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A helicopter is flying horizontally with a speed 'v' at an altitude 'h' has to drop a food packet for a man on the ground. What is the distance of helicopter from the man when the food packet is dropped?
(1) √((2ghv² + 1)/h²)
(2) √((2v²h)/g + h²)
(3) √((2gh)/v² + h²)
(4) √((2gh)/v² + h²)
Official Ans. by NTA (3)
Sol. R = √(2h/g)·v
D = √(R² + h²)
= √((2h/g·v)² + h²)
D = √(2hv²/g + h²)
Option (3) is correct
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In the following logic circuit the sequence of the inputs A, B are (0, 0), (0,1), (1, 0) and (1, 1). The output Y for this sequence will be :
(1) 1,0,1,0
(2) 0,1,0,1
(3) 1,1,1,0
(4) 0,0,1,1
Official Ans. by NTA (3)
Sol. Y = (A·B)·(A + B)
Y|(0,0) = 1
Y|(0,1) = 1
Y|(1,0) = 1
Y|(1,1) = 0
Option (3) is correct
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Two particles A and B having charges 20μC and –5μC respectively are held fixed with a separation of 5 cm. At what position a third charged particle should be placed so that it does not experience a net electric force?
20μC A 5cm B –5μC
(1) At 5 cm from 20μC on the left side of system
(2) At 5 cm from –5μC on the right side
(3) At 1.25 cm from –5μC between two charges
(4) At midpoint between two charges
Official Ans. by NTA (2)
Sol. 20μC –5μC
Null point is possible only right side of –5μC
E_N = + k(–5μC)/x² + k(20μC)/(5 + x)² = 0
x = 5 cm
∴ option (2) is correct
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A reversible engine has an efficiency of 1/4. If the temperature of the sink is reduced by 58°C, its efficiency becomes double. Calculate the temperature of the sink :
(1) 174°C
(2) 280°C
(3) 180.4°C
(4) 382°C
Official Ans. by NTA (1)
Official Ans. by ALLEN (Bonus)
Sol. T? = sink temperature
η = 1 – T?/T?
1/4 = 1 – T?/T?
T?/T? = 3/4 …(i)
1/2 = 1 – (T? – 58)/T?
T?/T? = 58/T? = 1/2
3/4 = 58/T? + 1/2
1/4 = 58/T? ⇒ T? = 232
T? = 3/4 × 232
T? = 174 K
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An object is placed at the focus of concave lens having focal length f. What is the magnification and distance of the image from the optical centre of the lens?
(1) 1, ∞
(2) Very high, ∞
(3) 1/2, f/2
(4) 1/4, f/4
Official Ans. by NTA (3)
Sol. U = –f
1/V – 1/U = 1/(–f) ⇒ 1/V = –2/f
V = –f/2
m = V/U = 1/2
distance = f/2
Option (3)
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A sample of a radioactive nucleus A disintegrates to another radioactive nucleus B, which in turn disintegrates to some other stable nucleus C. Plot of a graph showing the variation of number of atoms of nucleus B versus time is :
(Assume that at t = 0, there are no B atoms in the sample)
Official Ans. by NTA (2)
Sol. A B C (stable)
Initially no. of atoms of B = 0 after t = 0, no. of atoms of B will starts increasing & reaches maximum value when rate of decay of B = rate of formation of B.
After that maximum value, no. of atoms will starts decreasing as growth & decay both are exponential functions, so best possible graph is (2)
Option (2)
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A coil having N turns is wound tightly in the form of a spiral with inner and outer radii 'a' and 'b' respectively. Find the magnetic field at centre, when a current I passes through coil :
(1) μ?IN/(2(b – a)) log_e(b/a)
(2) μ?I/8 [(a + b)/(a – b)]
(3) μ?I/(4(a – b)) [1/a – 1/b]
(4) μ?I/8 ((a – b)/(a + b))
Official Ans. by NTA (1)
Sol. No. of turns in dx width = N/(b – a) dx
∫ dB = ∫_a^b (N/(b – a)) dx μ?i/(2x)
B = Nμ?i/(2(b – a)) ln(b/a)
Option (1)
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A body of mass M moving at speed V? collides elastically with a mass 'm' at rest. After the collision, the two masses move at angles θ? and θ? with respect to the initial direction of motion of the body of mass M. The largest possible value of the ratio M/m, for which the angles θ? and θ? will be equal, is :
(1) 4
(2) 1
(3) 3
(4) 2
Official Ans. by NTA (3)
Sol. given θ? = θ? = θ from momentum conservation in x-direction MV? = MV?cosθ + mV?cosθ in y-direction θ = MV?sinθ – mV?sinθ Solving above equations
V? = MV?/m, V? = 2V?cosθ
From energy conservation
1/2 MV?² = 1/2 MV?² + 1/2 MV?²
Substituting value of V? & V?, we will get
M/m + 1 = 4cos²θ ≤ 4
M/m ≤ 3
Option (3)
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The masses and radii of the earth and moon are (M?, R?) and (M?, R?) respectively. Their centres are at a distance 'r' apart. Find the minimum escape velocity for a particle of mass 'm' to be projected from the middle of these two masses:
V = 1/2√(4G(M? + M?)/r)
V = √(4G(M? + M?)/r)
V = 1/2√(2G(M? + M?)/r)
V = √(2G(M? + M?)/r)
Official Ans. by NTA (2)
Sol. 1/2 mV² – GM?m/(r/2) – GM?m/(r/2) = 0
1/2 mV² = 2Gm/r (M? + M?)
