PHYSICS
SECTION A
Q. 1: Match List-I with List-II:
| List-I | List-II |
| (A)h (Planck’s constant) | (i) [MLT–1] |
| (B)E (Kinetic energy) | (ii) [ML2T–1] |
| (C)V (Electric potential) | (iii) [ML2T–2] |
| (D)P (Linear momentum) | (iv) [ML2I–1T–3] |
Choose the correct answer from the options given below:
(A) (A) → (ii), (B) →(iii), (C) → (iv), (D) → (i)
(B) (A) →(i), (B) → (ii), (C) →(iv), (D) → (iii)
(C) (A) → (iii), (B) → (ii), (C) →(iv), (D) →(i)
(D) (A) → (iii), (B) → (iv), (C) →(ii), (D) →(i)
Answer: (A)
K.E. = [ML2T–2]
P (Linear momentum) = [MLT–1]
h (Planck’s constant) = [ML2T–1]
V (Electric potential) = [ML2T–3I–1]
Q. 2: The pitch of the screw gauge is 1 mm and there are 100 divisions on the circular scale. When nothing is put in between the jaws, the zero of the circular scale lines 8 divisions below the reference line. When a wire is placed between the jaws, the first linear scale division is clearly visible while the 72nd division on circular scale coincides with the reference line. The radius of the wire is:
(A) 1.64 mm
(B) 1.80 mm
(C) 0.82 mm
(D) 0.90 mm
Answer: (C)
Least count. = pitch/no. of div.
= 1 mm/100
= 0.01 mm
+ve zero error = 8 × L.C. = +0.08 mm
Measured reading = 1mm + 72 × L.C.
= 1mm + 0.72 mm
= 1.72 mm
True reading = 1.72 – 0.08
= 1.64 mm
Radius = 1.64/2 = 0.82 mm
Q. 3: If the time period of a two meter long simple pendulum is 2 s, the acceleration due to gravity at the place where pendulum is executing S.H.M. is:
(A) 2π2 ms–2
(B) 16 m/s2
(C) 9.8 ms–2
(D) π2 ms–2
Answer: (A)
T = 2π√(l/g)
T2 = 4π2l/g
g = 4π2l/T2
= 4π2×2/(2)2 = 2π2 ms–2
Q. 4: An α particle and a proton are accelerated from rest by a potential difference of 200 V. After this, their de Broglie wavelengths are λα and λp respectively. The ratio λp/λα is:
(A) 8
(B) 2.8
(C) 3.8
(D) 7.8
Answer: (B)
λ = h/p
= h/√(2mqV)
λp/λα = √(mαqα/mpqp)
= √(4×2/1×1)
= 2√2 = 2.8
Q. 5: Given below are two statements: one is labelled as Assertion A and the other is labelled as reason R.
Assertion A: The escape velocities of planet A and B are same. But A and B are of unequal masses.
Reason R: The product of their masses and radii must be same. M1R1 = M2R2
In the light of the above statements, choose the most appropriate answer from the options given below:
(A) Both A and R are correct but R is NOT the correct explanation of A
(B) A is correct but R is not correct
(C) Both A and R are correct and R is the correct explanation of A
(D) A is not correct but R is correct
Answer: (B)
Ve = escape velocity
ve = √(2GM/R)
So, for same ve, M1/R1 = M2/R2
A is true but R is false
Q. 6: A diatomic gas, having Cp = (7/2)R and Cv = (5/2)R, is heated at constant pressure. The ratio dU : dQ : dW
(A) 3 : 7 : 2
(B) 5 : 7 : 2
(C) 5 : 7 : 3
(D) 3 : 5 : 2
Answer: (B)
Since the gas is diatomic in nature and the process is isobaric, we have
Cp = (7/2)R
Cv = (5/2)R
dU = nCvdT
dQ = nCpdT
dW = nRdT
dU : dQ : dW
Cv : Cp : R
(5/2) R : (7/2) R : R
5 : 7 : 2
Q. 7: Statement I: A speech signal of 2 kHz is used to modulate a carrier signal of 1 MHz. The bandwidth requirement for the signal is 4 kHz.
Statement II: The side band frequencies are 1002 kHz and 998 kHz.
In the light of the above statements, choose the correct answer from the options given below:
(A) Both statement I and statement II are false
(B) Statement I is false but statement II is true
(C) Statement I is true but statement II is false
(D) Both statement I and statement II are true
Answer: (D)
Side band = (fc – fm) to (fc + fm)
= (1000 – 2) kHz to (1000 + 2) kHz
= 998 kHz to 1002 kHz
Band width = 2fm
= 2 ×2kHz
= 4 kHz
Both statements are true.
Q. 8: The current (i) at time t = 0 and t = ∞ respectively for the given circuit is:
(A) 18E/55,5E/18
(B) 5E/18,18E/55
(C) 5E/18,10E/33
(D) 10E/33,5E/18
Answer: (C)
At t = 0, inductor is open
Initial-state equivalent of the circuit shown in figure -1 is
Req = 6×9/(6+9) = 54/15
I (t = 0) = E×15/54 = 5E/18
At t = ∞, For steady state inductor is replaced by plane wire
Steady state equivalent of the circuit shown in figure-1 is
Equivalent circuit diagram is given by
Req =1×4/(1+4) + 5×5/(5+5)
= 4/5 + 5/2
= (8 + 25)/10
= 33/10
I = E/Req = 10E/33
Q. 9: Two satellites A and B of masses 200 kg and 400 kg are revolving round the earth at height of 600 km and 1600 km respectively.
