STAGE-1-Test Paper-7 Physics
Time : 1.00 Hr Max. Marks : 80
GENERAL INSTRUCTIONS
1. There are 20 questions in this paper.
2. For each question in Section I, you will be awarded 3 Marks if you give the correct answer and zero Mark if no answer is given. In all other cases, minus one (–1) Mark will be awarded.
3. For each question in Section II, you will be awarded 4 Marks if you give the correct answer and
zero
Mark if no answer is given. In all other cases, minus one (–1) Mark will be awarded.
4. For each question in Section III, you will be awarded 4 Marks if you give the correct answer and
zero
Mark if no answer is given. In all other cases, minus one (–1) Mark will be awarded.
5. For each question in Section IV, you will be awarded 2 Marks for each row in which you give the correct answer(s). Thus, each question in this section carries a maximum of 8 Marks. There is no negative marking for incorrect answer(s) for this section.
SECTION - I
Straight Objective Type
This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct.
1. A disc of radius a/4 having a uniformly distributed charge 6C is placed in the x-y plane with its centre at (–a/2, 0, 0). A rod of length a carrying a uniformly distributed charge 8C is placed on the x-axis from x = a/4 to x = 5a/4. Two point charges –7C and 3C are placed at (a/4, –a/4, 0) and (–3a/ 4, 3a/4, 0), respectively. Consider a cubical surface formed by six surfaces x = ± a/2, y = ± a/2, z = ±a/2. The electric flux through this cubical surface is
(A)
2C
0
2C
(B) 0
(C)
10C
0
(D)
12C
0
2. The x-t graph of a particle undergoing simple harmonic motion is shown 1 below. The acceleration of the particle at t = 4/3 s is
(A)
2 cm/s2 (B)
32
2
32
0
cm/s2
–1
t(s)
(C) 2
32
cm/s2 (D) –
3 2 cm/s2
32
3. Three concentric metallic spherical shells of radii R, 2R, 3R, are given charges Q1, Q2, Q3, respectively. It is found that the surface charge densities on the outer surfaces of the shells are equal. Then, the ratio of the charges given to the shells, Q1 : Q2 : Q3, is
(A) 1 : 2 : 3 (B) 1 : 3 : 5
(C) 1 : 4 : 9 (D) 1 : 8 : 18
4. A ball is dropped from a height of 20 m above the surface of water in a lake. The refractive index of water is 4/3. A fish inside the lake, in the line of fall of the ball, is looking at the ball. At an instant, When the ball is
12.8 m above the water surface, the fish sees the speed of ball as [Take g = 10 m/s2]
(A) 9 m/s (B) 12 m/s
(C) 16 m/s (D) 21.33 m/s
5. Look at the drawing given in the figure which has been drawn with ink of uniform line-thickness. The mass of ink used to draw each of the two inner circles and each of the two line segments is m. The mass of the ink used to draw the outer circle is 6m. The coordinates of the centres of the different parts are: outer circle (0, 0), left inner circle (–a, a), right inner circle (a, a), vertical line (0, 0) and horizontal line (0, –a). The y-coordinate of the
centre of mass of the ink in this drawing is
a
(A) 10
a
(B) 8
a
(C) 12
a
(D) 3
6. A block of base 10 cm × 10 cm and height 15 cm is kept on an inclined plane. The coefficient of friction
between them is from 0º. Then
. The inclination of this inclined plane from the horizontal plane is gradually increased
(A) at = 30º, the block will start sliding down the plane
(B) the block will remain at rest on the plane up to certain and then it will topple
(C) at = 60º, the block will start sliding down the plane and continue to do so at higher angles
(D) at = 60º, the block will start sliding down the plane and on further increasing , it will topple at certain
7. Two small particles of equal masses start moving in opposite directions from a point A in a horizontal circular orbit. Their tangential velocities are v and 2v, respectively, as shown in the figure. Between collisions, the par- ticles move with constant speeds. After making how many elastic colli sions, other than that at A, these two particles will again reach the point A?
(A) 4 (B) 3
(C) 2 (D) 1
8. The figure shows certain wire segments joined together to form a coplanar loop. The loop is placed in a perpendicular magnetic field in the direction going into the plane of the figure. The magnitude of the field increases with time. 1 and 2 are the currents in the segments ab and cd. Then,
(A) 1 > 2
(B) 1 < 2
(C) 1 is in the direction ba and 2 is in the direction cd
(D) 1 is in the direction ab and 2 is in the direction dc
v A 2v
SECTION - II
Multiple Correct Answer Type
This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) out of which ONE OR MORE may be correct.
9. A student performed the experiment of determination of focal length of a concave mirror by u-v method using an optical bench of length 1.5 meter. The focal length of the mirror used is 24 cm. The maximum error in the location of the image can be 0.2 cm. The 5 sets of (u, v) values recorded by the student (in cm) are : (42, 56), (48, 48), (60, 40), (66, 33), (78, 39). The data set(s) that cannot come from experiment and is (are) incor- rectly recorded, is (are)
(A) (42, 56) (B) (48, 48)
(C) (66, 33) (D) (78, 39)
10. If the resultant of all the external forces acting on a system of particles is zero, then from an inertial frame, one can surely say that
(A) linear momentum of the system does not change in time
(B) kinetic energy of the system does not changes in time
(C) angular momentum of the system does not change in time
(D) potential energy of the system does not change in time
11. Cv and Cp denotes the molar specific heat capacities of a gas at constant volume and constant pressure, respectively. Then
(A) Cp – Cv is larger for a diatomic ideal gas than for a monoatomic ideal gas
(B) Cp + Cv is larger for a diatomic ideal gas than for a monoatomic ideal gas
(C) Cp / Cv is larger for a diatomic ideal gas than for a monoatomic ideal gas
(D) Cp . Cv is larger for a diatomic ideal gas than for a monoatomic ideal gas
12. For the circuit shown in the figure
(A) the current I through the battery is 7.5 mA
(B) the potential difference across RL is 18 V
(C) ratio of powers dissipated in R1 and R2 is 3
(D) if R1 and R2 are interchanged, magnitude of the power dissipated in RL will decrease by a factor of 9
SECTION - III
Paragraph Type
This section contains 2 groups of questions. Each group has 3 multiple choice questions based on a paragraph. Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct.
Paragraph for Question Nos. 13 to 15
Scientists are working hard to develop nuclear fusion reactor. Nuclei of heavy hydrogen, 2H , known as deuteron and denoted by D, can be thought of as a candidate for fusion reactor. The D-D reaction is
2H 2 H 3 He n energy . In the core of fusion reactor, a gas of heavy hydrogen is fully ionized into
1 1 2
deuteron nuclei and electrons. This collection of 2H nuclei and electrons is known as plasma. The nuclei
move randomly in the reactor core and occasionally come close enough for nuclear fusion to take place. Usually, the temperatures in the reactor core are too high and no material wall can be used to confine the plasma. Special techniques are used which confine the plasma for a time t0 before the particles fly away from the core. If n is the density (number/volume) of deuterons, the product nt0 is called Lawson number. In one of the criteria, a reactor is termed successful if Lawson number is greater than 5×1014 s/cm3. It may be helpful
e2
to use the following: Boltzman constant k = 8.6×10–5 eV/K ; 4 = 1.44 × 10–9 eVm.
