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Question 1Physics (syllabus up to Class 12 CBSE India)

A body moves along a straight line with uniform acceleration. Which quantity is represented by the area under its velocity–time graph?

  1. Jerk of the motion
  2. Rate of change of velocity
  3. Displacement covered in that interval Correct answer.
  4. Acceleration of the body

Explanation

On a velocity–time graph the velocity is the height and time is the horizontal spread, so the product v·dt summed over the interval gives displacement. The slope of the same graph, not its area, gives acceleration. Area under an acceleration–time graph would give change in velocity.

Question 2Physics (syllabus up to Class 12 CBSE India)

A projectile is launched at speed u at angle θ to the horizontal on level ground, with air resistance neglected. At the highest point of its path, what is the magnitude of its acceleration?

  1. u² sin²θ / (2g)
  2. Zero, since vertical velocity vanishes
  3. g cos θ directed along the path
  4. g directed vertically downward Correct answer.

Explanation

Only gravity acts on the projectile throughout the flight, so the acceleration is g downward at every instant, including the top where the vertical velocity is momentarily zero. The horizontal velocity is constant, so the total acceleration equals g. A zero acceleration at the top would mean the velocity stops changing, which is false.

Question 3Physics (syllabus up to Class 12 CBSE India)

A block of mass m rests on a rough horizontal floor with coefficient of static friction μs. A horizontal force F is applied and the block remains at rest. What is the magnitude of the friction force acting on the block?

  1. Equal to μk m g
  2. Zero, because the block is not moving
  3. Equal to F Correct answer.
  4. Equal to μs m g

Explanation

Static friction is self-adjusting: it takes exactly the value needed to prevent relative sliding, up to a maximum of μs·N. While the block stays at rest, the horizontal force balance requires friction to equal the applied force F. Only at the verge of slipping does friction reach μs m g.

Question 4Physics (syllabus up to Class 12 CBSE India)

A satellite of mass m orbits Earth of mass M in a circular orbit of radius r. Which expression gives the total mechanical energy of the satellite?

  1. −GMm/r²
  2. −GMm/r
  3. GMm/(2r)
  4. −GMm/(2r) Correct answer.

Explanation

For a circular orbit the kinetic energy is GMm/(2r) and the potential energy is −GMm/r, so the total is their sum, −GMm/(2r). The negative sign reflects a bound orbit. The total energy is half the potential energy in magnitude, a standard circular-orbit result.

Question 5Physics (syllabus up to Class 12 CBSE India)

A solid sphere, a hollow sphere and a solid cylinder of the same mass and radius are released from rest and roll without slipping down the same incline. Which reaches the bottom first?

  1. The solid cylinder
  2. The hollow sphere
  3. The solid sphere Correct answer.
  4. All three arrive together

Explanation

Rolling acceleration down an incline is a = g sinθ/(1 + I/mR²), so the body with the smallest moment-of-inertia factor I/mR² wins. For a solid sphere I/mR² = 2/5, for a solid cylinder it is 1/2, and for a hollow sphere it is 2/3. The solid sphere therefore has the greatest acceleration and arrives first.

Question 6Physics (syllabus up to Class 12 CBSE India)

A particle executes simple harmonic motion with amplitude A. At what displacement from the mean position is its kinetic energy equal to its potential energy?

  1. A/√2 Correct answer.
  2. A√3/2
  3. A/4
  4. A/2

Explanation

In SHM, potential energy is (1/2)mω²x² and total energy is (1/2)mω²A². Setting kinetic equal to potential means potential is half the total, so x² = A²/2, giving x = A/√2. At this point each form of energy is half the total.

Question 7Physics (syllabus up to Class 12 CBSE India)

A uniform rod of mass M and length L is rotated about an axis through one end and perpendicular to its length. What is its moment of inertia about this axis?