If TA and TB are the time periods of A and B respectively then the value of TB – TA:
[Given: Radius of earth = 6400 km, mass of earth = 6×1024 kg]
(A) 4.24× 102 s
(B) 3.33 × 102 s
(C) 1.33 × 103 s
(D) 4.24 × 103 s
Answer: (C)
V = √(GMe/r)
T = 2𝜋r/ (√(GMe/r)) = 2𝜋r/ (√(r/GMe))
T =√(4𝜋2r3/GMe) = √(4𝜋2r3/GMe)
T2 – T1 = √(4𝜋2(8000×103)3/G×6×1024 ) – √(4𝜋2(7000×103)3/G×6×1024 )
≅ 1.33 ×103 s
Q. 10: An engine of a train, moving with uniform acceleration, passes the signal post with velocity u and the last compartment with velocity v. The velocity with which middle point of the train passes the signal post is:
(A) √(v2 – u2)/2
(B) (v – u)/2
(C) √(v2 + u2)/2
(D) (u + v)/2
Answer: (C)
a = uniform acceleration
u = velocity of first compartment
v = velocity of last compartment
l = length of train
v2 = u2 + 2as (3rd equation of motion)
v2 = u2 + 2al …..(1)
v2 middle = u2 + 2al/2
∴ v 2middle = u2 + al ….(2)
From equation (1) and (2)
v2middle = u2 + (v2 – u2)/2
= (v2 + u2)/2
∴ vmiddle = √(v2 + u2)/2
Q. 11: A 5V battery is connected across the points X and Y. Assume D1 and D2 to be normal silicon diodes. Find the current supplied by the battery if the +ve terminal of the battery is connected to point X.
(A) ~0.86 A
(B) ~0.5 A
(C) ~0.43 A
(D) ~1.5 A
Answer: (C)
Since silicon diode is used so 0.7 Volt is drop across it, only D1 will conduct so current through cell.
I = (5 – 0.7)/10 = 0.43 A
Q. 12: A solid sphere of radius R gravitationally attracts a particle placed at 3R from its centre with a force F1. Now a spherical cavity of radius R/2 is made in the sphere (as shown in figure) and the force becomes F2. The value of F1: F2 is:
(A) 41 : 50
(B) 36 : 25
(C) 50 : 41
(D) 25 : 36
Answer: (A)
Gravitational field intensity g1 = GM/(3R)2 = GM/9R2 …(1)
Gravitational field intensity g2 = GM/9R2 – G(M/8)/(5R/2)2
= GM/9R2 – GM/R250
= (41/9×50) GM/R2….(2)
Implies , g1/g2 = 41/50
⇒ F1/F2 = mg1/mg2 = 41/50
Q. 13: A student is performing the experiment of resonance column. The diameter of the column tube is 6 cm. The frequency of the tuning fork is 504 Hz. Speed of the sound at the given temperature is 336 m/s. The zero of the metre scale coincides with the top end of the resonance column tube. The reading of the water level in the column when the first resonance occurs is:
(A) 13 cm
(B) 14.8 cm
(C) 16.6 cm
(D) 18.4 cm
Answer: (B)
λ = v/f
= 336/504 = 66.66cm
λ/4 = l + e = l + 0.3d
= l + 1.8
16.66 = l + 1.8 cm
l = 14.86 cm
Q. 14: A proton, a deuteron and an α particle are moving with the same momentum in a uniform magnetic fiel(D) The ratio of magnetic forces acting on them is _______ and their speeds are in the ratio______.
(A) 2 : 1 : 1 and 4 : 2 : 1
(B) 1 : 2 : 4 and 2 : 1 :1
(C) 1 : 2 : 4 and 1 : 1 : 2
(D) 4 : 2 : 1 and 2 : 1 : 1
Answer: (A)
As v = p/m & F = qvB
∴ F = qpB/m
F1 = qpB/m, v1 = p/m
F2 = qpB/2m, v2 = p/2m
F3 = 2qpB/4m, v3 = p/4m
F1 : F2 : F3 & V1 : V2 : V3
1 : ½ : ½ & 1 : ½ :¼
2 : 1 : 1 & 4 : 2 : 1
Q. 15: Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: When a rod lying freely is heated, no thermal stress is developed in it.
Reason R: On heating, the length of the rod increases
In the light of the above statements, choose the correct answer from the options given below:
(A) A is true but R is false
(B) Both A and R are true and R is the correct explanation of A
(C) Both A and R are true but R is NOT the correct explanation of A
(D) A is false but R is true
Answer: (C)
If a rod is free and it is heated then there is no thermal stress produced in it.
The rod will expand due to increase in temperature.
So, both A & R are true.
Q. 16: In an octagon ABCDEFGH of equal side, what is the sum of
Answer: (A)
Q. 17: Two radioactive substances X and Y originally have N1 and N2 nuclei respectively. Half-life of X is half of the half-life of Y. After three half-lives of Y, numbers of nuclei of both are equal. The ratio N1/N2 will be equal to:
(A) 8/1
(B) 1/8
(C) 3/1
(D) 1/3
Answer: (A)
After n half-life no of nuclei undecayed = No/2n
Given, t(1/2)x= t(1/2y)/2
So 3 half-life of y = 6 half-life of x
Given, Nx = Ny after 3t(½)y
N1/26 = N2/23
N1/N2 = 26/23 = 23= 8/1
Q. 18: The angular frequency of alternating current in a L-C-R circuit is 100 rad/s. The components connected are shown in the figure. Find the value of inductance of the coil and capacity of condenser.
