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Tamil Nadu HSE First Year Physics study materials
Tamil Nadu HSE First Year Physics study resources

Tamil Nadu HSE First Year Physics practice questions

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Question 1Nature of Physical World and Measurement

In the SI system, which of the following is a supplementary unit whose value in radians is exactly 1?

  1. The mole, used for amount of substance
  2. The steradian, used for solid angle
  3. The degree, used for plane angle
  4. The radian, used for plane angle Correct answer.

Explanation

Plane angle and solid angle are the two supplementary quantities in the SI. The radian is defined as the angle subtended at the centre by an arc equal in length to the radius, so one radian is exactly 1 rad by definition and not a derived multiple of another unit.

Question 2Nature of Physical World and Measurement

A student writes the result of a length measurement as 4.520 m. How many significant figures does this value contain, and which digit is the least reliable?

  1. Five significant figures, with the digit 4 least reliable
  2. Two significant figures, with the digit 2 least reliable
  3. Four significant figures, with the final 0 least reliable Correct answer.
  4. Three significant figures, with the digit 5 least reliable

Explanation

All non-zero digits and any zeros written after the decimal point are significant, so 4.520 has four significant figures. The rightmost digit of a measured value carries the uncertainty of the instrument, so the trailing zero is the least reliable digit.

Question 3Nature of Physical World and Measurement

Which pair correctly matches a physical quantity with its dimensional formula?

  1. Pressure and [M L^-1 T^-2]
  2. Force and [M L T^-2] Correct answer.
  3. Work and [M L^2 T^-2]
  4. Power and [M L^2 T^-3]

Explanation

Force equals mass times acceleration, and acceleration has dimensions [L T^-2]. Multiplying by mass gives [M L T^-2], so this pairing is correct. Each of the other quantities has a different dimensional formula from the one listed.

Question 4Nature of Physical World and Measurement

The parallax method is used in astronomy to measure the distance of nearby stars. Which quantity is directly measured in this method?

  1. The wavelength shift of the star's spectral lines
  2. The drop in the star's brightness during a transit
  3. The time taken by light to travel from the star
  4. The shift in apparent position of the star against background stars Correct answer.

Explanation

Parallax relies on observing the same star from two positions of the Earth in its orbit and measuring the angle through which the star appears to shift against distant background objects. From that parallax angle and the baseline, the distance is calculated using geometry.

Question 5Nature of Physical World and Measurement

A physical quantity is given by Q = a^3 b^2 / c, where a, b and c are measured with percentage errors of 1%, 2% and 4% respectively. What is the maximum percentage error in Q?

  1. 11% Correct answer.
  2. 9%
  3. 7%
  4. 13%

Explanation

When quantities are multiplied or divided, their fractional errors add, with each error multiplied by the magnitude of its exponent. Here the error is 3(1%) + 2(2%) + 1(4%) = 3 + 4 + 4 = 11%.

Question 6Nature of Physical World and Measurement

A screw gauge has a pitch of 0.5 mm and 50 divisions on its circular scale. What is its least count?

  1. 0.1 mm
  2. 0.01 mm Correct answer.
  3. 0.001 mm
  4. 0.05 mm

Explanation

The least count of a screw gauge is the pitch divided by the number of circular scale divisions. Dividing 0.5 mm by 50 gives 0.01 mm, which is the smallest length the instrument can read directly.

Question 7Nature of Physical World and Measurement

Using dimensional analysis, which of the following expressions could represent a time period, where l is a length, g is the acceleration due to gravity and m is a mass?

  1. 2π sqrt(g/l)
  2. 2π (l g)
  3. 2π sqrt(m l / g)
  4. 2π sqrt(l/g) Correct answer.

Explanation

The dimensions of l/g are [L]/[L T^-2] = [T^2], so its square root has dimensions [T], which is correct for a time period. Mass cannot appear because no other quantity in the expression can cancel its dimension.

Question 8Nature of Physical World and Measurement

The radius of a sphere is measured as (2.1 ± 0.1) cm. What is the percentage error in the calculated volume of the sphere?

