This page is designed for students who want serious question practice with answer checking and clean step-by-step explanations.
How To Use
Attempt the MCQ first, select an option, then open the Answer Tab and Solution Tab. For diagrams, the source figure is kept with Kumar Physics Classes branding below it.
The layout is full-width, red-gold Jaipur/Jodhpur inspired, lightweight, left-aligned and ready for Elementor HTML widget use.
Class 11 And 12 Physics Formula Bank
Class 11
Units, Dimensions And Errors
Dimensional formula: [MaLbTcAdKemolfcdg]
Velocity [LT-1], acceleration [LT-2], force [MLT-2]
Rectifier ripple frequency half-wave = f, full-wave = 2f
Zener diode works in reverse breakdown
Transistor current: IE = IB + IC
α = IC/IE, β = IC/IB
β = α/(1 - α), α = β/(β + 1)
AND: Y = A.B
OR: Y = A + B
NOT: Y = A
NAND: Y = A.B
NOR: Y = A + B
Class 12
Communication Systems
Speed of EM wave: c = fλ
Wavelength: λ = c/f
Antenna minimum height: h = λ/4
Modulation index: μ = Am/Ac
AM wave: s(t) = Ac(1 + μcosωmt)cosωct
Bandwidth for AM = 2fm
Bandwidth for FM = 2(Δf + fm)
Range of TV tower: d = √(2Rh)
If receiver height hr is included: d = √(2Rht) + √(2Rhr)
Number of channels = total bandwidth/bandwidth per channel
Power in AM: Pt = Pc(1 + μ2/2)
Question 1ARBTS Paper 6 Q1
A solid sphere of radius R is uniformly charged with charge density ρ in its volume. A spherical cavity of radius R/2 is made in the sphere as shown. Then electric potential at the centre of sphere will be
Kumar Physics Classes +919955846145
Answer: Option (3)
Solution: Use superposition: potential of complete uniformly charged sphere at the centre minus the potential due to the removed smaller sphere/cavity. This gives option (3).
Question 2ARBTS Paper 6 Q2
Three short dipoles each of dipole moment of magnitude p are placed on a circle of radius R as shown. The magnitude of electric field intensity at centre will be
Kumar Physics Classes +919955846145
Answer: Option (2)
Solution: Resolve the three dipole fields at the centre. The vector sum of the symmetrically placed dipoles gives E = 2kp/R3.
Question 3ARBTS Paper 6 Q3
Equipotential surfaces corresponding to electric field due to infinite charged sheet are
Answer: Option (4)
Solution: An infinite sheet produces a uniform field normal to the sheet, so equipotential surfaces are parallel planes.
Question 4ARBTS Paper 6 Q4
A galvanometer has resistance 100 Ω. It gives full scale deflection on passing 10 mA current through it. To convert it into a voltmeter of range 0 - 10 V, the resistance to be added in series is
Answer: Option (2)
Solution: Total resistance needed is V/Ig = 10/0.01 = 1000 Ω. Series resistance = 1000 - 100 = 900 Ω.
Question 5ARBTS Paper 6 Q5
A charged particle q is shot from a large distance with speed v towards a fixed charged particle Q. It approaches Q upto a closest distance r and then returns. If velocity of q is halved, then the closest distance of approach would be
Kumar Physics Classes +919955846145
Answer: Option (4)
Solution: At closest approach, initial kinetic energy converts into electrostatic potential energy, so r ∝ 1/v2. If speed becomes v/2, closest distance becomes 4r.
Question 6ARBTS Paper 6 Q6
Select the correct option.
Answer: Option (3)
Solution: A neutral system can still have non-zero interaction energy because potential energy depends on pairwise charge products and separations.
Question 7ARBTS Paper 6 Q7
A particle having charge -q and mass m is released from rest on the axis of a fixed ring of total charge Q and radius R from a distance √3 R. Its kinetic energy when it reaches the centre of ring is
Kumar Physics Classes +919955846145
Answer: Option (2)
Solution: Potential on ring axis is kQ/√(R2 + x2). At x = √3R it is kQ/(2R), at centre it is kQ/R. Gain in kinetic energy = q[kQ/R - kQ/(2R)] = Qq/(8πε0R).