(A) 0.8 H and 250 μF
(B) 0.8 H and 150 μF
(C) 1.33 H and 250 μF
(D) 1.33 H and 150 μF
Answer: (A)
Since key is open, circuit is series L-C-R circuit
15 = irms (60)
∴irms = ¼ A
Now, 20 = ¼ XL = ¼ (ωL)
∴ L = 4/5 = 0.8 H
& 10 = ¼ 1/(100C)
C = (1/4000) F
= 250 μF
Q. 19: Two coherent light sources having intensities in the ratio 2x produce an interference pattern. The ratio (Imax – Imin)/(Imax + Imin) will be:
(A) 2√(2x)/(x + 1)
(B) √(2x)/(2x + 1)
(C) 2√(2x)/(2x + 1)
(D) √(2x)/(x + 1)
Answer: (C)
Let I1 = 2x
I2 = 1
Imax = (√I1 + √I2)2
Imin = (√I1 – √I2)2
(Imax-Imin)/ (Imax+ Imin) = [(√2x + 1)2 -(√2x – 1)2] / [(√2x + 1)2 +(√2x – 1)2]
= 4√(2x)/(2 + 4x)
= 2√(2x)/(2x + 1)
Q. 20: Magnetic fields at two points on the axis of a circular coil at a distance of 0.05 m and 0.2 m from the centre are in the ratio 8 : 1. The radius of coil is ______
(A) 0.15 m
(B) 0.2 m
(C) 0.1 m
(D) 1.0 m
Answer: (C)
B = μ0NiR2/2(R2 + x2)3/2
at x1 = 0.05m, B1 = μ0NiR2/2(R2 + (0.05)2)3/2
at x2 = 0.2m, B2 = μ0NiR2/2(R2 + (0.2)2)3/2
B1/B2 = (R2 + 0.04)3/2/(R2 + 0.0025)3/2
(8/1)2/3 = (R2 + 0.04)/(R2 + 0.0025)
4 (R2 + 0.0025) = R2 + 0.04
3R2 = 0.04 – 0.01
R2 = 0.03/3 = 0.01
R = √0.01 = 0.1 m
Section – B
Q. 1: The potential energy (U) of a diatomic molecule is a function dependent on r (interatomic distance) as U= α/r10– β/r5 – 3. Where, a and b are positive constants. The equilibrium distance between two atoms will (2α/β)a/(B) Where a =_______
Answer: 1
F = -dU/dr
F = -[-10α/r11 + 5β/r6]
for equilibrium, F = 0
10α/r11 = 5β/r6
2α/β = r5
r = (2α/β)1/5
a = 1
Q. 2: A small bob tied at one end of a thin string of length 1m is describing a vertical circle so that the maximum and minimum tension in the string are in the ratio 5 : 1. The velocity of the bob at the highest position is ______ m/s. (take g = 10 m/s2)
Answer: 5
By conservation of energy,
v2min = V2 – 4gl ….(1)
Tmax = mg + mv2/l ….(2)
Tmin = mv2min/l – mg ….(3)
from equation (1) and (3)
Tmin = (m/l) (v2 – 4gl) – mg
Tmax/Tmin = (v2/l + g)/(v2/l – 5g)
5/1 = (v2/1 + 10)/ (v2/1 – 50)
5v2 – 250 = v2 + 10
v2 = 65 ….(4)
from equation (4) and (1)
v2min = 65 – 40 = 25
vmin = 5 m/s
Q. 3: In a certain thermodynamic process, the pressure of a gas depends on its volume as kV3. The work done when the temperature changes from 1000 C to 3000 C will be ____ nR, where n denotes the number of moles of a gas.
Answer: 50
P = kv3
pv–3 = k
x = –3
w = nR(T1 – T2)/(x – 1)
= nR(100-300)/(-3 -1)
= nR(-200)/-4
= 50nR
Q. 4: In the given circuit of potentiometer, the potential difference E across AB (10 m length) is larger than E1 and E2 as well. For key K1 (closed), the jockey is adjusted to touch the wire at point J1 so that there is no deflection in the galvanometer. Now the first battery (E1) is replaced by the second battery (E2) for working by making K1 open and E2 close(D) The galvanometer gives then null deflection at J2. The value of E1/E2 is the smallest fraction of a/b, Then the value of a is ____.
Answer: 1
E1/E2 = l1/l2
= 3×100 cm + (100 – 20)cm)/(7×100 cm + 60 cm)
= 380/760
= ½
= a/b
a = 1
Q. 5: The same size images are formed by a convex lens when the object is placed at 20 cm or at 10 cm from the lens. The focal length of a convex lens is ______ cm.
Answer: 15
1/v – 1/u = 1/f …(1)
m = v/u ….(2)
from (1) and (2) we get
m = f/(f + u)
Given conditions
m1 = -m2
f/(f – 10) = -f/(f – 20)
f – 20 = –f + 10
2f = 30
f = 15 cm
Q. 6: A transmitting station releases waves of wavelength 960 m. A capacitor of 256 μF is used in the resonant circuit. The self inductance of coil necessary for resonance is _____ × 10–8H.
Answer: 10
At resonance
ωr = 1/√(Lc)
∴ 2πf = 1/√(Lc)
∴ 4π2c2/λ2 = 1/Lc
∴4π2×3×108×3×108/960×960 = 1/L×2.56×10-6
L = 375×960/10-6×4×π2×9×1016 = 103/1010
= 10–7 H
= 10 × 10–8 H
Q. 7: The electric field in a region is given by . The ratio of flux of reported field through the rectangular surface of area 0.2 m2 (parallel to y-z plane) to that of the surface of area 0.3 m2(parallel to x-z plane) is a : 2, where a = ________
[Here i, j and k are unit vectors along x, y and z-axes respectively]
Answer: 1
Therefore, a = 1
Q. 8: 512 identical drops of mercury are charged to a potential of 2 V each. The drops are joined to form a single drop. The potential of this drop is ____ V.
Answer: 128
Let charge on each drop = q
Radius = r
V = kq/r
2 = kq/r
Radius of bigger
4πR3/3 = 512 × (4/3) πr3
R = 8r
V = k(512)q/R = (512/8) (kq/r)
= (512/8) × 2
= 128 V
Q. 9: A monatomic gas of mass 4.0 u is kept in an insulated container. Container is moving with a velocity 30 m/s. If the container is suddenly stopped then the change in temperature of the gas (R=gas constant) is x/3R. Value of x is ______.
Answer: 3600
ΔKE = ΔU
ΔU = nCVΔT
½ mv2 = (3/2) nRΔT
mv2/3nR = ΔT
4×(30)2/3×1×R = ΔT
ΔT = 4×(30)2/3×1×R
x/3R = 1200/R
x = 3600
Q. 10: A coil of inductance 2 H having negligible resistance is connected to a source of supply whose voltage is given by V = 3t volt. (where t is in second). If the voltage is applied when t = 0, then the energy stored in the coil after 4 s is _______ J.