V = √(4G(M? + M?)/r)
Option (2)
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A small square loop of side 'a' and one turn is placed inside a larger square loop of side b and one turn (b >> a). The two loops are coplanar with their centres coinciding. If a current I is passed in the square loop of side 'b', then the coefficient of mutual inductance between the two loops is :
(1) μ?/4π 8√2 a²/b
(2) μ?/4π 8√2/a
(3) μ?/4π 8√2 b²/a
(4) μ?/4π 8√2/b
Official Ans. by NTA (1)
Sol. B = [μ?/4π 1/(b/2) × 2sin45] × 4
φ = 2√2 μ?/π 1/b × a²
∴ M = φ/I = 2√2 μ?a²/(πb) = μ?/4π 8√2 a²/b
Option (1)
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Choose the correct waveform that can represent the voltage across R of the following circuit, assuming the diode is ideal one:
Official Ans. by NTA (3)
Official Ans. by ALLEN (1)
Sol. When V_i > 3 volt, V_R > 0 Because diode will be in forward biased state When V_i ≤ 3 volt. V_R = 0 Because diode will be in reverse biased state.
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A uniform heavy rod of weight 10 kgms?², cross-sectional area 100 cm² and length 20 cm is hanging from a fixed support. Young modulus of the material of the rod is 2 × 10¹¹ Nm?². Neglecting the lateral contraction, find the elongation of rod due to its own weight.
(1) 2 × 10?? m
(2) 5 × 10?? m
(3) 4 × 10?? m
(4) 5 × 10?¹? m
Official Ans. by NTA (4)
Sol. We know,
Δ? = WL/(2AY)
Δ? = (10 × 1)/(2 × 5) × 100 × 10?? × 2 × 10¹¹
Δ? = 1/2 × 10?? = 5 × 10?¹? m
Option (4)
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Two plane mirrors M? and M? are at right angle to each other shown. A point source 'P' is placed at 'a' and '2a' meter away from M? and M? respectively. The shortest distance between the images thus formed is : (Take √5 = 2.3)
(1) 3a
(2) 4.6 a
(3) 2.3 a
(4) 2√10 a
Official Ans. by NTA (2)
Sol. Shortest distance is 2a between I? & I?
But answer given is for I? & I?
√((4a)² + (2a)²)
a√20
4.47 a
Option (2)
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Match List-I with List-II.
List-I List-II
(a) Torque (i) MLT?¹
(b) Impulse (ii) MT?²
(c) Tension (iii) MLT?²
(d) Surface Tension (iv) MLT?²
Choose the most appropriate answer from the option given below :
(1) (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
(2) (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
(3) (a)-(i), (b)-(iii), (c)-(iv), (d)-(ii)
(4) (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
Official Ans. by NTA (1)
Sol. torque τ → ML²T?² (III)
Impulse I ⇒ MLT?¹ (I)
Tension force ⇒ MLT?² (IV)
Surface tension ⇒ MT?² (II)
Option (1)
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For an ideal gas the instantaneous change in pressure 'p' with volume 'v' is given by the equation dp/dv = –ap. If p = p? at v = 0 is the given boundary condition, then the maximum temperature one mole of gas can attain is : (Here R is the gas constant)
(1) p?/(aeR)
(2) ap?/(eR)
(3) infinity
(4) 0°C
Official Ans. by NTA (1)
Sol. ∫_{p?}^{p} dp/p = –a∫_0^v dv
ln(p/p?) = –av
p = p?e^{-av}
For temperature maximum p-v product should be maximum
T = pv/(nR) = p?ve^{-av}/R
dT/dv = 0 ⇒ p?/R {e^{-av} + ve^{-av}(–a)}
p?e^{-av}/R {1 – av} = 0
v = 1/a, ∞
T = p?/(Rae) = p?/(Rae)
at v = ∞
T = 0
Option (1)
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Which of the following equations is dimensionally incorrect ?
Where t = time, h = height, s = surface tension, θ = angle, ρ = density, a, r = radius, g = acceleration due to gravity, v = volume, p = pressure, W = work done, Γ = torque, ε = permittivity, E = electric field, J = current density, L = length.