Question 8ARBTS Paper 6 Q8
In the network shown points A, B and C have potential of 70 V, zero and 10 V respectively. The ratio of current in the section AD, DB and DC are
Kumar Physics Classes +919955846145
Answer: Option (1)
Solution: Apply KCL at junction D and use I = potential difference/resistance for the 10 Ω, 20 Ω and 30 Ω branches. The branch currents come in the ratio 3 : 2 : 1.
Question 9ARBTS Paper 6 Q9
Find the current through 10 Ω resistor shown in figure.
Kumar Physics Classes +919955846145
Answer: Option (1)
Solution: The branch potentials make the 10 Ω branch balanced with the 3 V source; hence no net current flows through that resistor.
Question 10ARBTS Paper 6 Q10
The reading of voltmeter in the figure shown is
Kumar Physics Classes +919955846145
Answer: Option (3)
Solution: The voltmeter is across the 300 Ω and 600 Ω parallel combination. Reduce the circuit and use voltage division; the potential difference comes out as 40 V.
Question 11ARBTS Paper 6 Q11
In a region of space, suppose there exists an electric field E = 40x3 î. The potential difference VA - VO, where VO is the potential at origin and VA is the potential at x = 2 m, is
Answer: Option (1)
Solution: VA - VO = -∫0240x3dx = -40[x4/4]02 = -160 V.
Question 12ARBTS Paper 6 Q12
In the experiment of meter bridge if length corresponding to null deflection is x, what would be its value if the radius of meter bridge wire is doubled?
Answer: Option (1)
Solution: The null point depends on resistance ratio, not on the absolute resistance per unit length of the bridge wire. Doubling radius changes all wire resistances proportionally, so x remains unchanged.
Question 13ARBTS Paper 6 Q13
In the potentiometer circuit shown, internal resistance of the 12 V battery is 1 Ω and length of wire AB is 100 cm. When CB is 60 cm, the galvanometer shows no deflection. The emf of cell E is, given resistance of wire AB is 3 Ω.
Kumar Physics Classes +919955846145
Answer: Option (2)
Solution: Current in primary circuit = 12/(1 + 2 + 3) = 2 A, so potential across 100 cm wire = 2 x 3 = 6 V. Since CB = 60 cm, AC = 40 cm in the shown connection; hence E = 6 x 40/100 = 2.4 V.
Question 14ARBTS Paper 6 Q14
A cell of emf E and internal resistance r is connected in series with an external resistance nr. The ratio of terminal potential difference to emf is
Answer: Option (3)
Solution: Terminal voltage V = IR = E[nr/(r + nr)] = E n/(n + 1).
Question 15ARBTS Paper 6 Q15
A total charge of 20 µC is divided into two parts placed at some distance apart. If the charges experience maximum coulombian repulsion, then the charges should be
Answer: Option (2)
Solution: For fixed total charge, product q(20 - q) is maximum when the two charges are equal.
Question 16ARBTS Paper 6 Q16
The unit of Poynting vector is
Answer: Option (2)
Solution: Poynting vector represents energy flow per unit area per unit time, so unit is W/m2 = J/(m2s).
Question 17ARBTS Paper 6 Q17
If i1 = 3 sin ωt and i2 = 4 cos ωt, then i3 is
Kumar Physics Classes +919955846145
Answer: Option (1)
Solution: At the junction, i3 is the phasor sum of 3 sinωt and 4 cosωt. Resultant magnitude = 5 and phase lead = tan-1(4/3) = 53°.
Question 18ARBTS Paper 6 Q18
In an LCR circuit the voltage across each of the components L, C and R is 50 V each. Find the voltage across LC combination.
Answer: Option (4)
Solution: Voltages across L and C are equal and opposite in phase, so their vector sum is zero.
