IITJEE-2012-Paper 2-PCM-2
PAPER-2 [Code - 8]
IITJEE 2012

PART I : PHYSICS

SECTION I : Single Correct Answer Type
This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

Two identical discs of same radius R are rotating about their axes in opposite directions with the same constant angular speed ω. The discs are in the same horizontal plane. At time t = 0, the points P and Q are facing each other as shown in the figure. The relative speed between the two points P and Q is vr. In one time period (T) of rotation of the discs, vr as a function of time is best represented by

[Diagram: Two adjacent circular discs of radius R in horizontal plane; left disc rotates clockwise with angular speed ω with point P on rim, right disc rotates counter-clockwise with angular speed ω with point Q on rim facing P at t=0.]

(A) [Graph: vr vs t oscillating smoothly from 0 with period T, zero at t=0, T/2, T]
(B) [Graph: vr vs t starting at non-zero value, dropping to zero at T/2]
(C) [Graph: vr vs t with higher frequency oscillations]
(D) [Graph: vr vs t with multiple peaks and non-zero start]

Sol. (A)
In each rotation relative speed becomes zero twice and becomes maximum twice.

A loop carrying current I lies in the x-y plane as shown in the figure. The unit vector k̂ is coming out of the plane of the paper. The magnetic moment of the current loop is

[Diagram: Coordinate axes x and y. A closed loop carrying counter-clockwise current I formed by a square of side a in the third quadrant and three semicircles of radius a/2 outwards on the other three sides along positive axes.]

(A) a²I k̂
(B) (π/2 + 1)a²I k̂
(C) -(π/2 + 1)a²I k̂
(D) (2π + 1)a²I k̂

Sol. (B)
Magnetic moment, M̄ = I Ā = I (π/2 + 1)a² k̂

An infinitely long hollow conducting cylinder with inner radius R/2 and outer radius R carries a uniform current density along its length. The magnitude of the magnetic field, |B̄| as a function of the radial distance r from the axis is best represented by

(A) [Graph: |B| is 0 from 0 to R/2, rises linearly to R, then falls as 1/r for r > R]
(B) [Graph: |B| is constant from 0 to R/2, then non-zero]
(C) [Graph: |B| rises linearly from origin to R]
(D) [Graph: |B| is 0 for r < R/2, rises non-linearly/curved from R/2 to R, and decreases as 1/r for r > R]

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Sol. (D)
[Diagram: Cross-section of cylinder showing hollow region of inner radius a = R/2 and outer radius b = R with shaded current-carrying cross-section.]
Inside the cavity, B = 0
Outside the cylinder,
B = μ₀I / (2πr)
In the shaded region
B = (μ₀I / (2πr(b² - a²))) * (r - a²/r)
at r = a, B = 0
at r = b, B = μ₀I / (2πb)

A thin uniform cylindrical shell, closed at both ends, is partially filled with water. It is floating vertically in water in half-submerged state. If ρc is the relative density of the material of the shell with respect to water, then the correct statement is that the shell is
(A) more than half-filled if ρc is less than 0.5.
(B) more than half-filled if ρc is more than 1.0.
(C) half-filled if ρc is more than 0.5.
(D) less than half-filled if ρc is less than 0.5.

Sol. (A)
((Vw + Va + Vm) / 2) ρw g = Vm ρc ρw g + Vw ρw g
Va = Vm(2ρc - 1) + Vw
Vw = Vm(1 - 2ρc) + Va
if ρc > 1/2 ⇒ Vw < Va
if ρc < 1/2 ⇒ Vw > Va
where, Vw = volume occupied by water in the shell
Va = volume occupied by air in the shell
Vm = volume of the material in the shell

In the given circuit, a charge of +80 μC is given to the upper plate of the 4 μF capacitor. Then in the steady state, the charge on the upper plate of the 3 μF capacitor is

[Diagram: Parallel capacitor network. A 4 μF capacitor with +80 μC on top plate connected in parallel across a branch containing a 2 μF capacitor and a 3 μF capacitor in series/parallel combination.]