Answer: 144
L di/dt = ε
= 3t
L ∫di = 3 ∫t dt
Li = 3t2/2
i = 3t2/2L
Energy stored in the coil, E = ½ Li2
= ½ L (3t2/2L)2
= ½ × 9t4/4L
= 9/8 × (4)4/2
= 144 J
Chemistry
SECTION A
Q. 1. The hybridization and magnetic nature of [Mn(CN)6]4– and [Fe(CN)6]3–, respectively are:
(A) d2sp3 and paramagnetic
(B) sp3d2 and paramagnetic
(C) d2sp3 and diamagnetic
(D) sp3d2 and diamagnetic
Answer: (A)
Q. 2. Identify A and B in the chemical reaction.
Answer: (D)
Q. 3. Compound(s) which will liberate carbon dioxide with sodium bicarbonate solution is/are:
(A) 2 and 3 only
(B) 1 only
(C) 2 only
(D) 3 only
Answer: (A)
Compounds that are more acidic than H2CO3, gives CO2 gas in reaction with NaHCO3. Compound B i.e. Benzoic acid and compound C i.e. picric acid both are more acidic than H2CO3.
Q. 4. Ellingham diagram is a graphical representation of:
(A) ΔG vs T
(B) (ΔG – TΔS) vs T
(C) ΔH vs T
(D) ΔG vs P
Answer: (A)
Ellingham diagram tells us about the spontaneity of a reaction with temperature.
Q. 5. Which of the following equations depicts the oxidizing nature of H2O2?
(A) Cl2 + H2O2 → 2HCl + O2
(B) KlO4 + H2O2 → KlO3 + H2O + O2
(C) 2l– + H2O2 + 2H+ → I2 + 2H2O
(D) I2 + H2O2 + 2OH– → 2I– + 2H2O + O2
Answer: (C)
2l– + H2O2 + 2H+ → I2 + 2H2O
Oxygen reduces from –1 to –2
So, its reduction will take place. Hence it will behave as oxidising agent or it shows
oxidising nature.
Q. 6. In Freundlich adsorption isotherm at moderate pressure, the extent of adsorption (x/m) is directly proportional to Px. The value of x is:
(A) ∞
(B) 1
(C) zero
(D) 1/n
Answer: (D)
x / m = px
The formula is x / m = px
Hence, x= 1 / n. The value of ‘n’ is any natural number.
Q. 7. According to molecular orbital theory, the species among the following that does not exist is:
(A) He2–
(B) He2+
(C) O22-
(D) Be2
Answer: (D)
B.O. of Be2 is zero, so it does not exist.
Q. 8. In which of the following pairs, the outermost electronic configuration will be the same?
(A) Fe2+ and Co+
(B) Cr+ and Mn2+
(C) Ni2+ and Cu+
(D) V2+ and Cr+
Answer: (B)
Cr+ → [Ar]3d5
Mn2+ → [Ar]3d5
Q. 9. Given below are two statements:
Statement-I: An allotrope of oxygen is an important intermediate in the formation of reducing smog.
Statement-II: Gases such as oxides of nitrogen and Sulphur present in the troposphere contribute to the formation of photochemical smog.
(A) Statement I and Statement II are true
(B) Statement I is true about Statement II is false
(C) Both Statement I and Statement II are false
(D) Statement I is false but Statement II is true
Answer: (C)
Reducing smog acts as a reducing agent and the reducing character is due to presence of sulphur dioxide and carbon particles.
Q. 10. The plots of radial distribution functions for various orbitals of hydrogen atom against ‘r’ are given below:
(A) (4)
(B) (2)
(C) (1)
(D) (3)
Answer: (A)
3s orbital
Number of radial nodes = n – λ – 1
For 3s orbital n = 3, λ = 0
Number of radial nodes = 3 – 0 – 1 = 2
It is correctly represented in graph of option 4
Q. 11. Identify A in the given chemical reaction.
Answer: (D)
Aromatization reaction or hydroforming reaction.
Q. 12. Given below are two statements:
Statement-I: CeO2 can be used for oxidation of aldehydes and ketones.
Statement-II: Aqueous solution of EuSO4 is a strong reducing agent.
(A) Statement I is true, statement II is false
(B) Statement I is false, statement II is true
(C) Both Statement I and Statement II are false
(D) Both Statement I and Statement II are true
Answer: (D)
CeO2 can be used as an oxidizing agent like seO2. Similarly, EuSO4 is used as a reducing agent.
Q. 13. The major product of the following chemical reaction is:
(A) (CH3CH2CO)2O
(B) CH3CH2CHO
(C) CH3CH2CH3
(D) CH3CH2CH2OH
Answer: (B)
Q. 14. Complete combustion of 1.80 g of an oxygen-containing compound (CxHyOz) gave 2.64 g of CO2 and 1.08 g of H2O. The percentage of oxygen in the organic compound is:
(A) 63.53
(B) 51.63
(C) 53.33
(D) 50.33
Answer: (C)
n(CO2) = 2.64 / 44 = 0.06
nc = 0.06
weight of carbon = 0.06 × 12 = 0.72 gm
n(H2O) = 1.08 / 1.8 = 0.06
nH = 0.06 × 2 = 0.12
weight of H = 0.12 gm
∴ Weight of oxygen in CxHyOz
= 1.8 – (0.72 + 0.12)
= 0.96 gram
% weight of oxygen = 0.96/18 × 100 = 53.3%
Q. 15.The correct statement about B2H6 is:
(A) All B–H–B angles are 120°.
(B) Its fragment, BH3, behaves as a Lewis base.
(C) Terminal B–H bonds have less p-character when compared to bridging bonds.
(D) The two B–H–B bonds are not of the same length.