13. In the core of nuclear fusion reactor, the gas becomes plasma because of
(A) strong nuclear force acting between the deuterons
(B) Coulomb force acting between the deuterons
(C) Coulomb force acting between deuterons-electrons pairs
(D) the high temperature maintained inside the reactor core
14. Assume that two deuteron nuclei in the core of fusion reactor at temperature T are moving towards each other, each with kinetic energy 1.5 kT, when the separation between them is large enough to neglect Coulomb potential energy. Also neglect any interaction from other particles in the core. The minimum temperature T required for them to reach a separation of 4 × 10–15 m in the range.
(A) 1.0 × 109 K < T < 2.0 × 109 K (B) 2.0 × 109 K < T < 3.0 × 109 K
(C) 3.0 × 109 K < T < 4.0 × 109 K (D) 4.0 × 109 K < T < 5.0 × 109 K
15. Results of calculations for four different designs of a fusion reactor using D-D reaction are given below. Which of these is most promising based on Lawson criterion ?
(A) deuteron density = 2.0 × 1012 cm–3, confinement time = 5.0 × 10–3 s
(B) deuteron density = 8.0 × 1014 cm–3, confinement time = 9.0 × 10–1 s
(C) deuteron density = 4.0 × 1023 cm–3, confinement time = 1.0 × 10–11 s
(D) deuteron density = 1.0 × 1024 cm–3, confinement time = 4.0 × 10–12 s
Paragraph for Question Nos. 16 to 18
When a particle is restricted to move along x-axis between x = 0 and x = a, where a is of nanometer dimension, its energy can take only certain specific values. The allowed energies of the particle moving in such a restricted region, correspond to the formation of standing waves with nodes at its ends x = 0 and x = a. The wavelength of this standing wave is related to the linear momentum p of the particle according to
p2
the de-Broglie relation. The energy of the particle of mass m is related to its linear momentum as E = .
2m
Thus, the energy of the particle can be denoted by a quantum number ‘n’ taking values 1,2,3, , (n = 1,
called the ground state) corresponding to the number of loops in the standing wave.
Use the model described above to answer the following three questions for a particle moving in the line x = 0 to x = a. Take h = 6.6 × 10–34 J s and e = 1.6 × 10–19 C.
16. The allowed energy for the particle for a particular value of n is proportional to : (A) a–2 (B) a–3/2
(C) a–1 (D) a2
17. If the mass of the particle is m = 1.0 × 10–30 kg and a = 6.6 nm, the energy of the particle in its ground state is closest to :
(A) 0.8 meV (B) 8 meV
(C) 80 meV (D) 800 meV
18. The speed of the particle, that can take discrete values, is proportional to : (A) n–3/2 (B) n–1
(C) n1/2 (D) n
SECTION - IV
Matrix - Match Type
This section contains 2 questions. Each question contains statements given in two columns, which have to be matched. The statements in Column-I are labelled A, B, C and D, while the statements in Column-II are labelled p, q, r, s and t. Any given statement in Coloumn-I can have correct matching with ONE OR MORE statement(s) in Coloumn-II. The answers to these questions have to be appropriately marked as illustrated in the following example. If the correct matches are A-p, A-r, B-p, B-s, C-r, C-s, D-q and D-t then the answer should be written as : A p,r ; B p, s ; C r, s ; D q, t.
19. Six point charges, each of the same magnitude q, are arranged in different manners as shown in Column–
II. In each case, a point M and a line PQ passing through M are shown. Let E be the electric field and V be the electric potential at M (potential at infinity is zero) due to the given charge distribution when it is at rest. Now, the whole system is set into rotation with a constant angular velocity about the line PQ. Let B be the magnetic field at M and be the magnetic moment of the system in this condition. Assume each rotating charge to be equivalent to a steady current.
Column–I Column–II
(A) E = 0 (p) + – Q
– +
Charges are at the corners of a regular hexagon. M is at the centre of the hexagon. PQ is perpendicular to the plane of
(B) V 0 (q)
P + –
the hexagon.
Charges are on a line perpendicular to PQ at equal intervals. M is the midpoint between the two innermost charges.
Q
(C) B = 0 (r) + – + Q
– M –
Charges are placed on two coplanar insulating rings at equal intervals. M is the common centre of the rings. PQ is perpendicular to the plane of rings.
P
+
(D) 0 (s) – + –
– + –
Charges are placed at the corners of a rectangle of sides a and 2a and at the mid–points of the longer sides. M is at the centre of the rectangular. PQ is parallel to the longer sides.
(t)
Q
Charges are placed on two coplanar, identical insulating rings at equal intervals. M is the mid–point between the centres of the rings. PQ is perpendicular to the line joining the centres and coplanar to the rings.
20. Column II shows five systems in which two objects are labelled as X and Y. Also in each case a point P is shown. Column I gives some statements about X and/or Y. Match these statements to the appropriate system(s) from Column II.
Column I Column II
(A) The force exerted (p) by X on Y has a
magnitude Mg.
Block Y of mass M left on a fixed inclined plane X, slides on it with a constant velocity.
(B) The gravitational (q) Two ring magnets Y and Z, potential energy of each of mass M, are kept in
X is continuously frictionless vertical plastic
increasing, stand so that they repel each
other. Y rests on the base X and Z hangs in air in equilibrium. P is the topmost point of the stand on the common axis of the two rings. The whole system is in a lift that is going up with a constant velocity.
(C) Mechanical energy (r) A pulley Y of mass m0 is fixed of the system X + Y to a table through a clamp X.
is continuously A block of mass M hangs from
decreasing. a string that goes over the
pulley and is fixed at point P of the table. The whole system is kept in a lift that is going down with a constant velocity.
(D) The torque of the (s) A sphere Y of mass M is put weight of Y about in a nonviscous liquid X kept
point P is zero. in a container at rest. The
sphere is released and it moves down in the liquid.
(t) A sphere Y of mass M is falling with its terminal velocity in a viscous liquid X kept in a container.
PAPER - 2
Time : 1.00 Hr Max. Marks : 80
GENERAL INSTRUCTIONS
1. There are 19 questions in this paper.
2. For each question in Section I, you will be awarded 3 Marks if you give the correct answer and
zero Mark if no answer is given. In all other cases, minus one (–1) Mark will be awarded.
3. For each question in Section II, you will be awarded 4 Marks if you give the correct answer and
zero Mark if no answer is given. In all other cases, minus one (–1) Mark will be awarded.
4. For each question in Section III, you will be awarded 2 Marks for each row in which you give the correct answer(s). Thus, each question in this section carries a maximum of 8 Marks. There is no negative marking for incorrect answer(s) for this section.
5. For each question in Section IV, you will be awarded 4 Marks if you give the correct answer and
zero Mark if no answer is given. In all other cases, minus one (–1) Mark will be awarded.
SECTION - I
Straight Objective Type
This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct.
1. Photoelectric effect experiments are performed using three different metal plates p, q and r having work functions p = 2.0 eV, q = 2.5 eV and r = 3.0 eV respectively. A light beam containing wavelengths of 550 nm, 450 nm and 350 nm with equal intensities illuminates each of the plates. The correct -V graph for the
experiment is [Take hc = 1240 eV nm]
(A) (B)
(C) (D)
2. A uniform rod of length L and mass M is pivoted at the centre. Its two ends are attached to two springs of equal spring constants k. The springs are fixed to rigid supports as shown in the figure and the rod is free to oscillate in the horizontal plane. The rod is gently pushed through a small angle in one direction and released. The frequency of oscillation is :
(A) (B)
(C) (D)
3. A piece of wire is bent in the shape of a parabola y = kx2 (y-axis vertical) with a bead of mass m on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the x-axis with a constant acceleration a. The distance of the new equilibrium position of the bead, where the bead can stay at rest with respect to the wire, from the y-axis is
a
(A) gk
a
(B) 2gk
2a
(C) gk
a
(D) 4gk
4. The mass M shown in the figure oscillates in simple harmonic motion with amplitude A. The amplitude of the point P is
(A) k1A
2
k1A k1 k2
(B) k2A
1
k2A k1 k2
SECTION - II
Multiple Correct Answer Type
This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) out of which ONE OR MORE may be correct.
