1. A student measures the time period of 100 oscillations of a simple pendulum four times. The data set is 90 s, 91 s, 95 s and 92 s. If the minimum division in the measuring clock is 1 s, then the reported mean time should be :

(1) 92 ± 2 s
(2) 92 ± 5.0 s
(3) 92 ± 1.8 s
(4) 92 ± 3 s

1. (1)

x? = Σx_i / N = (90 + 91 + 95 + 92)/4 = 92
Mean deviation = Σ|x? - x_i| / N = (2 + 1 + 3 + 0)/4 = 1.5
L.C. = 1 s.
∴ Required value = 92 ± 2 s

L.C. = 1 s.

Required value = 92 ± 2 s

1. A particle of mass m is moving along the side of a square of side 'a', with a uniform speed v in the x-y plane as shown in the figure :

Which of the following statements is false for the angular momentum L about the origin?

(1) L = -mv/√2 R k when the particle is moving from A to B.
(2) L = mv[R/√2 - a] k when the particle is moving from C to D.
(3) L = mv[R/√2 + a] k when the particle is moving from B to C.
(4) L = mv/√2 R k when the particle is moving from D to A.

1. (2), (4)

Along CD, ⊥ distance from line of motion = (R/√2 + a). Magnitude of angular momentum = mv[R/√2 + a]
Hence (2) is incorrect.

In option (4) the direction of L is incorrect.

1. A point particle of mass m, moves along the uniformly rough track PQR as shown in the figure. The coefficient of friction, between the particle and the rough track equals μ. The particle is released, from rest, from the point P and it comes to rest at a point R. The energies, lost by the ball, over the parts, PQ and QR, of the track, are equal to each other, and no energy is lost when particle changes direction from PQ to QR.

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1. 0.2 and 6.5m
   (2) 0.2 and 3.5m
   (3) 0.29 and 3.5m
   (4) 0.29 and 6.5m

1. (3)

From work energy theorem and given condition

mgh - 2μmg cosθ h/sinθ = 0
∴ μ = 1/(2cot30) = 1/(2√3) = 0.29
again mgh/2 = μmg·QR
∴ QR = h/(2μ) = 2/(2×0.29) = 3.5m

1. A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up? Fat supplies 3.8×10^7 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g = 9.8 ms^-2 :

(1) 2.45×10^-3 kg
(2) 6.45×10^-3 kg
(3) 9.89×10^-3 kg
(4) 12.89×10^-3 kg

1. (4)

0.2×3.8×10^7×m = 10×g×1×1000
m = (10×9.8×1000)/(0.2×3.8×10^7) = 1.289×10^-2 kg = 12.89×10^-3 kg

1. A roller is made by joining together two cones at their vertices O. It is kept on two rails AB and CD which are placed asymmetrically (see figure), with its axis perpendicular to CD and its centre O at the centre of line joining AB and CD (see figure). It is given a light push so that it starts rolling with its centre O moving parallel to CD in the direction shown. As it moves, the roller will tend to :
   (1) turn left
   (2) turn right
   (3) go straight
   (4) turn left and right alternately

2. (1)

If we take 'r' as the distance of IAOR from the axis of rotation, then 'r' decreases on left side as the object moves forward.

So, for left v = ωr' < ωr (for right point)

So, the roller will turn to the left as it moves forward.

1. A satellite is revolving in a circular orbit at a height 'h' from the earth's surface (radius of earth R; h << R). The minimum increase in its orbital velocity required, so that the satellite could escape from the earth's gravitational field, is close to : (Neglect the effect of atmosphere).

(1) √2gR
(2) √gR
(3) √(gR/2)
(4) √(gR(√2 - 1))

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1. (4)

Orbital velocity v = √(GM/(R+h)) = √(GM/R) as h << R

Velocity required to escape

1/2 mv² = GMm/(R+h); v' = √(2GM/(R+h)) = √(2GM/R) (h << R)

Increase in velocity

v' - v = √(2GM/R) - √(GM/R) = √(2gR) - √(gR) = √(gR)(√2 - 1)

1. A pendulum clock loses 12 s a day if the temperature is 40°C and gains 4s a day if the temperature is 20°C. The temperature at which the clock will show correct time, and the co-efficient of linear expansion (α) of the metal of the pendulum shaft are respectively:

(1) 25°C; α = 1.85×10^-5 /°C
(2) 60°C; α = 1.85×10^-4 /°C
(3) 30°C; α = 1.85×10^-3 /°C
(4) 55°C; α = 1.85×10^-2 /°C

1. (1)

Time loss or gain is given by

Δt = (ΔT/T)t = 1/2 α·Δθ·t
∴ 12 = 1/2 α(40 - θ0)×Id ...(1)
4 = 1/2 α(θ0 - 20)×Id ...(2)
(1) gives (2)
3 = (40 - θ0)/(θ0 - 20)
Solving θ0 = 25°C and putting in (2)
α = 8/(5×29×60×60) =

1. An ideal gas undergoes a quasi static, reversible process in which its molar heat capacity C remains constant. If during this process the relation of pressure P and volume V is given by PV^n = constant, then n is given by (Here C_P and C_V are molar specific heat at constant pressure and constant volume, respectively):

(1) n = C_P/C_V
(2) n = (C - C_P)/(C - C_V)
(3) n = (C_P - C)/(C - C_V)
(4) n = (C - C_V)/(C - C_P)

1. (2)

C = C_v + R/(1-n) ⇒ 1 - n = R/(C - C_V)
n = 1 - R/(C - C_V) = (C - (C_V + R))/(C - C_V) = (C - C_P)/(C - C_V)

1. 'n' moles of an ideal gas undergoes a process A → B as shown in the figure. The maximum temperature of the gas during the process will be :

(1) 9P0V0/(4nR)
(2) 3P0V0/(2nR)
(3) 9P0V0/(2nR)
(4) 9P0V0/(nR)

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1. (1) Equation of line AB

y - y1 = (y2 - y1)/(x2 - x1)(x - x1)
P - P0 = (2P0 - P0)/(V0 - 2V0)(V - 2V0) = -P0/V0 (V - 2V0)
P = -P0/V0 V + 3P0
PV = -P0/V0 V² + 3P0V
nRT = -P0/V0 V² + 3P0V
T = 1/(nR)(-P0/V0 V² + 3P0V)
dT/dV = 0 (For maximum temperature)
-P0/V0 2V + 3P0 = 0
-P0/V0 2V = -3P0
V = 3/2 V0 (Condition for maximum temperature)
T_max = 1/(nR)(-P0/V0 × 9/4 V0² + 3P0 × 3/2 V0) = 1/(nR)(-9/4 P0V0 + 9/2 P0V0) = 9/4 P0V0/(nR)