(1) v = πpa?/(8ηL)
(2) h = 2scosθ/(ρrg)
(3) J = ε ∂E/∂t
(4) W = Γθ
Official Ans. by NTA (1)
Sol. (i) πpa?/(8ηL) = dv/dt = Volumetric flow rate (poiseuille's law)
(ii) hρg = 2s/r cosθ
(iii) RHS ⇒ ε × 1/(4πε?) a/r² × 1/ε = q/t × 1/r² = I/L² = IL?²
LHS T = I/A = IL?²
(iv) W = τθ
Option (1)
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Angular momentum of a single particle moving with constant speed along circular path :
(1) changes in magnitude but remains same in the direction
(2) remains same in magnitude and direction
(3) remains same in magnitude but changes in the direction
(4) is zero
Official Ans. by NTA (2)
Sol. |L| = mvr
And direction will be upward & remain constant
Option (2)
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In an ac circuit, an inductor, a capacitor and a resistor are connected in series with X_L = R = X_C. Impedance of this circuit is :
(1) 2R²
(2) Zero
(3) R
(4) R√2
Official Ans. by NTA (3)
Sol. Z = √((X_L – X_C)² + R²) = R ? X_L = X_C
Option (3)
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A moving proton and electron have the same de-Broglie wavelength. If K and P denote the K.E. and momentum respectively. Then choose the correct option :
(1) K_p < K_e and P_p = P_e
(2) K_p = K_e and P_p = P_e
(3) K_p < K_e and P_p < P_e
(4) K_p > K_e and P_p = P_e
Official Ans. by NTA (1)
Sol. λ_p = h/P_p, λ_e = h/P_e
? λ_p = λ_e
⇒ P_p = P_e
(K)_p = P_p²/(2m_p)
(K)_e = P_e²/(2m_e)
K_p < K_e as m_p > m_e
Option (1)
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Consider a galvanometer shunted with 5Ω resistance and 2% of current passes through it. What is the resistance of the given galvanometer ?
(1) 300 Ω
(2) 344 Ω
(3) 245 Ω
(4) 226 Ω
Official Ans. by NTA (3)
Sol. 0.02i Rg = 0.98i × 5
Rg = 245Ω
Option (3)
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When a rubber ball is taken to a depth of m in deep sea, its volume decreases by 0.5%.
(The bulk modulus of rubber = 9.8 × 10? Nm?² Density of sea water = 10³ kgm?³ g = 9.8 m/s²)
Official Ans. by NTA (500)
Sol. B = –ΔP/(ΔV/V) = –ρgh/(ΔV/V)
B ΔV/V = h
(9.8×10?×0.5)/(100×10³×9.8) = h
h = 500
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A particle of mass 1 kg is hanging from a spring of force constant 100 Nm?¹. The mass is pulled slightly downward and released so that it executes free simple harmonic motion with time period T. The time when the kinetic energy and potential energy of the system will become equal, is T/x. The value of x is
Official Ans. by NTA (8)
Sol. KE = PE
y = A/√2 = A sinωt
t = T/8 = T/x
x = 8
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If the sum of the heights of transmitting and receiving antennas in the line of sight of communication is fixed at 160 m, then the maximum range of LOS communication is km. (Take radius of Earth = 6400 km)
Official Ans. by NTA (64)
Sol. h_T = h_R = 160 …(i)
d = √(2Rh_T) + √(2Rh_R)
d = √(2R)[√h_T + √h_R]
d = √(2R)[√x + √(160 – x)]
d(d)/dx = 0
1/(2√x) + 1(–1)/(2√(160 – x)) = 0
1/√x = 1/√(160 – x)
x = 80 m
d_max = √(2×6400)[√(80/1000) + √(20/1000)]
= (80√2 × 2√80)/(10√10)
= 8×2×√2×2√2 = 64 km
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A square shaped wire with resistance of each side 3Ω is bent to form a complete circle. The resistance between two diametrically opposite points of the circle in unit of Ω will be ______.
Official Ans. by NTA (3)
Sol. R_eq = 3Ω
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A wire having a linear mass density 9.0 × 10?? kg/m is stretched between two rigid supports with a tension of 900 N. The wire resonates at a frequency of 500 Hz. The next higher frequency at which the same wire resonates is 550 Hz. The length of the wire is ______ m.
Official Ans. by NTA (10)
Sol. μ = 9.0×10?? kg/m
T = 900 N
V = √(T/μ) = √(900/(9×10??)) = 1000 m/s
f? = 500 Hz
f = 550
nV/(2?) = 500 …(i)
(n+1)V/(2?) = 500 …(ii)
(ii)-(i) V/(2?) = 50
? = 1000/(2×50) = 10
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The voltage drop across 15Ω resistance in the given figure will be ______ V.
Official Ans. by NTA (6)
Sol. ⇒ effective circuit diagram will be
i=1A, i?=2, 12V, 1Ω
Point drop across 6Ω = 1×6 = 6 = V_AB
⇒ Hence point drop across 15Ω = 6 volt = V_AB
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A block moving horizontally on a smooth surface with a speed of 40 ms?¹ splits into two equal parts. If one of the parts moves at 60 ms?¹ in the same direction, then the fractional change in the kinetic energy will be x:4 where x =
Official Ans. by NTA (1)
Sol. P_i = P_f
m×40 = m/2×v + m/2×60
40 = v/2 + 30
⇒ v = 20
(K.E.)_i = 1/2 m×(40)² = 800m
(K.E.)_f = 1/2 m/2 (20)² + 1/2 m/2 (60)² = 1000m
|ΔK.E.| = |1000m – 800m| = 200m
ΔK.E/(K.E.)_i = 200m/800m = 1/4 = x/4
x = 1
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The electric field in an electromagnetic wave is given by E = (50 N C?¹) sin ω(t – x/c)
The energy contained in a cylinder of volume V is 5.5 × 10?¹² J.
