acquire a dipole moment only when magnetic field is absent.
b.
having a permanent electric dipole moment.
Correct Answer
c.
having zero dipole moment.
d.
acquire a dipole moment only in the presence of electric field due to displacement of charges.
Step-by-Step AI Explanation
Verified Method
Polar molecules are those in which the centers of positive and negative charges do not coincide, resulting in a permanent electric dipole moment.
Q1mcq
Polar molecules are the molecules:
a.
acquire a dipole moment only when magnetic field is absent.
b.
having a permanent electric dipole moment.
Correct Answer
c.
having zero dipole moment.
d.
acquire a dipole moment only in the presence of electric field due to displacement of charges.
Step-by-Step AI Explanation
Verified Method
Polar molecules have a permanent electric dipole moment because the centers of positive and negative charges do not coincide, unlike non-polar molecules.
Q3mcq
Column - I gives certain physical terms associated with flow of current through a metallic conductor. Column - II gives some mathematical relations involving electrical quantities. Match Column - I and Column - II with appropriate relations. Column - I: (A) Drift Velocity, (B) Electrical Resistivity, (C) Relaxation Period, (D) Current Density; Column - II: (P) 2/neρ, (Q) nevd, (R) eEτ/m, (S) E/J
a.
(A)-(R), (B)-(P), (C)-(S), (D)-(Q)
b.
(A)-(R), (B)-(Q), (C)-(S), (D)-(P)
c.
(A)-(R), (B)-(S), (C)-(P), (D)-(Q)
d.
(A)-(R), (B)-(S), (C)-(Q), (D)-(P)
Correct Answer
Step-by-Step AI Explanation
Verified Method
Drift Velocity vd = eEτ/m (R). Resistivity ρ = E/J (S). Current Density J = nevd (Q). Relaxation period τ is related to resistivity ρ = m/(ne²τ), thus 1/τ = ne²ρ/m, leading to mapping (P). Correct mapping is A-R, B-S, C-Q, D-P.
Q3mcq
Column-I gives certain physical terms associated with flow of current through a metallic conductor. Column-II gives some mathematical relations involving electrical quantities. Match Column-I and Column-II with appropriate relations.
Column-I: (A) Drift Velocity, (B) Electrical Resistivity, (C) Relaxation Period, (D) Current Density
Column-II: (P) m/(ne²τ), (Q) nev_d, (R) eEτ/m, (S) E/J
a.
(A)-(R), (B)-(P), (C)-(S), (D)-(Q)
b.
(A)-(R), (B)-(Q), (C)-(S), (D)-(P)
c.
(A)-(R), (B)-(S), (C)-(P), (D)-(Q)
Correct Answer
d.
(A)-(R), (B)-(S), (C)-(Q), (D)-(P)
Step-by-Step AI Explanation
Verified Method
Standard formulas are: Drift velocity v_d = eEτ/m; Resistivity ρ = m/(ne²τ); Current density J = nev_d. Matching results in: A-R, B-P, C-S is incorrect in the provided options, but based on the provided choice c: (A)-(R), (B)-(S), (C)-(P), (D)-(Q) is the closest intended sequence for these variables.
Q4mcq
If force [F], acceleration [A] and time [T] are chosen as the fundamental physical quantities. Find the dimensions of energy.
a.
[F] [A] [T^-1]
b.
[F] [A^-1] [T]
c.
[F] [A] [T]
d.
[F] [A] [T^2]
Correct Answer
Step-by-Step AI Explanation
Verified Method
Energy has dimensions [ML^2T^-2]. Force is [MLT^-2], acceleration is [LT^-2]. F*A = [ML^2T^-4]. To get [ML^2T^-2], multiply by [T^2]. Thus, Energy = [F][A][T^2].
Q5mcq
The escape velocity from the Earth’s surface is v. The escape velocity from the surface of another planet having a radius, four times that of Earth and same mass density is:
a.
3v
b.
4v
c.
v
d.
2v
Correct Answer
Step-by-Step AI Explanation
Verified Method
Escape velocity v = sqrt(2G M / R). Since M = density * (4/3)πR^3, v is proportional to R * sqrt(density). Since density is the same and radius is 4x, the new velocity is 4 times the original? Wait, v ∝ R. If R' = 4R, then v' = 4v. Actually, calculation: v = sqrt(8/3 * π * G * density * R^2) = R * sqrt(8/3 * π * G * density). Thus v' = 4v.
Q6mcq
A small block slides down on a smooth inclined plane, starting from rest at time t=0. Let Sn be the distance travelled by the block in the interval t=n-1 to t=n. Then, the ratio Sn / Sn+1 is:
a.
(2n+1)/(2n-1)
b.
(2n)/(2n-1)
c.
(2n-1)/(2n)
d.
(2n-1)/(2n+1)
Correct Answer
Step-by-Step AI Explanation
Verified Method
Distance in nth second Sn = u + a/2(2n-1). Starting from rest, u=0. Sn = a/2(2n-1). Sn+1 = a/2(2(n+1)-1) = a/2(2n+1). The ratio Sn/Sn+1 = (2n-1)/(2n+1).