(A) +32 μC
(B) +40 μC
(C) +48 μC
(D) +80 μC

Sol. (C)
Let 'q' be the final charge on 3 μF capacitor then
(80 - q)/2 = q/3 ⇒ q = 48 μC

Two moles of ideal helium gas are in a rubber balloon at 30°C. The balloon is fully expandable and can be assumed to require no energy in its expansion. The temperature of the gas in the balloon is slowly changed to 35°C. The amount of heat required in raising the temperature is nearly (take R = 8.31 J/mol.K)
(A) 62 J
(B) 104 J
(C) 124 J
(D) 208 J

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Sol. (D)
ΔQ = n Cp ΔT (Isobaric process)
= 2 × (5/2)R × (35 - 30)
= 208 J

Consider a disc rotating in the horizontal plane with a constant angular speed ω about its centre O. The disc has a shaded region on one side of the diameter and an unshaded region on the other side as shown in the figure. When the disc is in the orientation as shown, two pebbles P and Q are simultaneously projected at an angle towards R. The velocity of projection is in the y-z plane and is same for both pebbles with respect to the disc. Assume that (i) they land back on the disc before the disc has completed 1/8 rotation, (ii) their range is less than half the disc radius, and (iii) ω remains constant throughout. Then

[Diagram: Disc in x-y plane centered at O with radius along y-axis connecting diametrically opposite points P, Q, O, R. Left half is unshaded, right half is shaded. Angular velocity ω indicated counter-clockwise about z-axis.]

(A) P lands in the shaded region and Q in the unshaded region.
(B) P lands in the unshaded region and Q in the shaded region.
(C) Both P and Q land in the unshaded region.
(D) Both P and Q land in the shaded region.

Sol. (C)
At t = (1/8) × (2π/ω) = π/(4ω)
x-coordinate of P = ω R (π/(4ω)) = πR/4 > R cos 45°
∴ Both particles P and Q land in unshaded region.
[Diagram: Geometry showing rotated disc at angle π/4 and landing positions of P and Q.]

A student is performing the experiment of resonance Column. The diameter of the column tube is 4 cm. The frequency of the tuning fork is 512 Hz. The air temperature is 38°C in which the speed of sound is 336 m/s. The zero of the meter scale coincides with the top end of the Resonance Column tube. When the first resonance occurs, the reading of the water level in the column is
(A) 14.0 cm
(B) 15.2 cm
(C) 16.4 cm
(D) 17.6 cm

Sol. (B)
L + e = λ/4
⇒ L = λ/4 - e
= 16.4 - 1.2 = 15.2 cm

IITJEE-2012-Paper 2-PCM-5

SECTION II : Paragraph Type
This section contains 6 multiple choice questions relating to three paragraphs with two questions on each paragraph. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.

Paragraph for Questions 9 and 10
The general motion of a rigid body can be considered to be a combination of (i) a motion of its centre of mass about an axis, and (ii) its motion about an instantaneous axis passing through the centre of mass. These axes need not be stationary. Consider, for example, a thin uniform disc welded (rigidly fixed) horizontally at its rim to a massless stick, as shown in the figure. When the disc-stick system is rotated about the origin on a horizontal frictionless plane with angular speed ω, the motion at any instant can be taken as a combination of (i) a rotation of the centre of mass of the disc about the z-axis, and (ii) a rotation of the disc through an instantaneous vertical axis passing through its centre of mass (as is seen from the changed orientation of points P and Q). Both these motions have the same angular speed ω in this case.

[Diagram: A horizontal disc with diameter PQ attached to a rod along y-axis rotating about vertical z-axis with angular velocity ω.]

Now consider two similar systems as shown in the figure: Case (a) the disc with its face vertical and parallel to x-z plane; Case (b) the disc with its face making an angle of 45° with x-y plane and its horizontal diameter parallel to x-axis. In both the cases, the disc is welded at point P, and the systems are rotated with constant angular speed ω about the z-axis.

[Diagram Case (a): Disc vertical, parallel to x-z plane, welded at P on rod along y-axis, rotating about z-axis at ω.]
[Diagram Case (b): Disc tilted at 45° to horizontal x-y plane, welded at P, rotating about z-axis at ω.]

Which of the following statements about the instantaneous axis (passing through the centre of mass) is correct?
(A) It is vertical for both the cases (a) and (b).
(B) It is vertical for case (a); and is at 45° to the x-z plane and lies in the plane of the disc for case (b).
(C) It is horizontal for case (a); and is at 45° to the x-z plane and is normal to the plane of the disc for case (b).
(D) It is vertical for case (a); and is 45° to the x-z plane and is normal to the plane of the disc for case (b).

Sol. (A)

Which of the following statements regarding the angular speed about the instantaneous axis (passing through the centre of mass) is correct?
(A) It is √2 ω for both the cases.
(B) It is ω for case (a); and ω/√2 for case (b).
(C) It is ω for case (a); and √2 ω for case (b).
(D) It is ω for both the cases.