(given ε? = 8.8 × 10?¹² C² N?¹ m?²)
Official Ans. by NTA (500)
Sol. E = 50 sin(ωt – ω/c x)
Energy density = 1/2 ε?E?²
Energy for volume V = 1/2 ε?E?²V = 5.5×10?¹²
1/2 8.8×10?¹²×2500V = 5.5×10?¹²
V = (5.5×2)/(2500×8.8) = .0005 m³
= .0005×10? (c.m)³
= 500 (c.m)³
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A capacitor of 50 μF is connected in a circuit as shown in figure. The charge on the upper plate of the capacitor is μC.
Official Ans. by NTA (100)
Sol. Pot. Diff. across each resistor = 2V
q = CV
= 50×10??×2 = 100×10?? = 100 μC
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A car is moving on a plane inclined at 30° to the horizontal with an acceleration of 10 ms?² parallel to the plane upward. A bob is suspended by a string from the roof of the car. The angle in degrees which the string makes with the vertical is (Take g = 10 ms?²)
Official Ans. by NTA (30)
Sol. tan(30 + θ) = (mg sin30° + ma)/(mg cos30°)
tan(30 + θ) = (5 + 10)/(5√3) = (1 + 2)/√3
(tanθ + 1/√3)/(1 – (1/√3)tanθ) = √3 tanθ + 1 = 3 – √3 tanθ
tanθ = 1/√3
θ = 30°
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The correct order of reactivity of the given chlorides with acetate in acetic acid is :
Official Ans. by NTA (1)
Sol. As it is example of SN¹.
so carbocation stability ↑, reaction rate ↑
-
Select the graph that correctly describes the adsorption isotherms at two temperatures T? and T? (T? > T?) for a gas :
(x - mass of the gas adsorbed ; m - mass of adsorbent ; P - pressure)
Official Ans. by NTA (4)
Sol. x/m α P^{1/n} (0 < 1/n < 1)
On Increasing temperature x/m decreases.
adsorption is generally exothermic
-
The major component/ingredient of Portland Cement is :
(1) tricalcium aluminate
(2) tricalcium silicate
(3) dicalcium aluminate
(4) dicalcium silicate
Official Ans. by NTA (2)
Sol. Major component of portland cement is "Tricalcium silicate (51%), 3CaO.SiO?.
-
In the structure of the dichromate ion, there is a :
(1) linear symmetrical Cr-O-Cr bond.
(2) non-linear symmetrical Cr-O-Cr bond.
(3) linear unsymmetrical Cr-O-Cr bond.
(4) non-linear unsymmetrical Cr-O-Cr bond.
Official Ans. by NTA (2)
Sol. dichromate ion contain non-linear symmetrical Cr–O–Cr Bond
-
Which one of the following compounds contains β-C?-C? glycosidic linkage?
(1) Lactose
(2) Sucrose
(3) Maltose
(4) Amylose
Official Ans. by NTA (1)
Sol. In Lactose it is β C? – C? glycosidic linkage.
In Maltose, Amylose α C? – C? glycosidic linkage is present
-
The major products A and B in the following set of reactions are :
Official Ans. by NTA (3)
Sol.
-
Which one of the following lanthanides exhibits +2 oxidation state with diamagnetic nature ? (Given Z for Nd = 60, Yb = 70, La = 57, Ce = 58)
(1) Nd
(2) Yb
(3) La
(4) Ce
Official Ans. by NTA (2)
Sol. Ytterbium shows +2 oxidation state with diamagnetic nature
So ans is 2
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Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : Aluminium is extracted from bauxite by the electrolysis of molten mixture of Al?O? with cryolite.
Reason (R) : The oxidation state of Al in cryolite is +3.
In the light of the above statements, choose the most appropriate answer from the options given below :
(1) (A) is true but (R) is false
(2) (A) is false but (R) is true
(3) Both (A) and (R) are correct and (R) is the correct explanation of (A)
(4) Both (A) and (R) are correct but (R) is not the correct explanation of (A)
Official Ans. by NTA (4)
Sol. (A) Aluminium is reactive metal so Aluminium is extracted by electrolysis of Alumina with molten mixture of Cryolite
(B) Cryolite, Na?AlF?
Here Al is in +3 O.S.
So Answer is 4
-
The major product formed in the following reaction is :
Official Ans. by NTA (2)
Sol.
-
Monomer of Novolac is :
(1) 3-Hydroxybutanoic acid
(2) phenol and melamine
(3) o-Hydroxymethylphenol
(4) 1,3-Butadiene and styrene
Official Ans. by NTA (3)
Sol. Monomer of Novolac is O-hydroxy methyl phenol
-
Given below are two statements :
Statement-I : The process of producing syn-gas is called gasification of coal.
Statement-II : The composition of syn-gas is CO + CO? + H? (1 : 1 : 1)
In the light of the above statements, choose the most appropriate answer from the options given below :
(1) Statement-I is false but Statement-II is true
(2) Statement-I is true but Statement-II is false
(3) Both Statement-I and Statement-II are false
(4) Both Statement-I and Statement-II are true
Official Ans. by NTA (2)
Sol. The process of producing syn-gas from coal is called gasification of coal.
Syn-gas having composition of CO & H? in 1 : 1
-
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : Treatment of bromine water with propene yields 1-bromopropan-2-ol.