  1. ML²
  2. ML²/12
  3. ML²/3 Correct answer.
  4. ML²/2

Explanation

The moment of inertia about the centre is ML²/12, and the parallel-axis theorem adds M(L/2)² = ML²/4. Adding these gives ML²/12 + 3ML²/12 = ML²/3. This is the standard result for a rod about an end.

Question 8Physics (syllabus up to Class 12 CBSE India)

Water flows steadily through a horizontal pipe that narrows from a wider section to a narrower section. Which statement is correct for the narrower section?

  1. The speed is lower and the pressure is lower
  2. The speed and pressure both stay unchanged
  3. The speed is higher and the pressure is lower Correct answer.
  4. The speed is higher and the pressure is higher

Explanation

By the equation of continuity Av = constant, a smaller cross-sectional area forces a larger speed. Bernoulli's principle for a horizontal pipe then requires the pressure to fall where the speed rises. Hence the narrow section has higher speed and lower pressure.

Question 9Physics (syllabus up to Class 12 CBSE India)

The absolute temperature of an ideal gas is doubled at constant volume. What happens to the root-mean-square speed of its molecules?

  1. It quadruples
  2. It increases by a factor of √2 Correct answer.
  3. It doubles
  4. It remains unchanged

Explanation

The rms speed is given by v_rms = √(3RT/M), so it varies as the square root of absolute temperature. Doubling T multiplies the speed by √2. The pressure, by contrast, would double at constant volume.

Question 10Physics (syllabus up to Class 12 CBSE India)

In an isothermal reversible expansion of an ideal gas, which quantity is zero for the gas?

  1. Change in internal energy Correct answer.
  2. Change in entropy
  3. Heat absorbed by the gas
  4. Work done by the gas

Explanation

The internal energy of an ideal gas depends only on temperature, so at constant temperature ΔU = 0. The first law then gives Q = W, meaning the heat absorbed equals the work done. Entropy, however, increases in a reversible isothermal expansion.

Question 11Physics (syllabus up to Class 12 CBSE India)

Two point charges +q and −q are separated by a distance 2a. What is the electric potential at the midpoint of the line joining them?

  1. kq/a²
  2. 2kq/a
  3. Zero Correct answer.
  4. kq/a

Explanation

Potential is a scalar, so the total is the algebraic sum of the potentials from each charge. At the midpoint each charge is at distance a, giving +kq/a and −kq/a, which cancel to zero. The electric field there, however, is not zero and points from +q toward −q.

Question 12Physics (syllabus up to Class 12 CBSE India)

A wire of resistance R is stretched uniformly until its length is doubled, its volume remaining constant. What is its new resistance?

  1. 4R Correct answer.
  2. R/2
  3. R
  4. 2R

Explanation

Resistance is R = ρL/A. Doubling the length at constant volume halves the cross-sectional area, so the new resistance is ρ(2L)/(A/2) = 4ρL/A = 4R. Both the length increase and the area decrease act to raise the resistance.

Question 13Physics (syllabus up to Class 12 CBSE India)

Three identical resistors, each of resistance R, are connected in parallel. What is the equivalent resistance of the combination?

  1. 3R
  2. 2R/3
  3. R/3 Correct answer.
  4. R

Explanation

For resistors in parallel, 1/R_eq = 1/R₁ + 1/R₂ + 1/R₃. With three equal resistors this gives 1/R_eq = 3/R, so R_eq = R/3. The equivalent resistance of a parallel network is always smaller than the smallest individual resistance.

Question 14Physics (syllabus up to Class 12 CBSE India)

A charged particle enters a uniform magnetic field with its velocity perpendicular to the field. Which quantity remains constant during its subsequent motion?

  1. Speed Correct answer.
  2. Kinetic energy and speed both change
  3. Velocity vector
  4. Magnetic force direction relative to the field

Explanation

The magnetic force qv×B is always perpendicular to the velocity, so it does no work on the particle. Hence the kinetic energy and therefore the speed stay constant while only the direction of velocity changes, producing circular motion. The velocity vector itself changes continuously.