  1. About 0.1%
  2. About 14.3% Correct answer.
  3. About 4.8%
  4. About 2.1%

Explanation

Volume depends on the cube of the radius, so the percentage error in volume is three times the percentage error in radius. The percentage error in radius is (0.1/2.1) × 100 ≈ 4.76%, and tripling it gives about 14.3%.

Question 9Nature of Physical World and Measurement

The length of a rod measured with a metre scale is reported as 25.4 cm. Which statement about the accuracy and precision of this measurement is most appropriate?

  1. The measurement is precise because it is written to 0.1 cm, which is the least count of the scale
  2. The measurement is precise but its accuracy cannot be judged without a standard value Correct answer.
  3. The measurement is accurate but not precise because no true value is stated
  4. The measurement is neither accurate nor precise because the scale has no zero error

Explanation

Precision refers to how finely and consistently a quantity is measured, while accuracy refers to closeness to the true value. Writing 25.4 cm shows the reading resolution, but accuracy can only be assessed by comparing with an accepted standard value.

Question 10Nature of Physical World and Measurement

Which of the following is a derived unit in the SI system?

  1. The ampere, defined through the force between current-carrying wires
  2. The newton, defined as kg m s^-2 Correct answer.
  3. The kelvin, defined through the triple point of water
  4. The candela, defined through the luminous intensity of a source

Explanation

Derived units are formed by combining base units, and the newton is defined as kg m s^-2 from the base units of mass, length and time. The kelvin, ampere and candela are base units with their own independent definitions.

Question 11Nature of Physical World and Measurement

A student measures the diameter of a wire five times and obtains 0.52, 0.53, 0.52, 0.54 and 0.52 mm. Which statement best describes the set of readings?

  1. The readings show only random error and no systematic error is possible
  2. The readings are precise but may not be accurate because a systematic zero error could shift them all Correct answer.
  3. The readings are accurate because the mean is close to 0.52 mm
  4. The readings are imprecise because they differ from one another

Explanation

The readings cluster closely around 0.52 mm, which shows good precision or repeatability. Precision alone does not guarantee accuracy, since a zero error in the instrument would shift every reading by the same amount and leave the spread unchanged.

Question 12Nature of Physical World and Measurement

A screw gauge with a positive zero error of +0.03 mm reads a main scale value of 4 mm and a circular scale reading of 35 divisions, with least count 0.01 mm. What is the corrected diameter of the wire?

  1. 4.35 mm
  2. 4.32 mm Correct answer.
  3. 4.40 mm
  4. 4.38 mm

Explanation

The observed reading is the main scale reading plus circular scale reading times least count, giving 4 mm + 0.35 mm = 4.35 mm. For a positive zero error the correction is subtracted, so 4.35 mm − 0.03 mm = 4.32 mm.

Question 13Kinematics

A body starts from rest and moves with uniform acceleration. Which graph of velocity against time represents this motion?

  1. A straight line with negative slope starting from the origin
  2. A straight line passing through the origin with positive slope Correct answer.
  3. A straight line parallel to the time axis
  4. A parabola opening upward

Explanation

For uniform acceleration, v = u + at. With u = 0, the velocity-time relation is v = at, which is a straight line through the origin with slope equal to the acceleration. A line parallel to the time axis would mean zero acceleration, and a parabola would mean changing acceleration.

Question 14Kinematics

The area under a velocity-time graph of a moving body gives which physical quantity?

  1. Jerk
  2. Rate of change of acceleration
  3. Displacement Correct answer.
  4. Acceleration

Explanation

Velocity is the rate of change of displacement, so the area under a velocity-time graph equals the displacement covered in that time interval. The slope of the same graph gives acceleration, not the area.

Question 15Kinematics

A particle moves along a straight line with uniform acceleration. Which equation relates displacement, initial velocity, time and acceleration?

  1. s = u + at
  2. s = ½ut + at²
  3. s = ut + ½at² Correct answer.
  4. s = ut − ½at²

Explanation

The second equation of motion for uniform acceleration is s = ut + ½at², obtained by integrating v = u + at with respect to time. It correctly combines the contribution of initial velocity and the contribution of acceleration to displacement.

Question 16Kinematics

Which of the following is a vector quantity?

