PHYSICS

SECTION-A

1. Consider the efficiency of Carnot's engine is given by η = (αβ/sinθ) log_e(βx/kT), where α and β are constants. If T is temperature, k is Boltzmann constant, θ is angular displacement and x has the dimensions of length. Then, choose the incorrect option.
(A) Dimensions of β is same as that of force.
(B) Dimensions of α?¹ x is same as that of energy.
(C) Dimensions of η?¹ sin θ is same as that of αβ
(D) Dimensions of α is same as that of β
Official Ans. by NTA (D)
Sol. [αβ] = [η] = [sin θ] = Dimensionless
[η?¹ sin θ] = [αβ] = DL.

2. At time t = 0 a particle starts travelling from a height 7z cm in a plane keeping z coordinate constant. At any instant of time it's position along the x and y directions are defined as 3t and 5t³ respectively. At t = 1s acceleration of the particle will be
(A) -30y
(B) 30y
(C) 3x + 15y
(D) 3x + 15y + 7z
Official Ans. by NTA (B)
Sol. r = 3t i + 5t³ j + 7 k
d²r/dt² = 30t j
At t = 1 ⇒ d²r/dt² = 30 j

TEST PAPER WITH SOLUTION

3. A pressure-pump has a horizontal tube of cross-sectional area 10 cm² for the outflow of water at a speed of 20 m/s. The force exerted on the vertical wall just in front of the tube which stops water horizontally flowing out of the tube, is: [given : density of water = 1000 kg/m³]
(A) 300 N
(B) 500 N
(C) 250 N
(D) 400 N
Official Ans. by NTA (D)
Sol. F = ρav² = 10³ × 10 × 10?? × 20 × 20
F = 400

4. A uniform metal chain of mass m and length 'L' passes over a massless and frictionless pulley. It is released from rest with a part of its length 'l' is hanging on one side and rest of its length 'L - l' is hanging on the other side of the pulley. At a certain point of time, when l = L/x, the acceleration of the chain is g/2. The value of x is ...
(A) 6
(B) 2
(C) 1.5
(D) 4
Official Ans. by NTA (D)
Sol. a = (m? - m?)/(m? + m?) g
g/2 = (λ(L - ?) - λ?)g/λL ⇒ L = L/4 = L/x
x = 4

5. A bullet of mass 200 g having initial kinetic energy 90 J is shot inside a long swimming pool as shown in the figure. If it's kinetic energy reduces to 40 J within 1s, the minimum length of the pool, the bullet has a to travel so that it completely comes to rest is
(A) 45 m
(B) 90 m
(C) 125 m
(D) 25 m
Official Ans. by NTA (A)
Sol. Using mv = √(2mk)
u = (1/0.2)√(2 × 0.2 × 90) = 30 m/s
v = (1/0.2)√(2 × 0.2 × 40) = 20 m/s
a = (20 - 30)/1 = -10 m/s²
s = -u²/2a = 45 m

6. Assume there are two identical simple pendulum Clocks-1 is placed on the earth and Clock-2 is placed on a space station located at a height h above the earth surface. Clock-1 and Clock-2 operate at time periods 4s and 6s respectively. Then the value of h is - (consider radius of earth R_E = 6400 km and g on earth 10 m/s²)
(A) 1200 km
(B) 1600 km
(C) 3200 km
(D) 4800 km
Official Ans. by NTA (C)
Sol. t ∝ 1/√g and g ∝ 1/(R + h)²
t?/t? = √(g/g) = √(R²/(R + h)²)
t?/t? = 4/6 = R/(R + h) ⇒ h = 3200 km

7. Consider a cylindrical tank of radius 1 m is filled with water. The top surface of water is at 15 m from the bottom of the cylinder. There is a hole on the wall of cylinder at a height of 5 m from the bottom. A force of 5 × 10? N is applied an the top surface of water using a piston. The speed of efflux from the hole will be : (given atmospheric pressure P_A = 1.01 × 10? Pa density of water ρ_w = 1000 kg/m³ and gravitational acceleration g = 10 m/s²)
(A) 11.6 m/s
(B) 10.8 m/s
(C) 17.8 m/s
(D) 14.4 m/s
Official Ans. by NTA (C)
Sol. Apply Bernoulli's theorem between Piston and hole P_A + ρgh = P? + 1/2 ρ v_e²
Assuming there is no atmospheric pressure on piston
(5 × 10?)/π + 10³ × 10 × 10 = 1.01 × 10? + 1/2 × 10³ × v_e²
v_e = 17.8 m/s

8. A vessel contains 14 g of nitrogen gas at a temperature of 27°C. The amount of heat to be transferred to the gap to double the r.m.s. speed of its molecules will be : (Take R = 8.32 J mol?¹ K?¹)
(A) 2229 J
(B) 5616 J
(C) 9360 J
(D) 13,104 J
Official Ans. by NTA (C)
Sol. V_rms ∝ √T
V_rms ∝ √(300 K)
V_rms = 2V_rms
V_rms ∝ √(1200 K)
T_f = 1200 K
T_i = 300 K
n = 14/28 = 1/2
Q = nC_v ΔT = 1/2 × 5R/2 × 900
Q = 9360 J

9. A slab of dielectric constant K has the same cross-sectional area as the plates of a parallel plate capacitor and thickness 3/4 d, where d is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be : (Given C? = capacitance of capacitor with air as medium between plates.)
(A) 4KC?/(3 + K)
(B) 3KC?/(3 + K)
(C) (3 + K)/(4KC?)
(D) K/(4 + K)
Official Ans. by NTA (A)
Sol. x + y + 3d/4 = d
x + y = d/4
Aε?/d = C?
ΔV = Ex + E/k × 3d/4 + Ey
= 3Ed/4k + E(x + y)
ΔV = E[3d/4k + d/4]
ΔV = σ/ε?[(3d + dk)/4k] = Qd/(Aε?)[(3 + k)/4k]
Q/ΔV = C = Aε?/d [4k/(3 + k)] = 4kC?/(k + 3)

10. A uniform electric field E = (8m/e) V/m is created between two parallel plates of length 1 m as shown in figure, (where m = mass of electron and e = charge of electron). An electron enters the field symmetrically between the plates with a speed of 2 m/s. The angle of the deviation (θ) of the path of the electron as it comes out of the field will be
(A) tan?¹(4)
(B) tan?¹(2)
(C) tan?¹(1/3)
(D) tan?¹(3)
Official Ans. by NTA (B)
Sol.
a_y = F_y/m = e(E)/m = e(8m/e)/m = 8 m/s²
s_x = u_x t
1 = 2 × t
t = 1/2 sec
v_y = u_y + a_y t
v_y = 0 + 8 × 1/2
v_y = 4 m/s
tanθ = v_y/v_x = 4/2 = 2 ⇒ θ = tan?¹(2)

11. Given below are two statements :
Statement I : A uniform wire of resistance 80Ω is cut into four equal parts. These parts are now connected in parallel. The equivalent resistance of the combination will be 5Ω.
Statement II : Two resistance 2R and 3R are connected in parallel in a electric circuit. The value of thermal energy developed in 3R and 2R will be in the ratio 3 : 2.
In the light of the above statements, choose the most appropriate answer from the options given below
(A) Both statement I and statement II are correct
(B) Both statement I and statement II are incorrect
(C) Statement I is correct but statement II is incorrect
(D) Statement I is incorrect but statement II is correct.
Official Ans. by NTA (C)
Sol. Statement 1 - R = 80Ω
R? = R? = R? = R? = 20Ω
In parallel R_eq = 20/4 = 5Ω
Statement 2 -
P_th = v²/R
P?/P? = (R?/R?) = 2/3 (where P is power)

12. A triangular shaped wire carrying 10A current is placed in a uniform magnetic field of 0.5T, as shown in figure. The magnetic force on segment CD is (Given BC = CD = BD = 5 cm).
(A) 0.126 N
(B) 0.312 N
(C) 0.216 N
(D) 0.245 N
Official Ans. by NTA (C)
Sol. F_M(CD) = BI?_eff
= 0.5 × (10) × (5 sin 60 × 10?²)
= 0.216 N

