1. Distance of the centre of mass of a solid uniform cone from its vertex is z0 . If the radius of its base is R and its height is h then z0 is equal to
(1) 5h/8
(2) 3h^2/8R
(3) h^2/4R
(4) 3h/4
Answer (4)
Sol. dm = πr^2 dyφ
y_CM = ∫ y dm / ∫ dm = ∫ πr^2 dy × ρ × y / (1/3 πR^2 hρ)
= 3h/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 1 m from the diode is
(1) 5.48 V/m
(2) 7.75 V/m
(3) 1.73 V/m
(4) 2.45 V/m
Answer (4)
Sol. I = P/(4πr^2) = U_av × c ...(1)
U_av = 1/2 ε0 E0^2 ...(2)
⇒ P/(4πr^2) = 1/2 ε0 E0^2 × c
⇒ E0 = √(2P/(4πr^2 ε0 c)) = 2.45 V/m
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, the time period changes to TM If the Young's modulus of the material of the wire is Y then 1/Y is equal to
(g = gravitational acceleration)
(1) [1 − (TM/T)^2] A/(Mg)
(2) [1 − (T/TM)^2] A/(Mg)
(3) [(TM/T)^2 − 1] A/(Mg)
(4) [(TM/T)^2 − 1] Mg/A
Answer (3)
Sol. T = 2π√(l/g)
TM = 2π√((l+Δl)/g)
Y = Fl/(AΔl) ⇒ Δl = Mgl/(AY)
⇒ 1/Y = A/(Mg)[(TM/T)^2 − 1]
4. 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)
Answer (4)
Sol. KE = 1/2 mω^2(A^2 − d^2)
PE = 1/2 mω^2 d^2
At d = ±A
PE = maximum while KE = 0.
5. A train is moving on a straight track with speed 20 ms^−1 . It is blowing its whistle at the frequency of 1000 Hz . The percentage 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%
Answer (4)
Sol. f1 = f[v/(v − vs)] = f[320/(320 − 20)] = f × 320/300 Hz
f2 = f[v/(v + vs)] = f × 320/340 Hz
100 × (f2/f1 − 1) = ((f2 − f1)/f1) × 100
100 × [300/340 − 1] = 12%
6. When 5 V potential difference is applied across a wire of length 0.1 m , the drift speed of electrons is 2.5 × 10^−4 ms^−1 . If the electron density in the wire is 8 × 10^28 m^−3 , the resistivity of the material is close to
(1) 1.6 × 10^−6 Ωm
(2) 1.6 × 10^−5 Ωm
(3) 1.6 × 10^−8 Ωm
(4) 1.6 × 10^−7 Ωm
Answer (2)
Sol. V = IR = Iρ l/A
⇒ ρ = VA/(Il) = VA/(IneAvd) = V/(I × n × e × vd)
⇒ ρ = 5/(0.1 × 2.5 × 10^−19 × 1.6 × 10^−19 × 8 × 10^28)
= 1.6 × 10^−5 Ωm
7. 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 θ with the vertical. If wires have mass λ per unit length then the value of I is (g = gravitational acceleration)
Answer (4)
Sol. T cosθ = λgl ...(1)
T sinθ = μ0/(2π) × I × Il/(2L sinθ) ...(2)
⇒ I = 2sinθ √(πλgL/(μ0 cosθ))
8. In the circuit shown, the current in the 1 Ω 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) 0 A
Sol. From KVL, 9 = 6I1 − I2 ...(1)
6 = 4I2 − I1 ...(2)
Solving, I1 − I2 = −0.13 A
9. Assuming human pupil to have a radius of 0.25 cm and a comfortable viewing distance of 25 cm the minimum separation between two objects that human eye can resolve at 500 nm wavelength is
(1) 100 μm
(2) 300 μm
(3) 1 μm
(4) 30 μm
Answer (4)
Sol. RP = 1.22λ/(2μ sinθ) = 1.22 × (500 × 10^−9 m)/(2 × 1 × (1/100))
= 3.05 × 10^−5 m
10. An inductor (L = 0.03 H) and a resistor (R = 0.15 kΩ) are connected in series to a battery of 15 V EMF in a circuit shown below. The key K1 has been kept closed for a long time. Then at t = 0 K1 is opened and key K2 is closed simultaneously. At t = 1 ms the current in the circuit will be (e^5 ≡ 150)
(1) 6.7 mA
(2) 0.67 mA
(3) 100 mA
(4) 67 mA
Answer (2)
Sol. I = I0 e^(t/τ), τ = L/R
= 15/150 e^(1 × 10^−3 / (1/5 × 10^3)) = 0.67 mA
11. An LCR circuit is equivalent to a damped pendulum. In an LCR circuit the capacitor is charged to Q0 and then connected to the L and R as shown below :
If a student plots graphs of the square of maximum charge (Q_Max^2) on the capacitor with time (t) for two different values L1 and L2 (L1 > L2) of L then which of the following represents this graph correctly? (Plots are schematic and not drawn to scale)
Answer (3)
Sol. For a damped pendulum, A = A0 e^−bt/2m
⇒ A = A0 e^−(R/2L)t
(Since L plays the same role as m )
12. In the given circuit, charge Q2 on the 2 μF capacitor changes as C is varied from 1 μF to 3 μF . Q2 as a function of C is given properly by : (Figures are drawn schematically and are not to scale)
Answer (4)
Sol. Caq = 3C/(3 + C) ...(i)
Total charges q = (3C/(3 + C))E ...(ii)
Charge upon capacitor 2 μF,
q′ = 2/3 × 3CE/(3 + C) = 2CE/(3 + C) = 2E/(1 + 3/C)
Now, dQ/dC > 0, d^2Q/dC^2 < 0
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 perpendicular to one of its faces is
(1) 4MR^2/(9√3π)
(2) 4MR^2/(3√3π)
(3) MR^2/(32√2π)
(4) 4MR^2/(9√3π)
Answer (1)
Sol. d = 2R = a√3
a = 2/√3 R
M = (4/3 πR^3)/((2/√3 R)^3) = √3/2 π