Question 19ARBTS Paper 6 Q19
Two short identical magnetic dipoles of magnetic moments 1 A m2 each are placed at a separation of 2 m with their axes perpendicular to each other. The resultant magnetic field at a point midway between the dipoles is
Answer: Option (2)
Solution: At the midpoint, axial and equatorial dipole fields are perpendicular, in the ratio 2 : 1. Resultant = √(22 + 12) x 10-7 = √5 x 10-7 T.
Question 20ARBTS Paper 6 Q20
A magnet of magnetic moment 50 î A m2 is placed in a magnetic field B = (0.5 î + 3 ĵ) T. The torque acting on the magnet is
Answer: Option (2)
Solution: Torque τ = M x B = 50 î x (0.5 î + 3 ĵ) = 150 k̂ N m.
Question 21ARBTS Paper 6 Q21
The inductance of a closely packed coil of 400 turns is 8 mH. When a current of 5 mA is passed through it, the magnetic flux per turn is (μ0 = permeability of free space)
Answer: Option (1)
Solution: Using L = NΦ/I, flux per turn Φ = LI/N = (8 x 10-3 x 5 x 10-3)/400 = 10-7 Wb = μ0/(4π) Wb.
Question 22ARBTS Paper 6 Q22
Find the time constant for the given RL circuit.
Kumar Physics Classes +919955846145
Answer: Option (2)
Solution: For the inductor, equivalent resistance is 6 Ω + (12 Ω || 6 Ω) = 10 Ω. Time constant τ = L/R = 2 mH/10 Ω = 0.2 ms.
Question 23ARBTS Paper 6 Q23
A coil having n turns and resistance R is connected with a galvanometer of resistance 4R. The combination is moved in time t such that magnetic flux φ1 per turn changes to flux φ2 per turn. The average induced current in the circuit is
Answer: Option (2)
Solution: Average emf = -n(φ2 - φ1)/t. Total resistance = R + 4R = 5R, so average current is option (2).
Question 24ARBTS Paper 6 Q24
At a place on earth, horizontal component of earth's magnetic field is B1 and vertical component is B2. If a magnetic needle is kept in a vertical plane making angle α with magnetic meridian, then square of time period of vibration needle in this plane is proportional to
Answer: Option (3)
Solution: For a magnetic needle, T2 ∝ 1/Beffective. In the given vertical plane the effective field is √[(B1cosα)2 + B22].
Question 25ARBTS Paper 6 Q25
A long wire carries a steady current. It is bent into a circle of one turn and magnetic field at the centre of coil is B. It is then bent into a circular loop of n-turns. The magnetic field at the centre of coil for same current is
Answer: Option (2)
Solution: For the same wire length, n turns reduce radius by factor n, and field at centre is proportional to n/R. Hence B' = n2B.
Question 26ARBTS Paper 6 Q26
A long straight wire along z-axis carries current I in the negative z direction. The magnetic field vector B at point (x, y) in z = 0 plane is
Answer: Option (1)
Solution: Using right-hand rule for current along negative z-axis and B = μ0I/(2πr), the vector form at (x, y) is option (1).
Question 27ARBTS Paper 6 Q27
The magnetic field due to current carrying circular loop of radius 3 cm at a point on the axis at a distance of 4 cm from centre is 54 µT. The value of magnetic field at centre of loop is
Answer: Option (1)
Solution: For axial field, Bx = B0R3/(R2 + x2)3/2. With R = 3 cm, x = 4 cm, factor = 27/125, so B0 = 54 x 125/27 = 250 µT.
Question 28ARBTS Paper 6 Q28
The length of the optical path of two media in contact of length d1 and d2, of refractive index μ1 and μ2 respectively, is
Answer: Option (1)
Solution: Optical path length is sum of μ times geometrical path in each medium.
Question 29ARBTS Paper 6 Q29
A ray of light falls on a transparent sphere with centre at C as shown. The ray emerges from the sphere parallel to line AB. The refractive index of the material of sphere is
Kumar Physics Classes +919955846145
Answer: Option (2)
Solution: Apply Snell's law at entry and use symmetry of refraction through the sphere. The given 60° geometry gives μ = √3.