13. The magnetic field at the center of current carrying circular loop is B?. The magnetic field at a distance of √3 times radius of the given circular loop from the center on its axis is B?. The value of B?/B? will be
(A) 9 : 4
(B) 12 : √5
(C) 8 : 1
(D) 5 : √3
Official Ans. by NTA (C)
Sol. B? = μ?I/2R
B? = μ?IR²/2(R² + 3R²)³/² = 1/8 (μ?I/2R) = B?/8
B?/B? = 8/1

14. A transformer operating at primary voltage 8 kV and secondary voltage 160 V serves a load of 80 kW. Assuming the transformer to be ideal with purely resistive load and working on unity power factor, the loads in the primary and secondary circuit would be
(A) 800Ω and 1.06Ω
(B) 10Ω and 500Ω
(C) 800Ω and 0.32Ω
(D) 1.06Ω and 500Ω
Official Ans. by NTA (C)
Sol. (8 × 10³)²/R_p = 80 × 10³
R_p = 800Ω
(160)²/R_s = 80 × 10³
R_s = 0.32Ω

15. Sun light falls normally on a surface of area 36 cm² and exerts an average force of 7.2 × 10?? N within a time period of 20 minutes. Considering a case of complete absorption, the energy flux of incident light is
(A) 25.92 × 10² W/cm²
(B) 8.64 × 10?? W/cm²
(C) 6.0 W/cm²
(D) 0.06 W/cm²
Official Ans. by NTA (D)
Sol. 1/C × area = force
1/C × 36 × 10?? = 7.2 × 10??
I = 7.2 × 10?? × 3 × 10? / (36 × 10?? × 10)
= 6 × 10?¹ / 10?³
I = 6 × 10² W/m²
= 0.06 W/cm²

16. The power of a lens (biconvex) is 1.25 m?¹ in particular medium. Refractive index of the lens is 1.5 and radii of curvature are 20 cm and 40 cm respectively. The refractive index of surrounding medium:
(A) 1.0
(B) 9/7
(C) 3/2
(D) 4/3
Official Ans. by NTA (B)
Sol. P = μ?/f = (μ? - μ?)(1/R? - 1/R?) (For this formula refer to NCERT Part-2, Chapter-9, Page no. 328, solved example 8)
(μ? is refractive index of lens and μ? is of surrounding medium)
1.25 = (1.5 - μ?)(1/0.2 + 1/0.4)
1.25 × 0.08/0.6 = (1.5 - μ?)
⇒ μ? = 4/3

17. Two streams of photons, possessing energies to five and ten times the work function of metal are incident on the metal surface successively. The ratio of the maximum velocities of the photoelectron emitted, in the two cases respectively, will be
(A) 1 : 2
(B) 1 : 3
(C) 2 : 3
(D) 3 : 2
Official Ans. by NTA (C)
Sol. 1/2 mv?² = 4φ
1/2 mv?² = 9φ
v?/v? = 2/3

18. A radioactive sample decays 7/4 times its original quantity in 15 minutes. The half-life of the sample is
(A) 5 min
(B) 7.5 min
(C) 15 min
(D) 30 min
Official Ans. by NTA (A)
Sol. Remaining = 1/8
3t?/? = 15 min
t?/? = 5 min

19. An n.p.n transistor with current gain β = 100 in common emitter configuration is shown in figure. The output voltage of the amplifier will be
(A) 0.1 V
(B) 1.0 V
(C) 10 V
(D) 100 V
Official Ans. by NTA (B)
Sol. V_in = β R_out/R_in
V_out = (100 × 10 × 10³)/10³ × 10?³
= 1 V

20. A FM Broad cast transmitter, using modulating signal of frequency 20 kHz has a deviation ratio of 10. The Bandwidth required for transmission is:
(A) 220 kHz
(B) 180 kHz
(C) 360 kHz
(D) 440 kHz
Official Ans. by NTA (D)
Sol. Given FM broadcast Modulating frequency = 20 kHz = f
Deviation ratio = frequency deviation/modulating frequency = Δf/f
⇒ frequency deviation - Δf = f × 10
= 20 kHz × 10 = 200 kHz
⇒ Bandwidth = 2(f + Δf)
= 2(20 + 200) kHz
= 440 kHz

SECTION-B

1. A ball is thrown vertically upwards with a velocity of 19.6 ms?¹ from the top of a tower. The ball strikes the ground after 6 s. The height from the ground up to which the ball can rise will be (k/5) m. The value of k is .... (use g = 9.8 m/s²)
Official Ans. by NTA (392)
Sol. t_a = u/g = 19.6/9.8 = 2s
t_d = 6 - 2s = √(2h_max/g)
⇒ h_max = 16 × 9.8/2 = 392/5

2. The distance of centre of mass from end A of a one dimensional rod (AB) having mass density ρ = ρ?(1 - x²/L²) kg/m and length L (in meter) is 3L/α m. The value of α is .... (where x is the distance form end A)
Official Ans. by NTA (8)
Sol. dm = λ · dx = λ?(1 - x²/?²)
X_cm = ∫ x dm / ∫ dm_?
= (?/0 (x - x³/?²) dx) / (? - ?³/3?²) = ?²/2 - ??/4?² / ? - ?³/3?² = 3?/8
= 3?/8

3. A string of area of cross-section 4 mm² and length 0.5 is connected with a rigid body of mass 2 kg. The body is rotated in a vertical circular path of radius 0.5 m. The body acquires a speed of 5 m/s at the bottom of the circular path. Strain produced in the string when the body is at the bottom of the circle is ... × 10??. (Use Young's modulus 10¹¹ N/m² and g = 10 m/s²)
Official Ans. by NTA (30)
Sol. Strain = F/AY
= (mg + mv²/R)/AY
= (20 + 2(5)²/0.5)/(4 × 10?? × 10¹¹) = 30 × 10??

4. At a certain temperature, the degrees of freedom per molecule for gas is 8. The gas performs 150 J of work when it expands under constant pressure. The amount of heat absorbed by the gas will be .... J.
Official Ans. by NTA (750)
Sol. W = nR ΔT = 150 J
Q = (f/2 + 1) nR ΔT = (8/2 + 1)150 = 750 J

5. The potential energy of a particle of mass 4 kg in motion along the x-axis is given by U = 4(1 - cos 4x) J. The time period of the particle for small oscillation (sin θ = θ) is (π/K) s. The value of K is .......
Official Ans. by NTA (2)
Sol. U = 4(1 - cos 4x)
F = -dU/dx = -4(+ sin 4x)4 = -16 sin(4x)
For small θ
sin θ ? θ
F = -64x
a = -64x/m = -16x
ω² = 16
T = 2π/ω = π/2

6. An electrical bulb rated 220V, 100W, is connected in series with another bulb rated 220V, 60W. If the voltage across combination is 220V, the power consumed by the 100W bulb will be about ....... W.
Official Ans. by NTA (14)
Sol. R? = V²/P = 220²/100 = 484
R? = V²/P = 220²/60 = 484(10/6)
I = 220/(484 + 484 × 10/6)
P? = I²R? = 14.06 W

7. For the given circuit the current through battery of 6V just after closing the switch 'S' will be ....... A.
Official Ans. by NTA (1)
Sol. Just after closing the switch S, inductor behaves like an open circuit.
I = 6/(2 + 4) = 1 A

8. An object 'o' is placed at a distance of 100 cm in front of a concave mirror of radius of curvature 200 cm as shown in the figure. The object starts moving towards the mirror at a speed 2 cm/s. The position of the image from the mirror after 10s will be at ....... cm.
Official Ans. by NTA (400)
Sol. After 10 sec.
u = -80 cm
f = -100 cm
1/v + 1/u = 1/f
v = 400 cm

9. In an experiment with a convex lens. The plot of the image distance (ν*) against the object distance (μ*) measured from the focus gives a curve ν*μ* = 225. If all the distances are measured in cm. The magnitude of the focal length of the lens is.
Official Ans. by NTA (15)
Sol. vu = f² (by Newton's formula)
f² = 225
f = 15 cm