5. Under the influence of the Coulomb field of charge +Q, a charge –q is moving around it in an elliptical orbit. Find out the correct statement(s).
(A) The angular momentum of the charge –q is constant
(B) The linear momentum of the charge –q is constant
(C) The angular velocity of the charge – q is constant
(D) The linear speed of the charge –q is constant
6. The figure shows the P-V plot of an ideal gas taken through a cycle ABCDA. The part ABC is a semi-circle and CDA is half of an ellipse. Then,
(A) the process during the path A B is isothermal
(B) heat flows out of the gas during the path B C D
(C) work done during the path A B C is zero
(D) positive work is done by the gas in the cycle ABCDA
7. A sphere is rolling without slipping on a fixed horizontal plane surface. In the figure, A is the point of contact, B is the centre of the sphere and C is its topmost point. Then
(A)
→ →
VC VA
→ →
2 VB VC
(B)
→ →
VC VB
→ →
= VB VA
→ → → →
→ → →
(C)
VC VA = 2 VB VC
(D)
VC VA = 4 VB
8. A student performed the experiment to measure the speed of sound in air using resonance air-column method. Two resonances in the air-column were obtained by lowering the water level. The resonance with the shorter air-column is the first resonance and that with the longer air-column is the second resonance. Then,
(A) the intensity of the sound heard at the first resonance was more than that at the second resonance
(B) the prongs of the tuning fork were kept in a horizontal plane above the resonance tube
(C) the amplitude of vibration of the ends of the prongs is typically around 1 cm
(D) the length of the air-column at the first resonance was somewhat shorter than 1/4th of the wavelength of the sound in air.
9. Two metallic rings A and B, identical in shape and size but having different A B resistivities A and B, are kept on top of two identical solenoids as shown in the
figure. When current is switched on in both the solenoids in identical manner,
the rings A and B jump to heights hA and hB, respectively, with hA > hB. The possible relation(s)
between their resistivities and their masses mA and mB is(are)
(A) A > B and mA = mB (B) A < B and mA = mB
(C) A > B and mA > mB (D) A < B and mA < mB
SECTION - III
Matrix - Match Type
This section contains 2 questions. Each question contains statements given in two columns, which have to be matched. The statements in Column-I are labelled A, B, C and D, while the statements in Column-II are labelled p, q, r, s and t. Any given statement in Coloumn-I can have correct matching with ONE OR MORE statement(s) in Coloumn-II. The answers to these questions have to be appropriately marked as illustrated in the following example. If the correct matches are A-p, A-r, B-p, B-s, C-r, C-s, D-q and D-t then the answer should be written as : A p,r ; B p, s ; C r, s ; D q, t.
10. Column I shows four situations of standard Young’s double slit arrangement with the screen placed far away
from the slits S
and S . In each of these cases S P = S P
, S P
– S P
= /4 and S P
– S P
= /3, where
1 2 1 0
2 0 1 1 2 1
1 2 2 2
is the wavelength of the light used. In the cases B,C and D, a transparent sheet of refractive index and thickness t is pasted on slit S2. The thicknesses of the sheets are different in different cases. The phase difference between the light waves reaching a point P on the screen from the two slits is denoted by (P) and the intensity by (P). Match each situation given in Column-I with the statement(s) in Column-II valid for that situation.
Column–I Column–II
(A) (p) (P ) = 0
(B) ( – 1)t = /4 (q) (P ) = 0
(C) ( – 1)t = /2 (r) (P ) = 0
(D) ( – 1)t = 3/4 (s) (P ) > (P )
0 1
(t) (P ) > (P )
2 1
11. Column II gives certain systems undergoing a process. Column I suggests changes in some of the parameters related to the system. Match the statements in Column-I to the appropriate process(es) from Column II.
Column–I Column–II
(A) The energy of the system is increased. (p) System: A capacitor, initially uncharged
Process: It is connected to a battery.
(B) Mechanical energy is provided to the system, (q) System: A gas in an adiabatic container fitted with which is converted into energy of random motion an adiabatic piston
of its parts Process: The gas is compressed by pushing the piston
(C) Internal energy of the system is converted (r) System: A gas in a rigid container
into its mechanical energy Process: The gas gets cooled due to colder atmosphere surrounding it
(D) Mass of the system is decreased (s) System: A heavy nucleus, initially at rest
Process: The nucleus fissions into two fragments of nearly equal masses and some neutrons are emitted
(t) System: A resistive wire loop
Process: The loop is placed in a time varying magnetic field perpendicular to its plane
SECTION - IV
Integer Answer Type
This section contains 8 questions. The answer to each of the questions is a single digit integer, ranging from 0 to 9.
12. A steady current goes through a wire loop PQR having shape of a right angle triangle with PQ = 3x, PR = 4x
k 0
and QR = 5x. If the magnitude of the magnetic field at P due to this loop is
48x , find the value of k.
13. A light inextensible string that goes over a smooth fixed pulley as shown in the figure connects two blocks of masses 0.36 kg and 0.72 kg. Taking g = 10 m/s2, find the work done (in joules) by the string on the block of mass 0.36 kg during the first second after the system is released from rest.
14. A solid sphere of radius R has a charge Q distributed in its volume with a charge density = kra, where k and
a are constants and r is the distance from its centre. If the electric field at r = R
2
the value of a.
1
is 8
times that at r = R, find
15. A metal rod AB of length 10 x has its one end A in ice at 0ºC and the other end B in water at 100ºC. If a point P on the rod is maintained at 400ºC, then it is found that equal amounts of water and ice evaporate and melt per unit time. The latent heat of evaporation of water is 540 cal/g and latent heat of melting of ice is 80 cal/g. If the point P is at a distance of x from the ice end A, find the value of . [ Neglect any heat loss to the surrounding
16. Two soap bubbles A and B are kept in a closed chamber where the air is maintained at pressure 8 N/m2. The radii of bubbles A and B are 2cm and 4cm, respectively. Surface tension of the soap-water used to make bubbles is 0.04 N/m. Find the ratio nB/nA, where nA and nB are the number of moles of air in bubbles A and B,
respectively. [Neglect the effect of gravity.]
17. A 20cm long string, having a mass of 1.0 g, is fixed at both the ends. The tension in the string is 0.5 N. The string is set into vibrations using an external vibrator of frequency 100 Hz. Find the separation (in cm) between the successive nodes on the string.
18. Three objects A,B and C are kept in a straight line on a frictionless horizontal surface. These have masses m, 2m and m, respectively. The object A moves towards B with a speed 9 m/s and makes an elastic collision with it. Thereafter, B makes completely inelastic collision with C. All motions occur on the same straight line. Find the final speed (in m/s) of the object C.
19. A cylindrical vessel of height 500 mm has an orifice (small hole) at its bottom. The orifice is initially closed and water is filled in it up to height H. Now the top is completely sealed with a cap and the orifice at the bottom is opened. Some water comes out from the orifice and the water level in the vessel becomes steady with height of water column being 200 mm. Find the fall in height (in mm) of water level due to opening of the orifice.