Question 15Physics (syllabus up to Class 12 CBSE India)

In a step-up transformer that is ideal, the secondary voltage is greater than the primary voltage. Which statement about the currents is correct?

  1. The secondary current is zero because no load is implied
  2. The secondary current is smaller than the primary current Correct answer.
  3. The secondary current equals the primary current
  4. The secondary current is larger than the primary current

Explanation

An ideal transformer conserves power, so V_p I_p = V_s I_s. If the secondary voltage is stepped up, the secondary current must be correspondingly smaller. The turns ratio sets both relations: N_s/N_p = V_s/V_p = I_p/I_s.

Question 16Physics (syllabus up to Class 12 CBSE India)

In a Young's double-slit experiment, the fringe width is β. If the separation between the slits is doubled while everything else is unchanged, what is the new fringe width?

  1. 2β
  2. β/2 Correct answer.
  3. 4β
  4. β

Explanation

Fringe width is β = λD/d, where d is the slit separation. Fringe width is inversely proportional to d, so doubling d halves the fringe width. Changing the screen distance or wavelength would scale it differently.

Question 17Physics (syllabus up to Class 12 CBSE India)

A convex lens of focal length f forms a real image of an object placed at distance 2f from the lens. Where is the image formed and what is its nature?

  1. At 2f on the other side, real and inverted, same size Correct answer.
  2. At 2f on the same side, virtual and erect
  3. At infinity, real and highly magnified
  4. At f on the other side, real and diminished

Explanation

Using 1/v − 1/u = 1/f with u = −2f gives 1/v = 1/f − 1/(2f) = 1/(2f), so v = 2f. The magnification m = v/u = −1, meaning the image is real, inverted and the same size as the object. This is the symmetric two-f configuration of a converging lens.

Question 18Physics (syllabus up to Class 12 CBSE India)

The work function of a metal is φ. Light of frequency ν, where hν > φ, falls on the metal. What is the maximum kinetic energy of the emitted photoelectrons?

  1. hν
  2. φ − hν
  3. hν − φ Correct answer.
  4. hν + φ

Explanation

Einstein's photoelectric equation states that the photon energy hν is used partly to overcome the work function and the remainder appears as maximum kinetic energy. Hence K_max = hν − φ. The threshold frequency corresponds to hν₀ = φ.

Question 19Physics (syllabus up to Class 12 CBSE India)

In the Bohr model of the hydrogen atom, how does the radius of the nth orbit depend on the principal quantum number n?

  1. Inversely proportional to n
  2. Proportional to n
  3. Proportional to 1/n²
  4. Proportional to n² Correct answer.

Explanation

The Bohr radius for the nth level is r_n = n²a₀, where a₀ is the Bohr radius of the ground state. The orbital radius therefore grows as the square of the quantum number. Energy, in contrast, varies as −13.6/n² eV.

Question 20Physics (syllabus up to Class 12 CBSE India)

A radioactive sample has a half-life of 20 minutes. What fraction of the original nuclei remains undecayed after 60 minutes?

  1. 1/6
  2. 1/3
  3. 1/8 Correct answer.
  4. 1/2

Explanation

Sixty minutes is three half-lives. After each half-life the remaining fraction is halved, so after three half-lives it is (1/2)³ = 1/8. The decay constant is λ = ln2/t½, and N = N₀e^(−λt) gives the same result.

Question 21Physics (syllabus up to Class 12 CBSE India)

A simple pendulum of length L is taken to the Moon, where the acceleration due to gravity is about one sixth of that on Earth. How does its period of small oscillations change?

  1. It increases by a factor of √6 Correct answer.
  2. It decreases by a factor of 6
  3. It stays the same
  4. It increases by a factor of 6

Explanation

The period of a simple pendulum is T = 2π√(L/g), so it is inversely proportional to the square root of g. Reducing g to one sixth multiplies the period by √6. The period does not depend on the mass of the bob.