10. In an experiment to find acceleration due to gravity (g) using simple pendulum, time period of 0.5 s is measured from time of 100 oscillation with a watch of 1s resolution. If measured value of length is 10 cm known to 1 mm accuracy. The accuracy in the determination of g is found to be x%. The value of x is
Official Ans. by NTA (5)
Sol. T = 2π√(?/g)
g = 1/4π² T²/?
Δg/g = 2ΔT/T + Δ?/?
Δg/g = 2 · 1/(100 × 0.5) + 1mm/10cm
Δg/g = 5/100

CHEMISTRY

SECTION-A

1. Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R
Assertion A : Zero orbital overlap is an out of phase overlap.
Reason : It results due to different orientation/direction of approach of orbitals.
In the light of the above statements. Choose the correct answer from the options given below
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Official Ans. by NTA (A)

2. The correct decreasing order for metallic character is
(A) Na > Mg > Be > Si > P
(B) P > Si > Be > Mg > Na
(C) Si > P > Be > Na > Mg
(D) Be > Na > Mg > Si > P
Official Ans. by NTA (A)
Sol. Across a period metallic character decreases

TEST PAPER WITH SOLUTION

3. Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R
Assertion A : The reduction of a metal oxide is easier if the metal formed is in liquid state than solid state.
Reason R : The value of ΔG° becomes more on negative side as entropy is higher in liquid state than solid state.
In the light of the above statements. Choose the most appropriate answer from the options given below
(A) Both A and R are correct and R is the correct explanation of A
(B) Both A and R are correct but R is NOT the correct explanation of A
(C) A is correct but R is not correct
(D) A is not correct but R is correct
Official Ans. by NTA (A)
Sol. ΔG = ΔH - TΔS
Entropy of liquid is more than solid on melting the entropy increases and ΔG becomes more negative and hence it becomes easier to reduce metal

4. The products obtained during treatment of hard water using Clark's method are:
(A) CaCO? and MgCO?
(B) Ca(OH)? and Mg(OH)?
(C) CaCO? and Mg(OH)?
(D) Ca(OH)? and MgCO?
Official Ans. by NTA (C)
Sol. In Clark's method lime water is used
Ca(HCO?)? + 2Ca(OH)? → 2CaCO? + 2H?O
Mg(HCO?)? + 2Ca(OH)? → 2CaCO? + Mg(OH)? + 2H?O

5. Statement I: An alloy of lithium and magnesium is used to make aircraft plates.
Statement II: The magnesium ions are important for cell-membrane integrity.
In the light the above statements, choose the correct answer from the options given below
(A) Both Statement I and Statement II are true
(B) Both Statement I and Statement II are false
(C) Statement I is true but Statement II is false
(D) Statement I is false but Statement II is true
Official Ans. by NTA (B)
Sol. Alloy of Li and Mg is used to make armour plates and not aircraft plates. Calcium plays important roles in neuromuscular function, interneuronal transmission and cell membrane integrity

6. White phosphorus reacts with thionyl chloride to give
(A) PCl? SO? and S?Cl?
(B) PCl? SO? and S?Cl?
(C) PCl? SO? and Cl?
(D) PCl? SO? and Cl?
Official Ans. by NTA (B)
Sol. P? + 8SOCl? → 4PCl? + 4SO? + 2S?Cl?

7. Concentrated HNO? reacts with Iodine to give
(A) HI NO? and H?O
(B) HIO? N?O and H?O
(C) HIO? NO? and H?O
(D) HIO? N?O and H?O
Official Ans. by NTA (C)
Sol. I? + 10HNO?(conc) ⇒ 2HIO? + 10NO? + 4H?O

8. Which of the following pair is not isoelectronic species? (At. no. Sm, 62; Er, 68: Yb, 70: Lu, 71; Eu, 63: Tb, 65; Tm, 69)
(A) Sm²? and Er³?
(B) Yb²? and Lu³?
(C) Eu²? and Tb??
(D) Tb²? and Tm??
Official Ans. by NTA (D)
Sol. Sm²? → electron = 60
Er³? → electron = 65 (not isoelectronic)
Tb²? → electron = 63
Tm?? → electron = 65

9. Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R
Assertion A: Permanganate titrations are not performed in presence of hydrochloric acid.
Reason R: Chlorine is formed as a consequence of oxidation of hydrochloric acid.
In the light of the above statements, choose the correct answer from the options given below
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Official Ans. by NTA (A)
Sol. 2KMnO? + 16HCl → 2MnCl? + 2KCl + 8H?O + Cl?
HCl gets oxidised by KMnO? into Cl?

10. Match List I with List II
A Ni(CO)? I sp³
B [Ni(CN)?]²? II sp³d²
C [Co(CN)?]³? III d²sp³
D [CoF?]³? IV dsp²
Choose the correct answer from the options given below:
(A) A-IV, B-I, C-III, D-II
(B) A-I, B-IV, C-III, D-II
(C) A-I, B-IV, C-II, D-III
(D) A-IV, B-I, C-II, D-III
Official Ans. by NTA (B)

11. Dinitrogen and dioxygen, the main constituents of air do not react with each other in atmosphere to form oxides of nitrogen because
(A) N? is unreactive in the condition of atmosphere.
(B) Oxides of nitrogen are unstable.
(C) Reaction between them can occur in the presence of a catalyst.
(D) The reaction is endothermic and require very high temperature.
Official Ans. by NTA (D)
Sol. N? + O? → (1483 - 2000 K) 2NO
(Endothermic and feasible at high temperature)

12. The major product in the given reaction is
Official Ans. by NTA (C)
Sol.

13. Arrange the following in increasing order of reactivity towards nitration
A. p-xylene
B. bromobenzene
C. mesitylene
D. nitrobenzene
E. benzene
Choose the correct answer from the options given below
(A) C < D < E < A < B
(B) D < B < E < A < C
(C) D < C < E < A < B
(D) C < D < E < B < A
Official Ans. by NTA (B)
Sol.
-NO? is strongly deactivating
-Br - deactivating
-CH? activating group
D < B < E < A < C

14. Compound I is heated with Conc. HI to give a hydroxy compound A which is further heated with Zn dust to give compound B. Identify A and B.
Official Ans. by NTA (D)
Sol.

15. Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R
Assertion A : Aniline on nitration yields ortho, meta & para nitro derivatives of aniline.
Reason R: Nitrating mixture is a strong acidic mixture.
In the light of the above statements, choose the correct answer from the options given below
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Official Ans. by NTA (A)
Sol.
Due to formation of anilinium ion in acidic medium meta product is also obtained in significant amount

16. Match List I with List II
List (Polymer) List II (Nature)
A. CH?-C=CH-CH? Cl I. Thermosetting polymer
B. H N-(CH?)-N-C-(CH?)-C Cl II. Fibers
C. CH?-CH Cl III. Elastomer
D. O-H CH?-CH Cl IV. Thermoplastic polymer
Choose the correct answer from the options given below:
(A) A-II, B-III, C-IV, D-I
(B) A-III, B-II, C-IV, D-I
(C) A-III, B-I, C-IV, D-II
(D) A-I, B-III, C-IV, D-II
Official Ans. by NTA (B)
Sol. Neoprene is elastomer
Nylon-6, 6 is fiber
PVC is thermoplastic
Novolac is thermosetting

17. Two statements in respect of drug-enzyme interaction are given below
Statement I : Action of an enzyme can be blocked only when an inhibitor blocks the active site of the enzyme.
Statement II : An inhibitor can form a strong covalent bond with the enzyme.
In the light of the above statements. Choose the correct answer from the options given below
(A) Both Statement I and Statement II are true
(B) Both Statement I and Statement II are false
(C) Statement I is true but Statement II is false
(D) Statement I is false but Statement II is true
Official Ans. by NTA (D)
Sol. Some drugs do not bind to active sites. These bind to different site of enzyme called allosteric sites.

18. Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R
Assertion A : Thin layer chromatography is an adsorption chromatography.
Reason : A thin layer of silica gel is spread over a glass plate of suitable size in thin layer chromatography which acts as an adsorbent.
In the light of the above statements, choose the correct answer from the options given below
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Official Ans. by NTA (A)
Sol. Theory based
Thin layer chromatography (TLC) is another type of adsorption chromatography, which involve separation of substance of a mixture over a thin layer of an adsorbent coated on glass plate.
A thin layer (about 0.2 mm thick) of an adsorbent (silica gel) or (Alumina) in spread over a glass plate of suitable size. Hence Assertion (A) is correct and Reason (R) is correct explanation of (A)

19. The formulas of A and B for the following reaction sequence are
(A) A = C?H??O?, B = C?H??
(B) A = C?H??O?, B = C?H??O
(C) A = C?H??O?, B = C?H??
(D) A = C?H??O?, B = C?H??O?
Official Ans. by NTA (A)
Sol.

20. Find out the major product for the above reaction.
Official Ans. by NTA (C)
Sol.

SECTION-B

1. 2L of 0.2M H?SO? is reacted with 2L of 0.1M NaOH solution, the molarity of the resulting product Na?SO? in the solution is millimolar. (Nearest integer).
Official Ans. by NTA (25)
Sol. H?SO? + 2NaOH → Na?SO? + 2H?O
0.4 mol 0.2 mol -
0.3 mol - 0.1 mol
Molarity of Na?SO? is 0.1/4 = 0.025 M = 25 mM

2. Metal M crystallizes into a FCC lattice with the edge length of 4.0 × 10?? cm. The atomic mass of the metal is g/mol. (Nearest integer). (Use : N_A = 6.02 × 10²³ mol?¹, density of metal, M = 9.03 g cm?³)
Official Ans. by NTA (87)
Sol. a = 4 × 10?? cm
d = 9.03 g/ml
d = ZM/(N_A a³)
M = 9.03 × 6.02 × 10²³ × 64 × 10?²?/4 = 86.97

3. If the wavelength for an electron emitted from H-atom is 3.3 × 10?¹? m, then energy absorbed by the electron in its ground state compared to minimum energy required for its escape from the atom, is times. (Nearest integer).
Given : h = 6.626 × 10?³? Js Mass of electron = 9.1 × 10?³¹
Official Ans. by NTA (2)
Sol. λ = h/√(2mK)
K = h²/(2mλ²)
K = h²/(2mλ³) = 43.9 × 10???/(2 × 9.1 × 10?³¹ × 10.89 × 10?²?)
K = 2.215 × 10?¹?
E_abs = E_req + K
E_abs/E_req = 1 + K/E_req = 1 + 2.215 × 10?¹?/(13.6 × 1.602 × 10?¹?) = 2.0166

4. A gaseous mixture of two substances A and B, under a total pressure of 0.8 atm is in equilibrium with an ideal liquid solution. The mole fraction of substance A is 0.5 in the vapour phase and 0.2 in the liquid phase. The vapour pressure of pure liquid A is atm. (Nearest integer)
Official Ans. by NTA (2)
Sol. Y_A = 0.5 ⇒ Y_B = 0.5
P_A = P_B = 0.4 atm
P_A = P_A? X_A
P_A? = 2

5. At 600 K, 2 mol of NO are mixed with 1 mol of O?.
2NO(g) + O?(g) ? 2NO?(g)
The reaction occurring as above comes to equilibrium under a total pressure of 1 atom. Analysis of the system shows that 0.6 mol of oxygen are present at equilibrium. The equilibrium constant for the reaction is . (Nearest integer).
Official Ans. by NTA (2)
Sol. 2NO + O? → 2NO?
2 - 2x 1 - x 2x
1.2 0.6 0.8
K_p = (0.8/2.6)²/((1.2/2.6)²(0.6/2.6)) = 1.925
K_p = (0.8/2.6)²/((1.2/2.6)²(0.6/2.6)) = 1.925

6. A sample of 0.125 g of an organic compound when analysed by Duma's method yields 22.78 mL of nitrogen gas collected over KOH solution at 280 K and 759 mmHg. The percentage of nitrogen in the given organic compound is . (Nearest integer). (a) The vapour pressure of water at 280 K is 14.2 mm Hg (b) R = 0.082 L atm K?¹ mol?¹
Official Ans. by NTA (22)
Sol. V = 22.78 ml
T = 280 K
P_total = 759 mmHg
P_N? = 759 - 14.2 = 744.8 mmHg
n_N? = 744.8 × 22.78/(760 × 1000 × 0.082 × 280) = 0.00097
W_Nitrogen = 0.02716
%N = 0.02716/0.125 × 1000 = 21.728

7. On reaction with stronger oxidizing agent like KIO? hydrogen peroxide oxidizes with the evolution of O?. The oxidation number of I in KIO? changes to
Official Ans. by NTA (5)
Sol. IO?? + H?O? → IO?? + O?

8. For a reaction, given below is the graph of ln k vs 1/T. The activation energy for the reaction is equal to cal mol?¹. (Nearest integer). (Given: R = 2 cal K?¹ mol?¹)
Official Ans. by NTA (8)
Sol. K = Ae^(-Ea/RT)
ln k = -Ea/RT + ln A
Slope = Ea/R = 20/5
Ea = 4R = 8 Cal/mol

9. Among the following the number of curves not in accordance with Freundlich adsorption isotherm is
Official Ans. by NTA (3)
Sol. X/m = K P^(1/n)
log(X/m) = 1/n log p + log k
log(X/m) = 1/n log p
log(X/m) = log k + 1/n log p

10. Among the following the number of state variable is
Internal energy (U)
Volume (V)
Heat (q)
Enthalpy (H)
Official Ans. by NTA (3)
Sol. Internal energy, volume enthalpy are state variable

MATHEMATICS

SECTION-A

1. Let S = {x ∈ [-6, 3] - {-2, 2} : (|x + 3| - 1)/(|x| - 2) ≥ 0} and T = {x ∈ Z : x² - 7|x| + 9 ≤ 0}. Then the number of elements in S ∩ T is
(A) 7
(B) 5
(C) 4
(D) 3
Official Ans. by NTA (D)
Sol. S ∩ T = {-5, -4, 3}

2. Let α, β be the roots of the equation x² - √2x + √6 = 0 and 1/α² + 1, 1/β² + 1 be the roots of the equation x² + ax + b = 0. Then the roots of the equation x² - (a + b - 2)x + (a + b + 2) = 0 are :
(A) non-real complex numbers
(B) real and both negative
(C) real and both positive
(D) real and exactly one of them is positive
Official Ans. by NTA (B)
Sol. a = -1/α² - 1/β² - 2
b = 1/α² + 1/β² + 1 + 1/α²β²
a + b = 1/(αβ)² - 1 = 1/6 - 1 = -5/6
x² - (-5/6 - 2)x + (2 - 5/6) = 0
6x² + 17x + 7 = 0
x = -7/3, x = -1/2 are the roots
Both roots are real and negative.

TEST PAPER WITH SOLUTION

3. Let A and B be any two 3 × 3 symmetric and skew symmetric matrices respectively. Then which of the following is NOT true?
(A) A? - B? is a symmetric matrix
(B) AB - BA is a symmetric matrix
(C) B? - A? is a skew-symmetric matrix
(D) AB + BA is a skew-symmetric matrix
Official Ans. by NTA (C)
Sol. Given that A? = A, B? = -B
C = A? - B?
C? = (A? - B?) = (A?)? - (B?)? = A? - B? = C
C = AB - BA
C? = (AB - BA)? = (AB)? - (BA)?
= B?A? - A?B? = -BA + AB = C
C = B? - A?
C? = (B? - A?)? = (B?)? - (A?)? = -B? - A?
C = AB + BA
C? = (AB + BA)? = (AB)? + (BA)?
= -BA - AB = -C
∴ Option C is not true.