1. A particle performs simple harmonic motion with amplitude A. Its speed is trebled at the instant that it is at a distance 2A/3 from equilibrium position. The new amplitude of the motion is :

(1) A/3 √41
(2) 3A
(3) A√3
(4) 7A/3

1. (4)

mω² = k
Total initial energy = 1/2 kA²
at x = 2A/3, potential energy = 1/2 k(2A/3)² = (1/2 kA²)(4/9)
Kinetic energy at (x = 2A/3) is = (1/2 kA²)(5/9)
If speed is tripled, new Kinetic energy = 1/2 kA²·5/9 = 5/2 kA²
New total energy = 5/2 kA² + 1/2 kA²(4/9) = kA²(49/9)/2
If next amplitude = A' then 1/2 kA'² = 1/2 kA²(49/9) ⇒ A' = 7/3 A

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1. A uniform string of length 20 m is suspended from a rigid support. A short wave pulse in introduced at its lowest end. It starts moving up the string. The time taken to reach the support is : (take g = 10 ms^-2)
   (1) 2π√2 s
   (2) 2 s
   (3) 2√2 s
   (4) √2 s

2. (3)

dy/dt = √(g y ρ A / μ)
dy/dt = ∫√(g y)
∫ dy/√y = √g dt
y^(-1/2+1)/(1/2+1)|_0^c = √g t|_0^t
t = 2√(20/10) = 2√2 sec

1. The region between two concentric spheres of radii 'a' and 'b', respectively (see figure), has volume charge density ρ = A/r, where A is a constant and r is the distance from the centre. At the centre of the spheres is a point charge Q. The value of A such that the electric field in the region between the spheres will be constant, is :

(1) Q/(2πa²)
(2) Q/(2π(b² - a²))
(3) 2Q/(π(a² - b²))
(4) 2Q/(πa²)

1. (1)

Charge in the shaded region = ∫_a^r 4×r² A/r · dr = 2πA(r² - a²)

Total field at P = 1/(4πε0) · Q/r² + 1/(4πε0) · 2πA(1 - a²/r²)

For field to be independent of r : Q = 2πAa²

1. A combination of capacitors is set up as shown in the figure. The magnitude of the electric field due to a point charge Q (having a charge equal to the sum of the charges on the 4μF and 9μF capacitors), at a point distant 30 m from it, would equal :

(1) 240 N/C
(2) 360 N/C
(3) 420 N/C
(4) 480 N/C

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1. (3)

Charges on 3μF = 3μF × 8V = 24μC : Charge on 4μF = Charge on 12μF = 24μC
Charge of 3μF = 3μF × 2V = 6μC
Charge of 9μF = 9μF × 2V = 18μC
Charge on 4μF + Charge on 9μF = (24 + 18)μC = 42μC
Electric field at 30 m = 9×10^9 × (42×10^-6)/(30×30) = 420 N/C

1. The temperature dependence of resistances of Cu and undoped Si in the temperature range 300 - 400 K, is best described by :

(1) Linear increase for Cu, linear increase for Si.
(2) Liner increase for Cu, exponential increase for Si.
(3) Linear increase for Cu, exponential decrease for Si.
(4) Linear decrease for Cu, linear decrease for Si.

1. (3) Fact based

2. Two identical wires A and B, each of length '?', carry the same current I. Wire A is bent into a circle of radius R and wire B is bent to form a square of side 'a'. If B_A and B_B are the values of magnetic field at the centres of the circle and square respectively, then the ratio B_A/B_B is :

(1) π²/8
(2) π²/(16√2)
(3) π²/16
(4) π²/(8√2)

1. (4)

For A
2πR = ?
R = ?/(2π)
B_A = μ0I/(2R) = μ0I/(2×?/(2π))
B_A = μ0πI/?

For B
4a = ?
a = ?/4
B_B = 4 × μ0I(?/4)/(2π(?/8)√((?/4)² + 4(?/8)²)) = 4μ0I/(π·√2 ?/4)

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1. Hysteresis loops for two magnetic materials A and B are given below : These materials are used to make magnets for electric generators, transformer core and electromagnet core. Then it is proper to use :
   (1) A for electric generators and transformers.
   (2) A for electromagnets and B for electric generators
   (3) A for transformers and B for electric generators
   (4) B for electromagnets and transformers

Conceptual (Requires low retentivity and low coercivity)

1. An arc lamp requires a direct current of 10 A at 80 V to function. It is connected to a 220 V (rms), 50 Hz AC supply, the series inductor needed for it to work is close to :

(1) 80 H
(2) 0.08 H
(3) 0.044 H
(4) 0.065 H

1. (4)

For dc
R = 80/10 = 8Ω

For ac
10 = 220/√(R² + ω²L²)
R² + ω²L² = (220/10)²
L² = (22² - 8²)/ω² ∴ L = √(30×14)/(2π×50) = √420/(100π) = 0.065 H

1. Arrange the following electromagnetic radiations per quantum in the order of increasing energy :

A : Blue light
B : Yellow light
C : X-ray
D : Radiowave

(1) D, B, A, C
(2) A, B, D, C
(3) C, A, B, D
(4) B, A, D, C

1. (1) D, B, A, C

2. An observer looks at a distant tree of height 10 m with a telescope of magnifying power of 20. To the observer the tree appears :

(1) 10 times taller
(2) 10 times nearer
(3) 20 times taller
(4) 20 times nearer

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1. (4)

2. The box of a pin hole camera, of length L, has a hole of radius a. It is assumed that when the hole is illuminated by a parallel beam of light of wavelength λ the spread of the spot (obtained on the opposite wall of the camera) is the sum of its geometrical spread and the spread due to diffraction. The spot would then have its minimum size (say b_min) when :

(1) a = λ²/L and b_min = (2λ²/L)
(2) a = √(λL) and b_min = √(4λL)
(3)
(4)

1. (3)

sinθ = λ/a
B = 2a + 2Lλ/a
∂B/∂a = 0 ⇒ 1 - Lλ/a² = 0
⇒ a = √(λL)
B_min = 2√(λL) + 2√(λL) [By substituting for a from (ii) in (i)]
= 4√(λL)
The radius of the spot = 1/2 4√(λL) = √(4λL)

1. Radiation of wavelength λ, is incident on a photocell. The fastest emitted electron has speed v. If the wavelength is changed to 3λ/4, the speed of the fastest emitted electron will be :

