Questions & explanations
1. A projectile is fired with speed 20 m/s at angle 30° above horizontal from the top of an incline of 30° downward. Find the range down the incline.
- 80√3 m
- 80 m
- 40 m
- 40√3 m
Answer: 80 m
Using ground frame: initial velocity components: u_x = 20 cos30° = 10√3 m/s, u_y = 20 sin30° = 10 m/s. Incline equation: y = -x tan30° = -x/√3. Projectile motion: x = 10√3 t, y = 10t - 5t². Substitute: 10t - 5t² = - (10√3 t)/√3 = -10t → 20t - 5t² = 0 → t = 4 s. Then x = 40√3 m, y = -40 m. Range = √(x²+y²) = √(4800+1600) = √6400 = 80 m.
2. A boat can move at 5 m/s in still water. The river flows at 3 m/s east. A log drifts with the current. The boat starts from the south bank aiming to reach the log. Find the minimum distance between the boat and the log if the boat's initial position is 40 m south of the log.
- 24 m
- 32 m
- 40 m
- 30 m
Answer: 24 m
The relative velocity of boat w.r.t. log is v_BL = v_B - v_L. The boat's speed relative to water is 5 m/s, river 3 m/s east, so v_BL magnitude = 4 m/s (perpendicular component). Minimum distance d_min = |r0| sin φ, where φ is angle between r0 (40 m south) and v_BL. sin φ = (3/5), so d_min = 40 × (3/5) = 24 m.
3. Rain is falling vertically at 5 m/s. A man walks horizontally at 2 m/s. At what angle to the vertical should he hold his umbrella?
- tan⁻¹(2/5) but backward
- tan⁻¹(5/2)
- tan⁻¹(2/5)
- tan⁻¹(5/2) but backward
Answer: tan⁻¹(2/5)
Relative velocity of rain w.r.t. man: v_rm = v_rg - v_mg. v_rg is downward (vertical), v_mg is horizontal. The horizontal component of v_rm is opposite to man's motion, magnitude 2 m/s. Vertical component is 5 m/s. Angle from vertical: tanθ = horizontal/vertical = 2/5, so θ = tan⁻¹(2/5) forward.
4. A projectile is fired with speed 20 m/s at angle 60° above horizontal up an incline of 30°. Find the range along the incline.
- 80/3 m
- 40/3 m
- 20/3 m
- 160/3 m
Answer: 80/3 m
Using rotated axes: x along incline, y perpendicular. u_x = 20 cos30° = 10√3 m/s, u_y = 20 sin30° = 10 m/s. a_y = -g cos30° = -5√3 m/s². Time of flight T = 2u_y / (-a_y) = 4/√3 s. Range R = u_x T + ½ a_x T² with a_x = -g sin30° = -5 m/s². R = (10√3)(4/√3) + ½(-5)(16/3) = 40 - 40/3 = 80/3 m.
5. Projectile A is fired from origin at 60° above horizontal (heading right) with speed 20 m/s. At the same instant projectile B is fired from (20, 0) m at 120° above horizontal (heading left) with speed 20 m/s. After what time do they collide?
- 0.5 s
- 4.0 s
- 1.0 s
- 2.0 s
Answer: 1.0 s
Both projectiles have the same vertical velocity component (20 sin 60° upward), so they stay at the same height at every instant. Horizontally, A moves right at 10 m/s and B moves left at 10 m/s. Relative speed = 20 m/s, gap = 20 m, so t = 20/20 = 1.0 s.
6. A passenger in a moving train drops a ball. Which statement about the ball's motion is correct?
- The ball appears to fall straight down to the passenger.
- The ball appears to fall straight down to a ground observer.
- The ball appears to move backward to the passenger.
- The ball appears to move forward to the ground observer.
Answer: The ball appears to fall straight down to the passenger.
According to the frame of reference, the passenger is at rest relative to the train. The ball has the same horizontal velocity as the train, so relative to the passenger it falls straight down. The ground observer sees a parabolic path.
7. In an experiment to determine resistivity of a wire, the percentage errors in voltage, current, length, and radius are 1%, 2%, 0.5%, and 1% respectively. What is the percentage error in resistivity?
- 6.0%
- 3.5%
- 5.5%
- 4.5%
Answer: 5.5%
Resistivity ρ = Vπr²/(IL). Relative errors add for products/quotients. Since r is squared, its relative error is doubled. So Δρ/ρ = ΔV/V + ΔI/I + ΔL/L + 2×(Δr/r) = 1% + 2% + 0.5% + 2×1% = 5.5%. Hence the percentage error in ρ is 5.5%.
8. What is 2.745 rounded to three significant figures?
- 2.74
- 2.75
- 2.7
- 2.745
Answer: 2.74
According to NCERT rounding rules, when the digit to be dropped is exactly 5, we look at the preceding digit. If it is even, we leave it unchanged. Here, the fourth digit is 5 and the third digit is 4 (even), so 2.745 rounds to 2.74.
9. The position of a particle is x(t) = 2t³ meters. What is its acceleration at t = 2 s?
- 48 m/s²
- 12 m/s²
- 6 m/s²
- 24 m/s²
Answer: 24 m/s²
Acceleration is the second derivative of position: a = d²x/dt². Given x = 2t³, first derivative v = 6t², second derivative a = 12t. At t = 2 s, a = 12 × 2 = 24 m/s². This uses the definition a = dv/dt from NCERT Class 11 Ch 3 §3.6.
10. A student measures the length of a table five times and gets values very close to each other but far from the true length. Which statement is correct?
- The measurements are precise but not accurate.
- The measurements are accurate but not precise.
- The measurements are both accurate and precise.
- The measurements are neither accurate nor precise.
Answer: The measurements are precise but not accurate.
Precision refers to the closeness of repeated measurements to each other; accuracy refers to closeness to the true value. Here, the values are close to each other (high precision) but far from the true value (low accuracy).
11. In an experiment, the time period of a pendulum is measured as 2.01 s, 2.00 s, 2.02 s, 1.99 s, and 2.00 s. What is the mean absolute error?
- 0.008 s
- 0.012 s
- 0.016 s
- 0.010 s
Answer: 0.010 s
The mean is (2.01+2.00+2.02+1.99+2.00)/5 = 2.004 s. Absolute deviations: 0.006, 0.004, 0.016, 0.014, 0.004. Mean absolute error = (0.006+0.004+0.016+0.014+0.004)/5 = 0.0088 s ≈ 0.010 s (rounded to one significant figure).
12. A student writes s = ut + at² for displacement under constant acceleration. Which of the following is true?
- The equation is dimensionally correct and physically correct.
- The equation is dimensionally incorrect because ut and at² have different dimensions.
- The equation is dimensionally correct but physically incorrect.
- The equation is dimensionally incorrect because s has dimension [L] while ut has [LT].
Answer: The equation is dimensionally correct but physically incorrect.
Dimensions: s is [L], ut is [LT⁻¹][T] = [L], at² is [LT⁻²][T²] = [L]. All terms have [L], so it is dimensionally consistent. However, the correct equation is s = ut + ½at², so the given equation is physically incorrect.