Question 30ARBTS Paper 6 Q30
Light is incident normally on face AB of a prism as shown. A liquid of refractive index μ is placed on face AC of the prism. The prism is made of glass of refractive index 3/2. The limit of μ for which total internal reflection takes place on face AC is
Kumar Physics Classes +919955846145
Answer: Option (2)
Solution: At face AC, use critical condition sin C = μliquid/μglass. The incidence angle from geometry gives the limiting value μ < 3√3/4.
Question 31ARBTS Paper 6 Q31
If the ratio of amplitudes of two waves of same frequency is 4 : 3, then the ratio of maximum and minimum intensities is
A thin mica sheet of thickness 2 x 10-6 m and refractive index 1.5 is introduced in the path of waves from upper slit in YDSE. The wavelength of the wave used is 5000 Å. The central bright maximum will shift
Answer: Option (1)
Solution: Number of fringe shifts = (μ - 1)t/λ = 0.5 x 2 x 10-6/(5 x 10-7) = 2. The shift is towards the slit covered by the sheet, i.e. upwards.
Question 33ARBTS Paper 6 Q33
A light has amplitude A after passing through the polariser. Angle between analyser and polariser is 60°. Light transmitted by analyser has amplitude
Answer: Option (4)
Solution: For the electric field amplitude after analyser, A' = A cos 60° = A/2.
Question 34ARBTS Paper 6 Q34
If the de-Broglie wavelengths for a proton and alpha-particle are equal, then the ratio of their velocities will be
Answer: Option (1)
Solution: For equal de Broglie wavelength, momenta are equal. Since p = mv and mα = 4mp, vp:vα = 4:1.
Question 35ARBTS Paper 6 Q35
The energy of a photon is E = hν and momentum of photon is P = h/λ, then the velocity of photon will be
Answer: Option (1)
Solution: E = hν and P = h/λ; hence E/P = νλ = c, the photon speed.
Question 36ARBTS Paper 6 Q36
Which of the following graphs is/are correct for photoelectric effect?
Kumar Physics Classes +919955846145
Answer: Option (4)
Solution: Photo current is proportional to intensity, saturation current varies with accelerating potential, and stopping potential varies linearly with frequency. Hence all shown graphs are correct.
Question 37ARBTS Paper 6 Q37
The first line in the Lyman series has wavelength λ. The wavelength of the first line in Balmer series is
Answer: Option (4)
Solution: Using Rydberg formula, Lyman first line is 2 to 1 and Balmer first line is 3 to 2. Ratio gives λBalmer = 27λ/5.
Question 38ARBTS Paper 6 Q38
The activity of a sample of radioactive material is A1 at time t1 and A2 at time t2 (t2 > t1). If its mean life is T, then
In the following circuit, if D1 and D2 are ideal diodes, then the value of I1 and I2 are respectively
Kumar Physics Classes +919955846145
Answer: Option (4)
Solution: With ideal-diode polarity, D1 is off and D2 conducts. Hence I1 = 0 and I2 = 5 mA.
Question 43ARBTS Paper 6 Q43
Name the gate represented by the following circuit.
Kumar Physics Classes +919955846145
Answer: Option (2)
Solution: The circuit combines OR and NAND paths to generate ĀB + AB̄, which is XOR.
Question 44ARBTS Paper 6 Q44
A radioactive element 23890X decays into 22283Y by α and β- emissions. The number of β- particles emitted is
Answer: Option (4)
Solution: Mass number decreases by 16, so 4 alpha particles are emitted. Atomic number after 4 alpha emissions is 90 - 8 = 82; to become 83, one beta-minus emission is needed.
Question 45ARBTS Paper 6 Q45
The ratio of the wavelength for 2 → 1 transition in Li++, He+ and H is
Answer: Option (3)
Solution: For hydrogen-like species, transition energy ∝ Z2, so wavelength ∝ 1/Z2. For Li++, He+, H, ratio is 1/9 : 1/4 : 1 = 4 : 9 : 36.