(1) > v(4/3)^(1/2)
(2) < v(4/3)^(1/2)
(3) = v(4/3)^(1/2)
(4) = v(3/4)^(1/2)

1. (1)

hc/λ - φ = 1/2 mv² ...(i)
4hc/(3λ) - φ = 1/2 mv'² ...(ii)
(ii) - (i) gives
hc/(3λ) = 1/2 m(v'² - v²) ⇒ v' = √(v² + 2hc/(3λm)) ...(iii)
also from (i) hc/λ = φ + mv²/2 ⇒ 2hc/(λm) = 2φ/m + v²
⇒ 2hc/(3λm) = 2φ/(3m) + v²/3 > v²/3 ...(iv)
combining (iii) & (iv)
v' > √(4v²/3)

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1. Half-lives of two radioactive elements A and B are 20 minutes and 40 minutes, respectively. Initially, the samples have equal number of nuclei. After 80 minutes, the ratio of decayed numbers of A and B nuclei will be :

(1) 1:16
(2) 4:1
(3) 1:4
(4) 5:4

1. (4)

80 minutes = 4 half-lives of A = 2 half-lives of B
Let the initial number of nuclei in each sample be N

N_A after 80 minutes = N/2^4 ⇒ Number of A nuclei decayed = 15/16 N
N_B after 80 minutes = N/2² ⇒ Number of B nuclei decayed = 3/4 N
Required ratio = (15/16)/(3/4) = 5/4

1. If a, b, c, d are inputs to a gate and x is its output, then, as per the following time graph, the gate is :

(1) NOT
(2) AND
(3) OR
(4) NAND

1. (3)

x is 1 when at least one of the inputs is 1. Hence x is an OR-Gate.

1. Choose the correct statement :

(1) In amplitude modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.
(2) In amplitude modulation the frequency of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.
(3) In frequency modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.
(4) In frequency modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the frequency of the audio signal.

1. (1)

2. A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement, it is found that when the two jaws of the screw gauge and brought in contact, the 45th division coincides with the main scale line and that the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is 0.5 mm and the 25th division coincides with the main scale line?

(1) 0.75 mm
(2) 0.80 mm
(3) 0.70 mm
(4) 0.50 mm

1. (2)

L.C. = pitch / No. of division on circular scale = 0.5/50 = 0.001 mm
-ve zero error = -5×L.C. = -0.005 mm
Measured value = main scale reading + screw gauge reading - zero error
= 0.5 mm + {25×0.001 - (-0.05)} mm = 0.80 mm

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1. A pipe open at both ends has a fundamental frequency f in air. The pipe is dipped vertically in water so that half of it is in water. The fundamental frequency of the air column is now :

(1) f/2
(2) 3f/4
(3) 2f
(4) f

1. (4)

2. A galvanometer having a coil resistance of 100Ω gives a full scale deflection, when a current of 1 mA is passed through it. The value of the resistance, which can convert this galvanometer into ammeter giving a full scale deflection for a current of 10 A is :

(1) 0.01Ω
(2) 2Ω
(3) 0.1Ω
(4) 3Ω

1. (1)

Maximum voltage that can be applied across the galvanometer coil = 100Ω × 10^-3 A = 0.1 V
If R_S is the shunt resistance :
R_S × 10 A = 0.1 V
⇒ R_S = 0.01Ω

1. In an experiment for determination of refractive index of glass of a prism by i - δ, plot, it was found that a ray incident at angle 35°, suffers a deviation of 40° and that it emerges at angle 79°. In that case which of the following is closest to the maximum possible value of the refractive index?

(1) 1.5
(2) 1.6
(3) 1.7
(4) 1.8

1. (1)

δ = i + e - A
40 = 35 + 79 - A
40 = 114 - A
A = 114 - 40 = 74 = r1 + r2
From this we get,
μ = 1.5
∴ δ_min < 40°
μ < sin((70 + 40)/2)/sin37
∴ μ_max = 1.44

1. Identify the semiconductor devices whose characteristics are given below, in the order (a), (b), (c), (d):

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1 Simple diode, Zener diode, Solar cell, Light dependent resistance.
2 Zener diode, Simple diode, Light dependent resistance, Solar cell.
3 Solar cell, Light dependent resistance, Zener diode, Simple diode.
4 Zener diode, Solar cell, Simple diode, Light dependent resistance.

1. (1)

2. For a common emitter configuration, if α and β have their usual meanings, the incorrect relationship between α and β is :

(1) 1/α = 1/β + 1
(2) α = β/(1 - β)
(3) α = β/(1 + β)
(4) α = β²/(1 + β²)

1. (2)

Standard Result

PART-B : CHEMISTRY

1. At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O2 by volume for complete combustion. After combustion the gases occupy 330 mL. Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is :

(1) C3H6
(2) C3H8
(3) C4H8
(4) C4H10

1. (2)

C_xH_y + (x + y/4)O2 → xCO2 + y/2 H2O
15 mL 15(x + y/4) mL 15x mL
V_O2 = 20/100 × 375 = 75 mL = 15(x + y/4)
⇒ x + y/4 = 5
⇒ C3H8

1. Two closed bulbs of equal volume (V) containing an ideal gas initially at pressure p_i and temperature T1 are connected through a narrow tube of negligible volume as shown in the figure below. The temperature of one of the bulbs is then raised to T2. The final pressure p_f is :

(1) p_i(T1T2/(T1+T2))
(2) 2p_i(T1/(T1+T2))
(3) 2p_i(T2/(T1+T2))
(4) 2p_i(T1T2/(T1+T2))

1. (3)

Initially
Later on
n1 = p_iV/(RT1), n2 = p_iV/(RT1)
n'1 = p_fV/(RT1), n'2 = p_fV/(RT2)

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1. A stream of electrons from a heated filament was passed between two charged plates kept at a potential difference V esu. If e and m are charge and mass of an electron, respectively, then the value of h/λ (where λ is wavelength associated with electron wave) is given by :

(1) meV
(2) 2meV
(3) √(meV)
(4) √(2meV)

1. (4)

λ = h/P = h/√(2mE) = h/√(2meV)
⇒ h/λ = √(2meV)

1. The species in which the N atom is in a state of sp hybridization is :

(1) NO2+
(2) NO2-
(3) NO3-
(4) NO2-

1. (1)

O = N = O
sp hybridisation

1. The heats of combustion of carbon and carbon monoxide are -393.5 and -283.5 kJ mol^-1, respectively. The heat of formation (in kJ) of carbon monoxide per mole is :

(1) 110.5
(2) 676.5
(3) -676.5
(4) -110.5

1. (4)

C(s) + O2(g) → CO2(g) ΔH1 = -393.5 kJ/mol
CO(g) + 1/2 O2(g) → CO2(g) ΔH2 = -283.5 kJ/mol
C(s) + 1/2 O2(g) → CO(g) ΔH = ΔH1 - ΔH2 = -110 kJ/mol

1. 18 g glucose (C6H12O6) is added to 178.2 g water. The vapor pressure of water (in torr) for this aqueous solution is :

(1) 7.6
(2) 76.0
(3) 752.4
(4) 759.0

1. (3)

n_H2O = 178.2/18 = 9.9 ; n_Glucose = 18/180 = 0.1
n_Total = 10
X_H2O = 0.99
P = P° X_solvent = 760 × 0.99 = 752.4 torr.