M′ = 2M/(√3π)
I = M′a^2/6 = 2M/(√3π) × 4/3 R^2 × 1/6
I = 4MR^2/(9√3π)
14. The period of oscillation of a simple pendulum is T = 2π√(L/g) . Measured value of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1 s resolution. The accuracy in the determination of g is
(1) 1%
(2) 5%
(3) 2%
(4) 3%
Answer (4)
Sol. g = 4π^2 · l/T^2
15. On a hot summer night, the refractive index of air is smallest near the ground and increases with height form the ground. When a light beam is directed horizontally, the Huygen's principle leads us to conclude that as it travels, the light beam
(1) Bends downwards
(2) Bends upwards
(3) Becomes narrower
(4) Goes horizontally without any deflection
Answer (2)
Sol. Consider a plane wavefront travelling horizontally. As it moves, its different parts move with different speeds. So, its shape will change as shown
⇒ Light bends upward
16. A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz . The frequencies of the resultant signal is/are
(1) 2005 kHz , 2000 kHz and 1995 kHz
(2) 2000 kHz and 1995 kHz
(3) 2 MHz only
(4) 2005 kHz and 1995 kHz
Answer (1)
Sol. Frequencies of resultant signal are
fe + fm, fe and fe − fm
(2000 + 5)kHz, 2000kHz, (2000 − 5)kHz,
2005kHz, 2000kHz, 1995kHz
17. A solid body of constant heat capacity 1 J/°C is being heated by keeping it in contact with reservoirs in two ways :
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same amount of heat.
(ii) Sequentially keeping in contact with 8 reservoirs such that 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) 2ln, 2ln2
(2) 2ln, 2ln2
(3) 2ln, 4ln2
(4) 2ln, 2ln2
Answer (None)
Sol. ds′ = dQ/T = ms dT/T
Δs′ = ∫ds′ = ms∫dT/T = 1 log_e(T2/T1) = log_e(473/373)
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 with internal energy per unit volume u = U/V ∝ T^4 and pressure P = 1/3 (U/V) . If the shell now undergoes an adiabatic expansion the relation between T and R is
(1) T ∝ 1/R
(2) T ∝ 1/R^3
(3) T ∝ e^−R
(4) T ∝ e^−3R
Answer (1)
Sol. P = 1/3 (U/V) = 1/3 kT^4 ...(i)
PV = μRT ...(ii)
μRT/V = 1/3 kT^4
⇒ V ∝ T^−3
R ∝ 1/T
19. Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial speed of 10 m/s and 40 m/s respectively. Which of the following graph best represents the time variation of relative position of the second stone with respect to the first? (Assume stones do not rebound after hitting the ground and neglect air resistance, take g = 10 m/s^2) (The figures are schematic and not drawn to scale)
Answer (2)
Sol. Till both are in air (From t = 0 to t = 8 sec)
Δx = x2 − x1 = 30t
⇒ Δx ∝ t
When second stone hits ground and first stone is in air Δx decreases.
20. A uniformly charged solid sphere of radius R has potential V0 (measured with respect to ∞ ) on its surface. For this sphere the equipotential surfaces with potentials 3V0/2, 5V0/4, 3V0/4 and V0/4 have radius R1, R2, R3 and R4 respectively. Then
(1) R1 = 0 and R2 < (R4 − R3)
(2) 2R < R4
(3) R1 = 0 and R2 > (R4 − R3)
(4) R1 ≠ 0 and (R2 − R1) > (R4 − R3)
Answer (1, 2)
Sol. Vol = kQ/R
Vol = kQ/(2R^3)(3R^2 − r^2)
Vol = 3/2 V0 ⇒ R1 = 0
Vol = 5kQ/(4R) = kQ((3R^2 − r^2)/(2R^3))
⇒ R2 = R/√2
Vol = 3kQ/(4R) = kQ/R^3
⇒ R3 = 4R/3
Vol = 1kQ/(4R) = kQ/R4
⇒ R4 = 4R ⇒ R4 > 2R
21. Monochromatic light is incident on a glass prism of angle A . If the refractive index of the material of the prism is μ , a ray, incident at an angle θ , on the face AB would get transmitted through the face AC of the prism provided.
(1) θ > cos^−1[μ sin(A + sin^−1(1/μ))]
(2) θ < cos^−1[μ sin(A + sin^−1(1/μ))]
(3) θ > sin^−1[μ sin(A − sin^−1(1/μ))]
(4) θ < sin^−1[μ sin(A − sin^−1(1/μ))]
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:
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?
(1) (b) and (d), respectively
(2) (b) and (c), respectively
(3) (a) and (b), respectively
(4) (a) and (c), respectively
Answer (1)
Stable equilibrium M || B
Unstable equilibrium M || (−B)
23. Two coaxial solenoids of different radii carry current I in the same direction. Let F1 be the magnetic force on the inner solenoid due to the outer one and F2 be the magnetic force on the outer solenoid due to the inner one. Then
(1) F1 is radially inwards and F2 = 0
(2) F1 is radially outwards and F2 = 0
(3) F1 = F2 = 0
(4) F1 is radially inwards and F2 is radially outwards
Answer (3)
Sol. Net force on each of them would be zero.
