Motion — Set 2
Physics · गति · Questions 11–20 of 60
Inertia is the property of a body to?
Correct Answer: A. Resist change in its state of rest or uniform motion
• **Resist change in its state of rest or uniform motion** = Inertia is the inherent property by which a body resists any change in its state of rest or uniform straight-line motion unless acted on by an external force. • **Newton's First Law** — this law is called the 'Law of Inertia'; the greater the mass, the greater the inertia. • 💡 Wrong-option analysis: Increase speed without force: impossible — forces are required to change speed; Change its motion easily: this is the opposite of inertia, which resists motion change; Move only in circles: circular motion requires centripetal force — it is not a property of inertia.
Two trains move in opposite directions with speeds 40 km/h and 60 km/h. What is their relative speed?
Correct Answer: D. 100 km/h
• **100 km/h** = For objects moving in opposite directions, relative speed = sum of speeds = 40 + 60 = 100 km/h. • **v_rel (opposite) = v₁ + v₂ = 100 km/h** — for same-direction motion it is |v₁ − v₂| = 20 km/h. • 💡 Wrong-option analysis: 20 km/h: this is the same-direction relative speed (difference), not opposite-direction; 60 km/h: this is just the speed of one train; 40 km/h: this is just the speed of the other train.
Which relation between angular speed (ω) and time period (T) is correct?
Correct Answer: A. ω = 2π/T
• **ω = 2π/T** = One complete revolution spans 2π radians in time period T, so angular speed ω = 2π/T. • **ω = 2π/T = 2πf** — ω has units rad/s; for T = 1 s, ω = 2π rad/s (standard rotation). • 💡 Wrong-option analysis: ω = T/2π: this is the reciprocal of the correct formula with wrong units; ω = 2πT: this incorrectly multiplies rather than divides by T, giving units of rad·s not rad/s; ω = π/T²: incorrect — has an extra power of T in the denominator.
A ball is thrown upward with initial speed 20 m/s. Taking g = 10 m/s^2, what maximum height does it reach?
Correct Answer: A. 20 m
• **20 m** = At maximum height v = 0; using v² = u² − 2gh: 0 = 400 − 2×10×h → h = 400/20 = 20 m. • **h = u²/(2g) = 20²/(2×10) = 400/20 = 20 m** — the standard result for maximum height with upward projection. • 💡 Wrong-option analysis: 25 m: would require u²/(2g) = 25, i.e. u = √500 ≈ 22.4 m/s; 10 m: error of computing h = u/(2g) = 20/20 = 1, scaled by 10; 15 m: does not match any standard formula with the given values.
Which is true about static friction?
Correct Answer: D. Its maximum value is μsN
• **Its maximum value is μsN** = Static friction adjusts to balance the applied force, up to a maximum of μsN (limiting friction); μs is the coefficient of static friction. • **fs ≤ μsN** — static friction is a self-adjusting force; it only reaches μsN just before the object starts sliding. • 💡 Wrong-option analysis: It always equals μkN: μkN is kinetic (sliding) friction — static friction is not always equal to μkN; It is always less than kinetic friction by definition: wrong — static friction is generally greater than kinetic friction; It acts only when there is relative motion: wrong — static friction acts precisely when there is no relative motion.
A projectile is launched with speed u at angle 30° to the horizontal. Which expression gives the horizontal component of its velocity?
Correct Answer: C. u cos 30°
• **u cos 30°** = The horizontal component of initial velocity is u cosθ; with θ = 30°, it equals u cos 30°. • **horizontal = u cosθ** — this component remains unchanged throughout the projectile's flight (no horizontal acceleration). • 💡 Wrong-option analysis: u/30: dividing by the angle in degrees is not physically meaningful; u tan 30°: tangent combines both components — it is not the horizontal component alone; u sin 30°: this is the vertical component of the initial velocity, not horizontal.
In uniform circular motion, the work done by the centripetal force on the body is?
Correct Answer: A. Zero
• **Zero** = Centripetal force is always perpendicular to the instantaneous velocity (displacement), so W = Fs cos90° = 0. • **W = 0 since θ = 90°** — force is radially inward, displacement is tangential; they are always perpendicular in uniform circular motion. • 💡 Wrong-option analysis: Maximum: perpendicular force does zero work, not maximum; Equal to mv²: mv² is twice the kinetic energy — unrelated to work done by centripetal force; Equal to mgh: mgh is gravitational PE — unrelated to centripetal force work.
A body moves in a circle of radius 2 m with speed 6 m/s. What is its centripetal acceleration?
Correct Answer: B. 18 m/s^2
• **18 m/s²** = Centripetal acceleration a = v²/r; with v = 6 m/s and r = 2 m: a = 36/2 = 18 m/s². • **a = v²/r = 36/2 = 18 m/s²** — centripetal acceleration increases with the square of speed and decreases with larger radius. • 💡 Wrong-option analysis: 12 m/s²: error of computing 6×2 = 12 instead of 6²/2 = 18; 6 m/s²: error of using v/r = 6/2 = 3, misapplied; 9 m/s²: arithmetic error — does not match any correct substitution.
The area under a velocity–time graph between two times represents?
Correct Answer: D. Displacement
• **Displacement** = The area under a velocity–time graph = ∫v dt, which by definition gives displacement. • **area = ∫v dt = displacement** — positive area above the time axis means positive displacement; area below means negative. • 💡 Wrong-option analysis: Acceleration: acceleration is the slope of the v–t graph, not the area; Jerk: jerk is the rate of change of acceleration and cannot be read from a v–t area; Force: force requires mass × acceleration — not the area under a v–t graph.
A body increases its speed from 10 m/s to 30 m/s in 5 s. What is the acceleration?
Correct Answer: A. 4 m/s^2
• **4 m/s²** = Using a = (v−u)/t with v = 30 m/s, u = 10 m/s, t = 5 s: a = 20/5 = 4 m/s². • **a = Δv/Δt = 20/5 = 4 m/s²** — straightforward first equation of motion application. • 💡 Wrong-option analysis: 3 m/s²: 3×5 = 15 m/s change, not 20 m/s; 2 m/s²: 2×5 = 10 m/s change, not 20 m/s; 5 m/s²: error of dividing final velocity 30 by time 5 ignoring initial velocity 10.