1. The equilibrium constant at 298 K for a reaction A + B ? C + D is 100. If the initial concentration of all the four species were 1M each, then equilibrium concentration of D (in mol L^-1) will be :

(1) 0.182
(2) 0.818
(3) 1.818
(4) 1.182

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1. (3)

A + B ? C + D
1 1 1 1
1-x 1-x 1+x 1+x
K_C = [C][D]/[A][B] = (1+x)²/(1-x)² = 100 ⇒ (1+x)/(1-x) = 10
⇒ 1 + x = 10 - 10x
⇒ 11x = 9
x = 9/11
[D] = 1 + x = 1 + 9/11
= 1.818 M

1. Galvanization is applying a coating of :

(1) Pb
(2) Cr
(3) Cu
(4) Zn

1. (4)

Galvanization is applying a coating of zinc.
E°_Pb2+/Pb = -0.13 V
E°_Cr3+/Cr = -0.74 V
E°_Cu2+/Cu = 0.34 V
E°_Zn2+/Zn = -0.76 V

1. Decomposition of H2O2 follows a first order reaction. In fifty minutes the concentration of H2O2 decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of H2O2 reaches 0.05 M, the rate of formation of O2 will be :

(1) 6.93×10^-2 mol min^-1
(2) 6.93×10^-4 mol min^-1
(3) 2.66 L min^-1 at STP
(4) 1.34×10^-2 mol min^-1

1. (2)

t_3/4 = 2×t_1/2 = 50 min
i.e. t_1/2 = 25 min
k = 0.693/t_1/2 = 0.693/25 min^-1
Rate of H2O2 decomposition = k[H2O2]
= 0.693/25 × 0.05 = -d[H2O2]/dt
H2O2 → H2O + 1/2 O2
d[H2O2]/dt = 2 d[O2]/dt
⇒ d[O2]/dt = 6.93×10^-4 mol min^-1

1. For a linear plot of log (x/m) versus log p in a Freundlich adsorption isotherm, which of the following statements is correct ? (k and n are constants)

(1) Both k and 1/n appear in the slope term.
(2) 1/n appears as the intercept.
(3) Only 1/n appears as the slope.
(4) log (1/n) appears as the intercept.

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1. (3)

2. Which of the following atoms has the highest first ionization energy ?

(1) Rb
(2) Na
(3) K
(4) Sc

1. (4)

Element Ionisation energy (kJ/mol)
Na 496
K 419
Rb 403
Sc 631
Scandium has the highest first Ionisation energy.

1. Which one of the following ores is best concentrated by froth flotation method ?

(1) Magnetite
(2) Siderite
(3) Galena
(4) Malachite

1. (3)

PbS, i.e. Galena is best concentrated by froth flotation method.

1. Which one of the following statements about water is FALSE ?

(1) Water is oxidized to oxygen during photosynthesis.
(2) Water can act both as an acid and as a base.
(3) There is extensive intramolecular hydrogen bonding in the condensed phase.
(4) Ice formed by heavy water sinks in normal water.

1. (3)

Water possess intermolecular hydrogen bonding in the condensed phase.

1. The main oxides formed on combustion of Li, Na and K in excess of air are, respectively :

(1) Li2O, Na2O and KO2
(2) LiO2, Na2O2 and K2O
(3) Li2O2, Na2O2 and KO2
(4) Li2O, Na2O2 and KO2

1. (4)

Li → Li2O

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1. (1)

Phosphorus acid series contain phosphorus in the oxidation state (+ III).
H H H
| | |
H-O-P-O-P-OH H-O-P-O-H
| | |
O O O
Pyrophosphorus acid orthophosphorus acid

1. Which of the following compounds is metallic and ferromagnetic ?
   (1) TiO2
   (2) CrO2
   (3) VO2
   (4) MnO2

2. (2)

CrO2 like metal conduct electricity and ferromagnetic.

1. The pair having the same magnetic moment is :
   [At. No.: Cr = 24, Mn = 25, Fe = 26, Co = 27]
   (1) [Cr(H2O)6]2+ and [CoCl4]2-
   (2) [Cr(H2O)6]2+ and [Fe(H2O)6]2+
   (3) [Mn(H2O)6]2+ and [Cr(H2O)6]2+
   (4) [CoCl4]2- and [Fe(H2O)6]2+

2. (2)

Cr2+ → (Ar) 3d4 4s0
Fe2+ → (Ar) 3d6 4s0

1. Which one of the following complexes shows optical isomerism ?
   (1) [Co(NH3)3Cl3]
   (2) cis[Co(en)2Cl2]Cl
   (3) trans[Co(en)2Cl2]Cl
   (4) [Co(NH3)4Cl2] Cl
   (en = ethylenediamine)

2. (2)

Cis-isomer

1. The concentration of fluoride, lead, nitrate and iron in a water sample from an underground lake was found to be 1000 ppb, 40 ppb, 100 ppm and 0.2 ppm, respectively. This water is unsuitable for drinking due to high concentration of :
   (1) Fluoride
   (2) Lead
   (3) Nitrate
   (4) Iron

2. (3)

Fluoride - 1000 ppm
Lead - 40 ppm
Nitrate - 100 ppm [> 40 ppm gives blue baby syndrome disease.]
Iron - 0.2 ppm

1. The distillation technique most suited for separating glycerol from spent-lye in the soap industry is :
   (1) Simple distillation
   (2) Fractional distillation
   (3) Steam distillation
   (4) Distillation under reduced pressure

2. (4)

Glycerol and spent-lye can be separated by distillation under reduced pressure.