24. A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving in the y direction with speed v . If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to
(1) 56%
(2) 62%
(3) 44%
(4) 50%
Answer (1)
Sol. m → 2v
= 2mv√2/(3m) = v′
KE loss = 1/2 m(2v)^2 + 1/2 (2m)v^2
− 1/2 × (3m)(2mv√2/(3m))^2 = 5/3 mv^2
Required% = (5/3 mv^2)/(2mv^2 + mv^2) × 100 = 56%
25. Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion, the average time of collision between molecules increases as V^q , where V is the volume of the gas. The value of q is
(γ = Cp/Cv)
(1) (γ + 1)/2
(2) (γ − 1)/2
(3) (3γ + 5)/6
(4) (γ − 1)/2
Answer (1)
Sol. τ = λ/v_rms = 1/(√2πd^2(N/V)√(3RT/M))
τ ∝ V/√T ...(iii)
T V^(γ−1) = k
⇒ τ ∝ V^((γ+1)/2)
26. From a solid sphere of mass M and radius R , a spherical portion of radius R/2 is removed, as shown in the figure. Taking gravitational potential V = 0 at r = ∞ , the potential at the centre of the cavity thus formed is
(G = gravitational constant)
(1) −2GM/(3R)
(2) −2GM/R
(3) −GM/(2R)
(4) −GM/R
Answer (4)
Sol. V = V1 − V2
V1 = −GM/(2R^3)[3R^2 − (R/2)^2]
V2 = −3G(M/8)/(2(R/2))
⇒ V = −GM/R
27. 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
Sol. Clearly fs = 120 N (for vertical equilibrium of the system)
28. A long cylindrical shell carries positive surface charge σ in the upper half and negative surface charge −σ in the lower half. The electric field lines around the cylinder will look like figure given in (figures are schematic and not drawn to scale)
Answer (3)
Sol. The field line should resemble that of a dipole.
29. As an electron makes a transition from an excited state to the ground state of a hydrogen-like atom/ion
(1) Kinetic energy decreases, potential energy increases but total energy remains same
(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
Answer (3)
Sol. PE = −27.2 z^2/n^2 eV
TE = −13.6 z^2/n^2 eV
KE = 13.6 z^2/n^2 eV
KE = 13.6/n^2 eV, As n decreases, KE ↑
PE = −27.2/n^2 eV, as n decreases, PE ↓
TE = −13.6/n^2 eV, as n decreases, TE ↓
30. Match List-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the choices given below the list:
List-I List-II
(A) Franck-Hertz experiment (i) Particle nature of light
(B) Photo-electric experiment (ii) Discrete energy levels of atom
(C) Davison-Germer experiment (iii) Wave nature of electron
(iv) Structure of atom
(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)
Answer (1)
Sol. Franck-Hertz exp.- Discrete energy level. Photo-electric effect- Particle nature of light Davison-Germer exp.- Diffraction of electron beam.
31. Let a, b and c be three non-zero vectors such that no two of them are collinear and (a × b) × c = 1/3 |b| |c| a . If θ is the angle between vectors b and c , then a value of sinθ is
(1) 2/3
(2) 2√3/3
(3) 2√2/3
(4) 2/3
Answer (3)
Sol. (a·c)b − (b·c)a = 1/3 |b| |c| a
∴ −(b·c) = 1/3 |b| |c|
∴ cosθ = −1/3
∴ sinθ = 2√2/3
32. Let O be the vertex and Q be any point on the parabola, x^2 = 8y . If the point P divides the line segment OQ internally in the ratio 1:3 , then the locus of P is
(1) y^2 = 2x
(2) x^2 = 2y
(3) x^2 = y
(4) y^2 = x
Answer (2)
Sol. x^2 = 8y
Let Q be (4t, 2t^2)
∴ P = (t, t^2/2)
∴ Let P be (h, k)
∴ h = t, k = t^2/2
∴ 2k = h^2
∴ Locus of (h, k) is x^2 = 2y.