Reason (R) : Attack of water on bromonium ion follows Markovnikov rule and results in 1-bromopropan-2-ol.
In the light of the above statements, choose the most appropriate answer from the options given below :
(1) Both (A) and (R) are true but (R) is NOT the correct explanation of (A)
(2) (A) is false but (R) is true.
(3) Both (A) and (R) are true and (R) is the correct explanation of (A)
(4) (A) is true but (R) is false
Official Ans. by NTA (3)
Sol. Its IUPAC name 1-bromopropan-2-ol
A and R are true and (R) is the correct explanation of (A)
-
The denticity of an organic ligand, biuret is :
(1) 2
(2) 4
(3) 3
(4) 6
Official Ans. by NTA (1)
Sol. Biuret :- Bidentate ligand
The denticity of organic ligand is 2.
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Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : Metallic character decreases and non-metallic character increases on moving from left to right in a period.
Reason (R) : It is due to increase in ionisation enthalpy and decrease in electron gain enthalpy, when one moves from left to right in a period.
In the light of the above statements, choose the most appropriate answer from the options given below :
(1) (A) is false but (R) is true.
(2) (A) is true but (R) is false
(3) Both (A) and (R) are correct and (R) is the correct explanation of (A)
(4) Both (A) and (R) are correct but (R) is not the correct explanation of (A)
Official Ans. by NTA (2)
Sol. From left to right in periodic table :-
Metallic character decreases Non-metallic character increases ⇒ It is due to increase in ionization enthalpy and increase in electron gain enthalpy.
-
Choose the correct name for compound given below :
(1) (4E)-5-Bromo-hex-4-en-2-yne
(2) (2E)-2-Bromo-hex-4-yn-2-ene
(3) (2E)-2-Bromo-hex-2-en-4-yne
(4) (4E)-5-Bromo-hex-2-en-4-yne
Official Ans. by NTA (3)
Sol. h.p. ⇒ higher priority l.p. ⇒ lower priority
2E- 2- bromo hex- 2- en- 4- yne
-
Which one of the following is the correct PV vs P plot at constant temperature for an ideal gas ? (P and V stand for pressure and volume of the gas respectively)
Official Ans. by NTA (1)
Sol. PV = nRT (n, T constant)
PV = constant
-
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R) :
Assertion (A) : A simple distillation can be used to separate a mixture of propanol and propanone.
Reason (R) : Two liquids with a difference of more than 20°C in their boiling points can be separated by simple distillations.
In the light of the above statements, choose the most appropriate answer from the options given below :
(1) (A) is false but (R) is true.
(2) Both (A) and (R) are correct but (R) is not the correct explanation of (A)
(3) (A) is true but (R) is false
(4) Both (A) and (R) are correct and (R) is the correct explanation of (A)
Official Ans. by NTA (4)
Sol. Both assertion & reason are correct & (R) is the correct explanation of (A)
-
Which one of the following 0.10 M aqueous solutions will exhibit the largest freezing point depression?
(1) hydrazine
(2) glucose
(3) glycine
(4) KHSO?
Official Ans. by NTA (4)
Sol. Van't Hoff factor is highest for KHSO?
colligative property (ΔT_f) will be highest for KHSO?
-
BOD values (in ppm) for clean water (A) and polluted water (B) are expected respectively:
(1) A > 50, B < 27
(2) A > 25, B < 17
(3) A < 5, B > 17
(4) A > 15, B > 47
Official Ans. by NTA (3)
Sol. BOD values of clean water (A) is less than 5 ppm
So A < 5
BOD values of polluted water (B) is greater than 17 ppm
So B > 17
So Ans.is 3
-
The structure of product C, formed by the following sequence of reactions is:
Official Ans. by NTA (1)
Sol.
-
Consider the following cell reaction :
Cd(s) + Hg?SO?(s) + 9/5 H?O(l) ? CdSO?·9/5 H?O(s) + 2Hg(l)
The value of E°_cell is 4.315 V at 25°C. If ΔH° = –825.2 kJmol?¹ the standard entropy change ΔS° in JK?¹ is ______. (Nearest integer) [Given : Faraday constant = 96487 Cmol?¹]
Official Ans. by NTA (25)
ΔG° = –nFE° = ΔH° – TΔS°
= (ΔH° + nFE°)/T
= (–825.2×10³ + 2×96487×4.315)/298
= (–825.2×10³ + 832.682×10³)/298
= 7.483×10³/298 = 25.11 JK?¹mol?¹
Nearest integer answer is 25
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The molarity of the solution prepared by dissolving 6.3 g of oxalic acid (H?C?O?·2H?O) in 250 mL of water in mol L?¹ is x × 10?². The value of x is ______. (Nearest integer) [Atomic mass : H : 1.0, C : 12.0, O : 16.0]
Official Ans. by NTA (20)
Sol. [H?C?O?·2H?O] = (weight/M_w)/V(L)
⇒ x×10?² = (6.3/126)/(250/1000)
x = 20
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Consider the sulphides HgS, PbS, CuS, Sb?S?, As?S? and CdS. Number of these sulphides soluble in 50% HNO? is
Official Ans. by NTA (4)
Sol. PbS, CuS, As?S?, CdS are soluble in 50% HNO?
HgS, Sb?S? are insoluble in 50% HNO?