Answer: (C)
The terminal bond angle is greater than that of bridge bond angle
Bond angle ∝ S-character
α 1/ p−character
Q. 16. Which of the glycosidic linkage galactose and glucose is present in lactose?
(A) C-1 of glucose and C-6 of galactose
(B) C-1 of galactose and C-4 of glucose
(C) C-1 of glucose and C-4 of galactose
(D) C-1 of galactose and C-6 of glucose
Answer: (B)
Q. 17. Which one of the following reactions will not form acetaldehyde?
Answer: (A)
Q. 18. Which of the following reactions will not give p-amino azobenzene?
(A) 2 only
(B) 1 and 2
(C) 3 only
(D) 1 only
Answer: (A)
Q. 19. The solubility of AgCN in a buffer solution of pH = 3 is x. The value of x is: [Assume: No cyano complex is formed; Ksp(AgCN) = 2.2 × 10–16 and Ka (HCN) = 6.2 × 10–10]
(A) 0.625 × 10-6
(B) 1.6 × 10-6
(C) 2.2 × 10-16
(D) 1.9 × 10-5
Answer: (D)
Let solubility is x
AgCN ⇌ Ag+ + CN–Ksp = 2.2 × 10–16
x x
H+ + CN– ⇌ HCN
K = 1/ Ka
K= 1/ 6.2 × 10–10
Ksp × 1/ka = [Ag+] [CN–] × [HCN] / [H+][CN–]
2.2 × 10-16 × 1 / 6.2 × 10-10 = [S][S] / 10-3
S2 = 2.2 / 6.2 × 10-9
S2 = 3.55 × 10–10
S = √3.55 × 10–10
S = 1.88 × 10–5 ⇒ 1.9 × 10–5
Q. 20. Which statement is correct?
(A) Buna-S is a synthetic and linear thermosetting polymer
(B) Neoprene is an addition copolymer used in plastic bucket manufacturing
(C) Synthesis of Buna-S needs nascent oxygen
(D) Buna-N is a natural polymer
Answer: (C)
Synthesis of Buna-S needs nascent oxygen.
SECTION B
Q. 1. A car tire is filled with nitrogen gas at 35 psi at 27°C. It will burst if pressure exceeds 40 psi. The temperature in °C at which the car tyre will burst is ______. (Rounded-off to the nearest integer)
Answer: 69.85°C ≃ 70°C
P1 / T1 = P2 / T2
35 / 300 = 40 / T2
T2 = 40 × 300 / 35
= 342.86 K
= 69.85°C ≃ 70°C
Q. 2. The reaction of cyanamide, NH2CN(s) with oxygen was run in a bomb calorimeter and ΔU was found to be –742.24 kJ mol–1. The magnitude of ΔH298 for the reaction NH2CN (s) + 3/2 O2 (g) → N2 (g) + O2 (g) + H2O(l) is______kJ. (Rounded off to the nearest integer). [Assume ideal gases and R = 8.314 J mol–1 K–1]
Answer: 741 kJ/mol
NH2CN(s) + 3/2 O2(g) → N2(g) + CO2(g) + H2O(λ)
Δng = (1 + 1) – 3/2 = ½
ΔH = ΔU + Δng RT
= -742.24 + ½ × 8.314 × 298 / 1000
= -742.24 + 1.24
= 741 kJ/mol
Q. 3. The ionization enthalpy of Na+ formation from Na(g) is 495.8 kJ mol–1, while the electron gain enthalpy of Br is –325.0 kJ mol–1. Given the lattice enthalpy of NaBr is –728.4 kJ mol–1. The energy for the formation of NaBr ionic solid is (–)_____ × 10–1 kJ mol–1.
Answer: 5576 kJ
Na(s) → Na+(g) ΔH = 495.8
½ Br2(λ) + e– → Br–(g) ΔH = 325
Na+(g) + Br–(g) → NaBr(s) ΔH = –728.4
Na(s) + ½ Βr2(λ) → NaBr(s). ΔH =?
ΔH = 495.8 – 325 – 728.4
–557.6 kJ = –5576 × 10–1 kJ
Q. 4. Consider the following chemical reaction.
Answer: 7
All carbon atoms in benzaldehyde are sp2 hybridized.
Q. 5. Among the following, the number of halide(s) which is/are inert to hydrolysis is ______.
(A) BF3
(B) SiCl4
(C) PCl5
(D) SF6
Answer: 1
Due to crowding, SF6 is not hydrolyzed.
Q. 6. In basic medium CrO42– oxidizes S2O32– to form SO24 and itself changes into Cr(OH)4–. The volume of 0.154 M CrO42– required to react with 40 mL of 0.25 M S2O32– is ______ mL. (Rounded-off to the nearest integer)
Answer: 173 mL
17H2O + 8CrO4 + 3S2O3 → 6SO4 + 8Cr(OH)4– + 2OH–
Applying mole-mole analysis
0.154 × v / 8 = 40 × 0.25 / 3
V = 173 mL
Q. 7. 1 molal aqueous solution of an electrolyte A2B3 is 60% ionise(D) The boiling point of the solution at 1 atm is _____ K. (Rounded-off to the nearest integer). [Given Kb for (H2O) = 0.52 K kg mol–1]
Answer: 375 K
A2B3 → 2A+3 + 3B–2
No. of ions = 2 + 3 = 5
wi = 1 + (n – 1) ∝
= 1 + (5 – 1) × 0.6
= 1 + 4 × 0.6 = 1 + 2.4 = 3.4
ΔTb = Kb × m × i
= 0.52 × 1 × 3.4 = 1.768°C
ΔTb = (Tb)solution – [(TbH2O)]solution
1.768 = (Tb)solution – 100
(Tb)solution = 101.768 °C
= 375 K
Q. 8. Using the provided information in the following paper chromatogram:
The calculated Rf value of A ______ × 10–1.
