📗 CDS General Knowledge20 Questions · No Negative Marking
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Question 1 of 20
A body moving with uniform velocity has:
Uniform velocity means constant speed AND constant direction. If velocity is constant, its rate of change (acceleration) is zero. Uniform circular motion has constant speed but changing direction — it has centripetal acceleration, so it is not uniform velocity.
Question 2 of 20
The slope of a velocity-time graph gives:
Slope of v-t graph = Δv/Δt = acceleration. Area under v-t graph = displacement. Slope of s-t graph = velocity. Area under a-t graph = change in velocity. These graphical interpretations are frequently tested in CDS Physics.
Question 3 of 20
A ball is dropped from a height. Its velocity when it reaches the ground (ignoring air resistance) is:
Using v² = u² + 2as: u = 0 (dropped), a = g, s = h. So v² = 2gh → v = √(2gh). This is derived from energy conservation: mgh = ½mv² → v = √(2gh). The mass cancels — all objects fall at the same rate regardless of mass.
Question 4 of 20
Newton's Third Law states that:
Newton's 3rd Law: every action has an equal and opposite reaction — but these forces act on different bodies (not on the same body, so they cannot cancel each other). Examples: walking (foot pushes ground back, ground pushes foot forward), rocket propulsion, swimming.
Question 5 of 20
The range of a projectile is maximum at an angle of:
R = u²sin(2θ)/g. Maximum when sin(2θ) = 1, i.e., 2θ = 90°, θ = 45°. The range at 30° equals that at 60° (complementary angles give same range). At 90°, range = 0 (vertical throw). At 45°, horizontal and vertical components of velocity are equal.
Question 6 of 20
Which equation of motion is INCORRECT for uniform acceleration?
The three standard equations of uniform motion are: v = u + at; s = ut + ½at²; v² = u² + 2as. Option D (s = vt - at²) is incorrect — the correct form using final velocity is s = vt - ½at². All three correct equations can be derived from the definition of uniform acceleration.
Question 7 of 20
The horizontal component of velocity of a projectile:
In projectile motion (ignoring air resistance): horizontal velocity = u cos θ = constant (no horizontal force). Vertical velocity: changes due to gravity (increases downward, decreases going up). At highest point: vertical component = 0, but horizontal = u cos θ (unchanged).
Question 8 of 20
A car accelerates uniformly from 0 to 72 km/h in 10 s. Its acceleration is:
72 km/h = 72 × (1000/3600) = 20 m/s. a = (v - u)/t = (20 - 0)/10 = 2 m/s². Always convert km/h to m/s first (divide by 3.6 or multiply by 5/18) before applying kinematic equations.
Question 9 of 20
Inertia of a body is measured by its:
Inertia is the tendency to resist changes in state of motion. It is measured by mass — greater mass means greater inertia. Weight is force (mg) not inertia. A body at rest or moving uniformly has inertia proportional to its mass. This is the basis of Newton's First Law.
Question 10 of 20
A bullet fired from a gun recoils. This demonstrates:
Gun recoil: bullet pushed forward (action) → gun pushed backward (reaction). By Newton's 3rd Law, forces are equal and opposite. By conservation of momentum: initial momentum = 0; final momentum of bullet + gun = 0 → gun recoils with momentum equal and opposite to bullet.
Question 11 of 20
The angle of projection for which horizontal range and maximum height are equal is:
R = u²sin2θ/g; H = u²sin²θ/2g. For R = H: sin2θ = sin²θ/2 → 2sinθcosθ = sin²θ/2 → 4cosθ = sinθ → tanθ = 4 → θ ≈ 76°. This specific angle (tanθ = 4) is a standard CDS question.
Question 12 of 20
The velocity of a body at any instant is called:
Instantaneous velocity: velocity at a specific instant = limit of (Δs/Δt) as Δt → 0 = ds/dt (derivative of displacement). Average velocity = total displacement/total time. Speed = distance/time (scalar). Instantaneous speed = magnitude of instantaneous velocity.
Question 13 of 20
A body is said to be in equilibrium when the net force acting on it is:
A body is in equilibrium (translational) when the vector sum of all forces acting on it is zero (ΣF = 0). This means the body either remains at rest or moves with constant velocity. For rotational equilibrium, the net torque must also be zero.
Question 14 of 20
The momentum of a body of mass 2 kg moving with velocity 5 m/s is:
Momentum p = mv = 2 × 5 = 10 kg·m/s. Momentum is a vector quantity in the direction of velocity. SI unit = kg·m/s = N·s. Newton's 2nd Law: F = dp/dt (rate of change of momentum). The impulse-momentum theorem: FΔt = Δp.
Question 15 of 20
A stone is thrown horizontally from a cliff. The time to hit the ground depends on:
Time of flight for horizontal projection: h = ½gt² → t = √(2h/g). This depends only on height h and g — NOT on horizontal speed. Horizontal speed only determines horizontal range = v × t. All horizontally projected stones from the same height take the same time to reach the ground.
Question 16 of 20
The area under an acceleration-time graph gives:
Area under a-t graph = ∫a dt = change in velocity (Δv). Area under v-t graph = displacement. Slope of v-t graph = acceleration. Slope of s-t graph = velocity. These fundamental graph relationships appear in every CDS Physics paper.
Question 17 of 20
At the highest point of a vertically projected body:
At the topmost point of vertical projection: velocity = 0 (momentarily at rest). Acceleration = g = 9.8 m/s² downward (gravity never stops acting). This is the most common misconception — students think acceleration is zero at the top. Only velocity is zero, not acceleration.
Question 18 of 20
A body of mass 5 kg is acted upon by a force of 20 N for 4 seconds. The impulse is:
Impulse = Force × time = 20 × 4 = 80 N·s. Impulse = change in momentum: Δp = 80 kg·m/s. Alternatively, impulse = m × Δv = 5 × 16 = 80 N·s (Δv = Ft/m = 80/5 = 16 m/s). Impulse is a vector in the direction of force.
Question 19 of 20
Two bodies of masses M and m (M > m) are dropped simultaneously from the same height. They reach the ground:
Galileo's law of free fall: all bodies fall with the same acceleration g regardless of mass (ignoring air resistance). Both reach the ground simultaneously. This was demonstrated at the Leaning Tower of Pisa. v = √(2gh) and t = √(2h/g) — both independent of mass.
Question 20 of 20
The motion of the hands of a clock is an example of:
Clock hands move at constant angular speed in circles — uniform circular motion. They have constant speed but changing direction (centripetal acceleration toward centre). The hour hand completes one revolution in 12 hours; minute hand in 60 minutes.