===== Page 18 =====

1. The product of the reaction given below is :

2. NBS/hv

3. H2O/K2CO3

(1)
(2)
(3)
(4)

1. (2)

2. The absolute configuration of

is :
(1) (2R, 3S)
(2) (2S, 3R)
(3) (2S, 3S)
(4) (2R, 3R)

1. (2)

(2 S 3 R)

1. 2-chloro-2-methylpentane on reaction with sodium methoxide in methanol yields :

(a)
(b)
(c)

(1) All of these
(2) (a) and (c)
(3) (c) only
(4) (a) and (b)

1. (1)

===== Page 19 =====

1. The reaction of propene with HOCl (Cl2 + H2O) proceeds through the intermediate :

(1) CH3 - CH+ - CH2 - OH
(2) CH3 - CH+ - CH2 - Cl
(3) CH3 - CH(OH) - CH2+
(4) CH3 - CHCl - CH2+

1. (2)

CH3 - CH = CH2 + HOCl → CH3 - CH - CH2 - Cl
CH3 - CH = CH2 + Cl+ → CH3 - CH - CH2 - Cl

1. In the Hofmann bromamide degradation reaction, the number of moles of NaOH and Br2 used per mole of amine produced are :

(1) One mole of NaOH and one mole of Br2
(2) Four moles of NaOH and two moles of Br2
(3) Two moles of NaOH and two moles of Br2
(4) Four moles of NaOH and one mole of Br2

1. (4)

R - C - NH2 + Br2 + 4NaOH → R - NH2 + 2NaBr + Na2CO3

1. Which of the following statements about low density polythene is FALSE ?

(1) Its synthesis requires high pressure.
(2) It is a poor conductor of electricity.
(3) Its synthesis requires dioxygen or a peroxide initiator as a catalyst.
(4) It is used in the manufacture of buckets, dust-bins etc.

1. (4)

n(H2C = CH2) → 200°C, 1500 atm (O2) → +H2C - CH2↑_n

1. Thiol group is present in :

(1) Cytosine
(2) Cystine
(3) Cysteine
(4) Methionine

1. (3)

HS - CH2 - CH(NH2)COOH

1. Which of the following is an anionic detergent ?

(1) Sodium stearate
(2) Sodium lauryl sulphate
(3) Cetyltrimethyl ammonium bromide
(4) Glyceryl oleate

1. (2)

CH3(CH2)11OSO3Na = (Sodium Lauryl Sulphate)

1. The hottest region of Bunsen flame shown in the figure below is :

(1) region 1
(2) region 2
(3) region 3
(4) region 4

1. (2)

Region - 2 → blue flame

===== Page 20 =====

1. If f(x) + 2f(1/x) = 3x, x ≠ 0, and S = {x ∈ R : f(x) = f(-x)} ; then S:

(1) is an empty set
(2) contains exactly one element
(3) contains exactly two elements.
(4) contains more than two elements.

1. (3)

f(x) + 2f(1/x) = 3x
Replace x by 1/x, f(1/x) + 2f(x) = 3/x
⇒ (3x - f(x))/2 = (3/x - 2f(x))/1
⇒ 3x - f(x) = 6/x - 4f(x)
⇒ f(x) = 2/x - x
f(x) = f(-x) ⇒ 2/x - x = -2/x + x
⇒ 4/x = 2x ⇒ x = ±√2

1. A value of θ for which (2 + 3i sinθ)/(1 - 2i sinθ) is purely imaginary, is:

(1) π/3
(2) π/6
(3) sin^-1(√3/4)
(4) sin^-1(1/√3)

1. (4)

(2 + 3i sinθ)/(1 - 2i sinθ) × (1 + 2i sinθ)/(1 + 2i sinθ) = ((2 - 6sin²θ) + i(7sinθ))/(1 + 4sin²θ)
To be purely imaginary if
(2 - 6sin²θ)/(1 + 4sin²θ) = 0 ⇒ 2 = 6sin²θ ⇒ sin²θ = 1/3 ⇒ θ = sin^-1(1/√3)

1. The sum of all real values of x satisfying the equation

(x² - 5x + 5)^(x² + 4x - 60) = 1 is

(1) 3
(2) -4
(3) 6
(4) 5

1. (1)

(x² - 5x + 5)^(x² + 4x - 60) = 1 = (x² - 5x + 5)^0
⇒ x² + 4x - 60 = 0 [a^x = a^y ⇒ x = y if a ≠ 1,0,-1]
x = -10, 6
& base x² - 5x + 5 = 0 or 1 or -1
If x² - 5x + 5 = 0
But it will not satisfy original equation.
x² - 5x + 5 = 1
∴ x = 4, 1

===== Page 21 =====

1. If A = [5a -b; 3 2] and A adj A = AA^T, then 5a + b is equal to:

(1) -1
(2) 5
(3) 4
(4) 13

1. (2)

A = [5a -b; 3 2]
A.adj A = A.A^T
Equate, 10a + 3b = 25a² + b²
& 10a + 3b = 13
& 15a - 2b = 0
a/2 = b/15 = k (let)
Solving a = 2/5, b = 3
So, 5a + b = 5×2/5 + 3 = 5

1. The system of linear equations

x + λy - z = 0
λx - y - z = 0
x + y - λz = 0

has a non-trivial solution for:

(1) infinitely many values of λ
(2) exactly one value of λ
(3) exactly two values of λ
(4) exactly three values of λ

1. (4)

x + λy - z = 0
λx - y - z = 0
x + y - λz = 0
For non-trivial solution ⇒ Δ = 0
λ + 1 - λ{-λ² + 1} - (λ + 1) = 0
λ(λ² - 1) = 0
λ = 0, ±1
Exactly 3 values of λ

1. If all the words (with or without meaning) having five letters, formed using the letters of the word SMALL and arranged as in a dictionary; then the position of the word SMALL is:

(1) 46th
(2) 59th
(3) 52nd
(4) 58th

===== Page 22 =====

1. (4)

A LL MS
A (LL MS) → 4!/2! = 24/2 = 12
L (AL MS) → 4! = 24
M (ALLS) → 4!/2! = 24/2 = 12
SA(MLL) → 3!/2! = 3
SL(ALM) → 3! = 6
Total words = 12 + 24 + 12 + 3 + 6 = 57
SMALL 58th
the position of the word SMALL is 58th

1. If the number of terms in the expansion of (1 - 2/x + 4/x²)^n, x ≠ 0, is 28, then the sum of the coefficients of all the terms in this expansion, is:
   (1) 64
   (2) 2187
   (3) 243
   (4) 729

2. (4)

Number of terms = (n + 1)(n + 2)/2 = 28
⇒ n = 6
∴ a0 + a1/x + a2/x² + ... + a2n/x^(2n) = (1 - 2/x + 4/x²)^n
Put x = 1, n = 6, a0 + a1 + a2 + ... + a2n = 3^6 = 729

