1995 kHz the ground and neglect air resistance, take g = 10 m/s 2 ) (The figures are schematic and not drawn to scale) Solution: (1) Frequencies of resultant signal are = Fc + Fs , Fc, Fc ­ Fs = 2005 KHz, 2000 KHz and 1995 KHz 1.
17. A solid body of constant heat capacity 1J/°C is being heated by keeping it in contact with reservoris in two ways. (i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same amount of heat. 2.
(ii) Sequentially keeping in contact with 8 reservoirs such th at each reservoir supplies same amount of heat. In both the cases body is brought from initial temperature 100°C to final temperature 200°C. Entropy change of the body in the two cases respectively is 1. ln2, 2ln2 2. 2ln2, 8ln2 3. ln2, 4ln2 4. ln2, ln2 3.
Solution : (2) 18. Consider a spherical shell of radius R at temperature T. The black body radiation inside it can be considered as an ideal gas of photons wi t h i n t ern al en ergy per u ni t vol u me 4. 1 æ U ö
U 4 u= µ T and pressure P = ç ÷ . If the shell Solution : (1) 3 è V ø
V
Till both stone are in air, they will have now undergoes an adiabatic expansion the constant relative velocity of 30m/s.
relation between t and R is: y 1 = 30t D y = y 2 ­y First stone will hit ground after 8 sec 1 1 1. T µ
2. T µ 3 ( Q ­240 = 10t­2t 2 ). R
R
After 8s, the relative velocity will be aqual to velocity of 2nd stone and the D x ­t graph will 3. T µ e - R 4. T µ e-3 R be inverted parabola. 20. A uniformly chaged solid sphere of radius R Solution : (1) has potential V 0 (measured with respect to
19. Two stones are thrown up simultaneously from ¥ )on i ts su rf ace. For t hi s sph ere th e the edge of a cliff 240 m high with initial speed equipotential surfaces wityh potentials of 10 m/s and 40 m/s respectively. Which of the following graph best represents the time 3V0 5V0 3 V0
V ,
,
and 0 have radius R 1 , R 2 , R 3 variation of relative position of the second 2
4
4
4 and R 4 respectively. Then
stone with respect to the first ? (Assume stones do not rebound after hitting 5 E­92, South Ex. Part 1, New Delhi 49. Ph: 011­41070321, 8510070321 8 www.niveditaclasses.com 1. R 1 = 0 and R 2 < (R 4 ­R 3 ) 2. 2R < R 4 3. R 1 = 0 and R 2 > (R 4 ­R 3 ) 4. R 1 ¹ 0 and (R 2 ­R 1 )> (R 4 ­R 3 ) Solution: (1),(2) For uniformly changed solid sphere
ì Kq for (r ³ R )
ï r ï
V = í
2 ï Kq éê3 - r ùú for (r < R) ïî 2R ë
R 2 û
Using fornula’s R 1 = 0; R 2 = R/ 2 ; R 3 = 4R/3 ; R 4 = 4R After checking the options, (1) and (2) satisfy. 21. Monochromatic light is incident on a glass prism of angle A. If the refractive index of the material of the prism is ix, a ray, incident at an angle 0, on t he f ace AB wou ld get transmitted through the face AC of the prism provided : r 2 a sin ­1 (1/ m ) for prism A = r 1 + r 2
­1 Þ A­r 1 < Sin (1/ m )
­1 Þ r 1 > A ­ sin (1/ m ) For face AB, Sin q = m sinr 1
­1 Þ q = sin ( m sinr 1 )
Þ q >sin­1 (m sin(A­sin­1 (1/m)))
22. A rectangular loop of sides 10 cm and 5 cm carrying a current I of 12 A is placed in different orientations as shown in the figures below : 1.
2.
3.
é
ù
æ
-1
-1 æ 1 ö ö
1. q > cos ê m sin ç A + sin ç m ÷ ÷ ú
êë
è ø ø úû
è
é
ù
æ
-1
-1 æ 1 ö ö
2. q < cos ê m sin ç A + sin ç m ÷ ÷ ú
è ø ø ûú
è
ëê
4.