So Answer is 4.
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The total number of reagents from those given below, that can convert nitrobenzene into aniline is (Integer answer)
I. Sn- HCl
II. Sn- NH?OH
III. Fe- HCl
IV. Zn- HCl
V. H? – Pd
VI. H? - Raney Nickel
Official Ans. by NTA (5)
Sol. Reagents used can be
(i) Sn + HCl
(ii) Fe + HCl
(iii) Zn + HCl
(iv) H? – Pd
(v) H? (Raney Ni)
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The number of halogen(s) forming halic (V) acid is
Official Ans. by NTA (3)
Sol. The number of halogen forming halic (V) acid
HClO?
HBrO?
HIO?
So Answer is 3
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For a first order reaction, the ratio of the time for 75% completion of a reaction to the time for 50% completion is (Integer answer)
Official Ans. by NTA (2)
Sol. k = 2.303/t log(a/(a – x))
2.303/t_50% log(100/(100 – 50)) = 2.303/t_75% log(100/(100 – 75))
t_75% = 2t_50%
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The number of hydrogen bonded water molecule(s) associated with stoichiometry CuSO?·5H?O is
Official Ans. by NTA (1)
Sol. One hydrogen bonded H?O molecule
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According to the following figure, the magnitude of the enthalpy change of the reaction A + B → M + N in kJ mol?¹ is equal to (Integer answer)
Official Ans. by NTA (45)
ΔH = E_{a_f} – E_{a_b}
= 20 – 65
= –45 KJ/mol
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Ge(Z = 32) in its ground state electronic configuration has x completely filled orbitals with m_l = 0. The value of x is
Official Ans. by NTA (7)
Sol. Completely filled orbital with m_l = 0 are
= 1 + 1 + 1 + 1 + 1 + 1
= 7
So Answer is 7
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A?B? is a sparingly soluble salt of molar mass M (g mol?¹) and solubility x g L?¹. The solubility product satisfies K_sp = a(x/M)?. The value of a is (Integer answer)
Official Ans. by NTA (108)
Sol. A?B?(s) ? 3A²? + 2B³?
3s 2s
K_sp = (3s)³(2s)²
K_sp = 108 s? & s = (x/M)
K_sp = 108(x/m)?
given K_sp = a(x/m)?
comparing a = 108
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Let *, ? ∈ {∧, ∨} be such that the Boolean expression (p * ~q) ⇒ (p ? q) is a tautology. Then :
(1) * = ∨, ? = ∨
(2) * = ∧, ? = ∧
(3) * = ∧, ? = ∨
(4) * = ∨, ? = ∧
Official Ans. by NTA (3)
Sol. (p ∧ ~q) → (p ∨ q) is tautology
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The number of real roots of the equation e^{4x} + 2e^{3x} – e^x – 6 = 0 is :
(1) 2
(2) 4
(3) 1
(4) 0
Official Ans. by NTA (3)
Sol. Let e^x = t > 0
f(t) = t? + 2t³ – t – 6 = 0
f'(t) = 4t³ + 6t² – 1
f(0) = –6, f(1) = –4, f(2) = 24
⇒ Number of real roots = 1
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The sum of 10 terms of the series
3/(1²×2²) + 5/(2²×3²) + 7/(3²×4²) + ... is :
(1) 1
(2) 120/121
(3) 99/100
(4) 143/144
Official Ans. by NTA (2)
Sol. S = (2² – 1²)/(1²×2²) + (3² – 2²)/(2²×3²) + (4² – 3²)/(3²×4²) + ...
= [1/1² – 1/2²] + [1/2² – 1/3²] + [1/3² – 1/4²] + ... + [1/10² – 1/11²]
= 1 – 1/121
= 120/121
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Let the equation of the plane, that passes through the point (1, 4, –3) and contains the line of intersection of the planes 3x – 2y + 4z – 7 = 0 and x + 5y – 2z + 9 = 0, be αx + βy + γz + 3 = 0, then α + β + γ is equal to :
(1) –23
(2) –15
(3) 23
(4) 15
Official Ans. by NTA (1)
Sol. Equation of plane is
3x – 2y + 4z – 7 + λ(x + 5y – 2z + 9) = 0
(3 + λ)x + (5λ – 2)y + (4 – 2λ)z + 9λ – 7 = 0
passing through (1, 4, –3)
⇒ 3 + λ + 20λ – 8 – 12 + 6λ + 9λ – 7 = 0
⇒ λ = 2/3
⇒ equation of plane is
–11x – 4y – 8z + 3 = 0
⇒ α + β + γ = –23