4. Let f(x) = ax² + bx + c be such that f(1) = 3, f(-2) = λ and f(3) = 4. If f(0) + f(1) + f(-2) + f(3) = 14 then λ is equal to
(A) -4
(B) 13/2
(C) 23/2
(D) 4
Official Ans. by NTA (D)
Sol. f(0) + 3 + λ + 4 = 14
∴ f(0) = 7 - λ = c
f(1) = a + b + c = 3 ...(i)
f(3) = 9a + 3b + c = 4 ...(ii)
f(-2) = 4a - 2b + c = λ ...(iii)
(ii)-(iii)
a + b = (4 - λ)/5
put in equation (i)
(4 - λ)/5 + 7 - λ = 3
6λ = 24; λ = 4

5. The function f : R → R defined by f(x) = lim_{n→∞} (cos(2πx) - x^{2n} sin(x - 1))/(1 + x^{2n+1} - x^{2n}) is continuous for all x in
(A) R - {-1}
(B) R - {-1, 1}
(C) R - {1}
(D) R - {0}
Official Ans. by NTA (B)
Note : n should be given as a natural number.
Sol. f(x) = { -sin(x - 1) x < -1
(x - 1)/(-(sin 2 + 1)) x = -1 }
Sol. f(x) = { -sin(x - 1) x > 1
(x - 1)/(-(sin 2 + 1)) x = -1 }
f(x) is discontinuous at x = -1 and x = 1

6. The function f(x) = x e^{x(1 - x)}, x ∈ R, is
(A) increasing in (-1/2, 1)
(B) decreasing in (1/2, 2)
(C) increasing in (-1, -1/2)
(D) decreasing in (-1/2, 1/2)
Official Ans. by NTA (A)
Sol. f(x) = x e^{x(1 - x)}
f'(x) = -e^{x(1 - x)}(2x + 1)(x - 1)
f(x) is increasing in (-1/2, 1)

7. The sum of the absolute maximum and absolute minimum values of the function f(x) = tan?¹(sin x - cos x) in the interval [0, π] is
(A) 0
(B) tan?¹(1/√2) - π/4
(C) cos?¹(1/√3) - π/4
(D) -π/12
Official Ans. by NTA (C)
Sol. f(x) = tan?¹(sin x - cos x)
f'(x) = (cos x + sin x)/((sin x - cos x)² + 1) = 0
∴ x = 3π/4
x: 0, 3π/4
f(x): -π/4, tan?¹√2
f(x)_max = tan?¹√2
f(x)_min = -π/4
sum = tan?¹√2 - π/4
= cos?¹(1/√3) - π/4

8. Let x(t) = 2√2 cos t √(sin 2t) and y(t) = 2√2 sin t √(sin 2t)
t ∈ (0, π/2)
(1 + (dy/dx)²)² / (d²y/dx²) at t = π/4 is equal to
(A) -2√2/3
(B) 2/3
(C) 1/3
(D) -2/3
Official Ans. by NTA (D)
Sol. x = 2√2 cos t √(sin 2t)
dx/dt = 2√2 cos 3t/√(sin 2t)
y(t) = 2√2 sin t √(sin 2t)
dy/dt = 2√2 sin 3t/√(sin 2t)
dy/dx = tan 3t
dy/dx = -1 at t = π/4
d²y/dx² = 3/(2√2) sec³ 3t · √(sin 2t) = -3 at t = π/4
∴ (1 + (dy/dx)²)/(d²y/dx²) = (1 + 1)/(-3) = -2/3

9. Let I_n(x) = ∫?? 1/(t² + 5)? dt, n = 1, 2, 3, .... Then
(A) 50I? - 9I? = xI?'
(B) 50I? - 11I? = xI?'
(C) 50I? - 9I? = I?'
(D) 50I? - 11I? = I?'
Official Ans. by NTA (A)
Sol. I_n(x) = ∫?? dt/(t² + 5)?
Applying integral by parts
I_n(x) = [t/(t² + 5)?]?? - ∫?? n(t² + 5)???¹ · 2t²
I_n(x) = x/(x² + 5)? + 2n ∫?? t²/(t² + 5)??¹ dt
I_n(x) = x/(x² + 5)? + 2n ∫?? (t² + 5) - 5/(t² + 5)??¹ dt
I_n(x) = x/(x² + 5)? + 2n I_n(x) - 10n I_{n+1}(x)
10n I_{n+1}(x) + (1 - 2n)I_n(x) = x/(x² + 5)?
Put n = 5

10. The area enclosed by the curves y = log_e(x + e²), x = log_e(2/y) and x = log_e 2, above the line y = 1 is
(A) 2 + e - log_e 2
(B) 1 + e - log_e 2
(C) e - log_e 2
(D) 1 + log_e 2
Official Ans. by NTA (B)
Sol. Required area is
= ∫?^{ln 2} ln(x + e²) - 1 dx + ∫_{ln 2}^{2e^{-1}} 2e^{-x} - 1 dx = 1 + e - ln 2

11. Let y = y(x) be the solution curve of the differential equation dy/dx + 1/(x² - 1) y = (x - 1)/(x + 1)^{1/2}, x > 1 passing through the point (2, √(1/3)). Then √7 y(8) is equal to
(A) 11 + 6 log_e 3
(B) 19
(C) 12 - 2 log_e 3
(D) 19 - 6 log_e 3
Official Ans. by NTA (D)
Sol. dy/dx + 1/(x² - 1) y = ((x - 1)/(x + 1))^{1/2},
dy/dx + Py = Q
IF = e^{∫Pdx} = ((x - 1)/(x + 1))^{1/2}
y((x - 1)/(x + 1))^{1/2} = ∫((x - 1)/(x + 1))^{1/2} dx
= x - 2 log_e |x + 1| + C
Curve passes through (2, √(1/3))
⇒ C = 2 log_e 3 - 5/3
at x = 8,
√7 y(8) = 19 - 6 log_e 3

12. The differential equation of the family of circles passing through the points (0, 2) and (0, -2) is
(A) 2xy dy/dx + (x² - y² + 4) = 0
(B) 2xy dy/dx + (x² + y² - 4) = 0
(C) 2xy dy/dx + (y² - x² + 4) = 0
(D) 2xy dy/dx - (x² - y² + 4) = 0
Official Ans. by NTA (A)
Sol. Equation of circle passing through (0, -2) and (0, 2) is
x² + (y² - 4) + λx = 0, (λ ∈ R)
Divided by x we get
(x² + (y² - 4))/x + λ = 0
Differentiating with respect to x
(x[2x + 2y · dy/dx] - [x² + y² - 4] · 1)/x² = 0
⇒ 2xy · dy/dx + (x² - y² + 4) = 0

13. Let the tangents at two points A and B on the circle x² + y² - 4x + 3 = 0 meet at origin O(0, 0). Then the area of the triangle of OAB is
(A) 3√3/2
(B) 3√3/4
(C) 3/(2√3)
(D) 3/(4√3)
Official Ans. by NTA (B)
Sol. C : (x - 2)² + y² = 1
Equation of chord AB : 2x = 3
OA = OB = √3
AM = √3/2
Area of triangle OAB = 1/2 (2AM)(OM)
= 3√3/4 sq. units

14. Let the hyperbola H : x²/a² - y²/b² = 1 pass through the point (2√2, -2√2). A parabola is drawn whose focus is same as the focus of H with positive abscissa and the directrix of the parabola passes through the other focus of H. If the length of the latus rectum of the parabola is e times the length of the latus rectum of H, where e is the eccentricity of H, then which of the following points lies on the parabola?
(A) (2√3, 3√2)
(B) (3√3, -6√2)
(C) (√3, -√6)
(D) (3√6, 6√2)
Official Ans. by NTA (B)
Sol. H : x²/a² - y²/b² = 1
Foci : S(ae, 0), S'(-ae, 0)
Foot of directrix of parabola is (-ae, 0)
Focus of parabola is (ae, 0)
Now, semi latus rectum of parabola = |SS'| = 2ae
Given, 4ae = e(2b²/a)
⇒ b² = 2a² ...(1)
Given, (2√2, -2√2) lies on H
⇒ 1/a² - 1/b² = 1/8 ...(2)
From (1) and (2)
a² = 4, b² = 8
∴ b² = a²(e² - 1)
∴ e = √3
⇒ Equation of parabola is y² = 8√3x

15. Let the lines (x - 1)/λ = (y - 2)/1 = (z - 3)/2 and (x + 26)/(-2) = (y + 18)/3 = (z + 28)/λ be coplanar and P be the plane containing these two lines. Then which of the following points does NOT lies on P?
(A) (0, -2, -2)
(B) (-5, 0, -1)
(C) (3, -1, 0)
(D) (0, 4, 5)
Official Ans. by NTA (B)
Sol. Given, L? : (x - 1)/λ = (y - 2)/1 = (z - 3)/2
and L? : (x + 26)/(-2) = (y + 18)/3 = (z + 28)/λ
are coplanar
Now, normal of plane P, which contains L? and L?
= -3i - 13j + 11k
⇒ Equation of required plane P :
3x + 13y - 11z + 4 = 0
(0, 4, 5) does not lie on plane P.