Answer: 4
Rr = Distance Travelled By Solvent / Distance Travelled By Compound
On chromatogram distance travelled by compound is → 2 cm
Distance travelled by solvent = 5 cm
So Rf = 2 / 5 = 4 × 10–1 = 0.4
Q. 9. For the reaction, aA + bB → cC + dD, the plot of log k vs 1/T is given below:
The temperature at which the rate constant of the reaction is 10–4s–1 is ________ K. [Rounded off to the nearest integer)
[Given: The rate constant of the reaction is 10–5 s–1 at 500 K]
Answer: 526 K
log10 K = log10 A – E0/2.303RT
Slope = E0/2.203 = -10000
log10 K1/K2 = E0/2.203 × (1/T1 – 1/T2)
log10 10-4/10-5 = 1000 × (1/500 -1/T)
1 = 1000 × (1/500 -1/T)
1/10000 = 1/500 = 1/T
1/T = 1/500 = 1/10000
1/T = 20-1 / 10000 = 19/10000
T = 10000 / 19
T= 526 K
Q. 10. A 0.4g mixture of NaOH, Na2CO3 and some inert impurities was first titrated with N / 10HCl using phenolphthalein as an indicator, 17.5 mL of HCl was required at the end point. After this methyl orange was added and titrate(D) 1.5 mL of the same HCl was required for the next end point. The weight percentage of Na2CO3 in the mixture is ______. (Rounded-off to the nearest integer)
Answer: 3%
1st end point reaction
NaOH + HCl → NaCl + H2O
nf = 1
NaCO3 + HCl → NaHCO3
nf = 1
Eq of HCl used = n(NaOH) × 1 + n(Na2CO3) × 1
17.5 × 1/10 × 10-3 = n(NaOH) + n(Na2CO3)
2nd end point
NaHCO3 + HCl → H2CO3
1.5 × 1/10 × 10-3 = n(NaHCO3) × 1 = n(NaHCO3)
0.15 mmol = n(Na2CO3)
0.15 = n(Na2CO3)
W(Na2CO3) = 0.15×106×10-3 × 100 × 10
= 3 × 106 × 10–2
= 3 × 1.06 = 3.18%
Maths
SECTION A
Q. 1: A man is observing, from the top of a tower, a boat speeding towards the tower from a certain point A, with uniform spee(D) At that point, the angle of depression of the boat with the man’s eye is 30° (Ignore man’s height). After sailing for 20 seconds towards the base of the tower (which is at the level of water), the boat has reached point B, where the angle of depression is 45°. Then the time taken (in seconds) by the boat from B to reach the base of the tower is :
(A) 10(√3-1)
(B) 10√3
(C) 10
(D) 10(√3+1)
Answer: (D)
x + y = √3h ….…….(1)
Also,
h/y = tan 45o
h = y …….(2)
put in (1)
x + y = √3y
x = (√3 – 1)y
x/20=v (speed)
Therefore, time taken to reach
Foot from B
= y/V
= x/(√3-1)x . 20
= 10 (√3 + 1)
Q. 2:
(A) xyz = 4
(B) xy – z = (x + y)z
(C) xy + yz + zx = z
(D) xy + z = (x+y)z
Answer: (D)
Q. 3: The equation of the line through the point (0,1,2) and perpendicular to the line (x-1)/2 = (y+1)/3 = (z-1)/-2 is :
(A) x/-3 = (y-1)/ 4 = (z-2)/3
(B) x/3 = (y-1)/ 4 = (z-2)/3
(C) x/3 = (y-1)/ -4 = (z-2)/3
(D) x/3 = (y-1)/ 4 = (z-2)/-3
Answer: (A)
(x-1)/2 = (y+1)/3 = (z-1)/-2 =λ
Any point on this line (2λ + 1, 3λ – 1, -2λ + 1)
Direction ratio of given line d1 ≡ (2, 3, -2)
Direction ratio of the line to be found d2 ≡(2λ+1,3λ−2,−2λ−1)
∴ d1 d2 = 0
λ = 2 / 17
Direction ratio of line (21, -28, -21) ≡ (3, -4, -3) ≡ (-3, 4, 3)
Q. 4: The coefficients a,b and c of the quadratic equation, ax2 + bx + c = 0 are obtained by throwing a dice three times. The probability that this equation has equal roots is :
(A) 1/54
(B) 1/72
(C) 1/36
(D) 5/216
Answer: (D)
ax2 + bx + c = 0
a,b,c ∈ {1,2,3,4,5,6}
n(s) = 6 × 6 × 6 = 216
D=0 ⇒ b2 = 4ac
ac = b2/4, If b = 2, ac = 1 ⇒ a = 1,c = 1
If b = 4, ac = 4 ⇒ a = 1, c = 4
a = 4, c = 1
a = 2, c = 2
If b = 6, ac = 9 ⇒ a = 3, c = 3
Therefore, probability = 5/216
Q. 5: Let α be the angle between the lines whose direction cosines satisfy the equations l +m – n = 0 and l2 + m2 – n2 = 0. Then the value of sin4 α + cos4 α is :
(A) 3/4
(B) 1/2
(C) 5/8
(D) 3/8
Answer: (C)
Given that l+m=n ….(1)
l2 + m2 – n2 = 0 ….(2)
Squaring equation (1)
l2 + m2 + 2lm = n2 ….(3)
From equations (2) and (3)
lm=0 ⇒ l = 0 or m = 0
Case (1) : l = 0
⇒ m=n
⇒ l2 + m2 + n2 = 0
⇒ m=n = ±1/(√2 )
∴(l,m,n) = (0, 1/√2, 1/√2) or (0, -1/√2, -1/√2)
Case (2): m = 0
⇒ l = n
⇒ l2 + m2 + n2 = 0
Q. 6: The value of the integral
(where c is a constant of integration)