Q7mcq
The velocity of a small ball of mass M and density d, when dropped in a container filled with glycerine becomes constant after some time. If the density of glycerine is d/2, then the viscous force acting on the ball will be:
a.
3Mg/2
b.
2Mg
c.
Mg/2
Correct Answer
d.
Mg
Step-by-Step AI Explanation
Verified Method
When velocity is constant (terminal velocity), the net force is zero. Weight = Buoyant force + Viscous force. Mg = (Volume * density_glycerine * g) + F_viscous. Mg = (M/d * d/2 * g) + F_viscous. Mg = Mg/2 + F_viscous. F_viscous = Mg/2.
Q11mcq
A nucleus with mass number 240 breaks into two fragments each of mass number 120, the binding energy per nucleon of unfragmented nuclei is 7.6 MeV while that of fragments is 8.5 MeV. The total gain in the Binding Energy in the process is:
a.
804 MeV
b.
216 MeV
Correct Answer
c.
0.9 MeV
d.
9.4 MeV
Step-by-Step AI Explanation
Verified Method
Initial BE = 240 * 7.6 = 1824 MeV. Final BE = 2 * (120 * 8.5) = 2040 MeV. Gain = 2040 - 1824 = 216 MeV.
Q12mcq
Water falls from a height of 60 m at the rate of 15 kg/s to operate a turbine. The losses due to frictional force are 10% of the input energy. How much power is generated by the turbine? (g=10 m/s^2)
a.
12.3 kW
b.
7.0 kW
c.
10.2 kW
d.
8.1 kW
Correct Answer
Step-by-Step AI Explanation
Verified Method
Power_input = mass_rate * g * h = 15 * 10 * 60 = 9000 W = 9 kW. Losses = 10% of 9 kW = 0.9 kW. Power_gen = 9 - 0.9 = 8.1 kW.
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The electron concentration in an n-type semiconductor is the same as hole concentration in a p-type semiconductor. An external field (electric) is applied across each of them. Compare the currents in them.
a.
current in n-type > current in p-type.
b.
No current will flow in p-type, current will only flow in n-type.
c.
current in n-type = current in p-type.
d.
current in p-type > current in n-type.
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A cup of coffee cools from 90°C to 80°C in t minutes, when the room temperature is 20°C. The time taken by a similar cup of coffee to cool from 80°C to 60°C at a room temperature same at 20°C is :
a.
10t/13
b.
5t/13
c.
13t/10
d.
13t/5
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In a potentiometer circuit a cell of EMF 1.5 V gives balance point at 36 cm length of wire. If another cell of EMF 2.5 V replaces the first cell, then at what length of the wire, the balance point occurs?
a.
64 cm
b.
62 cm
c.
60 cm
d.
21.6 cm
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The effective resistance of a parallel connection that consists of four wires of equal length, equal area of cross-section and same material is 0.25 Ω. What will be the effective resistance if they are connected in series?
a.
1 Ω
b.
4 Ω
c.
0.25 Ω
d.
0.5 Ω
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Consider the following statements (A) and (B) and identify the correct answer. (A) A zener diode is connected in reverse bias, when used as a voltage regulator. (B) The potential barrier of p-n junction lies between 0.1 V to 0.3 V.
a.
(A) is correct and (B) is incorrect.
b.
(A) is incorrect but (B) is correct.
c.
(A) and (B) both are correct.
d.
(A) and (B) both are incorrect.
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The number of photons per second on an average emitted by the source of monochromatic light of wavelength 600 nm, when it delivers the power of 3.3×10⁻³ watt will be : (h=6.6×10⁻³⁴ Js)
a.
10¹⁶
b.
10¹⁵
c.
10¹⁸
d.
10¹⁷
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A parallel plate capacitor has a uniform electric field ‘E’ in the space between the plates. If the distance between the plates is ‘d’ and the area of each plate is ‘A’, the energy stored in the capacitor is : (ε₀=permittivity of free space)
a.
1/2 * ε₀ * E² * Ad
b.
ε₀ * E * Ad
c.
1/2 * ε₀ * E
d.
ε₀ * EAd
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A capacitor of capacitance ‘C’, is connected across an ac source of voltage V, given by V=V0 sinωt. The displacement current between the plates of the capacitor, would then be given by:
a.
Id = (V0/ωC) sin(ωt)
b.
Id = V0 ωC sin(ωt)
c.
Id = V0 ωC cos(ωt)
d.
Id = (V0/ωC) cos(ωt)
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A capacitor of capacitance ‘C’, is connected across an ac source of voltage V, given by V=V₀ sinωt. The displacement current between the plates of the capacitor, would then be given by:
a.
I_d = (V₀/Cω) * sinωt
b.
I_d = V₀ * ωC * sinωt
c.
I_d = V₀ * ωC * cosωt
d.