  1. Distance
  2. Time
  3. Displacement Correct answer.
  4. Speed

Explanation

Displacement has both magnitude and direction, so it is a vector. Distance, speed and time are described completely by magnitude alone and are therefore scalars.

Question 17Kinematics

A body thrown vertically upward reaches its maximum height. At that instant, what is the acceleration of the body?

  1. Equal to g directed downward Correct answer.
  2. Equal to g directed upward
  3. Equal to 2g directed downward
  4. Zero

Explanation

At the highest point the velocity becomes zero, but gravity continues to act on the body throughout the motion. Hence the acceleration remains g directed downward at the topmost point.

Question 18Kinematics

Which of the following is the correct expression for the magnitude of relative velocity of two bodies moving with velocities v₁ and v₂ in the same direction?

  1. v₁ + v₂
  2. |v₁ − v₂| Correct answer.
  3. √(v₁² + v₂²)
  4. v₁v₂

Explanation

When two bodies move along the same direction, their relative velocity is the difference of their velocities, and its magnitude is |v₁ − v₂|. The sum applies when they move in opposite directions.

Question 19Kinematics

A car accelerates uniformly from rest at 2 m/s² for 5 s. What distance does it cover in this time?

  1. 25 m Correct answer.
  2. 10 m
  3. 50 m
  4. 20 m

Explanation

Using s = ut + ½at² with u = 0, a = 2 m/s² and t = 5 s gives s = ½ × 2 × 25 = 25 m. The initial velocity term vanishes because the car starts from rest.

Question 20Kinematics

A body moving with 20 m/s is brought to rest in 4 s by a constant retardation. What is the magnitude of the retardation?

  1. 0.2 m/s²
  2. 4 m/s²
  3. 80 m/s²
  4. 5 m/s² Correct answer.

Explanation

Retardation magnitude is |a| = (v − u)/t = (0 − 20)/4 = 5 m/s². The negative sign shows deceleration, but the magnitude asked is 5 m/s².

Question 21Kinematics

A stone is thrown vertically upward with a velocity of 20 m/s. Taking g = 10 m/s², what maximum height does it reach?

  1. 20 m Correct answer.
  2. 80 m
  3. 10 m
  4. 40 m

Explanation

At maximum height v = 0, so using v² = u² − 2gh gives 0 = 400 − 20h, hence h = 20 m. The upward velocity decreases under gravity until it momentarily vanishes at the top.

Question 22Kinematics

A body travels 4 m north and then 3 m east. What is the magnitude of its displacement?

  1. 7 m
  2. 1 m
  3. 12 m
  4. 5 m Correct answer.

Explanation

The two displacements are perpendicular, so the resultant magnitude is √(4² + 3²) = √25 = 5 m. Displacement uses vector addition, not simple arithmetic addition of path lengths.

Question 23Kinematics

A body covers the first half of a straight track with speed 30 m/s and the second half with speed 60 m/s. What is its average speed over the whole track?

  1. 45 m/s
  2. 90 m/s
  3. 40 m/s Correct answer.
  4. 50 m/s

Explanation

For equal distances covered at speeds v₁ and v₂, the average speed is the harmonic mean 2v₁v₂/(v₁ + v₂) = 2 × 30 × 60/90 = 40 m/s. The arithmetic mean is valid only for equal time intervals, not equal distances.

Question 24Kinematics

A particle starts from rest with uniform acceleration and attains a velocity of 40 m/s in 8 s. What is the distance travelled in this interval?

  1. 160 m Correct answer.
  2. 80 m
  3. 320 m
  4. 640 m

Explanation

The average velocity during uniformly accelerated motion is (u + v)/2 = 20 m/s, so the distance is 20 × 8 = 160 m. Alternatively, a = 5 m/s² and s = ½at² = 160 m.

Question 25Kinematics

A body moving with uniform acceleration covers 20 m in the 3rd second and 32 m in the 5th second. What is its acceleration?

  1. 12 m/s²
  2. 3 m/s²
  3. 4 m/s²
  4. 6 m/s² Correct answer.

Explanation

The distance covered in the nth second is sₙ = u + a(n − ½). Subtracting the two expressions gives 32 − 20 = a(5 − 3) = 2a, so a = 6 m/s². The difference of successive nth-second distances equals the acceleration.