Sol. (D)

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Paragraph for Questions 11 and 12
The β-decay process, discovered around 1900, is basically the decay of a neutron (n). In the laboratory, a proton (p) and an electron (e⁻) are observed as the decay products of the neutron. Therefore, considering the decay of a neutron as a two-body decay process, it was predicted theoretically that the kinetic energy of the electron should be a constant. But experimentally, it was observed that the electron kinetic energy has continuous spectrum. Considering a three-body decay process, i.e. n → p + e⁻ + ν̄ₑ, around 1930, Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino (ν̄ₑ) to be massless and possessing negligible energy, and the neutron to be at rest, momentum and energy conservation principles are applied. From this calculation, the maximum kinetic energy of the electron is 0.8 × 10⁶ eV. The kinetic energy carried by the proton is only the recoil energy.

If the anti-neutrino had a mass of 3 eV/c² (where c is the speed of light) instead of zero mass, what should be the range of the kinetic energy, K, of the electron?
(A) 0 ≤ K ≤ 0.8 × 10⁶ eV
(B) 3.0 eV ≤ K ≤ 0.8 × 10⁶ eV
(C) 3.0 eV ≤ K < 0.8 × 10⁶ eV
(D) 0 ≤ K < 0.8 × 10⁶ eV

Sol. (D)

What is the maximum energy of the anti-neutrino?
(A) Zero
(B) Much less than 0.8 × 10⁶ eV
(C) Nearly 0.8 × 10⁶ eV
(D) Much larger than 0.8 × 10⁶ eV

Sol. (C)

Paragraph for Questions 13 and 14
Most materials have the refractive index, n > 1. So, when a light ray from air enters a naturally occurring material, then by Snell's law, sin θ₁ / sin θ₂ = n₂ / n₁, it is understood that the refracted ray bends towards the normal. But it never emerges on the same side of the normal as the incident ray. According to electromagnetism, the refractive index of the medium is given by the relation, n = (c/v) = ±√(εᵣ μᵣ), where c is the speed of electromagnetic waves in vacuum, v its speed in the medium, εᵣ and μᵣ are the relative permittivity and permeability of the medium respectively.
In normal materials, both εᵣ and μᵣ are positive, implying positive n for the medium. When both εᵣ and μᵣ are negative, one must choose the negative root of n. Such negative refractive index materials can now be artificially prepared and are called meta-materials. They exhibit significantly different optical behavior, without violating any physical laws. Since n is negative, it results in a change in the direction of propagation of the refracted light. However, similar to normal materials, the frequency of light remains unchanged upon refraction even in meta-materials.

For light incident from air on a meta-material, the appropriate ray diagram is

(A) [Diagram: Ray enters meta-material and bends normally away across the boundary into the opposite side of normal]
(B) [Diagram: Ray reflects back into air]

IITJEE-2012-Paper 2-PCM-7

(C) [Diagram: Incident ray with angle θ₁ in Air reaches interface; refracted ray in Meta-material is on the same side of the normal with angle θ₂]
(D) [Diagram: Refracted ray continues straight without deflection]

Sol. (C)

Choose the correct statement.
(A) The speed of light in the meta-material is v = c|n|
(B) The speed of light in the meta-material is v = c / |n|
(C) The speed of light in the meta-material is v = c.
(D) The wavelength of the light in the meta-material (λₘ) is given by λₘ = λ_air |n|, where λ_air is wavelength of the light in air.

Sol. (B)

SECTION III : Multiple Correct Answer(s) Type
This section contains 6 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct.

In the given circuit, the AC source has ω = 100 rad/s. Considering the inductor and capacitor to be ideal, the correct choice(s) is (are)

[Diagram: AC source of 20 V connected across two parallel branches; top branch has 100 μF capacitor in series with 100 Ω resistor; bottom branch has 0.5 H inductor in series with 50 Ω resistor.]

(A) The current through the circuit, I is 0.3 A.
(B) The current through the circuit, I is 0.3√2 A.
(C) The voltage across 100 Ω resistor = 10√2 V.
(D) The voltage across 50 Ω resistor = 10 V.

Sol. (A, C)
I_upper = 20 / (100√2) ; +π/4 ahead of voltage
I_lower = 20 / (50√2) ; -π/4 behind voltage
I = √(I₁² + I₂²) = √(1/10) ≈ 0.3 A
V_100 Ω = (20 / (100√2)) × 100 = 10√2 V
[Phasor diagrams: Upper branch capacitive with impedance 100√2 at π/4; lower branch inductive with impedance 50√2 at π/4.]

IITJEE-2012-Paper 2-PCM-8

Six point charges are kept at the vertices of a regular hexagon of side L and centre O, as shown in the figure. Given that K = (1 / (4πε₀)) * (q / L²), which of the following statement(s) is (are) correct?