[Take atmospheric pressure = 1.0 × 105 N/m2 , density of water = 1000 kg/m3 and g = 10 m/s2. Neglect any effect of surface tension]
A nswers
PAPER - 1
1. (A) 2. (D) 3. (B) 4. (C) 5. (A)
6. (B) 7. (C) 8. (D) 9. (C) 10. (A)
11. (B), (D) 12. (A), (D) 13. (D) 14. (A) 15. (B)
16. (A) 17. (B) 18. (D)
19. (A) (p), (r), (s); (B) (r), (s); (C) (p), (q), (t); (D) (r), (s)
20. (A) (p), (t); (B) (q), (s), (t); (C) (p), (r), (t); (D) (q)
PAPER - 2
1. (A) 2. (C) 3. (B) 4. (D) 5. (A)
6. (B), (D) 7. (B), (C) 8. (A), (D) 9. (B), (D)
10. (A) p, s; (B) q; (C) t ; (D) r,s,t 11. (A) p, q, t; (B) q, t (C) s, (D) s 12. k = 7
13. 8J 14. a = 2 15. = 9 16.
nB 6 nA
17. 5 cm
18. VC = 4 m/sec. 19. 6 mm
STAGE SOLUTIONS TO TEST PAPERS (PHYSICS)
2
1. From Gauss law
Qin
0
= (8C / 4) – 7C (6C / 2) = –
0
2C .
0
2. T = 8 second., A = 1 cm,
2
x = A sint. = 1sin 8 t.
2 2
a = – 2x = – 8
2
sin 8 t cm/s2
4 2 2
At, t = 3 second. a = – 8 sin 3 = –
cm/s2.
32
3. The charge distribution on the surfaces of the shells are given. As per the given condition.
Q1
4R2
Q1 Q2
= 4(2R)2 =
Q1 Q2 Q3
4(3R)2
Q1 Q2 Q3 .
1 3 5
4. x' =
x v
, v' = =
2 10 (20 – 12.8) 4
= 16 m/s.
nrel nrel 1 3
5. ycm =
m1y1 m2y2 m3y3 m4y4 m5y5
m1 m2 m3 m4 m5
cm =
6m(0) m(a) m(0) m(a) m(–a) m m m m 6m
a
= 10 .
6. Angle of repose 0 = tan–1 = tan–1
= 60º
tan =
5
15 / 2
2
= 3 .
< 45º.
Block will topple before it starts to slide down.
7. Since masses of particles are equal, collisons are elastic,so parwill exchange velocities after each collision. The first collision will be at a point P and second at point Q again and before third collision the par ticles will reach at A.
8. Current 1 = 2 , Since magnetic field increases with time So induced net flux should be outward i.e. current will flow from a to b
9. 1 1 1 or
1 1 1
|v| =
| u | | f |
v u f
| v | | u | | f |
| u | | f |
For |u| = 42, |f| = 24 ; |v| =
(42)(24)
42 24
= 56 cm so (42, 56) is correct observation
For |u| = 48 or |u| = 2f or |v| = 2f so (48, 48) is correct observation For |u| = 66 cm ; |f| = 24 cm
|v| =
(66)(24)
66 24
36 cm which is not in the permissible limit
so (66, 33), is incorrect recorded For |u| = 78, |f| = 24 cm
|v| =
(78)(24)
78 24
32 cm which is also not in the permissible limit.
so (78, 39), is incorrect recorded.
10. (A) Since there in no resultant external force, linear momentum of the system remains constant.
(B) Kinetic energy of the system may change.
(C) Angular momentum of the system may change as in case of couple, net force is zero but torque is not zero.Hence angular momentum of the system is not constant.
(D) Potential energy may also change.
5 3
11. For monoatomic gas, Cp = 2 R, Cv = 2 R. Cp – Cv = R
7 5
For diatomic gas Cp = 2 R, Cv = 2 R. Cp – Cv = R
Cp – Cv is same for both Cp + Cv = 6R (for diatomic) Cp + Cv = 4R (for mono) so (Cp + Cv)dia > (Cp + Cv)mono
Cp 7
= 1.4 (for diatomic)
Cv 5
Cp 5
= 1.66 (for monoatomic)
Cv 3
35
(C ) (C ) = R2 = 8.75 R2 (for diatomic)
p v 4
(C ) (C ) = 15 R2 (for monoatomic)
p v 4
so (Cp . Cv)diatomic > (Cp . Cv)monoatomic
6 1.5 2 6 16
12. Req =
7.5
= + 2 = k
V
= Req =
24
16 × 5 mA =
3
2 × 5 = 7.5 mA
for potential difference across R1 V1 = 7.5 × 2 = 15 V for potential difference across R2 V2 = 24 – 15 = 9 V
V2 V2
(15)2 92 25
for power :
PR : PR
= 1 : 2 = : =
1 2 R1 R2
2 6 3
V2
P 2 =
L RL
92
1.5
= 54 mW
If R1 and R2 are interchanged
(2)(1.5) 3
R´ = R1 || R2 =
2 1.5 =
3.5
V´L =
R´ R2 R´
× 24 V = 3 V
Now power dissipated in RL
is P´L
V ´2
= L
RL
= 32
1.5
= 6 mW
13. Plasma state is achieved at high temperature.
14. From energy conservation
3 1 e2
2 2 kT 4 r
0
T =
(1.44 109 )
5 ×
1
15
= 1.39 × 109 K
(8.6 10 ) 3 4 10
15. nt > 5 × 1014 s/cm3
for deuteron density = 8.0 × 1014 cm–3, confinement time = 9.0 × 10–1 s nt = 7.2 × 1014 s/cm3
16.
h
n 2 = a and p =
nh
= 2a
Energy E = p2
2m
n2h2
= 8ma2
E 1 a2
17. E =
n2h2 8 ma2
For ground state n = 1 E =
(1)2(6.6 1034 )2
8 1030 (6.6 109 )2 1.6 1019
= 8 meV
nh
18. P = 2a
= mv v n
→ → →
19. E = E1 E2 + ......... v = v1 + v2 + v3 + ........
→ →
B = B1 +
→ + .......... → = →
2
..........
20.
(p)
(A) Net force on Y due to X = = mg
(B) As the inclined is fixed. So, gravitational P.E. of X is constant
(C) As K.E. is constant and P.E. of Y is decreasing. So mechanical energy of (X + Y) is decreasing.
(q) (A) force on Y due to X will be greater than Mg which is equal to (Mg + repulsion force)
(B) As the system is moving up, P.E. of X is increasing.
(C) Mechanical energy of (X + Y) is increasing
(D) Torque of the weight of Y about point P = 0
(r)
(A) force on Y due to X =
(B) As the system moves down, gravitational P.E. of X decreases
(C) As the system moves down, total mechanical energy of (X + Y) also decreases
(D) 0
(s)
(A) force on Y due to X = Buoyancy force which is less than mg
(B) As the sphere moves down, that volume of water comes up, so gravitational P.E. of X increases.
(C) As there is no non–conservative force, so total mechanical energy of X + Y remains conserved.
(D) 0
(t)
(A) As the sphere is moving with constant velocity B + fv = Mg
so force on Y due to X is B + fv = Mg
(B) As the sphere moves down, that volume of water comes up, so gravitational P.E. of X will increase
(C) Increase in mechanical energy = wfr = –ve
(D) = 0
1. E 550 nm 1240 eV 2.25 eV E 450 nm 1240 eV 2.8 eV
1 550 2 450
E 350 nm 1240 eV 3.5 eV
3 350
For metal r, only 3 is able to generate photoelectron.