1. If the 2nd, 5th and 9th terms of a non-constant A.P. are in G.P., then the common ratio of this G.P. is:

(1) 8/5
(2) 4/3
(3) 1
(4) 7/4

1. (2)

t2 = a + d
t5 = a + 4d
t9 = a + 8d
Given t2, t5, t9 are in G.P.
(a + 4d)² = (a + d)(a + 8d)
a² + 16d² + 8ad = a² + 8d² + 9ad
8d² - ad = 0
d(8d - a) = 0
As given non-constant AP. ⇒ d ≠ 0
∴ d = a/8 ⇒ a = 8d
so, A.P. is 8d, 9d, 10d, ....
Common ratio of G.P. = t5/t2 = (a + 4d)/(a + d) = 12d/9d = 4/3

===== Page 23 =====

1. If the sum of the first ten terms of the series

(1/5)² + (2 2/5)² + (3 1/5)² + 4² + (4 4/5)² + ... , is 16/5 m, then m is equal to:

(1) 102
(2) 101
(3) 100
(4) 99

1. (2)

(1/5)² + (2 2/5)² + (3 1/5)² + 4² + (4 4/5)² + ... upto 10 terms
= (8/5)² + (12/5)² + (16/5)² + (20/5)² + (24/5)² + ... upto 10 terms.
(8)² + (12)² + (16)² + ... up to 10 terms
T_n = [4(n + 1)]² where n varies from 1 to 10.
= 16(n² + 2n + 1)
ΣT_n = Σ_{n=1}^{10} 16(n² + 2n + 1)
= 16[385 + 55(2) + 10]
= 16(505)
Σ_{n=1}^{10} n² = n(n+1)(2n+1)/6 = 10×11×21/6 = 385
Σ_{n=1}^{10} n = n(n+1)/2 = 10×11/2 = 55
Σ_{n=1}^{10} 1 = n = 10
∴ (8/5)² + (12/5)² + (16/5)² + ... upto 10 terms = 16×505/25
It is given that 16×505/25 = 16/5 m
∴ m = 505/5 = 101

1. Let p = lim_{x→0+} (1 + tan²√x)^(1/(2x)) then log p is equal to:

(1) 2
(2) 1
(3) 1/2
(4) 1/4

1. (3)

p = lim_{x→0+} {1 + tan²√x}^{1/tan²√x × tan²√x/(2x)}
= e^{tan²√x/(√x)² × 1/2} = e^{1/2}
log_e p = 1/2

===== Page 24 =====

1. For x ∈ R, f(x) = |log 2 - sin x| and g(x) = f(f(x)), then:

(1) g is not differentiable at x = 0
(2) g'(0) = cos(log 2)
(3) g'(0) = -cos(log 2)
(4) g is differentiable at x = 0 and g'(0) = -sin(log 2)

1. (2)

g(x) = |log_e2 - sin(|log_e2 - sin x|)|
At x = 0, g(x) = log_e(2) - sin(log_e2 - sin x)
∴ g'(x) = cos(log_e(2) - sin x) × cos(x)
⇒ g'(0) = cos(log_e(2))

1. Consider

f(x) = tan^-1(√((1 + sin x)/(1 - sin x))), x ∈ (0, π/2).

A normal to y = f(x) at x = π/6 also passes through the point:

(1) (0,0)
(2) (0, 2π/3)
(3) (π/6, 0)
(4) (π/4, 0)

1. (2)

f(x) = tan^-1(tan(π/4 + x/2)) = π/4 + x/2
f at x = π/6, f(π/6) = π/4 + π/12 = 4π/12 = π/3
f'(x) = 1/2
Normal y = π/3 = -2(x - π/6)
So, (0, 2π/3) Satisfies

1. A wire of length 2 units is cut into two parts which are bent respectively to form a square of side = x units and a circle of radius = r units. If the sum of the areas of the square and the circle so formed is minimum, then:

(1) 2x = (π + 4)r
(2) (4 - π)x = πr
(3) x = 2r
(4) 2x = r

1. (3)

Let length of two parts be 'a' and '2 - a'
As per condition given, we write
a = 4x and 2 - a = 2πr
∴ x = a/4 and r = (2 - a)/(2π)
∴ A(square) = (a/4)² = a²/16 and
A(circle) = π[(2 - a)/(2π)]² = π(4 + a² - 4a)/(4π²)
= (a² - 4a + 4)/(4π)

===== Page 25 =====

f(a) = a²/16 + (a² - 4a + 4)/(4π)
∴ f(a) = (a²π + 4a² - 16a + 16)/(16π)
∴ f'(a) = 1/(16π)[2aπ + 8a - 16]
f'(a) = 0 ⇒ 2aπ + 8a - 16 = 0 ⇒ 2aπ + 8a = 16
∴ 2a(π + 4) = 16 ⇒ a = 8/(π + 4)
x = a/4 = 2/(π + 4) and r = (2 - a)/(2π) = (2 - 8/(π + 4))/(2π)
= (2π + 8 - 8)/(2π(π + 4)) = 1/(π + 4)
∴ x = 2/(π + 4) and r = 1/(π + 4) ⇒ x = 2r

1. The integral ∫ (2x^12 + 5x^9)/(x^5 + x^3 + 1)^3 dx is equal to:

(1) -x^5/(x^5 + x^3 + 1)² + C
(2) x^10/(2(x^5 + x^3 + 1)²) + C
(3) x^5/(2(x^5 + x^3 + 1)²) + C
(4) -x^10/(2(x^5 + x^3 + 1)²) + C

1. (2)

∫ (2x^12 + 5x^9)/(x^5 + x^3 + 1)^3 dx = ∫ (2x^12 + 5x^9)/(x^15(1 + 1/x² + 1/x^5)^3) dx
Dividing numerator and denominator by x^15 we get,
= ∫ (2/x^3 + 5/x^6)/(1 + 1/x² + 1/x^5)^3 dx
Put (1 + 1/x² + 1/x^5) = t
= ∫ -dt/t³
= -t^(-3+1)/(-3+1) + C = 1/2 × 1/t² + C
= 1/2 1/(1 + 1/x² + 1/x^5)² + C
= 1/2 x^10/(x^5 + x^3 + 1)² + C

===== Page 26 =====

1. lim_{n→∞} (((n + 1)(n + 2)...3n)/n^(2n))^(1/n) is equal to:
   (1) 18/e^4
   (2) 27/e^2
   (3) 9/e^2
   (4) 3log3 - 2

2. (2)