Question 46ARBTS Paper 7 Q1
In an experiment four quantities a, b, c and d are measured with percentage error 2%, 3%, 1% and 0.5% respectively. A quantity Q is defined as Q = a√b/(c3/2d4). Maximum percentage error in calculation of Q will be
Solution: The acceleration is initially positive, then negative, then zero, so slope of x-t first increases, then decreases smoothly and finally becomes nearly constant. This matches graph (2).
Question 49ARBTS Paper 7 Q4
A particle has initial velocity u = (4î - 5ĵ) m/s and acceleration a = (1/4 î + 1/5 ĵ) m/s2. Velocity of the particle at t = 2 second is
Answer: Option (2)
Solution: v = u + at = (4, -5) + 2(1/4, 1/5) = (4.5, -4.6) m/s.
Question 50ARBTS Paper 7 Q5
Three blocks A, B and C are placed on a rough horizontal surface. Friction coefficient between blocks and surface is 0.6. Acceleration of block C in given situation is (g = 10 m/s2)
Kumar Physics Classes +919955846145
Answer: Option (4)
Solution: Maximum friction on A and B together is sufficient to balance the pull due to 5.5 kg hanging block, so the system remains at rest.
Question 51ARBTS Paper 7 Q6
Three blocks each of mass m are hanged vertically with inextensible strings and an ideal spring. Initially the system was in equilibrium. At any instant lower most string is cut. Acceleration of block B just after cutting the string is
Kumar Physics Classes +919955846145
Answer: Option (1)
Solution: Spring force cannot change instantaneously. Just after cutting, forces on B give net downward force mg, so acceleration is g.
Question 52ARBTS Paper 7 Q7
A block P of mass m is moving with velocity v0 and collides elastically with identical block Q as shown. If spring constant is K, maximum compression in subsequent motion is
Kumar Physics Classes +919955846145
Answer: Option (3)
Solution: After elastic collision, Q moves with v0. During compression, reduce Q and 2m to a two-body spring system with reduced mass 2m/3. Energy conservation gives x = v0√(2m/3K).
Question 53ARBTS Paper 7 Q8
Given diagram represents potential energy curve of particle in a field. Particle will be in equilibrium at position
Kumar Physics Classes +919955846145
Answer: Option (2)
Solution: Equilibrium occurs where dU/dx = 0. That happens at the extrema A and C.
Question 54ARBTS Paper 7 Q9
Three thin rods each of mass m and length L are joined to form the shown shape. Moment of inertia about axis xx' passing through rod PQ is
Kumar Physics Classes +919955846145
Answer: Option (3)
Solution: Rod on the axis contributes negligible moment; two perpendicular rods contribute by rod moment and parallel-axis terms. Total = 4mL2/3.
Question 55ARBTS Paper 7 Q10
A cylinder of mass m and radius R rolls purely over an inclined surface while moving up on it. Correct statement regarding friction acting on it is
Answer: Option (3)
Solution: As the rolling cylinder moves up, static friction acts upward to reduce angular speed consistently with pure rolling.
Question 56ARBTS Paper 7 Q11
Two satellites are in parking orbits around the earth. Mass of one is 10 times that of the other. The ratio of their periods of revolution is
Answer: Option (1)
Solution: Orbital period depends on orbital radius and central mass, not on satellite mass.
Question 57ARBTS Paper 7 Q12
Four particles each of mass m move along a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is
Answer: Option (4)
Solution: Resolve gravitational attraction of the other three masses towards the centre and equate net inward force to mv2/R.
Question 58ARBTS Paper 7 Q13
When an elastic material with Young's modulus Y is subjected to stretching stress S, elastic energy stored per unit volume is
Answer: Option (2)
Solution: Energy density = 1/2 x stress x strain = 1/2 x S x (S/Y) = S2/(2Y).