For metal q, only and are able to generate photoelectron. For metal p, all wavelengths are able to generate photoelectron.
Hence photoelectric current will be maximum for p and least for r.
2. Torque about P = (kx)
L
2 + (kx)
L
2 = kxL = k
L2
2
L
For small angle , x = 2
=
– KL2 ML2
6K =
2
=
12 M
and f = 2 =
3. ma cos = mg cos (90 – )
a tan g
a dy
g = dx
d (kx2) = a
dx
x =
a
2gk = D
4.
Extensions in springs are x1 and x2 then k1x1 = k2x2 and x1 + x2 = A
x1 +
k1x1 k2
= A x1 =
k2A k1 k2
5. Torque about Q of charge –q is zero, so angular momentum charge –q is constant, but distance between charges is changing, so force is changing, so speed and velocity are changing.
6. (A) process is not isothermal
(B) volume decreases and temperature decreases
U = negative
W = negative
so Q = negative
(C) Work done in process A B C is positive
(D) Cycle is clockwise, so work done by the gas is positive.
7. → ˆ ˆ ; →
ˆ ; → ˆ ˆ
VA V( i) R(i)
VB Vi VC Vi Ri
→ → ˆ
VC VA 2Ri
→ →
VB VC
2[ V(ˆi ) V(ˆi) R(ˆi )] = –2R( ˆi )
→ → → →
Hence
VC VA = 2(VB VC)
→ → → →
so | VC VA | = | 2(VB VC) |
→ → ˆ
VC VB = R( i )
→ →
VB VA
= R( ˆi )
→ → → →
VC VB VB VA
Hence
→ →
VC VA
2R(ˆi )
→ → → →
VC VB VB VA
→ = 4V( ˆi ) = 4R ( ˆi )
→ → →
Hence
VC VA 2(VB )
8. (A) The intensity of sound decreases with increases of order. The intensity of sound is maximum for first resonance.
(B) The prongs vibrate in vertical plane.
(C*) The prongs do not vibrate in amplitude of that order.
(D) Consider end correction, the length of air column is slightly less than /4 So ans are (A) and (D)
9. The horizontal component of magnetic field due to solenoid will exert force on ring in vertical direction
F = BHi (2r)
Ft = MV
( / t)
i = (2r)
A
BH i (2r) t = MV
V = BHA K
M M
V2 K2
h = 2g
2M2
hA > hB
K2
2 M2
K2
2 M2
A A B B
BMB > AMA Using this we get
10. (A) (P ) = + + 2
2
cos 4
2
= + + 2 . = (2 + )
1
(P ) = + + 2 .
(P ) > (P )
= 2 1 2 . cos 3
= + + 2 .
0 0 0
=
(B) (P ) = ( – 1)t . 2
= . 2
0 4 2
2
(P1) = ( 1)t 4 . 0
2
(P ) = ( 1)t .
2 3 6
(P ) = + + 2 . cos (P )
= 0
+ 0
+ 2
. cos 2
= 2
(P ) = 4
1 0
(P ) =
+ 0
+ 2
cos
= (2 + )
(C). (P ) = ( – 1)t . 2
= . 2
0 2
( 1)t. 2 2
(P ) =
1
4 2
2
(P ) = ( 1)t .
2 3 6
. 2 2
(P ) =
0 2 3 6 3
(D) (P ) = 3 2 3
0 4 2
(P ) = 0
(P2) > 0, (P0) > 0
11. (A) (p) : Capacitor is charged, hence its energy is increased
(q) : The temperature is increased, hence its energy is increased or as the external positive work is done, hence energy increases
(r) : The temperature decreases, its energy is decreased
(s) : All natural process, energy of the system decreases
(t) : The current is produced. Hence energy of the system increases
(B) (p), (r), (s) no mechanical energy is provided to the system
(q) the mechanical energy is provided which increases the temperature and hence random motion of molecules
(t) Mechanical work is done to change the magnetic field, which increases the mechanical energy of electron and these electrons strike with stationary positive charge and energy is converted in random motion.
(C) (s) Internal binding energy is converted into mechanical energy
(D) (s) Mass changes only in nuclear process.
12. B =
0i
4 12x
(sin 37º + sin 53º)
5
0i
4 3
0i
B = 48x
5
5 5
B = 7 48x
k = 7
13. T = 2m1m2 g =
1 2
T = 4.8 N
m1 m2
2 0.72 0.36
0.72 0.36
g
× 10
1 2
1 g 10
a = m m
g = 3
s = at
2
= 2 3 (1)2 = 6
1 2
10
Work done by T = (T) (S) = (4.8) × 6 = 8J
rR
14. Total charge Q = d V (Kra )(4r2dr) 4k (Ra3 )
rR / 2
r0
4k R a3
Q´ = d V (Kra )(4r2dr)
r0
a 3 2
1 Q´ 1 1 Q
According to question
4 2 = 2
0 (R / 2) 8 40 R
Putting the value of Q and Q´ we get a = 2
dm L
i dt f L
400 0
400 100
1 f
15. i1 =
(x / kA) , i2 =
(10 )x / kA
i dm L
dt v
400 / x 80
= 9 .
300 /(10 )x 540
4T
16. PA = P0 + rA
P = 8 +
4 0.04
0.02
P = 16 N/m2
P = P + 4T
rB
= 8 +
4 0.04
0.04
P = 12 N/m2
for bubble A, PV = nRT
4
(16) 3 (0.02)3 = n RT (1)
4 (0.04)3
for bubble B (12) = n RT (2)
3 B
dividing equation (1) and (2)
nA 1
nB 6
nB 6 nA
17. v = =
v = f
10 = (100)
= 0.1 m = 10 cm
= 10 m/sec.
distance between two successive nodes = 2
= 5 cm
18.
from momentum conservation : 9m = (2m) V1 – (m)V2
9 = 2V – V (1)
2
e = V1 V2 1 9
......(2)
from eqn(1) and eqn(2) V1 = 6 m/sec.
for second collision between second block and third block : (2m) 6 + m(0) = (2m + m) VC
V = 4 m/sec.
19. P + 200 × 10–3 × 1000 × 10 = P (1)
P0 (500 – H) = P . (300 mm)
P =
P0(500 H)mm
300 mm
....(2)
from (1) and (2)
P0 (500 H)
300
105(500 H)
300
+ 2000 = P0
+ 2000 = 105
5 × 107 – H × 105 + 6 = 300
H = 206 mm
fall in height = 6 mm
STAGE
SIMILAR TEST PAPERS (PHYSICS)
3
Time : 1.00 Hr Max. Marks : 80
GENERAL INSTRUCTIONS
1. There are 20 questions in this paper.
2. For each question in Section I, you will be awarded 3 Marks if you give the correct answer and zero Mark if no answer is given. In all other cases, minus one (–1) Mark will be awarded.
3. For each question in Section II, you will be awarded 4 Marks if you give the correct answer and
zero
Mark if no answer is given. In all other cases, minus one (–1) Mark will be awarded.
4. For each question in Section III, you will be awarded 4 Marks if you give the correct answer and
zero
Mark if no answer is given. In all other cases, minus one (–1) Mark will be awarded.