16. A plane P is parallel to two lines whose direction ratios are -2, 1, -3, and -1, 2, -2 and it contains the point (2, 2, -2). Let P intersect the co-ordinate axes at the points A, B, C making the intercepts α, β, γ. If V is the volume of the tetrahedron OABC, where O is the origin and p = α + β + γ then the ordered pair (V, p) is equal to
(A) (48, -13)
(B) (24, -13)
(C) (48, 11)
(D) (24, -5)
Official Ans. by NTA (B)
Sol. Normal of plane P :
Equation of plane P which passes through (2, 2, -2) is 4x - y - 3z - 12 = 0
Now, A(3, 0, 0), B(0, -12, 0), C(0, 0, -4)
⇒ α = 3, β = -12, γ = -4
⇒ p = α + β + γ = -13
Now, volume of tetrahedron OABC
V = |1/6 OA · (OB × OC)| = 24
(V, p) = (24, -13)

17. Let S be the set of all a ∈ R for which the angle between the vectors u = a(log_e b)i - 6j + 3k and v = (log_e b)i + 2j + 2a(log_e b)k, (b > 1) is acute. Then S is equal to
(A) (-∞, -4/3)
(B) Φ
(C) (-4/3, 0)
(D) (12/7, ∞)
Official Ans. by NTA (C)
Sol. For angle to be acute
u · v > 0
⇒ a(log_e b)² - 12 + 6a(log_e b) > 0
∀ b > 1
let log_e b = t ⇒ t > 0 as b > 1
y = at² + 6at - 12 & y > 0, ∀ t > 0
⇒ a ∈ Φ

18. A horizontal park is in the shape of a triangle OAB with AB = 16. A vertical lamp post OP is erected at the point O such that ∠PAO = ∠PBO = 15° and ∠PCO = 45°, where C is the midpoint of AB. Then (OP)² is equal to
(A) 32/√3 (√3 - 1)
(B) 32/√3 (2 - √3)
(C) 16/√3 (√3 - 1)
(D) 16/√3 (2 - √3)
Official Ans. by NTA (B)
Sol. OP/OA = tan 15°
⇒ OA = OP cot 15°
OP/OC = tan 45° ⇒ OP = OC
Now, OP = √(OA² - 8²)
⇒ OP² = (OP)² cot² 15° - 64
⇒ OP² = 32/√3 (2 - √3)

19. Let A and B be two events such that P(B|A) = 2/5
P(A|B) = 1/7 and P(A ∩ B) = 1/9. Consider
(S1) P(A' ∪ B) = 5/6,
(S2) P(A' ∩ B') = 1/18.
(A) Both (S1) and (S2) are true
(B) Both (S1) and (S2) are false
(C) Only (S1) is true
(D) Only (S2) is true
Official Ans. by NTA (A)
Sol. P(A|B) = 1/7 ⇒ P(A ∩ B)/P(B) = 1/7
⇒ P(B) = 7/9
P(B|A) = 2/5 ⇒ P(A ∩ B)/P(A) = 2/5
⇒ P(A) = 5/18
Now, P(A' ∪ B) = 1 - P(A ∪ B) + P(B)
= 1 - P(A) + P(A ∩ B) = 5/6
P(A' ∩ B') = 1 - P(A ∪ B)
= 1 - P(A) - P(B) + P(A ∩ B) = 1/18
⇒ Both (S1) and (S2) are true.

20. Let
p : Ramesh listens to music.
q : Ramesh is out of his village
r : It is Sunday
s : It is Saturday
Then the statement "Ramesh listens to music only if he is in his village and it is Sunday or Saturday" can be expressed as
(A) ((~q) ∧ (r ∨ s)) ⇒ p
(B) (q ∧ (r ∨ s)) ⇒ p
(C) p ⇒ (q ∧ (r ∨ s))
(D) p ⇒ ((~q) ∧ (r ∨ s))
Official Ans. by NTA (D)
Sol. p ≡ Ramesh listens to music
~q ≡ He is in village.
r ∨ s ≡ Saturday or sunday
p ⇒ ((~q) ∧ (r ∨ s))

SECTION-B

1. Let the coefficients of the middle terms in the expansion of (1/√6 + βx)?, (1 - 3βx)² and (1 - β/2 x)?, β > 0, respectively form the first three terms of an A.P. If d is the common difference of this A.P., then 50 - 2d/β² is equal to
Official Ans. by NTA (57)
Sol. ?C? × β²/6, -6β, -?C? × β³/8 are in A.P
β² - 5/2 β³ = -12β
β = 12/5 or β = -2 : β = 12/5
d = -72/5 - 144/25 = -504/25
∴ 50 - 2d/β² = 57

2. A class contains b boys and g girls. If the number of ways of selecting 3 boys and 2 girls from the class is 168, then b + 3g is equal to
Official Ans. by NTA (17)
Sol. bC? × gC? = 168
b(b - 1)(b - 2)(g)(g - 1) = 8 × 7 × 6 × 3 × 2
b + 3g = 17

3. Let the tangents at the points P and Q on the ellipse x²/2 + y²/4 = 1 meet at the point R(√2, 2√2 - 2). If S is the focus of the ellipse on its negative major axis, then SP² + SQ² is equal to
Official Ans. by NTA (13)
Sol. Ellipse is
x²/2 + y²/4 = 1; e = 1/√2; S ≡ (0, -√2)
Chord of contact is
x/√2 + (2√2 - 2)y/4 = 1
⇒ x/√2 = 1 - (√2 - 1)y/2 solving with ellipse
⇒ y = 0, √2 ∴ x = √2, 1
P ≡ (1, √2), Q ≡ (√2, 0)
∴ (SP)² + (SQ)² = 13

4. If 1 + (2 + ??C? + ??C? + ... + ??C??)(??C? + ??C? + ... + ??C??) is equal to 2?, where m is odd, then n + m is equal to
Official Ans. by NTA (99)
Sol. 1 + (1 + 2??)(2?? - 1) = 2??
m = 1, n = 98
m + n = 99

5. Two tangent lines l? and l? are drawn from the point (2, 0) to the parabola 2y² = -x. If the lines l? and l? are also tangent to the circle (x - 5)² + y² = r, then 17r is equal to
Official Ans. by NTA (9)
Sol. y² = -x/2
y = mx - 1/8m
this tangent pass through (2, 0)
m = ±1/4 i.e., one tangent is x - 4y - 2 = 0
17r = 9

6. If 6/3¹² + 10/3¹¹ + 20/3¹? + 40/3? + ... + 10240/3 = 2? · m, where m is odd, then m·n is equal to
Official Ans. by NTA (12)
Sol. 6/3¹² + 10(1/3¹¹ + 2/3¹? + 2²/3? + 2³/3? + ... + 2¹?/3)
6/3¹² + 10/3¹¹ ((6¹¹ - 1)/(6 - 1))
= 2¹² · 1; m·n = 12

7. Let S = [-π, π/2) - (π/2, -π/4, -3π/4, π/4). Then the number of elements in the set
A = {θ ∈ S : tan θ (1 + √5 tan(2θ)) = √5 - tan(2θ)}
Official Ans. by NTA (5)
Sol. tan θ + √5 tan 2θ tan θ = √5 - tan 2θ
tan 3θ = √5
θ = nπ/3 + α/3; tan α = √5
Five solution

8. Let z = a + ib, b ≠ 0 be complex numbers satisfying z² = z? · 2^{1 - |z|}. Then the least value of n ∈ N, such that z? = (z + 1)?, is equal to
Official Ans. by NTA (6)
Sol. |z²| = |z?| · 2^{1 - |z|} ⇒ |z| = 1
z² = z? ⇒ z³ = 1 : ∴ z = ω or ω²
ω? = (1 + ω)? = (-ω²)?
Least natural value of n is 6.