Let P = lim_{n→∞} [(n + 1)/n · (n + 2)/n · ... (n + 2n)/n]^(1/n)
Taking log
Log P = lim_{n→∞} 1/n Σ_{r=1}^{2n} log(1 + r/n)
= ∫_0^2 log(1 + x)dx
= log(1 + x)·x|_0^2 - ∫_0^2 x/(1 + x) dx
= 2ln3 - ∫_0^2 (1 - 1/(1 + x))dx = 2ln3 - [x_0^2 - ln(1 + x)]_0^2
= 2ln3 - [2 - ln3] = 3ln3 - 2
= ln3³ - ln e²
= ln(27/e²)
P = 27/e²

1. The area (in sq. units) of the region {(x,y): y² ≥ 2x and x² + y² ≤ 4x, x ≥ 0, y ≥ 0} is:

(1) π - 4/3
(2) π - 8/3
(3) π - 4√2/3
(4) π/2 - 2√2/3

1. (2)

x² + y² ≤ 4x & y² ≥ 2x
To find point of intersection,
x² + y² = 4x ⇒ x² + 2x = 4x
⇒ x² - 2x = 0 ⇒ x(x - 2) = 0
⇒ x = 0 or x = 2
∴ y = 0 or y = 2
Solve (x,y) = (0,0) & (x,y) = (2,2)
Area = ∫_0^2 (y1 - y2)dx = ∫_0^2 y_circle dx - ∫_0^2 y_parabola dx
= π×r²/4 - ∫_0^2 √2·x^(1/2) dx
= π×4/4 - √2·2/3 × x^(3/2)|_0^2
= π - 2√2/3·(2^(3/2) - 0) = π - 8/3

===== Page 27 =====

1. If a curve y = f(x) passes through the point (1, -1) and satisfies the differential equation, y(1 + xy) dx = x dy, then f(-1/2) is equal to:

(1) -2/5
(2) -4/5
(3) 2/5
(4) 4/5

1. (4)

y/x(1 + xy) = dy/dx
y = vx ⇒ y/x = v
dy/dx = v + x dv/dx
v(1 + vx²) = v + x dv/dx
v²x² = x dv/dx
v²x = dv/dx
∫ x dx = ∫ 1/v² dv
x²/2 = -1/v + c
x²/2 = -x/y + c
Put (1, -1)
1/2 = 1/1 + c ⇒ c = -1/2
x²/2 = -x/y - 1/2
We have to find f(-1/2)
Put x = -1/2
(-1/2)²/2 = -(-1/2)/y - 1/2
1/8 = 1/(2y) - 1/2
y = 4/5

1. Two sides of a rhombus are along the lines, x - y + 1 = 0 and 7x - y - 5 = 0. If its diagonals intersect at (-1, -2), then which one of the following is a vertex of this rhombus?

(1) (-3, -9)
(2) (-3, -8)
(3) (1/3, -8/3)
(4) (-10/3, -7/3)

===== Page 28 =====

1. (3) Coordinates of A ≡ (1, 2) : Slope of AE = 2
   ⇒ Slope of BD = -1/2
   ⇒ Eq. of BD is (y + 2)/(x + 1) = -1/2
   ⇒ x + 2y + 5 = 0 : Co-ordinates of D = (1/3, -8/3)

2. The centres of those circles which touch the circle, x² + y² - 8x - 8y - 4 = 0, externally and also touch the x-axis, lie on:

(1) a circle
(2) an ellipse which is not a circle
(3) a hyperbola
(4) a parabola

1. (4)

x² + y² - 8x - 8y - 4 = 0
Centre (4,4)
Radius = √(4² + 4² + 4) = 6
Let centre of the circle is (h,k)
√((h - 4)² + (k - 4)²) = (6 + k)
(h - 4)² + (k - 4)² = (6 + k)²
h² - 8h + 16 + k² - 8k + 16 = 36 + k² + 12k
h² - 8h - 20k - 4 = 0
x² - 8x - 20y - 4 = 0
Which is an equation of parabola

1. If one of the diameters of the circle, given by the equation, x² + y² - 4x + 6y - 12 = 0, is a chord of a circle S, whose centre is at (-3, 2), then the radius of S is:

(1) 5√2
(2) 5√3
(3) 5
(4) 10

1. (2)

x² + y² - 4x + 6y - 12 = 0
Centre (2, -3)
Radius √(4 + 9 + 12) = 5
Distance b/w two centres c1(2, -3) and c2(-3, 2)
d = √((2 + 3)² + (-3 - 2)²) = √50
Radius of (S) = √(5² + (√50)²) = √75 = 5√3

1. Let P be the point on the parabola, y² = 8x which is at a minimum distance from the centre C of the circle, x² + (y + 6)² = 1. Then the equation of the circle, passing through C and having its centre at P is:

x² + y² - 4x + 8y + 12 = 0
x² + y² - x/4 + 2y - 24 = 0

===== Page 29 =====

1. (1)

Normal at P(at², 2at) is y + tx = 2at + at³
Given it passes (0, -6)
⇒ -6 = 2at + at³ (a = 2)
-6 = 4t + 2t³
t³ + 2t + 3 = 0
t = -1
so, P(a, -2a) = (2, -4). [a = 1]
radius of circle = CP = √(2² + (-4 + 6)²) = 2√2
Circle is (x - 2)² + (y + 4)² = (2√2)²
x² + y² - 4x + 8y + 12 = 0

1. The eccentricity of the hyperbola whose length of the latus rectum is equal to 8 and the length of its conjugate axis is equal to half the distance between its foci, is:

(1) 4/3
(2) 4/√3
(3) 2/√3
(4) √3

1. (3)

? = 2b²/a = 8 ⇒ b² = 4a ...(1)
2b = 1/2(2ae)
2b = ae ...(2)
Squaring eqn. (2), we get
4b² = a²e² ⇒ 4b²/a² = e² and we know that e² = 1 + b²/a² ⇒ b²/a² = e² - 1
4(e² - 1) = e²
4e² - e² = 4
3e² = 4
e = 2/√3

1. The distance of the point (1, -5, 9) from the plane x - y + z = 5 measured along the line x = y = z is:

(1) 3√10
(2) 10√3
(3) 10/√3
(4) 20/3

1. (2)

Let Q (1, -5, 9)
(x - 0)/1 = (y - 0)/1 = (z - 0)/1
Line is (x - 1)/1 = (y + 5)/1 = (z - 9)/1 = r (say)
Any pt on line we can take P(r + 1, r - 5, r + 9)
So, Pt satisfy Plane
⇒ (r + 1) - (r - 5) + (r + 9) = 5
r = -10
So, Point P = (-9, -15, -1)
Distance is PQ = √((10)² + (10)² + (10)²) = 10√3

===== Page 30 =====

1. If the line, (x - 3)/2 = (y + 2)/-1 = (z + 4)/3 lies in the plane, ?x + my - z = 9, then ?² + m² is equal to:

(1) 26
(2) 18
(3) 5
(4) 2

1. (4)

Point on line is P = (3, -2, -4)
'P' lies on ?x + my - z = 9
⇒ 3? - 2m + 4 = 9
3? - 2m = 5 ...(1)
As line lies on plane ⇒ 2×? + m×(-1) + 3×(-1) = 0
2? - m = 3 ...(3)
Solving ? = 1, m = -1
So, ?² + m² = 2.