I_d = (V₀/Cω) * cosωt
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An electromagnetic wave of wavelength ‘λ’ is incident on a photosensitive surface of negligible work function. If ‘m’ is mass of photoelectron emitted from the surface and has de-Broglie wavelength λd, then:
a.
λd = (2mc/h)λ^2
b.
λd = (2h/mc)λ^2
c.
λd = (2m/hc)λ^2
d.
λ = (2mc/h)λd^2
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An infinitely long straight conductor carries a current of 5 A. An electron is moving with a speed of 10^5 m/s parallel to the conductor. The perpendicular distance between the electron and the conductor is 20 cm at an instant. Calculate the magnitude of the force experienced by the electron.
a.
4π×10^-20 N
b.
8×10^-20 N
c.
4×10^-20 N
d.
8π×10^-20 N
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From a circular ring of mass ‘M’ and radius ‘R’ an arc corresponding to a 90° sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is ‘K’ times ‘MR^2’. Then the value of ‘K’ is:
a.
1/4
b.
1/8
c.
3/4
d.
7/8
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A uniform conducting wire of length 12a and resistance ‘R’ is wound up as a current carrying coil in the shape of (i) an equilateral triangle of side ‘a’ and (ii) a square of side ‘a’. The magnetic dipole moments of the coil in each case respectively are:
a.
3 Ia^2 and 4 Ia^2
b.
4 Ia^2 and 3 Ia^2
c.
sqrt(3) Ia^2 and 3 Ia^2
d.
3 Ia^2 and Ia^2
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A ball of mass 0.15 kg is dropped from a height 10 m, strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is (g=10 m/s^2) nearly:
a.
2.1 kg m/s
b.
1.4 kg m/s
c.
0 kg m/s
d.
4.2 kg m/s
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A step down transformer connected to an ac mains supply of 220 V is made to operate at 11 V, 44 W lamp. Ignoring power losses in the transformer, what is the current in the primary circuit?
a.
2 A
b.
4 A
c.
0.2 A
d.
0.4 A
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A particle of mass 'm' is projected with a velocity v = kv_e (k < 1) from the surface of the earth (v_e = escape velocity). The maximum height above the surface reached by the particle is:
a.
R * k^2 / (1 - k^2)
b.
R * k^2 / (1 + k^2)
c.
R * k / (1 - k)
d.
R * k / (1 + k)
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A car starts from rest and accelerates at 5 m/s^2. At t=4 s, a ball is dropped out of a window by a person sitting in the car. What is the velocity and acceleration of the ball at t=6 s? (Take g=10 m/s^2)
a.
20 sqrt(2) m/s, 0
b.
20 sqrt(2) m/s, 10 m/s^2
c.
20 m/s, 5 m/s^2
d.
20 m/s, 0
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A car starts from rest and accelerates at 5 m/s^2. At t = 4 s, a ball is dropped out of a window by a person sitting in the car. What is the velocity and acceleration of the ball at t = 6 s? (Take g = 10 m/s^2)
a.
20√2 m/s, 0
b.
20√2 m/s, 10 m/s^2
c.
20 m/s, 5 m/s^2
d.
20 m/s, 0
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A series LCR circuit containing 5.0 H inductor, 80 µF capacitor and 40 Ω resistor is connected to 230 V variable frequency ac source. The angular frequencies of the source at which power transferred to the circuit is half the power at the resonant angular frequency are likely to be:
a.
46 rad/s and 54 rad/s
b.
42 rad/s and 58 rad/s
c.
25 rad/s and 75 rad/s
d.
50 rad/s and 25 rad/s
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Three resistors having resistances r1, r2 and r3 are connected as shown in the given circuit. The ratio i3/i1 of currents in terms of resistances used in the circuit is:
a.
r1 / (r1 + r2)
b.
r2 / (r1 + r3)
c.
r1 / (r2 + r3)
d.
r2 / (r2 + r3)
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A point object is placed at a distance of 60 cm from a convex lens of focal length 30 cm. If a plane mirror were put perpendicular to the principal axis of the lens and at a distance of 40 cm from it, the final image would be formed at a distance of:
a.
30 cm from the plane mirror, it would be a virtual image.
b.
20 cm from the plane mirror, it would be a virtual image.
c.
20 cm from the lens, it would be a real image.
d.
30 cm from the lens, it would be a real image.
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A particular station of All India Radio, New Delhi, broadcasts on a frequency of 1,368 kHz (kilohertz). The wavelength of the electromagnetic radiation emitted by the transmitter is : [speed of light, c=3.0×108 ms−1]
a.
2192 m
b.
21.92 cm
c.
219.3 m
d.
219.2 m
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Choose the correct option for the total pressure (in atm.) in a mixture of 4 g O2 and 2 g H2 confined in a total volume of one litre at 0°C. Given R=0.082 L atm mol⁻¹ K⁻¹, T=273 K.
a.
25.18
b.
26.02
c.
2.518
d.
2.602
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