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Let f be a non-negative function in [0, 1] and twice differentiable in (0, 1). If ∫_0^x √(1 – (f'(t))²) dt = ∫0^x f(t) dt, 0 ≤ x ≤ 1 and f(0) = 0, then lim{x→0} 1/x² ∫_0^x f(t) dt :
(1) equals 0
(2) equals 1
(3) does not exist
(4) equals 1/2
Official Ans. by NTA (4)
Sol. ∫_0^x √(1 – (f'(t))²) dt = ∫0^x f(t) dt 0 ≤ x ≤ 1
differentiating both the sides
√(1 – (f'(x))²) = f(x)
⇒ 1 – (f'(x))² = f²(x)
f'(x)/√(1 – f²(x)) = 1
sin?¹ f(x) = x + C
? f(0) = 0 ⇒ C = 0 ⇒ f(x) = sinx
Now lim{x→0} ∫_0^x sin t dt / x² (0/0) = 1/2
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Let a and b be two vectors such that |2a + 3b| = |3a + b| and the angle between a and b is 60°. If (1/8)a is a unit vector, then |b| is equal to :
(1) 4
(2) 6
(3) 5
(4) 8
Official Ans. by NTA (3)
Sol. |3a + b|² = |2a + 3b|²
(3a + b)·(3a + b) = (2a + 3b)·(2a + 3b)
9a·a + 6a·b + b·b = 4a·a + 12a·b + 9b·b
5|a|² – 6a·b = 8|b|²
5(8)² – 6·8·|b| cos60° = 8|b|² (? 1/8|a| = 1 ⇒ |a| = 8)
40 – 3|b| = |b|²
⇒ |b|² + 3|b| – 40 = 0
|b| = –8, |b| = 5 (rejected)
|b| = 5
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The function f(x) = |x² – 2x – 3|·e^{|x² – 12x + 4|} is not differentiable at exactly :
(1) four points
(2) three points
(3) two points
(4) one point
Official Ans. by NTA (3)
Sol. f(x) = |(x – 3)(x + 1)|·e^{|3x – 2|²}
f(x) = { (x – 3)(x + 1)e^{|3x – 2|²} ; x ∈ (3, ∞)
–(x – 3)(x + 1)e^{|3x – 2|²} ; x ∈ [–1, 3]
(x – 3)(x + 1)e^{|3x – 2|²} ; x ∈ (–∞, –1) }
Clearly, non-differentiable at x = –1 & x = 3.
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Three numbers are in an increasing geometric progression with common ratio r. If the middle number is doubled, then the new numbers are in an arithmetic progression with common difference d. If the fourth term of GP is 3 r², then r² – d is equal to :
(1) 7 – 7√3
(2) 7 + √3
(3) 7 – √3
(4) 7 + 3√3
Official Ans. by NTA (2)
Sol. Let numbers be a/r, a, ar → G.P
a/r, 2a, ar → A.P ⇒ 4a = a/r + ar ⇒ r + 1/r = 4
r = 2 ± √3
4th form of G.P = 3r² ⇒ ar³ = 3r² ⇒ a = 3
r = 2 + √3, a = 3, d = 2a – a/r = 3√3
r² – d = (2 + √3)² – 3√3
= 7 + 4√3 – 3√3
= 7 + √3
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Which of the following is not correct for relation R on the set of real numbers ?
(1) (x, y) ∈ R ⇔ 0 < |x| – |y| ≤ 1 is neither transitive nor symmetric.
(2) (x, y) ∈ R ⇔ 0 < |x – y| ≤ 1 is symmetric and transitive.
(3) (x, y) ∈ R ⇔ |x| – |y| ≤ 1 is reflexive but not symmetric.
(4) (x, y) ∈ R ⇔ |x – y| ≤ 1 is reflexive and symmetric.
Official Ans. by NTA (2)
Sol. Note that (1,2) and (2,3) satisfy 0 < |x – y| ≤ 1 but (1,3) does not satisfy it so 0 ≤ |x – y| ≤ 1 is symmetric but not transitive So, (2) is correct.
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The integral ∫ 1/(4√((x – 1)³(x + 2)?)) dx is equal to :
(where C is a constant of integration)
(1) 3/4 ((x + 2)/(x – 1))^{1/4} + C
(2) 3/4 ((x + 2)/(x – 1))^{5/4} + C
(3) 4/3 ((x – 1)/(x + 2))^{1/4} + C
(4) 4/3 ((x – 1)/(x + 2))^{5/4} + C
Official Ans. by NTA (3)
Sol. ∫ dx/((x – 1)^{3/4}(x + 2)^{5/4})
= ∫ dx/(((x + 2)/(x – 1))^{5/4}·(x – 1)²)
put (x + 2)/(x – 1) = t
= –1/3 ∫ dt/t^{5/4}
= 4/3·1/t^{1/4} + C
= 4/3 ((x – 1)/(x + 2))^{1/4} + C
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If p and q are the lengths of the perpendiculars from the origin on the lines,
x cos α – y sec α = k cot 2α and
x sin α + y cos α = k sin 2α
respectively, then k² is equal to :
(1) 4p² + q²
(2) 2p² + q²
(3) p² + 2q²
(4) p² + 4q²
Official Ans. by NTA (1)
Sol. First line is x/sinα – y/cosα = k cos2α/sin2α
⇒ x cos α – y sin α = k/2 cos 2α
⇒ p = |k/2 cos 2α| ⇒ 2p = |k cos 2α| ...(i)
second line is x sinα + y cosα = k sin2α
⇒ q = |k sin2α| ...(ii)
Hence 4p² + q² = k² (From (i) & (ii))