Question 22Physics (syllabus up to Class 12 CBSE India)

A parallel-plate capacitor with air between its plates is charged by a battery and then the battery is disconnected. A dielectric of constant K is now inserted to fill the gap. What happens to the stored energy?

  1. It remains unchanged
  2. It increases by a factor of K²
  3. It increases by a factor of K
  4. It decreases by a factor of K Correct answer.

Explanation

With the battery disconnected the charge Q is fixed, and energy is U = Q²/(2C). Inserting a dielectric raises the capacitance to KC, so the energy becomes Q²/(2KC), which is the original energy divided by K. The energy is reduced because the dielectric is drawn into the field.

Question 23Physics (syllabus up to Class 12 CBSE India)

A charged particle of charge q and mass m moves in a circle of radius r in a uniform magnetic field B, with v perpendicular to B. Which expression gives the time period of its circular motion?

  1. 2πr/(qB)
  2. qB/(2πm)
  3. 2πm/(qB) Correct answer.
  4. 2πqB/m

Explanation

The magnetic force provides the centripetal force, qvB = mv²/r, giving r = mv/(qB). The period is T = 2πr/v, and substituting r gives T = 2πm/(qB), independent of speed. This cyclotron result is why the frequency depends only on q, m and B.

Question 24Physics (syllabus up to Class 12 CBSE India)

An AC voltage source of angular frequency ω drives a series circuit containing only an inductor of inductance L. How does the current behave relative to the voltage?

  1. It leads the voltage by π/2
  2. It lags the voltage by π/2 Correct answer.
  3. It lags the voltage by π/4
  4. It is in phase with the voltage

Explanation

For a pure inductor the voltage is v = L di/dt, so the current is the integral of a sinusoid, which is a cosine shifted back by a quarter cycle. Hence the current lags the applied voltage by π/2. The inductive reactance is ωL and rises with frequency.

Question 25Physics (syllabus up to Class 12 CBSE India)

A block of mass m is pushed up a rough incline of angle θ with initial speed v and slides back down to its starting point. Compared with the upward journey, what is true of the downward journey?

  1. Friction acts up the incline, so the downward acceleration is smaller in magnitude than the upward one Correct answer.
  2. The downward acceleration equals g sinθ exactly, as friction vanishes
  3. Friction acts down the incline, making the downward acceleration larger
  4. Friction is absent on the way down because motion reverses

Explanation

Friction always opposes relative sliding, so it points down the incline during the ascent and up the incline during the descent. Going up, the deceleration is g sinθ + μg cosθ; coming down, the acceleration is g sinθ − μg cosθ. The downward acceleration is therefore smaller in magnitude than the upward deceleration.

Question 26Physics (syllabus up to Class 12 CBSE India)

A projectile is launched at speed u at an angle θ to the horizontal on level ground, with air resistance neglected. Which expression gives the maximum height reached above the launch point?

  1. u² cos²θ / (2g)
  2. u² sin²θ / (2g) Correct answer.
  3. u² sin²θ / g
  4. u² sin 2θ / g

Explanation

The vertical component of velocity is u sinθ, and the vertical motion is uniformly decelerated by g. Using v² = u² − 2gh with v = 0 at the top gives h = u² sin²θ/(2g).

Question 27Physics (syllabus up to Class 12 CBSE India)

A body starts from rest and moves with uniform acceleration along a straight line. The ratio of the distance covered in the first 2 s to that covered in the first 4 s is

  1. 1 : 2
  2. 1 : 8
  3. 1 : 16
  4. 1 : 4 Correct answer.

Explanation

For motion from rest with uniform acceleration, distance is proportional to the square of time: s = ½at². Hence the ratio is (2²) : (4²) = 4 : 16 = 1 : 4.