Question 26Kinematics

A boat can move at 5 m/s in still water. It crosses a river 200 m wide flowing at 3 m/s, heading straight across. How long does the crossing take?

  1. 25 s
  2. 50 s
  3. 67 s
  4. 40 s Correct answer.

Explanation

The component of the boat's velocity perpendicular to the bank is 5 m/s, unaffected by the current. The crossing time is therefore width/velocity = 200/5 = 40 s. The current only causes drift along the bank.

Question 27Kinematics

A projectile is launched with speed u at an angle θ to the horizontal. Which expression gives its horizontal range on level ground?

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

Explanation

The range is R = u² sin 2θ / g, obtained by multiplying the horizontal velocity u cos θ by the time of flight 2u sin θ / g. It is maximum at θ = 45°, where sin 2θ = 1.

Question 28Kinematics

A body is projected at 45° to the horizontal with a speed of 20 m/s. Taking g = 10 m/s², what is its horizontal range?

  1. 10 m
  2. 20 m
  3. 40 m Correct answer.
  4. 80 m

Explanation

Using R = u² sin 2θ / g with θ = 45° gives R = 400 × 1/10 = 40 m. At 45° the range is maximum for a given launch speed.

Question 29Kinematics

Two trains, each 100 m long, move on parallel tracks in opposite directions with speeds 10 m/s and 15 m/s. How long do they take to cross each other completely?

  1. 10 s
  2. 8 s Correct answer.
  3. 4 s
  4. 20 s

Explanation

When moving in opposite directions, the relative speed is 10 + 15 = 25 m/s, and the total distance to be covered is 100 + 100 = 200 m. The time is therefore 200/25 = 8 s.

Question 30Kinematics

A particle starts from rest and moves with uniform acceleration. The ratio of the distances covered in the first, second and third seconds is

  1. 1 : 4 : 9
  2. 1 : 2 : 3
  3. 1 : 1 : 1
  4. 1 : 3 : 5 Correct answer.

Explanation

For a body starting from rest, the distance in the nth second is proportional to (2n − 1). Hence for n = 1, 2, 3 the ratio is 1 : 3 : 5, the well-known odd-number ratio.

Question 31Kinematics

A ball is dropped from a height h and simultaneously another ball is thrown upward from the ground with speed u. They meet after a time t. Which relation must hold at the meeting point?

  1. ½gt² − ut = h
  2. ut + ½gt² = h
  3. ut = h
  4. ½gt² + (ut − ½gt²) = h Correct answer.

Explanation

The falling ball descends ½gt² from the top, while the rising ball ascends ut − ½gt² from the ground. Since they meet, the sum of these two distances must equal the initial separation h, giving ½gt² + ut − ½gt² = h, i.e. ut = h.

Question 32Kinematics

A particle moves such that its position is given by x = 3t² − 2t + 5 (in metres, t in seconds). What is its acceleration at t = 2 s?

  1. 3 m/s²
  2. 10 m/s²
  3. 12 m/s²
  4. 6 m/s² Correct answer.

Explanation

Velocity is dx/dt = 6t − 2 and acceleration is d²x/dt² = 6 m/s², which is constant. The value of t does not affect this constant acceleration.

Question 33Laws of Motion

A body of mass 5 kg is acted upon by a constant net force of 20 N. What is the acceleration of the body?

  1. 0.25 m s⁻²
  2. 25 m s⁻²
  3. 4 m s⁻² Correct answer.
  4. 100 m s⁻²

Explanation

Newton's second law states that the net force equals mass times acceleration, F = ma. Rearranging gives a = F/m = 20/5 = 4 m s⁻². The acceleration is in the direction of the applied net force.

Question 34Laws of Motion

Which quantity is a measure of the inertia of a body?

  1. Its mass Correct answer.
  2. Its weight
  3. Its momentum
  4. Its velocity

Explanation

Inertia is the tendency of a body to resist change in its state of rest or uniform motion. Mass is the quantitative measure of inertia, so a larger mass means greater resistance to change in motion. Weight depends on gravity and momentum depends on velocity, so neither measures inertia itself.