33. If the angles of elevation of the top of a tower from three collinear points A, B and C , on a line leading to the foot of the tower, are 30° , 45° and 60° respectively, then the ratio, AB:BC , is
(1) 1:√3
(2) 2:3
(3) √3:1
(4) √3:√2
Answer (3)
Sol. AO = h cot30°
∴ h = √3
BO = h
CO = h/√3
∴ AB/BC = (AO − BO)/(BO − CO)
∴ (h√3 − h)/(h − h/√3)
= √3
34. The number of points, having both co-ordinates as integers, that lie in the interior of the triangle with vertices (0, 0), (0, 41) and (41, 0) , is
(1) 820
(2) 780
(3) 901
(4) 861
Answer (2)
Sol. Total number of integral coordinates as required
= 39 + 38 + 37 + … + 1
∴ 39 × 40/2 = 780
35. The equation of the plane containing the line 2x − 5y + z = 3; x + y + 4z = 5 , and parallel to the plane, x + 3y + 6z = 1 , is
(1) x + 3y + 6z = 7
(2) 2x + 6y + 12z = −13
(3) 2x + 6y + 12z = 13
(4) x + 3y + 6z = −7
Answer (1)
Sol. Required plane is
(2x − 5y + z − 3) + λ(x + y + 4z − 5) = 0
It is parallel to x + 3y + 6z = 1
∴ (2 + λ)/1 = (−5 + λ)/3 = (1 + 4λ)/6
Solving λ = −11/2
Required plane is
(2x − 5y + z − 3) − 11/2 (x + y + 4z − 5) = 0
∴ x + 3y + 6z − 7 = 0
36. Let A and B be two sets containing four and two elements respectively. Then the number of subsets of the set A × B , each having at least three elements is
(1) 275
(2) 510
(3) 219
(4) 256
Answer (3)
Sol. n(A) = 4 n(B) = 2
n(A × B) = 8
= 2^8 − (8C0 + 8C1 + 8C2)
= 256 − 37
= 219
37. Locus of the image of the point (2, 3) in the line (2x − 3y + 4) + k(x − 2y + 3) = 0 , k ∈ R , is a
(1) Circle of radius √2
(2) Circle of radius √3
(3) Straight line parallel to x-axis
(4) Straight line parallel to y-axis
Answer (1)
Sol. After solving equation (i) & (ii)
2x − 3y + 4 = 0 ...(i)
2x − 4y + 6 = 0 ...(i)
x = 1 and y = 2
Slope of AB × Slope of MN = −1
(b − 3)/(a − 2) × (b + 3)/(a + 2) = −1
(y − 3)(y − 1) = −(x − 2)x
y^2 − 4y + 3 = −x^2 + 2x
x^2 + y^2 − 2x − 4y + 3 = 0
Circle of radius = √2
38. lim_{x→0} ((1 − cos2x)(3 + cosx))/tan4x is equal to
(1) 2
(2) 1/2
(3) 4
(4) 3
Answer (1)
Sol. lim_{x→0} (2sin^2x · (3 + cosx))/(x^2 tan4x) × x^2/(4x) = 2
39. The distance of the point (1, 0, 2) from the point of intersection of the line (x − 2)/3 = (y + 1)/4 = (z − 2)/12 and the plane x − y + z = 16 , is
(1) 3√21/2
(2) 13
(3) 2√14/4
(4) 8
Answer (2)
Sol. (x − 2)/3 = (y + 1)/4 = (z − 2)/12 = λ
P(3λ + 2, 4λ − 1, 12λ + 2)
Lies on plane x − y + z = 16
Then,
3λ + 2 − 4λ + 1 + 12λ + 2 = 16
11λ + 5 = 16
λ = 1 P(5, 3, 14)
Distance = √(16 + 9 + 144) = √169 = 13
40. The sum of coefficients of integral powers of x in the binomial expansion of (1 − 2√x)^50 is
(1) 1/2 (3^50 − 1)
(2) 1/2 (2^50 + 1)
(3) 1/2 (3^50 + 1)
(4) 1/2 (3^50)
Answer (3)
Sol. (1 − 2√x)^50 = 50C0 − 50C1(2√x)^1 + 50C2(2√x)^2 + …
+ 50C50(−2√x)^50
Sum of coefficient of integral power of x
= 50C0 2^0 + 50C2 · 2^2 + 50C4 · 2^4 + … + 50C50 · 2^50
We know that
(1 + 2)^50 = 50C0 + 50C1 · 2 + … + 50C50 · 2^50
Then,
50C0 + 50C2 · 2^2 + … + 50C50 · 2^50 = (3^50 + 1)/2
41. The sum of first 9 terms of the series
1^3/1 + (1^3 + 2^3)/(1 + 3) + (1^3 + 2^3 + 3^3)/(1 + 3 + 5) + …
(1) 142
(2) 192
(3) 71
(4) 96
Answer (4)
Sol. tn = [n(n + 1)]^2/n^2
= (n + 1)^2/4
= 1/4 [n^2 + 2n + 1]
= 1/4 [n(n + 1)(2n + 1)/6 + 2(n)(n + 1)/2 + 1]
= 1/4 [9 × 10 × 19/6 + 9 × 10 + 9]
42. The area (in sq. units) of the region described by {(x, y): y^2 ≤ 2x and y ≥ 4x − 1} is
(1) 15/64
(2) 9/32
(3) 7/32
(4) 5/64
Answer (2)
After solving y = 4x − 1 and y^2 = 2x
y = 4 · y^2/2 − 1
2y^2 − y − 1 = 0
y = (1 ± √(1 + 8))/4 = (1 ± 3)/4 y = 1, −1/2
A = ∫_{−1/2}^{1} ((y + 1)/4)dy − ∫_{−1/2}^{1} y^2/2 dy
= 1/4 [y^2/2 + y]_{−1/2}^{1} − 1/2 [y^3/3]_{−1/2}^{1}
= 1/4 [(4 + 8 − 1 + 4)/8] − 1/2 [(8 + 1)/24]
= 1/4 [15/8] − 9/48
= 15/32 − 6/32 = 9/32
43. The set of all values of λ for which the system of linear equations
2x1 − 2x2 + x3 = λx1
2x1 − 3x2 + 2x3 = λx2
−x1 + 2x2 = λx3
has a non-trivial solution
(1) Contains two elements
(2) Contains more than two elements
(3) Is an empty set
(4) Is a singleton
Answer (1)
Sol. x1(2 − λ) − 2x2 + x3 = 0
2x1 + x2(−λ − 3) + 2x3 = 0
−x1 + 2x2 − λx3 = 0
|2 − λ −2 1|
|2 −λ − 3 2| = 0
|−1 2 −λ|
(2 − λ)(λ^2 + 3λ − 4) + 2(−2λ + 2) + (4 − λ − 3) = 0
2λ^2 + 6λ − 8 − λ^3 − 3λ^2 + 4λ − 4λ + 4 − λ + 1 = 0
⇒ −λ^3 − λ^2 + 5λ − 3 = 0
⇒ λ^3 + λ^2 − 5λ + 3 = 0
λ^3 − λ^2 + 2λ^2 − 2λ − 3λ + 3 = 0
λ^2(λ − 1) + 2λ(λ − 1) − 3(λ − 1) = 0
(λ − 1)(λ^2 + 2λ − 3) = 0
(λ − 1)(λ + 3)(λ − 1) = 0
⇒ λ = 1, 1, −3
Two elements.