Question 28Physics (syllabus up to Class 12 CBSE India)

Two bodies of masses m and 3m moving towards each other with equal speeds v collide head-on and stick together. The speed of the combined body after the collision is

  1. zero
  2. v towards the lighter body's original direction
  3. v/4 towards the heavier body's original direction
  4. v/2 towards the heavier body's original direction Correct answer.

Explanation

Taking the heavier body's direction as positive, total momentum is 3mv − mv = 2mv. The combined mass is 4m, so the final speed is 2mv/4m = v/2 in the heavier body's direction.

Question 29Physics (syllabus up to Class 12 CBSE India)

A solid sphere, a solid cylinder and a thin ring, all of the same mass and radius, roll without slipping down the same incline starting together. Which reaches the bottom last?

  1. All three arrive simultaneously
  2. The solid cylinder
  3. The solid sphere
  4. The thin ring Correct answer.

Explanation

Rolling acceleration is a = g sinθ/(1 + I/MR²), so the body with the largest moment of inertia factor is slowest. The ring has I = MR², giving the largest factor 1, so it accelerates least and arrives last.

Question 30Physics (syllabus up to Class 12 CBSE India)

A satellite is in a circular orbit of radius r around the Earth. If it is moved to a circular orbit of radius 4r, the orbital speed changes by a factor of

  1. 2
  2. 1/4
  3. 4
  4. 1/2 Correct answer.

Explanation

For a circular orbit, gravitational force supplies the centripetal force: GMm/r² = mv²/r, so v = √(GM/r). Thus v ∝ 1/√r, and increasing r by a factor of 4 reduces v by a factor of 2.

Question 31Physics (syllabus up to Class 12 CBSE India)

A wire of length L and cross-sectional area A carries a steady current I under a potential difference V. Which expression gives the number of free electrons crossing any cross-section per second?

  1. n e A, where n is the free-electron density
  2. V A / (e L)
  3. I/e Correct answer.
  4. I L / (e A)

Explanation

Current is the rate of flow of charge, I = Ne, where N is the number of electrons per second. Therefore N = I/e, independent of the wire's geometry and material.

Question 32Physics (syllabus up to Class 12 CBSE India)

An ideal voltmeter and an ideal ammeter are used to measure a resistor. Which statement correctly describes their ideal properties?

  1. The voltmeter has zero resistance and the ammeter has infinite resistance
  2. Both instruments have infinite resistance
  3. The voltmeter has infinite resistance and the ammeter has zero resistance Correct answer.
  4. Both instruments have zero resistance

Explanation

An ideal voltmeter draws no current, so it must have infinite resistance when connected in parallel. An ideal ammeter must not develop a voltage drop, so it has zero resistance in series.

Question 33Physics (syllabus up to Class 12 CBSE India)

A charged capacitor of capacitance C is discharged through a resistor of resistance R. The time constant of the discharge is

  1. C/R
  2. R/C
  3. RC Correct answer.
  4. 1/(RC)

Explanation

During discharge the charge decays as q = q₀e^(−t/RC), so the time constant is the product RC. It has units of seconds and does not depend on the initial charge.

Question 34Physics (syllabus up to Class 12 CBSE India)

A long straight wire carries a current I. At a perpendicular distance d from the wire, the magnetic field magnitude is B. At distance 2d the field magnitude is

  1. 2B
  2. B/2 Correct answer.
  3. 4B
  4. B/4

Explanation

The field of a long straight wire falls off as 1/d, since B = μ₀I/(2πd). Doubling the distance therefore halves the field.

Question 35Physics (syllabus up to Class 12 CBSE India)

A bar magnet is dropped through a horizontal copper ring held fixed. As the magnet approaches the ring from above, the induced current in the ring

  1. alternates direction continuously as the magnet falls
  2. attracts the magnet, producing an attractive force
  3. opposes the approaching magnet, producing a repulsive force Correct answer.
  4. is zero because the ring is not connected to a circuit

Explanation

By Lenz's law, the induced current opposes the change in magnetic flux that produces it. As the magnet approaches, the flux through the ring increases, so the induced current creates a field that repels the magnet.