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cosec 18° is a root of the equation :
(1) x² + 2x – 4 = 0
(2) 4x² + 2x – 1 = 0
(3) x² – 2x + 4 = 0
(4) x² – 2x – 4 = 0
Official Ans. by NTA (4)
Sol. cosec18° = 1/sin18° = 4/(√5 – 1) = √5 + 1
Let cosec18° = x = √5 + 1
⇒ x – 1 = √5
Squaring both sides, we get
x² – 2x + 1 = 5
⇒ x² – 2x – 4 = 0
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If the following system of linear equations
2x + y + z = 5
x – y + z = 3
x + y + az = b
has no solution, then :
(1) a = –1/3, b ≠ 7/3
(2) a ≠ 1/3, b = 7/3
(3) a ≠ –1/3, b = 7/3
(4) a = 1/3, b ≠ 7/3
Official Ans. by NTA (4)
Sol. Here D = |2 1 1; 1 –1 1; 1 1 a| = 2(–a – 1) – 1(a – 1) + 1(1 + 1) = 1 – 3a
D? = |2 1 5; 1 –1 3; 1 1 b| = 2(–b – 3) – 1(b – 3) + 5(1 + 1) = 7 – 3b
for a = 1/3, b ≠ 7/3, system has no solutions
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The length of the latus rectum of a parabola, whose vertex and focus are on the positive x-axis at a distance R and S (>R) respectively from the origin, is :
(1) 4(S + R)
(2) 2(S – R)
(3) 4(S – R)
(4) 2(S + R)
Official Ans. by NTA (3)
Sol. V → Vertex
F → focus
VF = S – R
So latus rectum = 4(S – R)
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If the function f(x) = { (1/x log_e((1 + x)/(1 – x)), x < 0
k, x = 0
(cos²x – sin²x – 1)/(√(x² + 1) – 1), x > 0 }
is continuous at x = 0, then 1/a + 1/b + 4/k is equal to :
(1) –5
(2) 5
(3) –4
(4) 4
Official Ans. by NTA (1)
Sol. If f(x) is continuous at x = 0, RHL = LHL = f(0)
lim_{x→0+} f(x) = lim_{x→0+} (cos²x – sin²x – 1)/(√(x² + 1) – 1) (Rationalisation)
lim_{x→0+} –2sin²x/x² (√(x² + 1) + 1) = –4
lim_{x→0–} f(x) = lim_{x→0–} 1/x ln((1 + x)/(1 – x))
= 1/a + 1/b
So 1/a + 1/b = –4 = k
⇒ 1/a + 1/b + 4/k = –4 – 1 = –5
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If dy/dx = (2^{x+y} – 2^x)/2^y, y(0) = 1, then y(1) is equal to :
(1) log?(2 + e)
(2) log?(1 + e)
(3) log?(2e)
(4) log?(1 + e²)
Official Ans. by NTA (2)
Sol. dy/dx = (2^x 2^y – 2^x)/2^y
2^y dy/dx = 2^x(2^y – 1)
∫ 2^y/(2^y – 1) dy = ∫ 2^x dx
ln(2^y – 1)/ln2 = 2^x/ln2 + C
⇒ log?(2^y – 1) = 2^x log?e + C
? y(0) = 1 ⇒ 0 = log?e + C
C = –log?e
⇒ log?(2^y – 1) = (2^x – 1) log?e
put x = 1, log?(2^y – 1) = log?e
2^y = e + 1
y = log?(e + 1) Ans.
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lim_{x→0} sin²(π cos?x)/x? is equal to :
(1) π²
(2) 2π²
(3) 4π²
(4) 4π
Official Ans. by NTA (3)
Sol. lim_{x→0} sin²(π cos?x)/x?
lim_{x→0} (1 – cos(2π cos?x))/2x?
lim_{x→0} (1 – cos(π – 2π cos?x))/[2π(1 – cos?x)]² 4π²·sin?x/2x? (1 + cos²x)²
= 1/2·4π²·1/2(2)² = 4π²
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A vertical pole fixed to the horizontal ground is divided in the ratio 3 : 7 by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a point on the ground 18 m away from the base of the pole, then the height of the pole (in meters) is :
(1) 12√15
(2) 12√10
(3) 8√10
(4) 6√10
Official Ans. by NTA (2)
Sol. Let height of pole = 10?
tan α = 3?/18 = ?/6
tan 2α = 10?/18
2tanα/(1 – tan²α) = 10?/18
tanα = ?/6 ⇒ ? = √(72/5)
height of pole = 10? = 12√10
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If a_r = cos(2πr/9) + i sin(2πr/9), r = 1,2,3,..., i = √–1, then the determinant |a? a? a?; a? a? a?; a? a? a?| is equal to :
(1) a?a? – a?a?
(2) a?
(3) a?a? – a?a?
(4) a?
Official Ans. by NTA (3)
Sol. a_r = e^{i2πr/9}, r = 1,2,3,... a?, a?, a?, ... are in G.P.
Now a?a? – a?a? = a?^{10} – a?^{10} = 0
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The line 12x cos θ + 5y sin θ = 60 is tangent to which of the following curves?
(1) x² + y² = 169
(2) 144x² + 25y² = 3600
(3) 25x² + 12y² = 3600
(4) x² + y² = 60
Official Ans. by NTA (2)
Sol. 12x cos θ + 5y sin θ = 60
x cosθ/5 + y sinθ/12 = 1
144x² + 25y² = 3600