(A) 1/18[9 – 2 sin6θ – 3 sin4θ – 6 sin2θ ](3/2) + c
(B) 1/18[11 – 18 sin2θ + 9 sin4θ – 2 sin6θ ](3/2) + c
(C) 1/18[11 – 18 cos2θ + 9 cos4θ – 2 cos6θ ](3/2) + c
(D) 1/18[9 – 2 cos6θ – 3 cos4θ – 6 cos2θ ](3/2) + c
Answer: (C)
Using trig identities, sin2A = 2sinAcosA and 1-cos2A = 2sin2A
Q. 7: The statement A→ (B → A) is equivalent to:
(A) A → (A ᐱ B)
(B) A → (A ᐯ B)
(C) A → (A → B)
(D) A → (A ↔ B)
Answer: (B)
A → (B → A)
⇒ A → (∼ B ᐯ A)
⇒ ∼ A ᐯ (∼ B ᐯ A)
⇒ ∼ B ᐯ (∼ A ᐯ A)
⇒ ∼ B ᐯ t
= t (tautology)
From options:
(B) A → (A ᐯ B)
⇒ ∼ A ᐯ (A ᐯ B)
⇒ (∼ A ᐯ A) ᐯ B
⇒ t ᐯ B
⇒ t
Q. 8: The integer k, for which the inequality x2 – 2(3k – 1)x + 8k2 – 7 >0 is valid for every x in R is :
(A) 3
(B) 2
(C) 4
(D) 0
Answer: (A)
D < 0
(2(3k-1))2 – 4(8k2 – 7) < 0
4(9k2 – 6k + 1) – 4(8k2 – 7) < 0
k2 – 6 k + 8 < 0
(k–4)(k–2)<0
2< k < 4
then k = 3
Q. 9: All possible values of θ ∈ [0, 2π] for which sin2θ + tan2θ > 0 lie in
a.(0,π/2)∪(π,3π/2)
(B) (0,π/4)∪(π/2,3π/4)∪(π,5π/4)∪(3π/2,7π/4)
(C) (0,π/2)∪(π/2,3π/4)∪(π,7π/4)
(D) (0,π/4)∪(π/2,3π/4)∪(3π/2,11π/6)
Answer: (B)
Q. 10: The image of the point (3,5) in the line x – y + 1 = 0, lies on :
(A) (x – 2)2 + (y – 4) 2 =4
(B) (x – 4) 2 + (y + 2)2 =16
(C) (x – 4)2 + (y – 4)2 = 8
(D) (x – 2)2 + (y – 2)2 =12
Answer: (A)
Image of P(3, 5) on the line x – y + 1 = 0 is
(x-3)/1 = (y-5)/-1 = (-2(3-5+1))/ 2 = 1
x = 4, y = 4
Image is (4, 4)
Which lies on (x – 2)2 + (y – 4)2 = 4
Q. 11: If Rolle’s theorem holds for the function f(x) = x3 – ax2 + bx – 4, x ∈ [1, 2] with f'(4/3) = 0, then ordered pair (a, b) is equal to :
(A) (–5, 8)
(B) (5, 8)
(C) (5, –8)
(D) (–5, –8)
Answer: (B)
f(1) = f(2)
⇒ 1 – a + b –4 = 8 – 4a + 2b –4
3a – b = 7 …(1)
f’(x) = 3x2 – 2ax + b
⇒f’(4/3) = 0 ⇒ 3 x 16/9 – 8a/3 + b =0
⇒–8a + 3b = –16 …(2)
a = 5, b = 8
Q. 12: If a curve passes through the origin and the slope of the tangent to it at any point (x, y) is (x2-4x+y+8)/(x-2), then this curve also passes through the point :
(A) (4, 5)
(B) (5, 4)
(C) (4, 4)
(D) (5, 5)
Answer: (D)
dy/dx = [(x-2)2 +y+4]/(x-2) = (x-2) + (y+4)/(x-2)
Let x – 2 = t ⇒ dx = dt
and y + 4 = u ⇒dy = du
dy/dx = du/dt
du/dt = t + u/t ⇒ du/dt – u/t = t
Here, IF = 1/t
u. (1/t) = ∫ t.(1\t) dt
⇒ u/t = t + c
⇒ (y+4)/(y-2) = (x – 2) + c
Passing through (0, 0)
c = 0
⇒ (y + 4) = (x – 2)2
Q. 13: The value of ∫1-1 x2 ex^3 dx where [t] denotes the greatest integer ≤ t,is :
(A) (e+1)/3
(B) (e-1)/3e
(C) (e+1)/3e
(D) 1/3e
Answer: (C)
Q. 14: When a missile is fired from a ship, the probability that it is intercepted is 1/3 and the probability that the missile hits the target, given that it is not intercepted, is 3/4. If three missiles are fired independently from the ship, then the probability that all three hit the target, is:
(A) 1/8
(B) 1/27
(C) 3/4
(D) 3/8
Answer:(A)
Probability of not getting intercepted = 2/3
Probability of missile hitting target = 3/4
Probability that all 3 hit the target = (2/3 x3/4)3 = ⅛
Q. 15: If the curves, x2/a + y2/b and x2/c + y2/d intersect each other at an angle of 90°, then which of the following relations is true ?
(A) a + b = c + d
(B) a- b = c – d
(C) ab = (c+d)/(a+b)
(D) a-c = b+d
Answer: (B)
x2/a + y2/b ………(1)
diff :
2x/a +2y/b dx/dy =0
y/b . dx/dy = -x/a
dx/dy = -bx/ay……(2)
x2/c + y2/d………(3)
Diff :
dy/dx = -dx/cy ………(4)
m1m2 = –1
⇒ -bx/ay × -dx/cy =−1
⇒ bdx2 = – acy2…….(5)
(1)–(3) ⇒ (1/a – 1/c)x2 + (1/b – 1/d) y2 = 0
Solve above equation using 5
⇒ (c – a) – (d – b) = 0
⇒ c – a = d – b
⇒ c – d = a – b
Q. 16: A tangent is drawn to the parabola y2 = 6x which is perpendicular to the line 2x + y =1. Which of the following points does NOT lie on it?
(A) (0,3)
(B) (-6,0)
(C) (4,5)
(D) (5,4)
Answer: (D)
Equation of tangent : y = mx + 3/(2m)
mT = (1/2) (Because perpendicular to line 2x + y = 1)
Therefore, tangent is: y = x/2 + 3 ⇒ x – 2y + 6 = 0
Q. 17: Let f, g: N→N such that f(n + 1)= f(n)+ f(1) for all n ∈ N and g be any arbitrary function. Which of the following statements is NOT true ?