é
ù
æ
-1 æ 1 ö ö
3. q > sin ê m sin ç A - sin ç m ÷ ÷ ú
è ø ø ûú
è
ëê
-1
é
ù
æ
-1
- 1 æ 1 ö ö
4. q < sin ê m sin ç A - sin ç m ÷ ÷ ú
êë
è ø ø úû
è
If there is a uniform magnetic field of 0.3 T in the positive z direction, in which orientations the loop would be in (i) stable equilibrium and (ii) unstable equilibrium ? Solution: (3)
1. (b) and (d), respectively A 2. (b) and (c), respectively A q
3. (a) and (b), respectively r 1 r 2 4. (a) and (c), respectively
B C T avoid T 1 R at face AC 6 E­92, South Ex. Part 1, New Delhi 49. Ph: 011­41070321, 8510070321 8 www.niveditaclasses.com Solution: (1) # For stable equilibrium ­ magnetic dipole moment and magnetic field should be in same direction. (figure (b)) r
r
# For unstable equilibrium ­ m
and b should æ 2 3n ö
1 ÷÷
= 3m n 2 ­ . 3m çç
2 è 3 ø
= 2 5 2 m n
3
be anti­parallel. (figure (d)) 23. Two coaxial solenoids of different radii carry 25. Consider an ideal gas confined in an isolated uur
closed chamber. As the gas undergoes an current I in the same direction. Let F1 be the adiabatic expansion, the average time of magnetic force on the inner solenoid due to the collision between molecules increases as V q , uur where V is the volume of the gas. The value outer one and F2 be the magnetic force on the outer of q is solenoid due to the inner one. Then : uur
uur 1. F1 is radially inwards and F2 =0 uur
uur æ
C P ö
çg =
÷
Cv ø
è
2. F1 is radially outwards and F2 =0 3. g + 1 1. uur
uur F1 = F2 =0 uur
uur 4. F1 is radially inwards and F2 is radially outwards 2
3g + 5 6
3. 2. 4. g - 1 2
3g - 5 6
26. From a solid sphere of mass M and radius R, a Solution: (3) R spherical portion of radiusis removed, as 2 shown in the figure. Taking gravitational potential V=0 at r= ¥ the potential at the centre of the cavity thus formed is : S 2 F 2 S 1 F 1 Consider the front view. # The net force F 1 will be zero as it will be in all directions # There will be no field outside the solenoid due r
to S 1 . Therefore the force F2 = 0 (G = gravitational constant) 24. A Particle of mass 2m moving in the x direction with speed 2v. is hit by another particle of mass 2 m moving in the y direction with speed v. If the collisioin is perfectly inelastic, the percentage loss in the energy during the collision is close to: 1. 56% 2. 62% 3. 44% 4. 50% Solution: (1) m12 n
3m 1nc
1. -2 GM 3 R
2. -2GM R
3. -GM 2 R
4. -GM R
Solution: (4) V = V s ­ V c 2 ù
R (
GM é
2 )
ê3 - 2 ú
V s = 2R ê
R ú
ë
û
= -
11 GM 8 R
2m 1n
Using conservation of momentum
ur
ur
2 2 3mvc = 2mv $i + $j Þ v c =
v 3
KE loss = KE i ­ KE f ( )
7 V c = -
3GM1 3G(M 8 )
3 GM ==R 2r
2( 2 )
8 R
V = V S ­ V C = - GM R
E­92, South Ex. Part 1, New Delhi 49. Ph: 011­41070321, 8510070321 8 www.niveditaclasses.com Solution: (3) We can treat the given configration as an electric dipole. 29. As an electron makes a transition from an excited state of a hydrogen – like atom/ion: 27. 1. kinetic energy decreases, potential energy increases but total energy remains same Given in the figure are two blocks A and B of weight 20 N and 100 N, respectively. These are being pressed against a wall by a force F as shown. If the coefficient of friction between the blocks is 0.1 and between block B and the wall is 0.15, the frictional force applied by the wall on block B is : 1. 120 N 2. 150 N 3. 100 N 4. 80 N Solution: (1) fr F AB N 2. kinetic energy and total energy decrease but potential energy increases 3. its kinetic energy increases but potential energy and total energy decrease 4. kinetic energy, potential energy and total energy decrease Solution: (3) KE = ­TE = -
PE
2
KE a
1 , KE - as n ¯
n 2 PE a
-2 , PE ¯ as n ¯
n2 120N -1 TE a 2 , TE ¯ as n ¯
For equilibrium F r = 120N n
28. A long cylindrical shell carries positive surface 30. Match List ­ I (Fundamental Experiment) with charge s in the upper half and negative e List ­ II (its conclusion) and select the correct surface charge — s in the lower half. The option from the choices given below the list : electric field lines around the cylinder will look like figure given in : (figures are schematic and not drawn to scale) 1.
2.
3.
1. (A)­ (ii) (B)­(i) (C)­(iii) 2. (A)­ (iv) (B)­(iii) (C)­(ii) 3. (A)­ (i) (B)­(iv) (C)­(iii) 4. (A)­ (ii) (B)­(iv) (C)­(iii) Solution: (1)
4. 8 E­92, South Ex. Part 1, New Delhi 49. Ph: 011­41070321, 8510070321 8 www.niveditaclasses.com