Question 59ARBTS Paper 7 Q14
The normal density of gold is ρ0 and its bulk modulus is B. The increase in density of sphere of gold when pressure P is applied uniformly is (P << B)
Answer: Option (3)
Solution: Bulk modulus B = -P/(ΔV/V). Since density increases by approximately Δρ/ρ0 = P/B, Δρ = ρ0P/B.
Question 60ARBTS Paper 7 Q15
A large tank filled with water to height h is emptied through a small hole at the bottom. Ratio of time for level to fall from h to h/2 and from h/2 to zero is
Answer: Option (1)
Solution: Emptying time from height H to h is proportional to √H - √h. Ratio = (√h - √(h/2))/√(h/2) = √2 - 1.
Question 61ARBTS Paper 7 Q16
Specific heat S of a 1 kg container varies with temperature T as S = A + BT, where A = 100 cal kg-1K-1 and B = 2 x 10-2 cal kg-1K-2. Heat required from 27°C to 227°C is
Answer: Option (2)
Solution: Q = ∫m(A + BT)dT from 300 K to 500 K = 21600 cal.
Question 62ARBTS Paper 7 Q17
A spherical body of emissivity e = 0.7 and surface area A is placed inside a perfect black body maintained at temperature T. Energy radiated per second by black body will be
Answer: Option (4)
Solution: A perfect black body has emissivity 1, so emitted power is σAT4.
Question 63ARBTS Paper 7 Q18
An ideal gas expands from volume V to 2V according to law VP2 = constant. If initial temperature is T, final temperature will be
Answer: Option (2)
Solution: From VP2 = constant and PV = nRT, P ∝ V-1/2, so T ∝ PV ∝ V1/2. Doubling V gives T' = T√2.
Question 64ARBTS Paper 7 Q19
In a cyclic process ABCA, V-T graph is shown. P-V graph corresponding to the given process can be best depicted by
Kumar Physics Classes +919955846145
Answer: Option (3)
Solution: On V-T graph, AB has V ∝ T so pressure is constant; BC is isochoric; CA is isothermal. The matching P-V cycle is graph (3).
Question 65ARBTS Paper 7 Q20
A refrigerator with coefficient of performance 7 extracts heat from low temperature compartment at 250 J/cycle. Work done per cycle required is nearly
Answer: Option (2)
Solution: COP = QL/W = 7, so W = 250/7 = 35.7 J ≈ 36 J.
Question 66ARBTS Paper 7 Q21
The ratio of velocity of sound in oxygen to that in argon at same temperature is
Answer: Option (1)
Solution: v = √(γRT/M). Use γO2 = 7/5, M = 32 and γAr = 5/3, M = 40. Ratio = √(21/20).
Question 67ARBTS Paper 7 Q22
The frequency of second overtone of open pipe is equal to first overtone of closed pipe. The ratio of lengths of closed pipe to open pipe is
Answer: Option (3)
Solution: Second overtone open pipe frequency = 3v/(2Lo); first overtone closed pipe = 3v/(4Lc). Equating gives Lc/Lo = 1/2.
Question 68ARBTS Paper 7 Q23
A simple pendulum performs SHM about x = 0 with amplitude A and period T. Speed at x = 3A/4 will be
A beam of light converges towards point O, 10 cm behind a concave mirror of focal length 20 cm. Magnification produced by the mirror is
Answer: Option (2)
Solution: Treat the converging point as a virtual object for mirror and use mirror formula. Magnification comes out 2/3.
Question 70ARBTS Paper 7 Q25
A circular beam of light having diameter 4 cm falls on a plane glass slab at angle of incidence 60°. If refractive index of slab is μ = 3/2, diameter of refracted beam is
Answer: Option (3)
Solution: Use Snell's law to find refraction angle, then projected width of the oblique beam inside slab. The refracted beam diameter is 8√(2/3) cm.
Question 71ARBTS Paper 7 Q26
A compound microscope has magnifying power 30. Focal length of eyepiece is 5 cm. If final image is at least distance of distinct vision, magnification produced by objective is
Answer: Option (3)
Solution: Eyepiece magnification for final image at D is 1 + D/fe = 1 + 25/5 = 6. Objective magnification = 30/6 = 5.