Question 35Laws of Motion

A book lies at rest on a horizontal table. According to Newton's third law, the reaction to the gravitational force exerted by the Earth on the book is

  1. the normal force exerted by the table on the book
  2. the gravitational force exerted by the book on the Earth Correct answer.
  3. the frictional force between the book and the table
  4. the weight of the table acting downward

Explanation

Action and reaction pairs act on two different bodies and are of the same type. The Earth pulls the book downward, so the reaction is the book pulling the Earth upward with an equal gravitational force. The normal force is a contact force and pairs with the book's push on the table, not with gravity.

Question 36Laws of Motion

The SI unit of impulse is the same as the SI unit of

  1. force
  2. acceleration
  3. power
  4. momentum Correct answer.

Explanation

Impulse is force multiplied by time, so its unit is N s. Since impulse equals change in momentum, its unit is also kg m s⁻¹, which is the unit of momentum. Force has the unit N, power the unit W, and acceleration m s⁻², so only momentum matches.

Question 37Laws of Motion

A force of 10 N acts on a body for 3 s. What is the impulse delivered to the body?

  1. 30 N s Correct answer.
  2. 13 N s
  3. 3.33 N s
  4. 0.3 N s

Explanation

Impulse is the product of force and the time for which it acts, J = F × t. Here J = 10 × 3 = 30 N s. This impulse equals the change in momentum produced in the body.

Question 38Laws of Motion

A cricketer moves his hands backward while catching a fast-moving ball. This action

  1. decreases the change in momentum of the ball
  2. increases the rate of change of momentum
  3. increases the time of impact and so reduces the force on the hands Correct answer.
  4. increases the impulse delivered to the hands

Explanation

The change in momentum of the ball is fixed by its initial and final velocities. By increasing the contact time, the average force F = Δp/Δt is reduced, protecting the hands. The impulse itself stays the same because Δp is unchanged.

Question 39Laws of Motion

A block of mass 4 kg rests on a horizontal surface where the coefficient of static friction is 0.5. Taking g = 10 m s⁻², the minimum horizontal force needed to just start moving the block is

  1. 20 N Correct answer.
  2. 2 N
  3. 8 N
  4. 40 N

Explanation

The limiting static friction equals μs times the normal force. On a horizontal surface the normal force equals the weight, mg = 4 × 10 = 40 N, so f = 0.5 × 40 = 20 N. A horizontal force slightly above this value starts the block moving.

Question 40Laws of Motion

The coefficient of static friction between two surfaces is generally

  1. less than the coefficient of kinetic friction
  2. equal to the coefficient of kinetic friction
  3. independent of the nature of the surfaces
  4. greater than the coefficient of kinetic friction Correct answer.

Explanation

For a given pair of surfaces, more force is needed to break the interlocking at rest than to maintain sliding. Hence limiting static friction exceeds kinetic friction, so μs > μk. Both coefficients, however, depend on the nature and roughness of the surfaces in contact.

Question 41Laws of Motion

A body of mass 2 kg is pulled along a horizontal floor by a horizontal force of 12 N against a constant kinetic friction of 4 N. The acceleration of the body is

  1. 2 m s⁻²
  2. 6 m s⁻²
  3. 8 m s⁻²
  4. 4 m s⁻² Correct answer.

Explanation

The net horizontal force is the applied force minus kinetic friction, that is 12 − 4 = 8 N. Applying F = ma gives a = 8/2 = 4 m s⁻². The friction always opposes the relative sliding motion.

Question 42Laws of Motion

A body is in equilibrium under the action of several coplanar forces. Which condition must hold?

  1. The forces must all be equal in magnitude
  2. The algebraic sum of the magnitudes of the forces must be zero
  3. The forces must all act along the same line
  4. The vector sum of all the forces must be zero Correct answer.

Explanation

For translational equilibrium the vector sum of all forces acting on the body must be zero. This is equivalent to the sum of the components along each direction being zero. The magnitudes need not be equal, and the forces need not be collinear as long as they cancel vectorially.