(A) f is one-one
(B) If fog is one-one, then g is one-one
(C) If g is onto, then fog is one-one
(D) If f is onto, then f(n) = n for all n ∈ N
Answer: (C)
f(n + 1) = f(n) +f(1)
⇒ f(n + 1)- f(n)= f(1) → A.P. with common difference = f(1)
General term = Tn = f(1) + (n-1)f(1) = nf(1)
⇒f(n) = nf(1)
Clearly f(n) is one-one.
For fog to be one-one, g must be one-one.
For f to be onto, f(n) should take all the values of natural numbers.
As f(x) is increasing, f(1)=1
⇒f(n)=n
If g is many one, then fog is many one. So “if g is onto, then fog is one-one” is incorrect.
Q. 18: Let the lines (2 – i)z = (2 + i)zˉ and (2 + i)z + (i – 2)zˉ– 4i = 0, (here i2 = –1) be normal to a circle C. If the line iz + zˉ+ 1 + i = 0 is tangent to this circle C, then its radius is :
(A) 3/√2
(B) 3√2
(C) 3/(2√2)
(D) 1/(2√2)
Answer: (C)
(2–i)z = (2+i)zˉ
⇒(2–i)(x+ iy)=(2+i)(x–iy)
⇒2x–ix + 2iy + y=2x+ ix–2iy+y
⇒2ix –4iy=0
L1 ∶ x–2y=0
⇒(2+i)z+(i –2)zˉ–4i=0.
⇒(2+i) (x + iy)+(i –2)(x–iy)– 4i = 0.
⇒2x +ix + 2iy – y + ix – 2x + y +2iy – 4i =0
⇒2ix + 4iy – 4i =0
L2 ∶ x+2y–2 = 0
Solve L1 and L2
4y=2 or y=1/2
∴ x = 1
Centre(1, 1/2)
L3∶ iz + zˉ+ 1 + i = 0
⇒i(x+iy)+x–iy+1+i=0
⇒ix–y+x–iy+1+i=0
⇒(x–y+1)+i(x–y+1)=0
Radius = distance from (1, 1/2) to x – y + 1 = 0
r = (1-1/2+1)/ √2
r = 3/2√2
Q. 19:
(A) 1/2
(B) 1/e
(C) 1
(D) 0
Answer: (C)
Q. 20: The total number of positive integral solutions (x, y, z) such that xyz = 24 is
(A) 36
(B) 45
(C) 24
(D) 30
Answer: (D)
x.y.z = 24
x.y.z= 23 × 31
x=2(a1) ⋅3(b1)
y = 2(a2) ⋅3(b2)
z = 2(a3)⋅3(b3)
a1, a2, a3 ∈ {0,1,2,3}
b1, b2, b3 ∈ {0,1}
Case 1: a1 + a2 + a3 =3
Non negative solution = (3+3-1) C(3-1) = 5C2 = 10
Case 2: b1 + b2 + b3 = 1
Non negative solution = (1+3-1) C(3-1) = 3C2 = 3
∴ Total solutions =10×3=30
Section B
Q. 1: Let A is a matrix where x, y and z are real numbers such that x + y + z > 0 and xyz = 2. If A2 = l3, then the value of x3 + y3 + z3 is ______.
Answer: 7
Q. 2: Let A1, A2, A3,…. be squares such that for each n ≥ 1, the length of the side of An equals the length of diagonal of An+1. If the length of A1 is 12 cm, then the smallest value of n for which area of An is less than one is ____________.
Answer: (9)
Q. 3: The locus of the point of intersection of the lines (√3)kx + ky – 4√3 = 0 and √3x – y – 4√3 k = 0 is a conic, whose eccentricity is _________.
Answer: (2)
(√3)kx + ky – 4√3 = 0 …(1)
√3kx -ky= 4√3 k2 …(2)
Adding equation (1) & (2)
2√3 kx = 4√3(k^2 + 1)
x = 2 (k + 1/k) ….(3)
Subtracting equation (1) & (2)
y = 2√3(1/k – k) ………(4)
x2/4 – y2/12 = 4
x2/16 – y2/48 = 1 – Hyperbola
e2 = 1 + 48/16
or e = 2
Q. 4:
Answer: 13
|I2 + A|=|I2 – A|
From equation (1)
∴a2 + b2 = 1
⇒13(a2 + b2) = 13
Q. 5: The graphs of sine and cosine functions, intersect each other at a number of points and between two consecutive points of intersection, the two graphs enclose the same area A. Then A4 is equal to ___________
Answer: (64)
Q. 6: The number of points, at which the function f(x) = |2x + 1| – 3|x+2|+|x2 + x–2|, x ∈ R is not differentiable, is ________.
Answer: 2
Check at 1, –2 and -1/2
Non-Differentiable at x = 1 and -1/2
Q. 7: If the system of equations
kx + y + 2z = 1
3x – y – 2z = 2
–2x – 2y – 4z = 3
has infinitely many solutions, then k is equal to __________.
Answer: 21
D = D1 = D2 = D3 = 0
Choose D2 = 0
k(–8+6)–1(–12–4)+2(9+4)=0
–2k + 16 + 26 = 0
2k = 42
k = 21
Q. 8: The total number of numbers, lying between 100 and 1000 that can be formed with the digits 1, 2, 3, 4,5, if the repetition of digits is not allowed and numbers are divisible by either 3 or 5 is ________.
Answer: 32
Q. 9: Let f(x) be a polynomial of degree 6 in x, in which the coefficient of x6 is unity and it has extrema at x = –1 and x = 1. If limx→0 f(x)/x3 =1, then 5.f(2) is equal to _________.
Answer: (144)
f(x) = x6 + ax5 + bx4 + x3
f’(x) = 6x5 + 5ax4 + 4bx3 + 3x2
Roots 1 & – 1
Therefore, 6 + 5a + 4b + 3 = 0 & – 6 + 5a – 4b + 3 = 0 solving
a = -3/5 and b = -3/2
f(x) = x6 – (3/5)x5 – (3/2)x4 + x3
5.f(2) = 5 [64 – 96/5 – 24 + 8] = 144
Q. 10: Let 
→r.→a =0 is equal to _______
Answer: 12