1. Let a, b and c be three unit vectors such that a × (b × c) = √3/2 (b + c). If b is not parallel to c, then the angle between a and b is:

(1) 3π/4
(2) π/2
(3) 2π/3
(4) 5π/6

1. (4)

a × (b × c) = √3/2 (b + c)
(a·c)b - (a·b)c = √3/2 b + √3/2 c
Equate a·c = √3/2 & -(a·b) = √3/2
|a| |b| cosθ = -√3/2
cosθ = -√3/2 as a & b unit vectors
θ = 5π/6

1. If the standard deviation of the numbers 2, 3, a and 11 is 3.5, then which of the following is true?

(1) 3a² - 26a + 55 = 0
(2) 3a² - 32a + 84 = 0
(3) 3a² - 34a + 91 = 0
(4) 3a² - 23a + 44 = 0

1. (2)

S.D. = √(Σd²/N - (Σd/N)²)
⇒ (4 + 9 + a² + 121)/4 - ((2 + 3 + a + 11)/4)² = 49/4
⇒ (4a² + 536) - (a² + 32a + 256) = 196
⇒ 3a² - 32a + 84 = 0

===== Page 31 =====

1. Let two fair six-faced dice A and B be thrown simultaneously. If E1 is the event that die A shows up four, E2 is the event that die B shows up two and E3 is the event that the sum of numbers on both dice is odd, then which of the following statements is NOT true?

(1) E1 and E2 are independent
(2) E2 and E3 are independent
(3) E1 and E3 are independent
(4) E1, E2 and E3 are independent

1. (4)

P(E1) = 1/6
P(E2) = 1/6
P(E3) = 1/2
P(E1 ∩ E2) = 1/36
P(E2 ∩ E3) = 1/12
P(E3 ∩ E1) = 1/12
P(E1 ∩ E2 ∩ E3) = 0
∴ E1, E2, E3 are not independent

1. If 0 ≤ x < 2π, then the number of real values of x, which satisfy the equation cos x + cos 2x = cos 3x + cos 4x = 0, is:

(1) 3
(2) 5
(3) 7
(4) 9

1. (3)

We have, cos x + cos 2x + cos 3x + cos 4x = 0
(cos x + cos 4x) + (cos 2x + cos 3x) = 0
2cos(5x/2)·cos(3x/2) + 2cos(5x/2)·cos(x/2) = 0
2cos(5x/2)[cos(3x/2) + cos(x/2)] = 0
cos(5x/2) = 0 ⇒ 5x/2 = (2n + 1)π/2 ⇒ x = (2n + 1)π/5 ⇒ x = π/5, 3π/5, π, 7π/5, 9π/5
Or
cos(3x/2) + cos(x/2) = 0
4cos³(x/2) - 3cos(x/2) + cos(x/2) = 0 ⇒ 4cos³(x/2) - 2cos(x/2) = 0 ⇒ 2cos(x/2)[2cos²(x/2) - 1] = 0
2cos(x/2)[cos x] = 0
cos(x/2) = 0 ⇒ x/2 = (2n + 1)π/2 ⇒ x = (2n + 1)π or cos x = 0 ⇒ x = (2n + 1)π/2
x = π or x = π/2, 3π/2
Solution are x = π/5, 3π/5, π, 7π/5, 9π/5, π/2, 3π/2 ... (0 ≤ x < 2π)

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1. A man is walking towards a vertical pillar in a straight path, at a uniform speed. At a certain point A on the path, he observes that the angle of elevation of the top of the pillar is 30°. After walking for 10 minutes from A in the same direction, at a point B, he observes that the angle of elevation of the top of the pillar is 60°. Then the time taken (in minutes) by him, from B to reach the pillar, is:

(1) 6
(2) 10
(3) 20
(4) 5

1. (4)

Let AB = x, BQ = y, PQ = z
tan30° = z/(x + y) ⇒ z = (x + y)/√3
tan60° = z/y ⇒ z = √3 y
∴ (x + y)/√3 = √3 y ⇒ x + y = 3y ⇒ x = 2y
∴ y = x/2
To go x, it takes 10 minutes.
To go y, it takes 5 minutes.

1. The Boolean Expression (p ∧ ~q) ∨ q ∨ (~p ∧ q) is equivalent to:

(1) ~p ∧ q
(2) p ∧ q
(3) p ∨ q
(4) p ∨ ~q

1. (3)

(p ∧ ~q) ∨ q ∨ (~p ∧ q)

p q ~p ~q p∧~q ~p∧q (p∧~q)∨q (p∧~q)∨q∨(~p∧q) p∨q
T T F F F F T T T
T F F T T F T T T
F T T F F T T T T
F F T T F F F F F

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1. The test is of 3 hours duration.

2. The Test Booklet consists of 90 questions. The maximum marks are 360.

3. There are three parts in the question paper A, B, C consisting of Mathematics, Physics and Chemistry having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for each correct response.

4. Candidates will be awarded marks as stated above in Instructions No. 3 for correct response of each question. 1/4 (one-fourth) marks will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer sheet.

5. There is only one correct response for each question. Filling up more than one response in each question will be treated as wrong response and marks for wrong response will be deducted accordingly as per instruction 4 above.

6. For writing particulars/marking responses on Side-1 and Side-2 of the Answer Sheet use only Blue/Black Ball Point Pen provided by the Board.

7. No candidate is allowed to carry any textual material, printed or written, bits of papers, pager, mobile phone, any electronic device, etc. except the Admit Card inside the examination hall/room.

FOR COMPLETE QUESTION PAPER WITH DIAGRAM AND IMAGE REFER PAGE NO 381 to 413 from Given PDF link 

JEE MAINS Previous Year Question paper solwed question paper free pdf download link

JEE MAINS PREVIOUS YEAR QUESTION PAPER 015

JEE MAINS PREVIOUS YEAR QUESTION PAPER 2015 CODE C