9. A bag contains 4 white and 6 black balls. Three balls are drawn at random from the bag. Let X be the number of white balls, among the drawn balls. If σ² is the variance of X, then 100σ² is equal to
Official Ans. by NTA (56)
Sol. σ² = Σ X² P(X) - (Σ X P(X))² = 56/100

10. ∫?^{π/2} sin(6x) sin(6x) sin(6x) sin(6x) sin(6x) dx
Official Ans. by NTA (104)
Sol. I = 60∫?^{π/2} ((sin 6x - sin 4x)/sin x + (sin 4x - sin 2x)/sin x + sin 2x/sin x) dx
I = 60∫?^{π/2} (2cos 5x + 2cos 3x + 2cos x) dx
I = 60(2/5 sin 5x + 2/3 sin 3x + 2 sin x)∫?^{π/2} = 104

PHYSICS

SECTION-A

1. The dimensions of (B²/μ?) will be :
(if μ? : permeability of free space and B : magnetic field)
(A) [M L² T?²]
(B) [M L T?²]
(C) [M L?¹ T?²]
(D) [M L² T?² A?¹]
Official Ans. by NTA (C)
Sol. u = B²/2μ?
u → Energy per unit volume
[B²/μ?] = [u] = [M L² T?²]/[L³] = [M L?¹ T?²]

2. A NCC parade is going at a uniform speed of 9 km/h under a mango tree on which a monkey is sitting at a height of 19.6 m. At any particular instant, the monkey drops a mango. A cadet will receive the mango whose distance from the tree at time of drop is :
(Given g = 9.8 m/s²)
(A) 5 m
(B) 10 m
(C) 19.8 m
(D) 24.5 m
Official Ans. by NTA (A)
Sol. Monkey
Time taken by mango = √(2n/g)
= √(2 × 19.6/9.8) = 2 second
Distance = vt
= 9 × 5/18 × 2 = 5 m

TEST PAPER WITH SOLUTION

3. In two different experiments, an object of mass 5 kg moving with a speed of 25 m s?¹ hits two different walls and comes to rest within (i) 3 second, (ii) 5 seconds, respectively. Choose the correct option out of the following :
(A) Impulse and average force acting on the object will be same for both the cases.
(B) Impulse will be same for both the cases but the average force will be different.
(C) Average force will be same for both the cases but the impulse will be different.
(D) Average force and impulse will be different for both the cases.
Official Ans. by NTA (B)
Sol. Impulse = change in momentum
I = ΔP
F_avg = ΔP/Δt
Δt? = 3, Δt? = 5
ΔP? = ΔP?
I? = I?
F_avg in case (i) is more than (ii)

4. A balloon has mass of 10 g in air. The air escapes from the balloon at a uniform rate with velocity 4.5 cm/s. If the balloon shrinks in 5 s completely. Then, the average force acting on that balloon will be (in dyne).
(A) 3
(B) 9
(C) 12
(D) 18
Official Ans. by NTA (B)
Sol. F = dm/dt v
= 10 g/5 s (4.5 cm/s) = 9 g cm/s² = 9 dyne

5. If the radius of earth shrinks by 2% while its mass remains same. The acceleration due to gravity on the earth's surface will approximately :
(A) decrease by 2%
(B) decrease by 4%
(C) increase by 2%
(D) increase by 4%
Official Ans. by NTA (D)
Sol. g = GM/R²
M = constant g < 1/R²
100 Δg/g = -2 ΔR/R 100
% change = -2(-2)
% change in g = 4%
increase by 4%

6. The force required to stretch a wire of cross-section 1 cm² to double its length will be :
(Given Yong's modulus of the wire = 2 × 10¹¹ N/m²)
(A) 1 × 10? N
(B) 1.5 × 10? N
(C) 2 × 10? N
(D) 2.5 × 10? N
Official Ans. by NTA (C)
Sol. F = γA Δ?/?
= 2 × 10¹¹ × 10?? ((2? - ?)/?)
= 2 × 10? N

7. A Carnot engine has efficiency of 50%. If the temperature of sink is reduced by 40°C its efficiency increases by 30%. The temperature of the source will be :
(A) 166.7 K
(B) 255.1 K
(C) 266.7 K
(D) 367.7 K
Official Ans. by NTA (C)
Sol. η = 1 - T_L/T_H
1/2 = 1 - T_L/T_H
1/2 (1.3) = 1 - (T_L - 40)/T_H
1/2 (1.3) = 1/2 + 40/T_H
T_H = 266.7 K

8. Given below are two statements :
Statement I : The average momentum of a molecule in a sample of an ideal gas depends on temperature.
Statement II : The rms speed of oxygen molecules in a gas is v. If the temperature is doubled and the oxygen molecules dissociate into oxygen atoms, the rms speed will become 2v.
In the light of the above statements, choose the correct answer from the options given below :
(A) Both Statement I and Statement II are true
(B) Both Statement I and Statement II are false
(C) Statement I is true but Statement II is false
(D) Statement I is false but Statement II is true
Official Ans. by NTA (D)
Sol. [P_avg = 0] (due to random motion)
v_rms = √(3RT/M)
T_new = 2T
M_new = M/2
v_new = √(2T/(M/2))/v
v_new = 2v
v_new = 2v

9. In the wave equation
y = 0.5 sin(2π/λ)(400t - x) m
the velocity of the wave will be :
(A) 200 m/s
(B) 200√2 m/s
(C) 400 m/s
(D) 400√2 m/s
Official Ans. by NTA (C)
Sol. y = 0.5 sin((2π/λ)400t - (2π/λ)x)
ω = (2π/λ)400
K = 2π/λ
v = ω/k [v = 400 m/s]

10. Two capacitors, each having capacitance 40 μF are connected in series. The space between one of the capacitors is filled with dielectric material of dielectric constant K such that the equivalence capacitance of the system became 24 μF. The value of K will be :
(A) 1.5
(B) 2.5
(C) 1.2
(D) 3
Official Ans. by NTA (A)
Sol. C → KC
C_eq = C(KC)/(C + KC) = KC/(K + 1)
24 = K40/(K + 1)
[K = 1.5]

11. A wire of resistance R? is drawn out so that its length is increased by twice of its original length. The ratio of new resistance to original resistance is:
(A) 9 : 1
(B) 1 : 9
(C) 4 : 1
(D) 3 : 1
Official Ans. by NTA (A)
Sol. R? = ρ L?/A?
R? = ρ(3L?/(A?/3)) = 9ρ L?/A?
∴ R?/R? = 9

12. The current sensitivity of a galvanometer can be increased by :
(A) decreasing the number of turns
(B) increasing the magnetic field
(C) decreasing the area of the coil
(D) decreasing the torsional constant of the spring
Choose the most appropriate answer from the options given below :
(A) (B) and (C) only
(B) (C) and (D) only
(C) (A) and (C) only
(D) (B) and (D) only
Official Ans. by NTA (D)
Sol. i = (K/NAB)θ
∴ dθ/di = NAB/K