Question 72ARBTS Paper 7 Q27
In YDSE, if slit separation is halved and distance between slits and screen is doubled, fringe width becomes
Answer: Option (2)
Solution: Fringe width β = λD/d. If D doubles and d halves, β becomes 4 times.
Question 73ARBTS Paper 7 Q28
A beam of light is incident on a glass plate at angle 60°. The reflected ray is completely polarized. Refractive index of glass plate is
Answer: Option (3)
Solution: At Brewster angle, μ = tan iB = tan 60° = √3 = 1.732.
Question 74ARBTS Paper 7 Q29
Two charges each +q are placed at vertices A and B of a right angled isosceles triangle. Charge to be placed at vertex C so that net electrostatic energy is zero is
Kumar Physics Classes +919955846145
Answer: Option (4)
Solution: Write total potential energy of three-charge system and set it zero. Since AB = r√2 and AC = BC = r, solving gives QC = -√2q/(√2 + 1).
Question 75ARBTS Paper 7 Q30
Two conducting charged spheres having different radii are connected by a conducting wire. Which statement is true?
Answer: Option (2)
Solution: Connected conductors come to equal potential. For spheres V = kQ/R, so larger radius sphere gets larger charge.
Question 76ARBTS Paper 7 Q31
Plate A of a parallel plate air-filled capacitor is connected to a spring of force constant K and plate B is fixed. If +q is given to A and -q to B, extension of spring in equilibrium is
Kumar Physics Classes +919955846145
Answer: Option (2)
Solution: Attractive pressure between plates is σ2/(2ε0), so force = q2/(2ε0A). Equate with Kx.
Question 77ARBTS Paper 7 Q32
When the key is pressed at t = 0, charge on the capacitor after a very long time in given circuit will be
Kumar Physics Classes +919955846145
Answer: Option (2)
Solution: At steady state the capacitor branch is open. The voltage across it is the voltage across the parallel 2 Ω resistor, equal to 4 V. Q = CV = 4µF x 4V = 16µC.
Question 78ARBTS Paper 7 Q33
A thin rectangular magnet suspended freely has period 4 s. If the magnet is broken into two halves, and one piece oscillates in same field, its time period becomes
Answer: Option (1)
Solution: For half-length magnet, magnetic moment and moment of inertia change so that T becomes half. Hence 2 s.
Question 79ARBTS Paper 7 Q34
Soft iron is used in many parts of electrical machines because it exhibits
Answer: Option (4)
Solution: Soft iron is preferred because it magnetizes easily and has low energy loss per cycle.
Question 80ARBTS Paper 7 Q35
The wire loop carries current I as shown. Magnetic field at centre O will be
Kumar Physics Classes +919955846145
Answer: Option (1)
Solution: Straight radial parts give zero field at O. The two circular arcs contribute in the same direction, giving option (1).
Question 81ARBTS Paper 7 Q36
A charged particle is projected into a region where there may be electric field E and/or magnetic field B. If it goes unaccelerated, then it is not possible that
Answer: Option (2)
Solution: With only non-zero electric field, electric force qE cannot be balanced by magnetic force, so unaccelerated motion is impossible.
Question 82ARBTS Paper 7 Q37
The working of a dynamo is based on the principle of
Answer: Option (1)
Solution: A dynamo converts mechanical energy to electrical energy by electromagnetic induction.
Question 83ARBTS Paper 7 Q38
Power factor of the AC circuit shown is
Kumar Physics Classes +919955846145
Answer: Option (4)
Solution: For the shown AC circuit, calculate impedance Z and use cosφ = R/Z. The value is 0.8.
Question 84ARBTS Paper 7 Q39
The Maxwell equation ∮B.dl = μ0(I + ε0dφE/dt) is a statement of
Answer: Option (2)
Solution: The displacement-current term added to Ampere's law gives Maxwell-Ampere, or modified Ampere's law.