Question 36Physics (syllabus up to Class 12 CBSE India)

In a series LCR circuit driven at angular frequency ω, resonance occurs when

  1. ω = 1/√(LC) Correct answer.
  2. ω = √(LC)
  3. ω = R/L
  4. ω = L/C

Explanation

Resonance occurs when the inductive and capacitive reactances are equal: ωL = 1/(ωC). Solving gives ω = 1/√(LC), at which the impedance is purely resistive and minimum.

Question 37Physics (syllabus up to Class 12 CBSE India)

An electromagnetic wave travels in vacuum. Which statement about its electric and magnetic field vectors is correct?

  1. They oscillate in phase but at right angles to the direction of propagation only for the magnetic field
  2. They are mutually perpendicular and both perpendicular to the direction of propagation Correct answer.
  3. They are mutually perpendicular and one of them is along the direction of propagation
  4. They are parallel to each other and perpendicular to the direction of propagation

Explanation

In a plane electromagnetic wave, E and B are perpendicular to each other and to the propagation direction, forming a right-handed triad. They oscillate in phase, with E/B = c in vacuum.

Question 38Physics (syllabus up to Class 12 CBSE India)

A parallel-plate capacitor is connected to a battery that keeps the potential difference fixed. A dielectric of relative permittivity 3 is inserted to fill the space between the plates. The stored energy becomes

  1. nine times the original value
  2. one third of the original value
  3. three times the original value Correct answer.
  4. unchanged

Explanation

With the voltage fixed, capacitance increases by the factor 3, and energy stored is U = ½CV². Since C triples while V is constant, the stored energy triples.

Question 39Physics (syllabus up to Class 12 CBSE India)

In a plane-convex lens of refractive index 1.5 and radius of curvature 20 cm for the curved surface, the focal length in air is

  1. 20 cm
  2. 40 cm Correct answer.
  3. 10 cm
  4. 30 cm

Explanation

Using the lens maker's formula 1/f = (n − 1)(1/R₁ − 1/R₂) with the plane surface having infinite radius: 1/f = 0.5 × (1/20) = 1/40. Hence f = 40 cm.

Question 40Physics (syllabus up to Class 12 CBSE India)

Light of wavelength λ in vacuum passes from air into a medium of refractive index n. Inside the medium, the wavelength and frequency are respectively

  1. λ/n and unchanged Correct answer.
  2. λn and unchanged
  3. λ/n and reduced by a factor n
  4. λ and increased by a factor n

Explanation

The frequency of light is set by the source and does not change on refraction. The speed becomes c/n, so the wavelength becomes λ/n, since v = fλ.

Question 41Physics (syllabus up to Class 12 CBSE India)

A double-convex lens of focal length 20 cm is placed in contact with a concave lens of focal length 30 cm. The focal length of the combination is

  1. +60 cm Correct answer.
  2. +12 cm
  3. −60 cm
  4. +50 cm

Explanation

For thin lenses in contact, 1/F = 1/f₁ + 1/f₂ = 1/20 − 1/30 = 1/60, giving F = +60 cm. The combination is converging because the convex lens is stronger.

Question 42Physics (syllabus up to Class 12 CBSE India)

Photons of energy 5 eV strike a metal surface whose work function is 2 eV. The stopping potential for the emitted photoelectrons is

  1. 7 V
  2. 5 V
  3. 3 V Correct answer.
  4. 2 V

Explanation

By Einstein's photoelectric equation, maximum kinetic energy is 5 − 2 = 3 eV. The stopping potential in volts equals the maximum kinetic energy in electron-volts, so it is 3 V.