44. A complex number z is said to be unimodular if |z| = 1 . Suppose z1 and z2 are complex numbers such that (z1 − 2z2)/(2 − z1 z2¯) is unimodular and z2 is not unimodular. Then the point z1 lies on a
(1) Circle of radius 2
(2) Circle of radius √2
(3) Straight line parallel to x-axis
(4) Straight line parallel to y-axis
Answer (1)
Sol. |(z1 − 2z2)/(2 − z1 z2¯)| = 1
((z1 − 2z2)/(2 − z1 z2¯))((z1¯ − 2z2¯)/(2 − z1¯ z2)) = 1
z1 z1¯ − 2z1 z2¯ − 2z2 z1¯ + 4z2 z2¯
= 4 − 2z1¯ z2 − 2z1 z2¯ + z1 z1¯ z2 z2¯
z1 z1¯ + 4z2 z2¯ = 4 + z1 z1¯ z2 z2¯
z1 z1¯(1 − z2 z2¯) − 4(1 − z2 z2¯) = 0
(z1 z1¯ − 4)(1 − z2 z2¯) = 0
⇒ z1 z1¯ = 4
|z| = 2, i.e. z lies on circle of radius 2.
45. The number of common tangents to the circles x^2 + y^2 − 4x − 6y − 12 = 0 and x^2 + y^2 + 6x + 18y + 26 = 0, is
(1) 3
(2) 4
(3) 1
(4) 2
Answer (1)
Sol. x^2 + y^2 − 4x − 6y − 12 = 0
C1(center) = (2, 3), r = √(2^2 + 3^2 + 12) = 5
x^2 + y^2 + 6x + 18y + 26 = 0
C2(center) = (−3, −9), r = √(9 + 81 − 26)
= √64 = 8
C1C2 = 13, C1C2 = r1 + r2
Number of common tangent is 3.
46. The number of integers greater than 6,000 that can be formed, using the digits 3, 5, 6, 7 and 8, without repetition, is
(1) 120
(2) 72
(3) 216
(4) 192
Answer (4)
Sol. 4 digit numbers
3, 5, 6, 7, 8
|6|7|8| |
3 4 5 2 = 72
5 digit numbers
| | | | | |
5
5 × 4 × 3 × 2 × 1 = 120
Total number of integers = 72 + 120 = 192
47. Let y(x) be the solution of the differential equation (x log x)dy/dx + y = 2x log x, (x ≥ 1). Then y(e) is equal to
(1) 2
(2) 2e
(3) e
(4) 0
Answer (1)*
Sol. It is best option. Theoretically question is wrong, because initial condition is not given.
x log x dy/dx + y = 2x log x If x = 1 then y = 0
dy/dx + y/(x log x) = 2
I.F. = e^∫(1/(x log x))dx = e^(log log x) = log x
Solution is y · log x = ∫2 log x dx + c
y log x = 2(x log x − x) + c
x = 1, y = 0
Then, c = 2, y(e) = 2
48. If A = [1 2 2; 2 1 −2; a 2 b] is a matrix satisfying the equation AA^T = 9I , where I is 3 × 3 identity matrix, then the ordered pair (a, b) is equal to
(1) (2, 1)
(2) (−2, −1)
(3) (2, −1)
(4) (−2, 1)
Answer (2)
Sol. [1 2 2; 2 1 −2; a 2 b][1 2 a; 2 1 2; 2 −2 b] = [9 0 0; 0 9 0; 0 0 9]
a + 4 + 2b = 0
2a + 2 − 2b = 0
a + 1 − b = 0
2a − 2b = −2
a + 2b = −4
3a = −6
a = −2
−2 + 1 − b = 0
b = −1
a = −2
(−2, −1)
49. If m is the A.M. of two distinct real numbers l and n(l, n > 1) and G1, G2 and G3 are three geometric means between l and n , then G1^4 + 2G2^4 + G3^4 equals.
(1) 4lmn^2
(2) 4l^2m^2n^2
(3) 4l^2mn
(4) 4lmn^2
Answer (4)
Sol. (l + n)/2 = m
l + n = 2m ...(i)
G1 = l(n/l)^(1/4)
G2 = l(n/l)^(2/4)
G3 = l(n/l)^(3/4)
Now G1^4 + 2G2^4 + G3^4
l^4 · n/l + 2 · (l^2)(n/l)^2 + l^4(n/l)^3
= nl^3 + 2n^2l^2 + n^3l
= 2n^2l^2 + nl(n^2 + l^2)
= 2n^2l^2 + nl((n + l)^2 − 2nl)
= nl(n + l)^2
= nl · (2m)^2
= 4nlmn^2
50. The negation of ∼s ∨ (∼r ∧ s) is equivalent to
(1) s ∨ (r ∨ ∼s)
(2) s ∧ r
(3) s ∧ ∼r
(4) s ∧ (r ∧ ∼s)
Answer (2)
Sol. ∼(∼s ∨ (∼r ∧ s))
= s ∧ (r ∨ ∼s)
= (s ∧ r) ∨ (s ∧ ∼s)