Question 43Laws of Motion

A block of weight 50 N hangs from a ceiling by a single vertical rope. The tension in the rope is

  1. 25 N
  2. 100 N
  3. 50 N Correct answer.
  4. zero

Explanation

The block is at rest, so the net force on it is zero. The upward tension must exactly balance the downward weight, giving T = 50 N. A single vertical rope carries the whole weight when the body is stationary.

Question 44Laws of Motion

A monkey of mass 10 kg climbs up a vertical rope with an acceleration of 2 m s⁻². Taking g = 10 m s⁻², the tension in the rope is

  1. 100 N
  2. 120 N Correct answer.
  3. 20 N
  4. 80 N

Explanation

The net upward force on the monkey is tension minus weight, which equals mass times acceleration. So T − mg = ma, giving T = m(g + a) = 10 × (10 + 2) = 120 N. When accelerating upward, the tension exceeds the weight.

Question 45Laws of Motion

A rocket moves forward in space because

  1. the rocket is lighter than the surrounding air
  2. the momentum of the ejected gases backward is balanced by forward momentum of the rocket Correct answer.
  3. gravity pulls the rocket forward
  4. the exhaust gases push against the atmosphere

Explanation

This is an application of conservation of momentum and Newton's third law. The rocket pushes gases backward, and the gases push the rocket forward with an equal and opposite force. The motion does not require an atmosphere, so it works in the vacuum of space.

Question 46Laws of Motion

A body of mass 3 kg moving at 4 m s⁻¹ collides with and sticks to a stationary body of mass 1 kg. The velocity of the combined body after collision is

  1. 3 m s⁻¹ Correct answer.
  2. 1 m s⁻¹
  3. 2 m s⁻¹
  4. 4 m s⁻¹

Explanation

In a perfectly inelastic collision, total momentum is conserved. Initial momentum is 3 × 4 = 12 kg m s⁻¹, and the combined mass is 4 kg, so the final velocity is 12/4 = 3 m s⁻¹. Kinetic energy is not conserved in this collision, but momentum is.

Question 47Laws of Motion

A body of mass m moves in a horizontal circle of radius r with constant speed v. The centripetal force required is

  1. mv²r directed toward the centre
  2. mv²/r directed toward the centre Correct answer.
  3. mv²/r directed along the tangent
  4. mv/r directed toward the centre

Explanation

Uniform circular motion requires a net force directed toward the centre, called the centripetal force. Its magnitude is mv²/r, obtained from the centripetal acceleration v²/r. The tangential direction has no net force because the speed is constant.

Question 48Laws of Motion

A car rounds a flat circular curve of radius 20 m. If the coefficient of static friction between the tyres and road is 0.5 and g = 10 m s⁻², the maximum safe speed is

  1. 100 m s⁻¹
  2. 5 m s⁻¹
  3. 10 m s⁻¹ Correct answer.
  4. 20 m s⁻¹

Explanation

On a flat curve the centripetal force is provided by static friction, so mv²/r = μmg. Cancelling m and solving gives v = √(μgr) = √(0.5 × 10 × 20) = √100 = 10 m s⁻¹. Above this speed the friction is insufficient and the car skids.

Question 49Laws of Motion

A block of mass 5 kg is pulled up a rough incline of angle 30° by a force parallel to the incline. If the coefficient of kinetic friction is 0.2 and g = 10 m s⁻², the force needed to move it up at constant speed is closest to

  1. 16 N
  2. 25 N
  3. 50 N
  4. 34 N Correct answer.

Explanation

At constant speed the applied force balances the component of weight down the incline plus kinetic friction. The weight component is mg sin30° = 5 × 10 × 0.5 = 25 N, and friction is μ mg cos30° = 0.2 × 50 × 0.866 ≈ 8.7 N. The total is about 33.7 N, closest to 34 N.

Question 50Laws of Motion

A lift of mass 500 kg is moving upward with a uniform velocity of 2 m s⁻¹. Taking g = 10 m s⁻², the tension in the supporting cable is

  1. 5000 N Correct answer.
  2. 6000 N
  3. 1000 N
  4. 4000 N

Explanation

Uniform velocity means zero acceleration, so the net force on the lift is zero. The tension must equal the weight, T = mg = 500 × 10 = 5000 N. The velocity value does not affect the tension when the motion is uniform.