13. As shown in the figure, a metallic rod of linear density 0.45 kg m?¹ is lying horizontally on a smooth incline plane which makes an angle of 45° with the horizontal. The minimum current flowing in the rod required to keep it stationary, when 0.15 T magnetic field is acting on it in the vertical upward direction, will be :
Use g = 10 m/s²
(A) 30 A
(B) 15 A
(C) 10 A
(D) 3 A
Official Ans. by NTA (A)
Sol. mg sin 45° = ILB cos 45°
∴ I = (m/L) g/B
= (0.45)(10)/0.15 = 30 A

14. The equation of current in a purely inductive circuit is 5 sin(49πt - 30°). If the inductance is 30 mH then the equation for the voltage across the inductor, will be :
{Let π = 22/7}
(A) 1.47 sin(49πt - 30°)
(B) 1.47 sin(49πt + 60°)
(C) 23.1 sin(49πt - 30°)
(D) 23.1 sin(49πt + 60°)
Official Ans. by NTA (D)
Sol. v? = i?x_L
i?(wL)
= (5)(49π)(30 × 10?³)
= 23.1
Voltage will lead current by 90°
∴ V = 23.1 sin(49πt + 60°)

15. As shown in the figure, after passing through the medium 1. The speed of light v? in medium 2 will be :
(Given c = 3 × 10? m s?¹)
(A) 1.0 × 10? ms?¹
(B) 0.5 × 10? ms?¹
(C) 1.5 × 10? ms?¹
(D) 3.0 × 10? ms?¹
Official Ans. by NTA (A)
Sol. μ?/μ_air = C/v?
∴ √(μ?ε?/(1)) = C/v?
∴ √(1)(9) = C/v?
∴ v? = C/3
∴ v? = C/3

16. In normal adjustment, for a refracting telescope, the distance between objective and eye piece is 30 cm. The focal length of the objective, when the angular magnification of the telescope is 2, will be:
(A) 20 cm
(B) 30 cm
(C) 10 cm
(D) 15 cm
Official Ans. by NTA (A)
Sol. f? + f_c = 30
m = f?/f_c
2 = f?/f_c ⇒ f? = 2f_c
So f? + f?/2 = 30
f? = 20 cm

17. The equation λ = 1.227/x nm can be used to find the de-Broglie wavelength of an electron. In this equation x stands for : Where, m = mass of electron P = momentum of electron K = Kinetic energy of electron V = Accelerating potential in volts for electron
(A) √(mK)
(B) √P
(C) √K
(D) √V
Official Ans. by NTA (D)
Sol. λ = h/mv (de-Broglie's wavelength)
λ = h/√(2m(K·E))
h = h/√(2mV)
Putting the values of m ; q
We get λ = 1.22/√V nm

18. The half life period of a radioactive substance is 60 days. The time taken for 7/8 th of its original mass to disintegrate will be :
(A) 120 days
(B) 130 days
(C) 180 days
(D) 20 days
Official Ans. by NTA (C)
Sol. 7/8 disintegrates means 1/8 remains
Or (1/2)³
3 half lives
= 180 days

19. Identify the solar cell characteristics from the following options :
Official Ans. by NTA (B)
Sol. Conceptual / theory

20. In the case of amplitude modulation to avoid distortion the modulation index (μ) should be :
(A) μ ≤ 1
(B) μ ≥ 1
(C) μ = 2
(D) μ = 0
Official Ans. by NTA (A)
Sol. μ = A_m/A_c
μ ≤ 1 to avoid distortion
because μ > 1 will result in interference between career frequency & message frequency.

SECTION-B

1. If the projection of 2i + 4j - 2k on i + 2j + αk is zero. Then, the value of α will be
Official Ans. by NTA (5)
Sol. a · b = 0
∴ a · b = 0
∴ 2 × 1 + 4 × 2 - 2 × α = 0
∴ α = 5

2. A freshly prepared radioactive source of half life 2 hours 30 minutes emits radiation which is 64 times the permissible safe level. The minimum time, after which it would be possible to work safely with source, will be hours.
Official Ans. by NTA (15)
Sol. A = A? × 2^{-t/T}
A?/64 = A? × 2^{-t/t}
∴ t = 6T = 6 × 2.5 = 15 hours

3. In a Young's double slit experiment, a laser light of 560 nm produces an interference pattern with consecutive bright fringes' separation of 7.2 mm. Now another light is used to produce an interference pattern with consecutive bright fringes' separation of 8.1 mm. The wavelength of second light is nm.
Official Ans. by NTA (630)
Sol. β ∝ λ
λ? = 9/8 λ?
∴ β? = 9/8 β? = 9/8 × 560 = 630 nm

4. The frequencies at which the current amplitude in an LCR series circuit becomes 1/√2 times its maximum value, are 212 rad s?¹ and 232 rad s?¹. The value of resistance in the circuit is R = 5Ω. The self inductance in the circuit is mH.
Official Ans. by NTA (250)
Sol. Band width = 232 - 212 = R/L
∴ L = 5/20 = 250 mH

5. As shown in the figure, a potentiometer wire of resistance 20Ω and length 300 cm is connected with resistance box (R.B.) and a standard cell of emf 4 V. For a resistance 'R' of resistance box introduced into the circuit, the null point for a cell of 20mV is found to be 60 cm. The value of 'R' is Ω
Official Ans. by NTA (780)
Sol. E = AC/AB (V_A - V_B)
∴ 20 × 10?³ = 60/300 × (4 × 20)/(R + 20)
∴ R = 780 Ω

6. Two electric dipoles of dipole moments 1.2 × 10?³? cm and 2.4 × 10?³? cm are placed in two difference uniform electric fields of strengths 5 × 10? NC?¹ and 15 × 10? NC?¹ respectively. The ratio of maximum torque experienced by the electric dipoles will be 1/x. The value of x is
Official Ans. by NTA (6)
Sol. |τ|_max = PE
τ?/τ? = P?E?/P?E? = (1.2 × 10?³? × 5 × 10?)/(2.4 × 10?³? × 15 × 10?) = 1/6
Hence x = 6

7. The frequency of echo will be Hz if the train blowing a whistle of frequency 320 Hz is moving with a velocity of 36 km/h towards a hill from which an echo is heard by the train driver. Velocity of sound in air is 330 m/s.
Official Ans. by NTA (340)
Sol. The hill will be a secondary source.
f? = frequency of the car w.r.t. the hill
f? = (v/(v - v_s))f = (330/320) × 320 = 330 Hz
f? = Frequency of the sound reflected by hill w.r.t. the car (echo)
f? = ((v + v?)/v)f? = (330 + 10)/330 × 330 = 340 Hz

8. The diameter of an air bubble which was initially 2 mm, rises steadily through a solution of density 1750 kg m?³ at the rate of 0.35 cm s?¹. The coefficient of viscosity of the solution is ______. poise (in nearest integer). (the density of air is negligible).
Official Ans. by NTA (11)
Sol. As the bubble is rising steadily the net force acting on it will be zero
(Because of density of air the value of mg can be neglected)
So B = F ⇒ 4π/3 R³ρg = 6πηRv
Putting R = 1 mm = 10?³ m
ρ = 1.75 × 10³ kg/m³
g = 10 m/s²
v = 0.35 × 10?² m/s
η = 10/9 = 1.11 SI unit = 11 poise (CGS)

9. A block of mass 'm' (as shown in figure) moving with kinetic energy E compresses a spring through a distance 25 cm when, its speed is halved. The value of spring constant of used spring will be nE Nm?¹ for n =
Official Ans. by NTA (24)
Sol. Using work-energy theorem
W_net = (K_f - K_i)
⇒ -1/2 Kx² = 1/2 m(v/2)² - 1/2 mv² = E/4 - E
⇒ 1/2 Kx² = 3E/4 ⇒ K = 3E/2x²
⇒ K = 3E/(2 × (1/4)²) = 24E
n = 24

10. Four identical discs each of mass 'M' and diameter 'a' are arranged in a small plane as shown in figure. If the moment of inertia of the system about OO' is x/4 Ma². Then, the value of x will be
Official Ans. by NTA (3)
Sol. I? = I? = MR²/4
I? = MR²/4 + MR² = 5/4 MR² = I?

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