5. For each question in Section IV, you will be awarded 2 Marks for each row in which you give the correct answer(s). Thus, each question in this section carries a maximum of 8 Marks. There is no negative marking for incorrect answer(s) for this section.
SECTION - I
Straight Objective Type
This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct.
1. Consider a closed surface formed by two hexagonal parallel surfaces located at z = ± a (are parallel to x y
2
plane). Other six surfaces have square shape. A disc of radius a having uniformly distributed charge 6C is
– 3 a a
– a
3a
placed at , ,0 Two point charges of 2C each are placed at 0, , 0 and , ,0 . A rod
of length 10a having total charge C uniformly distributed over its length, is placed along line
y = x – a as shown in figure. Rod lie in (xy) plane. find total flux through this closed surface.
5 6 10 1– 7
3
(A) 0 C (B) C (C) C (D) C
0 0 0
2. The y – t graph of a particle having mass 0.5 kg, undergoing S. H. M. is shown. The accleration and K.E. of the particle at t = 4/3 are.
(A)
(C)
64
– 2
16
2
,
64
32
,
64
2
(B) 64 ,
2
(D) 64 ,
64
– 2
4
3. If surface charge densties on outer shell and inner most shell are equal. Charge on middle shell is Q and some charge is present on inner most shell. Then the charges on inner and outer shells respectively are.
(A)
–Q , –90
(B)
Q , 90
10 10 10 10
(C)
–70 , 30
(D)
–70 , –30
10 10 10 10
4. A bird dives to catch a fish with speed 30 m/s. Bird is at 10m height above water surface. If refractive index of water is 4/3, then find the speed of bird as seen by fish.
(A) 22.5 m/s (B) 40 m/s
(C) 20 ms (D) none of these
5. Look at the drawing in the figure which has been drawn with ink of uniform line thickness. Masses of ink used to draw each straight lines (outer boundries of the house) and each smaller rectangles are m. there are five straight lines. Mass of ink used to draw bigger rectangle is 2m. Coordinates of centers of different points are
– a , a a , a
Two vertical lines are 2
2 and 2 2
– a ,a a ,a
Two inclined lines are 2
and 2
– a , 3a a , 3a
Two small rectangles 4
4 and 4 4
0, 5a
Horizontal line is (0, 0 ) Bigger rectangle
8
y–coordinates of centre of mass of the drawing is.
a 23 a
(A) 36 (B) 36
25 a
(C) 36
(D) a 12
6. A block of base 6cm × 6cm and height 10 cm is kept on an plane. The coefficient of friction between them is 4/3. The inclination of this plane from horizontal plane is gradually increased from 0º then.
(A) at = 30º, the block will start sliding down the plane.
(B) the block will remain at rest on the plane up to certain and then it will topple.
(C) at = 53º the block will start sliding down the plane and continue to do so at higher angles.
(D) at = 53º the block will start sliding down the plane and on further increasing , it will topple at certain .
7. Two small particles of equal masses start moving in opposite direction from a point A in a horizontal. circular orbit. Their tangential velocities are v and 3v as shown in figure. Between collisions particles move with constant speeds. After how many elastic collisions including collision at A, these two particles will again reach the
point A.
(A) 4 (B) 5
(C) 3 (D) 2
8. The figure shows certain wire segment joined together to form a coplanar loop the loop is placed in a perpendicular magnetic field in the direction going into the plane of the figure. The magnitude of the field increases with time. I1 and I2 are the currents in the segment ab and cd then.
(A) I1 > I2
(B) I1 < I2
(C) I1 is in the direction ba and I2 is in the direction cd
(D) I1 is in the direction ab and I2 is in direction dc.
SECTION - II
Multiple Correct Answer Type
This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) out of which ONE OR MORE may be correct.
9. A student performed the experiment of determination of focal length of a concave mirror by u – v method using an optical bench of length 2.5 metre. the focal length of the mirror is 20 cm. the maximum. error in the location of the image can be 0.2 cm the 5 sets of (u, v) values recorded by the student
(in cm) are (40, 40), (45, 36), (50, 35), (36, 48) (60, 30)
The data set(s) that a cannot come from experiment and is (are) incorrectly recorded (A) 50, 35 (B) 45, 36
(C) 60, 30 (D) 36, 48
10. If the resultant of all the external torques about an axis on a system is zero. Then from an inertial frame one can surely say that.
(A) linear momentum of the system does not changes in time
(B) Kinetic energy of the system does not changes in time about that axis.
(C) Angular momentum of the system does not changes in time about that axis
(D) potential energy of the system may changes in time.
11. Cp and Cv denote the molar specific heat capacities of a gas at constant pressure and at constant volume respectively. Assume H2 and He can be treatad as ideal gas. Then:
(A) C 2 – C 2 is smaller for H than He (B) C 2 + C 2 is smaller for H
than He
p v 2 p v 2
(C) (C / C )2 is smaller for H than He (D) (C 2 . C 2 ) is smaller for H
than He
p v 2 p v 2
12. For the circuit shown in the figure :
(A) The current I through the battery is 12 A
(B) Poiential diffrence across R3 is 9V
(C) Ratio of powers dissipated in R3 and R2 is 81
(D) Ratio of power dissipated in R2 and R1 is 2
SECTION - III
Paragraph Type
This section contains 2 groups of questions. Each group has 3 multiple choice questions based on a paragraph. Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct.
Paragraph for Question Nos. 13 to 15
THE ENGINE OF THE STAR
One of the important features of nuclear reactions is that much more energy is released (i.e., converted from mass-energy) than in normal chemical reactions, such as the burning of fossil fuels. Let us look at the binding energy curve as shown here.
Examining the left end of the curve, we see that we could increase the binding energy of the system and release energy by combining two light nuclei. This is the process of fusion and is made difficult by the fact that the nuclei must overcome a very strong Coulomb repulsion to come close enough to fuse. One way to assist the nuclei in overcoming this repulsion is to cause them to move to very high kinetic energy by raising the system of nuclei to very high temperature. This situation is difficult to achieve on earth but it is achieved in the core of the stars, where temperature (~106 K) is sufficient to provide the required thermal kinetic energy. This process is the source of energy of stars including our Sun.
For stars with hotters cores (T > 15 × 106 K) a nuclear reaction called carbon cycle dominates. The step wise reaction is given below, which completes in five steps.
1H +
1H +
1H +
12C
13C
14N
13C + e+ +
14N +
15O +
15O
15N + e+ +
1H +
15N
12C +
4He
e2
It may be useful to use the following 4 0
= 1.44 × 10–9 eV-m. The radius of a nucleus of mass number A,
r = r0 A1/3 , r0 = 1.44 × 10–15 m.
13. The region of greatest stability is :
(A) 1 A 10 (B) 20 A 30
(C) 50 A 70 (D) 200 A 230
14. Consider 27A𝑙 coming towards a fixed 127. For fusion to occur, 27A𝑙 must have kinetic energy of the order:
(A) 12 MeV (B) 33 MeV
(C) 59 MeV (D) 86 MeV
15. The energy released in the carbon cycle depends upon
(A) the binding energy of 12C (B) the binding energy of 1H
(C) the binding energy of 4He (D) all the above
Paragraph for Question Nos. 16 to 18 RESONANCE of particle in between two walls.
When we blow a flute of given length, sound of certain frequencies produce high intensity sound and others die off. Only those frequencies which are able to form standing waves in the flute survive. Those frequencies are called resonance frequencies and the phenomenon is called resonance. This occurs for any wave and for particle waves also.