Question 43Physics (syllabus up to Class 12 CBSE India)

An electron and a proton have the same kinetic energy. The de Broglie wavelength of the electron is

  1. equal to that of the proton
  2. smaller than that of the proton
  3. larger than that of the proton Correct answer.
  4. zero, because the electron is charged

Explanation

De Broglie wavelength is λ = h/√(2mK) for a given kinetic energy K. Since the electron has much smaller mass, its momentum is smaller and its wavelength is larger.

Question 44Physics (syllabus up to Class 12 CBSE India)

In a nuclear reaction, the mass of the products is slightly less than the mass of the reactants. The energy released is given by

  1. Δm / c², where Δm is the mass defect
  2. Δm c, where Δm is the mass defect
  3. ½ Δm c², where Δm is the mass defect
  4. Δm c², where Δm is the mass defect Correct answer.

Explanation

Einstein's mass-energy relation states that a mass defect Δm corresponds to energy E = Δm c². This is the origin of the energy released in fission and fusion.

Question 45Physics (syllabus up to Class 12 CBSE India)

A p-n junction diode is connected in forward bias with a series resistor. As the forward voltage is increased beyond the knee voltage, the current

  1. decreases linearly with voltage
  2. remains constant at the reverse saturation value
  3. increases rapidly and almost exponentially Correct answer.
  4. becomes zero until breakdown occurs

Explanation

In forward bias, once the applied voltage exceeds the barrier potential, the depletion region narrows and the current rises steeply. The diode equation gives an approximately exponential rise with voltage.

Question 46Physics (syllabus up to Class 12 CBSE India)

A gas undergoes an adiabatic expansion in which it does work W on the surroundings. Which statement about the gas is correct?

  1. Its internal energy increases by W and its temperature rises
  2. Heat flows into the gas to compensate for the work done
  3. Its internal energy stays constant because the process is adiabatic
  4. Its internal energy decreases by W and its temperature falls Correct answer.

Explanation

For an adiabatic process Q = 0, so the first law gives ΔU = −W. The gas loses internal energy equal to the work it does, and for an ideal gas a fall in internal energy means a fall in temperature.

Question 47Physics (syllabus up to Class 12 CBSE India)

A black body at absolute temperature T radiates with peak wavelength λ. If the temperature is raised to 2T, the peak wavelength becomes

  1. λ/2 Correct answer.
  2. 2λ
  3. 4λ
  4. λ/4

Explanation

Wien's displacement law states that the peak wavelength is inversely proportional to absolute temperature: λ_max T = constant. Doubling T therefore halves the peak wavelength.

Question 48Physics (syllabus up to Class 12 CBSE India)

A body of mass m is attached to a spring of force constant k and executes simple harmonic motion on a frictionless horizontal surface. Its angular frequency is

  1. √(m/k)
  2. k/m
  3. √(k/m) Correct answer.
  4. m/k

Explanation

For a mass-spring oscillator the restoring force is −kx, giving acceleration −(k/m)x. Comparing with a = −ω²x gives ω = √(k/m).

Question 49Physics (syllabus up to Class 12 CBSE India)

A wave of frequency 500 Hz travels along a stretched string with speed 200 m/s. The phase difference between two points on the string separated by 0.1 m is

  1. π/4 rad
  2. 2π rad
  3. π rad
  4. π/2 rad Correct answer.

Explanation

The wavelength is v/f = 200/500 = 0.4 m. A separation of 0.1 m is one quarter of a wavelength, giving a phase difference of 2π × (0.1/0.4) = π/2 rad.

Question 50Physics (syllabus up to Class 12 CBSE India)

A galvanometer of resistance G gives full-scale deflection for current I_g. To convert it into a voltmeter of range V, the resistance connected in series with it must be

  1. V/I_g − G Correct answer.
  2. I_g G / V
  3. V G / I_g
  4. V/I_g + G

Explanation

The total series resistance needed for full-scale deflection is R_total = V/I_g, since the same current I_g must flow through the whole path. The series resistance to be added is therefore V/I_g − G.