= s ∧ r
51. The integral ∫ dx/(x^2(x^4 + 1)^(3/4)) equals.
(1) −(x^4 + 1)^(1/4) + c
(2) −((x^4 + 1)/x^4)^(1/4) + c
(3) ((x^4 + 1)/x^4)^(3/4) + c
(4) (x^4 + 1)^(1/4) + c
Answer (2)
Sol. I = ∫ dx/(x^2(x^4 + 1)^(3/4)) = ∫ dx/(x^5(1 + 1/x^4)^(3/4))
Let 1 + 1/x^4 = t ⇒ −4/x^5 dx = dt
So, I = −1/4 ∫ dt/t^(3/4) = −1/4 ∫ t^(−3/4)dt
= −1/4 (t^(1/4)/(1/4)) + c
= −(1 + 1/x^4)^(1/4) + c
So, option (2).
52. The normal to the curve, x^2 + 2xy − 3y^2 = 0 at (1,1):
(1) Meets the curve again in the third quadrant
(2) Meets the curve again in the fourth quadrant
(3) Does not meet the curve again
(4) Meets the curve again in the second quadrant
Answer (2)
Sol. Curve is x^2 + 2xy − 3y^2 = 0
Differentiate w.r.t. x, 2x + 2[dy/dx + y] − 6y dy/dx = 0
⇒ (dy/dx)_{(1,1)} = 1
So equation of normal at (1,1) is
y − 1 = −1(x − 1)
⇒ y = 2 − x
Solving it with the curve, we get
x^2 + 2x(2 − x) − 3(2 − x)^2 = 0
⇒ −4x^2 + 16x − 12 = 0
⇒ x^2 − 4x + 3 = 0
⇒ x = 1, 3
So points of intersections are (1,1) & (3, −1) i.e. normal cuts the curve again in fourth quadrant.
53. Let tan^−1y = tan^−1x + tan^−1(2x/(1 − x^2))
where |x| < 1/√3 . Then a value of y is
(1) (3x − x^3)/(1 + 3x^2)
(2) (3x − x^3)/(1 − 3x^2)
(3) (3x − x^3)/(1 − 3x^2)
(4) (3x − x^3)/(1 + 3x^2)
Answer (3)
Sol. tan^−1y = tan^−1x + tan^−1(2x/(1 − x^2))
3tan^−1x = tan^−1((3x − x^3)/(1 − 3x^2))
y = (3x − x^3)/(1 − 3x^2)
54. If the function.
g(x) = { k√(x + 1), 0 ≤ x ≤ 3
mx + 2, 3 < x ≤ 5
is differentiable, the value of k + m is
(1) 10/3
(2) 2
(3) 2
(4) 4
Answer (3)
Sol. g(x) = { k√(x + 1), 0 ≤ x ≤ 3
mx + 2, 3 < x ≤ 5
R.H.D.
lim_{h→0} (g(3 + h) − g(3))/h
= lim_{h→0} (m(3 + h) + 2 − 2k)/h
= lim_{h→0} ((3m − 2k) + mh + 2)/h = m
and 3m − 2k + 2 = 0
L.H.D.
lim_{h→0} (k√((3 − h) + 1) − 2k)/(−h)
lim_{h→0} (−k[√(4 − h) − 2])/h
lim_{h→0} −k × (4 − h − 4)/(h(√(4 − h) + 2)) = k/4
From above,
k/4 = m and 3m − 2k + 2 = 0
m = 2/5 and k = 8/5
k + m = 8/5 + 2/5 = 10/5 = 2
Alternative Answer
g(x) = { k√(x + 1), 0 ≤ x ≤ 3
mx + 2, 3 < x ≤ 5
g is constant at x = 3
k√4 = 3m + 2
2k = 3m + 2 ...(i)
Also (k/(2√(x + 1)))_{x = 3} = m
k/4 = m
k = 4m ...(ii)
m = 2/5, k = 8/5
m + k = 2/5 + 8/5 = 2
55. The mean of the data set comprising of 16 observations is 16. If one of the observation valued 16 is deleted and three new observations valued 3, 4 and 5 are added to the data, then the mean of the resultant data, is
(1) 15.8
(2) 14.0
(3) 16.8
(4) 16.0
61. Which compound would give 5-keto-2-methyl hexanal upon ozonolysis?
Answer (4)
Sol. 5-keto-2-methylhexanal is
62. Which of the vitamins given below is water soluble?
(1) Vitamin E
(2) Vitamin K
(3) Vitamin C
(4) Vitamin D
Answer (3)
Sol. Vitamin C is water soluble vitamin.
63. Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater than its lattice enthalpy?
(1) BaSO4
(2) SrSO4
(3) CaSO4
(4) BeSO4
Answer (4)
Sol. BeSO4 has hydration energy greater than its lattice energy.
64. In the reaction
the product E is
Answer (1)
65. Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 4.29 Å . The radius of sodium atom is approximately
(1) 5.72 Å
(2) 0.93 Å
(3) 1.86 Å
(4) 3.22 Å
Answer (3)
Sol. Edge length of BCC is 4.29 Å .