A particle is restricted between two walls separated by a distance a = 1Å. Between the walls, particle can have certain kinetic energies only. The kinetic energy should be such that the corresponding particle waves will form nodes at the positions of the walls. From de–Broglie hypothesis, the wavelength of the particle wave
h
is = p =
, where p = linear momentum and E is kinetic energy.
Use the passage above to answer the following three questions for a particle moving in a line perpendicular to the wall. (Take hc = 12400 eV–Å, If m is the mass of the electron then m c2 = 0.511 MeV)
16. The wavelength of particle between the wall will :
(A) have maximum value 1Å (B) have maximum value 2Å
(C) have minimum value 1 Å (D) have minimum value 2Å
17. If mass of the particle is m = 20 me(me = mass of electron) the possible kinetic energy of the particle is :
(A) 5 eV (B) 10 eV
(C) 15 eV (D) 20 eV
18. The speed of the particle between the walls depends on a as :
(A) a2 (B) a
(C) a–1 (D) a–2
SECTION - IV
Matrix - Match Type
This section contains 2 questions. Each question contains statements given in two columns, which have to be matched. The statements in Column-I are labelled A, B, C and D, while the statements in Column-II are labelled p, q, r, s and t. Any given statement in Coloumn-I can have correct matching with ONE OR MORE statement(s) in Coloumn-II. The answers to these questions have to be appropriately marked as illustrated in the following example. If the correct matches are A-p, A-r, B-p, B-s, C-r, C-s, D-q and D-t then the answer should be written as : A p,r ; B p, s ; C r, s ; D q, t.
19. Point charges, each of the same magnitude q, are arranged in different manners as shown in
Column–II. In each case, a point M and a line PQ passing through M are shown. Let E be the electric field and V be the electric potential at M (potential at infinity is zero) due to the given charge distribution when it is at rest. Now, the whole system is set into rotation with a constant angular velocity about the line PQ. Let B be the magnetic field at M and be the magnetic moment of the system in this condition. Assume each rotating charge to be equivalent to a steady current.
Column–I Column–II
(A) E = 0 (p) Charges are at the corners of a regular pentagon. M is at the centre of the pentagon. PQ is perpendicular to the plane of the pentagon.
(B) V 0 (q)
Charges are on a line perpendicular to PQ at equal intervals. M is the midpoint between the two innermost charges.
Q
(C) B = 0 (r) Charges are placed on two coplanar insulating triangles at equal intervals. M is the common centre of the triangles. PQ is perpendicular to the plane of triangles.
(D) 0 (s) Charges are placed on ring. M is at
the center of ring. PQ is along a diameter
(t) Charges are placed on two coplanar, identical insulating triangles at equal intervals. M is the mid–point between the centres of the triangles. PQ is perpendicular to the line joining
the centres and coplanar to the triangles.
20. Column II shows five systems in which two objects are labelled as X and Y. Also in each case a point P is shown. Column I gives some statements about X and and/or Y. Match these statements to the appropriate system(s) from Column II.
Column I Column II
(A) The force exerted (p) by X on Y has a
magnitude Mg.
A block X of mass M is moving over a smooth and fixed circular track of radius R, as shown in figure. At the given instant
velocity v = & = 60º.
(B) The gravitational (q) Two ring magnets Y and Z, potential energy of each of mass M, are kept in
X is continuously frictionless vertical plastic
increasing, stand so that they repel each
other. Y rests on the base X and Z hangs in air in equilibrium. P is the topmost point of the stand on the common axis of the two rings.
The whole system is in a lift
that is going down with a constant acceleration. (a < g)
(C) Mechanical energy of (r) Two blocks of same mass M are
the system X + Y is placed on a smooth and fixed
continuously decreasing. wedge. Pulley is also smooth and massless. The whole system is kept in a lift that is going upward with constant velocity.
(D) The torque of the (s) Two hollow balls of mass 2M,M weight of Y about and volume 2V, V respectively
point P is zero. are moving with constant
velocity v, in a non–viscous liquid as shown in figure. Pulley is fixed at point P.
(t) Two identical spherical balls of mass M are falling with terminal velocities in a viscus X kept in a container.
PAPER - 2
Time : 1.00 Hr Max. Marks : 80
GENERAL INSTRUCTIONS
1. There are 19 questions in this paper.
2. For each question in Section I, you will be awarded 3 Marks if you give the correct answer and
zero
Mark if no answer is given. In all other cases, minus one (–1) Mark will be awarded.
3. For each question in Section II, you will be awarded 4 Marks if you give the correct answer and
zero
Mark if no answer is given. In all other cases, minus one (–1) Mark will be awarded.
4. For each question in Section–III, you will be awarded 2 Marks for each row in which you give the correct answer(s). Thus, each question in this section carries a maximum of 8 Marks. There is no negative marking for incorrect answer(s) for this section.
5. For each question in Section IV, you will be awarded 4 Marks if you give the correct answer and
zero
Mark if no answer is given. In all other cases, minus one (–1) Mark will be awarded.
SECTION - I
Straight Objective Type
This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct.
1. Photoelectric effect experiments are performed using three different metal plates p, q and r having work
functions p = 2.0 eV, q = 2.5 eV and r = 3.0 eV respectively. A light beam containing wavelengths of 500 nm, 400 nm and 310 nm with equal intensities illuminates each of the plates. The correct -V graph for the experiment is [Take hc = 1240 eV nm]
(A) (B)
(C) (D)
2. A rod of mass M and length L is pivoted at a point O which divides the length in ratio 1 : 2. The two ends are connected to two springs of spring constants shown. The rod is free to oscillate in horizontal plane about
O. The rod is gently pushed through a small angle in one direction and
released. The frequency of oscillation is :
(A) (B)
(C) (D)
3. A piece of wire is bent in the shape of a parabola y = kx2 (y-axis vertical) with a bead of mass m on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now rotated about the y-axis with constant angular velocity . The distance of the new equilibrium position of the bead, where the bead can stay at rest with respect to the wire, from the y-axis is
(A)
2
gk2
(B)
gk2
2
2
(C) g
(D) any value.
4. A pointer is attached to a spring of spring constant k at point P which divides the spring in the ratio n1 : n2. The mass M shown in the figure oscillates in simple harmonic motion with amplitude A. The amplitude of the point P is :
(A)
n1 .A
n2
(B) n2 .A
1
n1A n1 n2
n2A n1 n2
SECTION - II
Multiple Correct Answer Type
This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) out of which ONE OR MORE may be correct.
5. Under the influence of the Coulomb field of charge +Q, a charge –q is moving around it in an elliptical orbit. Find out the correct statement(s).
(A) Torque on–q about Q is zero
(B) Angular momentum of –q about Q is zero
(C) Angular momentum of –q about Q is constant
(D) The magnitude of linear momentum of –q is constant.
6. The figure shows the P – V plot of an ideal gas taken through a cycle ADCBA. The part ADC is semi circle and part CBA is rectangular hyper- bolic. D is the state of maximum volume and B and D are at same pres sure. Then
(A) the heat flows out in the process CBA.
(B) the heat flows into the system in the process ADC.
(C) the work done by the gas in the process DCB is negative.
(D) the temperatures of states A and C are same.