[Diagram: Regular hexagon with vertices labeled counter-clockwise: F(+q), E(+q), D(-2q), C(-q), B(+q), A(-q) or with charges +q, -q, +2q, -2q, +q, -q along the perimeter, lines PR and ST passing through center O.]

(A) The electric field at O is 6K along OD.
(B) The potential at O is zero.
(C) The potential at all points on the line PR is same.
(D) The potential at all points on the line ST is same.

Sol. (A, B, C)
Line PR is perpendicular bisector of all the dipoles.
[Diagram: Vector addition of electric field components at O giving resultant 6K along OD.]
At point O : (1 / (4πε₀)) ∑ (Qᵢ / rᵢ) = 0

Two spherical planets P and Q have the same uniform density ρ, masses M_P and M_Q and surface areas A and 4A respectively. A spherical planet R also has uniform density ρ and its mass is (M_P + M_Q). The escape velocities from the planets P, Q and R are V_P, V_Q and V_R, respectively. Then
(A) V_Q > V_R > V_P
(B) V_R > V_Q > V_P
(C) V_R / V_P = 3
(D) V_P / V_Q = 1/2

Sol. (B, D)
By calculation, if Mass of P = M and Radius of P = R
Then Mass of Q = 8M and radius of Q = 2R
and Mass of R = 9M and radius of R = 9^(1/3) R
V_P = √(2GM / R)
V_Q = √(2G(8M) / (2R)) = 2 V_P
V_R = √(2G(9M) / (9^(1/3) R)) = 9^(1/3) V_P
V_R > V_Q > V_P
V_P / V_Q = 1/2

IITJEE-2012-Paper 2-PCM-9

The figure shows a system consisting of (i) a ring of outer radius 3R rolling clockwise without slipping on a horizontal surface with angular speed ω and (ii) an inner disc of radius 2R rotating anti-clockwise with angular speed ω/2. The ring and disc are separated by frictionless ball bearings. The point P on the inner disc is at a distance R from the origin, where OP makes an angle of 30° with the horizontal. Then with respect to the horizontal surface,

[Diagram: Outer ring of radius 3R rolling on ground at ω clockwise; inner concentric disc of radius 2R rotating at ω/2 counter-clockwise with bearings in between; point P at distance R from O at angle 30°.]

(A) the point O has linear velocity 3Rω î
(B) the point P has linear velocity (11/4)Rω î + (√3/4)Rω k̂
(C) the point P has linear velocity (13/4)Rω î - (√3/4)Rω k̂
(D) the point P has linear velocity (3 - √3/4)Rω î + (1/4)Rω k̂

Sol. (A, B)
v̄_0 - (3R)ω î = 0
∴ v̄_0 = 3Rω î
v̄_P,0 = -(Rω/4) î + (Rω√3/4) ĵ
∴ v̄_P = v̄_P,0 + v̄_0 = (11/4)Rω î + (√3/4)Rω ĵ
[Diagram: Velocity vector diagram for point P at angle 30° with tangential velocity Rω/2.]

Two solid cylinders P and Q of same mass and same radius start rolling down a fixed inclined plane from the same height at the same time. Cylinder P has most of its mass concentrated near its surface, while Q has most of its mass concentrated near the axis. Which statement(s) is (are) correct?
(A) Both cylinders P and Q reach the ground at the same time.
(B) Cylinders P has larger linear acceleration than cylinder Q.
(C) Both cylinders reach the ground with same translational kinetic energy.
(D) Cylinder Q reaches the ground with larger angular speed.

Sol. (D)
a = (Mg sin θ) / (M + I/R²)

IITJEE-2012-Paper 2-PCM-10

a_P = (Mg sin θ) / (M + MR²/R²) = g/2
a_Q ≈ g sin θ as I_Q ~ 0
∴ ω_P = √(2 * (g/2) * l) / R
ω_Q = √(2 * g * l) / R
∴ ω_Q > ω_P

A current carrying infinitely long wire is kept along the diameter of a circular wire loop, without touching it, the correct statement(s) is(are)
(A) The emf induced in the loop is zero if the current is constant.
(B) The emf induced in the loop is finite if the current is constant.
(C) The emf induced in the loop is zero if the current decreases at a steady rate.
(D) The emf induced in the loop is infinite if the current decreases at a steady rate.

Sol. (A, C)
ϕ = zero
∴ dϕ/dt = zero
∴ A, C are correct.
[Diagram: Circular wire loop with a straight vertical wire carrying current I along its diameter, showing magnetic field into the plane ⊗ on one side and out of the plane ⊙ on the other side.]

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