Question 85ARBTS Paper 7 Q40
The ratio of de-Broglie wavelength of an alpha-particle and a proton of same kinetic energy is
Answer: Option (1)
Solution: For same kinetic energy, λ ∝ 1/√m. Since mα = 4mp, λα:λp = 1:2.
Question 86ARBTS Paper 7 Q41
If 10% of a material decays in 5 days, then original material left after 20 days is approximately
Answer: Option (2)
Solution: After every 5 days, 90% remains. After 20 days: (0.9)4 = 0.6561, about 65%.
Question 87ARBTS Paper 7 Q42
For stability of any nucleus
Answer: Option (1)
Solution: Greater binding energy per nucleon means a more stable nucleus.
Question 88ARBTS Paper 7 Q43
When a P-N junction is forward biased, current across the junction is mainly due to
Answer: Option (2)
Solution: Forward bias lowers the barrier and majority carriers diffuse across the junction.
Question 89ARBTS Paper 7 Q44
The Boolean equation for the given circuit is
Kumar Physics Classes +919955846145
Answer: Option (4)
Solution: A passes through NOT gate and B, C pass through OR gate; final AND gate gives y = Abar.(B + C), so option (4) is correct.
Question 90ARBTS Paper 7 Q45
The peak voltage in the output of a half wave diode rectifier fed with a sinusoidal signal without filter is 10 V. The DC component of output voltage is
Answer: Option (2)
Solution: For half-wave rectifier, Vdc = Vm/π = 10/π V.
Question 91Original NEET Practice Q1
A 2 kg block moves on a smooth horizontal surface with speed 6 m/s. A constant opposing force of 12 N acts on it. Distance travelled before stopping is
Answer: Option (2)
Solution: Initial kinetic energy = 1/2 x 2 x 62 = 36 J. Work done by retarding force = Fs, so s = 36/12 = 3 m.
Question 92Original NEET Practice Q2
A capacitor of 6 µF is charged to 10 V and then connected in parallel with an uncharged 3 µF capacitor. Final common potential is
Answer: Option (2)
Solution: Total charge remains 6µF x 10V = 60µC. Total capacitance = 9µF, so V = 60/9 = 6.67 V.
Question 93Original NEET Practice Q3
A wire of resistance R is stretched to double its length without change in volume. New resistance is
Answer: Option (4)
Solution: When length doubles, area becomes half. Since R = ρL/A, new resistance = 4R.
Question 94Original NEET Practice Q4
A photon of wavelength 400 nm has energy approximately (take hc = 1240 eV nm)
Answer: Option (3)
Solution: E = hc/λ = 1240/400 = 3.10 eV.
Question 95Original NEET Practice Q5
In a uniform magnetic field, a charged particle moves in a circle of radius r. If its speed is doubled, radius becomes
Answer: Option (3)
Solution: r = mv/(qB), so radius is directly proportional to speed.
Question 96Original NEET Practice Q6
For a convex lens, object is placed at 2f. The image is formed at
Answer: Option (2)
Solution: For a convex lens, an object at 2f forms a real inverted image at 2f on the other side.
Question 97Original NEET Practice Q7
If rms value of AC current is 5 A, its peak value is
Answer: Option (3)
Solution: I0 = √2 Irms = 5√2 A.
Question 98Original NEET Practice Q8
A radioactive sample has half-life 10 days. Fraction remaining after 30 days is
Answer: Option (3)
Solution: 30 days equals 3 half-lives, so remaining fraction = (1/2)3 = 1/8.
Question 99Original NEET Practice Q9
The dimensional formula of Planck's constant is
Answer: Option (2)
Solution: h has dimension of energy x time = (ML2T-2)T = ML2T-1.
Question 100Original NEET Practice Q10
A transformer has primary turns 500 and secondary turns 2500. If primary voltage is 220 V, secondary voltage for an ideal transformer is
Answer: Option (4)
Solution: Vs/Vp = Ns/Np = 2500/500 = 5, so Vs = 1100 V.