In BCC,
edge length = 4/√3 r
4.29 = 4/√3 r
r = 4.29/4 √3 ≈ 1.86 Å
66. Which of the following compounds is not colored yellow?
(1) (NH4)3[As(Mo3O10)4]
(2) BaCrO4
(3) Zn2[Fe(CN)6]
(4) K3[Co(NO2)6]
Answer (3)
Sol. (NH4)3[As(Mo3O10)4] BaCrO4 and K3[Co(NO2)6] are yellow colored compounds but Zn2[Fe(CN)6] is not yellow colored compound.
67. Which of the following is the energy of a possible excited state of hydrogen?
(1) −3.4 eV
(2) +6.8 eV
(3) +13.6 eV
(4) −6.8 eV
Answer (1)
Sol. Energy of excited state is negative and correspond to n > 1 .
68. Which of the following compounds is not an antacid?
(1) Phenelzine
(2) Ranitidine
(3) Aluminium Hydroxide
(4) Cimetidine
Answer (1)
Sol. Phenelzine is not antacid, it is anti-depressant.
69. The ionic radii (in Å^3 of N^3− O^2− and F^− are respectively
(1) 1.71, 1.40 and 1.36
(2) 1.71, 1.36 and 1.40
(3) 1.36, 1.40 and 1.71
(4) 1.36, 1.71 and 1.40
Answer (1)
Sol. Radius of N^3− O^2− and F^− follow order N^3− > O^2− > F^−
As per inequality only option (1) is correct that is 1.71 A, 1.40 A and 1.36 A
70. In the context of the Hall-Heroult process for the extraction of Al, which of the following statement is false?
(1) Al^3+ is reduced at the cathode to form Al
(2) Na3AlF6 serves as the electrolyte
(3) CO and CO2 are produced in this process
(4) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity
Answer (2)
Sol. In Hall-Heroult process Al2O3 (molten) is electrolytic.
71. In the following sequence of reactions :
Toluene —KMnO4→ A —SOCl2→ B —H2/Pd→ C,
the product C is
(1) C6H5CH2OH
(2) C6H5CHO
(3) C6H5COOH
(4) C6H5CH3
Answer (2)
72. Higher order (>3) reactions are rare due to
(1) Shifting of equilibrium towards reactants due to elastic collisions
(2) Loss of active species on collision
(3) Low probability of simultaneous collision of all the reacting species
(4) Increase in entropy and activation energy as more molecules are involved
Answer (3)
Sol. Higher order greater than 3 for reaction is rare because there is low probability of simultaneous collision of all the reacting species.
73. Which of the following compounds will exhibit geometrical isomerism?
(1) 2-Phenyl-1-butene
(2) 1,1-Diphenyl-1-propane
(3) 1-Phenyl-2-butene
(4) 3-Phenyl-1-butene
Answer (3)
Sol. For geometrical isomerism doubly bonded carbon must be bonded to two different groups which is only satisfied by 1-Phenyl-2-butene.
74. Match the catalysts to the correct processes :
Catalyst Process
a. TiCl3 (i) Wacker process
b. PdCl2 (ii) Ziegler-Natta polymerization
c. CuCl2 (iii) Contact process
d. V2O5 (iv) Deacon's process
(1) a(ii), b(ii), c(iv), d(i)
(2) a(ii), b(i), c(ii), d(iv)
(3) a(ii), b(ii), c(iv), d(i)
(4) a(ii), b(i), c(iv), d(ii)
Answer (4)
Sol. TiCl3 - Ziegler-Natta polymerization V2O5 - Contact process PdCl2 - Wacker process CuCl2 - Deacon's process
75. The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is
(1) London force
(2) Hydrogen bond
(3) Ion-ion interaction
(4) Ion-dipole interaction
Answer (2)
Sol. H-bond is one of the dipole-dipole interaction and dependent on inverse cube of distance between the molecules.
76. The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na mol.wt.206). What would be the maximum uptake of Ca^2+ ions by the resin when expressed in mole per gram resin?
Answer (2)
Sol. Co^2+ + 2C8H7SO3^−Na^+ → Ca(C8H7SO3)2^− + 2Na^+
The maximum uptake = 1/(206 × 2) = 1/412 mol/g
77. Two faraday of electricity is passed through a solution of CuSO4 . The mass of copper deposited at the cathode is (at. mass of Cu = 63.5 amu)
(1) 2 g
(2) 127 g
(3) 0 g
(4) 63.5 g
Answer (4)
Sol. Cu^2+ + 2e → Cu SO2 F charge deposit 1 mol of Cu. Mass deposited = 63.5 g
78. The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]^+ is (py = pyridine)
(1) 4
(2) 6
(3) 2
(4) 3
Answer (4)
as per question a = Cl b = py c = NH3 and d = NH2OH are assumed.
79. In Carius method of estimation of halogens, 250 mg of an organic compound gave 141 mg of AgBr . The percentage of bromine in the compound is
(At. mass Ag = 108 Br = 80)
(1) 48
(2) 60
(3) 24
(4) 36
Answer (3)
Sol. Percentage of Br
= (Weight of AgBr)/(Mol. mass of AgBr) × (Mol. mass of Br)/(Weight of O.C.) × 100
= 141/188 × 80/250 × 100
= 24%
80. The color of KMnO4 is due to
(1) L → M charge transfer transition
(2) σ − σ* transition
(3) M → L charge transfer transition
(4) d-d transition
Answer (1)
Sol. Charge transfer spectra from ligand (L) to metal (M) is responsible for color of KMnO4
81. The synthesis of alkyl fluorides is best accomplished by
(1) Finkelstein reaction
(2) Swarts reaction
(3) Free radical fluorination
(4) Sandmeyer's reaction
Answer (2)
Sol. Swart's reaction
CH3 − Cl + AgF —Δ→ CH3F + AgCl
82. 3 g of activated charcoal was added to 50 mL of acetic acid solution (0.06 N) in a flask. After an hour it was filtered and the strength of the filtrate was found to be 0.042 N. The amount of acetic acid adsorbed (per gram of charcoal) is
(1) 42 mg
(2) 54 mg
(3) 18 mg
(4) 36 mg
Answer (3)
Sol. Number of moles of acetic acid adsorbed
= (0.06 × 50/1000 − 0.042 × 50/1000)
= 0.9/1000 moles
Weight of acetic acid adsorbed = 0.9 × 60 mg
= 54 mg
Hence, the amount of acetic acid adsorbed per g of
charcoal = 54/3 mg
= 18 mg
Hence, option (3) is correct.