7. A sphere is rolling without slipping on a fixed horizontal plane surface. In the figure, A is the point of contact, B is the centre of the sphere and C is its topmost point D and E are points on same horizontal line passing through B. Sphere Mmoves with speed v. Then,
→ → → → → → → →
(A)
VD VA = VC VE
(B) VC VD = VE VB
→ → → →
→ → → →
(C)
VE VD = VC VA
(D) VE VB = VB VD
8. A student performed the experiment to measure the speed of sound in air using resonance air-column method. Two resonances in the air-column were obtained by lowering the water level. The resonance with the shorter air-column is the first resonance and that with the longer air-column is the second resonance. Then,
(A) The displacement antinode at both the resonances is inside the tube.
(B) The longer air column length is nearly three times the shorter air column length.
(C) The frequency with which resonance occurs in the second case is three times that in the first case.
(D) Displacement node is observed at the water surface.
9. A metallic ring is kept on a solenoid as shown. When a current is switched on in the solenoid the ring jumps vertically upto height h. Then,
(A) The ring gets thrust only due to radial component of magnetic field of the solenoid.
(B) Smaller is the resistivity of the ring, more is the height h jumped.
(C) More is the mass of the ring, less is the height jumped.
(D) More is the cross section area of the solenoid, more is the height jumped.
SECTION - III
Matrix - Match Type
This section contains 2 questions. Each question contains statements given in two columns, which have to be matched. The statements in Column-I are labelled A, B, C and D, while the statements in Column-II are labelled p, q, r, s and t. Any given statement in Coloumn-I can have correct matching with ONE OR MORE statement(s) in Coloumn-II. The answers to these questions have to be appropriately marked as illustrated in the following example. If the correct matches are A-p, A-r, B-p, B-s, C-r, C-s and D-q, D-t then the answer should be written as : A p,r ; B p, s ; C r, s ; D q, t.
10. Column I shows four situations of standard Young’s double slit arrangement with the screen placed far away
from the slits S and S . In each of these cases S P
= S P , S P – S P = /6 and S P
– S P
= 3/4, where
1 2 1 0
2 0 1 1 2 1
1 2 2 2
is the wavelength of the light used. In the cases B,C and D, a transparent sheet of refractive index and thickness t is pasted on slit S2. The thicknesses of the sheets are different in different cases. The intensity of light coming out of the each slit is . The phase difference between the light waves reaching a point P on the screen from the two slits is denoted by (P) and the intensity by (P). Match each situation given in Column-I with the statement(s) in Column-II valid for that situation.
Column–I Column–II
(A)
(p) (P ) =
0
(B)
( – 1)t = /4
(q)
(P ) = 0
2
(C)
( – 1)t = /2
(r)
(P ) = 2
0 0
(D)
( – 1)t = 3/4
(s)
(t)
(P ) > (P )
0 1
(P ) > (P )
2 1
11. Column II gives certain systems undergoing a process. Column I suggests changes in some of the parameters related to the system. Match the statements in Column-I to the appropriate process(es) from Column II.
Column–I Column–II
(A) The energy of the system is decreased . (p) System: An inductor with resistor in series.
Process: It is connected to a battery.
(B) No mechanical energy is provided to the , (q) System: A gas in perfectly conducting container system fitted with a conducting moveable piston.
Process: The container is placed in atmosphere at lower temperature.
(C) Energy is released from the system at the (r) System: A gas in an adiabatic container. expense of decrease of internal energy Process: The gas under goes free expansion. of the system
(D) Mass of the system is decreased (s) System: Two light nuclei moving in opposite
directions with same momentum.
Process: Two nuclei fuses. The reaction is exoergic
(t) System: A non-resistive wire loop
Process: The loop is placed in a time varying magnetic field perpendicular to its plane
SECTION - IV
Integer Answer Type
This section contains 8 questions. The answer to each of the questions is a single digit integer, ranging from 0 to 9.
12. A steady current = 10 A goes through a wire loop PQRS. The part PQR is semicircle of radius r = 1.0 m. RS = SP and RSP = 90º. Find the magnetic field at O in 10–6 T in nearest integer. Take = 4 × 10–7 tesla meter
ampere–1.
13. A light inextensible string that goes over a smooth fixed pulley as shown in the figure connects two blocks of masses 0.06 kg and 0.12 kg. Taking g = 10 m/s2, find the negative of the work done (in joules) by the earth on the block of mass 0.06 kg during the fifth second after the system is released
from rest.
14. A Long solid cylinder of radius R has a charge Q distributed in its volume with a charge density = kra, where
k and a are constants and r is the distance from its axis. If the electric field at r = r = R, find the value of a.
R is
2
1
4 times that at
15. A metal rod AB of length 9.1 m has its one end A in ice at 0ºC and the other end B in water at 100ºC. If a point P on the rod is maintained at 600ºC, then it is found that equal amounts of water and ice evaporate and melt per unit time. The latent heat of evaporation of water is 540 cal/g and latent heat of melting of ice is 80 cal/g.
If the thermal conductivity K and cross section area S of the rod be such that K.S = 3.24 cal-m/s/kelvin, find the mass of (in g) ice melting per second.[ Neglect any heat loss to the surrounding ]
16. Two soap bubbles A containing Ne gas and B containing O2 gas are kept in a closed chamber where the air is maintained at pressure 6 N/m2. The radii of bubbles A and B are 6 cm and 3 cm, respectively. Surface tension of the soap-water used to make bubbles is 0.03 N/m. Find the ratio mB/mA, where mA
and mB are the masses of gases in bubbles A and B, respectively. [Neglect the effect of gravity.] Molar
mass of Ne = 20 and that of O2 = 32.
17. A 50 cm long string, having a mass of 2.5g, is fixed at one end The other end is connected with a vibrator which oscillates that end normal to the string with frequency 50 Hz. The tension in the string is 2 N. Find the number of nodes excluding at the ends if any.
18. Three objects A, B, and C are kept in a straight line on a frictionless horizontal surface. They have masses 2m, m and 2m respectively. The object A moves towards B with a velocity vo and makes an
elastic collision with it. All collisions are elastic. All motions
occur on the same straight line. Find the number of collisions.
19. A vessel of height H = 0.5 m is filled with water up to top and sealed. It has one inlet at the top to pass an ideal gas and at the bottom a small orifice for outlet. The gas is let in at a rate such that the rate of flow of water from the orifice is constant. Find the number of moles of the gas in per unit m3 of the vessel when water just flows out completely.
Take atmospheric pressure = 1.0 × 105 N/m2, density of water = 1000 Kg/m3, g = 10 m/s2 universal gas
25
constant, R = 3 J/mol/kelvin, Temperature of the gas = 300 kelvin,
A nswers
PAPER - 1
1. (A) 2. (C) 3. (A) 4. (B) 5. (B)
6. (B)
(D) 7. (A) 8. (D) 9. (A), (D) 10. ( C ) ,
11. (C) 12. (A), (D) 13. (C) 14. (D) 15. (C)
16. (A) 17. (C) 18. (C)
19. (A) (r), (s); (B) (p), (r), (s); (C) (q), (t); (D) (p), (r), (s)
20. (A) (p), (r), (s), (t); (B) (p), (s), (t); (C) (q), (t) ; (D) (q), (r).
PAPER - 2
1. (A) 2. (C) 3. (D) 4. (C) 5. (A,C)
6. (B), (C), (D) 7. (A), (C), (D) 8. (B), (D) 9. (A), (B), (C), (D)
10. (A) s; (B) r; (C) p, t ; (D) q, r, s, t 11. (A) q, s (B) p, r, s, t (C) q, s (D) s 12. 7
13.
9
14.
1
15.
3
16.
4
17.
2
18.
3
19.
2
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