83. The vapour pressure of acetone at 20°C is 185 torr. When 1.2 g of a non-volatile substance was dissolved in 100 g of acetone at 20°C its vapour pressure was 183 torr. The molar mass (g mol^−1) of the substance is
(1) 128
(2) 488
(3) 32
(4) 64
Answer (4)
Sol. Vapour pressure of pure acetone P_A^0 = 185 torr
Vapour pressure of solution, P_S = 183 torr
Molar mass of solvent, M_A = 58 g/mole
as we know (P_A^0 − P_S)/P_S = n_B/n_A
⇒ (185 − 183)/183 = W_B/M_B × M_A/W_A
⇒ 2/183 = 1.2/M_B × 58/100
⇒ M_B = 1.2/2 × 58/100 × 183
84. Which among the following is the most reactive?
(1) I2
(2) ICl
(3) Cl2
(4) Br2
Answer (2)
Sol. Because of polarity and weak bond interhalogen compounds are more reactive.
85. The standard Gibbs energy change at 300 K for the reaction 2A ? B + C is 2494.2 J . At a given time, the composition of the reaction mixture is [A] = 1/2, [B] = 2 and [C] = 1/2 . The reaction proceeds in the : [R = 8.314 J/K/mol, e = 2.718]
(1) Forward direction because Q < KC
(2) Reverse direction because Q < KC
(3) Forward direction because Q > KC
(4) Reverse direction because Q > KC
Answer (4)
Sol. 2A ? B + C, ΔG° = 2494.2 J
As we know ΔG° = −2.303RT log KC
⇒ 2494.2 = −2.303 × 8.314 × 300 lg KC
⇒ −0.434 = log KC
⇒ KC = antilog(−0.434)
⇒ KC = 0.367
Now [A] = 1/2, [B] = 2 and [C] = 1/2
Now QC = [C][B]/[A]^2 = (1/2)(2)/(1/2)^2 = 4
as QC > KC hence reaction will shift in backward direction.
86. Assertion : Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen.
Reason : The reaction between nitrogen and oxygen requires high temperature.
(1) The assertion is incorrect, but the reason is correct
(2) Both the assertion and reason are incorrect
(3) Both assertion and reason are correct, and the reason is the correct explanation for the assertion
(4) Both assertion and reason are correct, but the reason is not the correct explanation for the assertion
Answer (3)
Sol. N2 + O2 → 2NO
Required temperature for above reaction is around 3000°C which is a quite high temperature. This reaction is observed during thunderstorm.
87. Which one has the highest boiling point?
(1) Kr
(2) Xe
(3) He
(4) Ne
Answer (2)
Sol. Down the group strength of van der Waal's force of attraction increases hence Xe have highest boiling point.
88. Which polymer is used in the manufacture of paints and lacquers?
(1) Polypropylene
(2) Poly vinyl chloride
(3) Bakelite
(4) Glyptal
Answer (4)
Sol. Glyptal is used in manufacture of paints and lacquers.
89. The following reaction is performed at 298 K.
2NO(g) + O2(g) ? 2NO2(g)
The standard free energy of formation of NO(g) is 86.6 kJ/mol at 298 K. What is the standard free energy of formation of NO2(g) at 298 K?
(Kp = 1.6 × 10^12)
(1) 86600 − ln(1.6 × 10^12)/(R(298))
(2) 0.5[2 × 86,600 − R(298)ln1.6 × 10^12]
(3) R(298)ln(1.6 × 10^12) − 86600
(4) 86600 + R(298)ln(1.6 × 10^12)
Answer (2)
Sol. 2NO(g) + O2(g) ? 2NO2(g)
(ΔG°)_reaction = [(ΔG°)_formation]_product
− [(ΔG°)_formation]_reactant
⇒ −RT lnKp = 2 × (ΔG°)_NO2 − 2(ΔG°)_NO
⇒ (ΔG°)_NO2 = 2(ΔG°)_NO − RT lnKp
⇒ (ΔG°)_NO2 = (2 × 86600 − R(298)lnKp)/2
= (2 × 86600 − R(298)ln1.6 × 10^12)/2
= 0.5[2 × 86,600 − R(298)ln1.6 × 10^12]
90. From the following statement regarding H2O2 choose the incorrect statement
(1) It has to be stored in plastic or wax lined glass bottles in dark.
(2) It has to be kept away from dust
(3) It can act only as an oxidizing agent
(4) It decomposes on exposure to light
Answer (3)
Sol. H2O2 can be reduced or oxidised. Hence, it can act as reducing as well as oxidising agent.
refer page number 210 to 231 from given PDF
PART 2 JEE MAINS 2014
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