Class 11 Physics NCERT Solutions

Step-by-step solution

1. 1(a) 1 cm = 10⁻² m, so (1 cm)³ = (10⁻²)³ m³ = 10⁻⁶ m³.
2. 2(b) Total surface area of a cylinder of radius r and height h is 2πr(h + r). With r = 2.0 cm and h = 10.0 cm: A = 2π(2.0)(12.0) = 48π ≈ 1.508 × 10² cm².
3. 31 cm² = (10 mm)² = 10² mm², so A = 1.508 × 10² × 10² = 1.508 × 10⁴ mm².
4. 4(c) 18 km h⁻¹ = 18 × (1000 m)/(3600 s) = 5 m s⁻¹, so the vehicle covers 5 m in 1 s.
5. 5(d) Relative density equals density in g cm⁻³, so density = 11.3 g cm⁻³. Since 1 g cm⁻³ = 10³ kg m⁻³ (multiply by 10⁻³/10⁻⁶ = 10³), density = 11.3 × 10³ kg m⁻³.

Step-by-step solution

1. 1(a) 1 kg = 10³ g and 1 m² = 10⁴ cm², so 1 kg m² s⁻² = 10³ × 10⁴ g cm² s⁻² = 10⁷ g cm² s⁻².
2. 2(b) 1 ly = 9.46 × 10¹⁵ m, so 1 m = 1/9.46 × 10⁻¹⁵ ≈ 1.06 × 10⁻¹⁶ ly.
3. 3(c) 3.0 m s⁻² = 3.0 × 10⁻³ km s⁻². Since 1 h = 3600 s, 1 s⁻² = (3600)² h⁻² = 1.296 × 10⁷ h⁻².
4. 4So 3.0 × 10⁻³ × 1.296 × 10⁷ = 3.888 × 10⁴ ≈ 3.9 × 10⁴ km h⁻².
5. 5(d) G = 6.67 × 10⁻¹¹ N m² kg⁻² = 6.67 × 10⁻¹¹ (kg m s⁻²) m² kg⁻² = 6.67 × 10⁻¹¹ m³ s⁻² kg⁻¹.
6. 61 m³ = (10² cm)³ = 10⁶ cm³ and 1 kg = 10³ g: G = 6.67 × 10⁻¹¹ × 10⁶/10³ = 6.67 × 10⁻⁸ cm³ s⁻² g⁻¹.

Step-by-step solution

1. 1Energy has dimensions [ML²T⁻²].
2. 2In SI, 1 calorie = 4.2 kg m² s⁻².
3. 3The new units are: mass α kg, length β m, time γ s. A numerical value transforms as n′ = n (u₁/u₁′)ᵃ (u₂/u₂′)ᵇ (u₃/u₃′)ᶜ, where a = 1, b = 2, c = −2 for [ML²T⁻²].
4. 4n′ = 4.2 × (1/α)¹ × (1/β)² × (1/γ)⁻² = 4.2 α⁻¹ β⁻² γ², which is exactly what had to be shown.

Step-by-step solution

1. 1A statement like ‘large’ or ‘small’ has meaning only relative to a chosen standard; the same quantity can be large compared with one standard and small compared with another.
2. 2(a) Reframe: atoms are very small compared with objects of everyday size (a few tenths of a nanometre).
3. 3(b) Reframe: a jet plane moves with great speed compared with a bicycle or car (about 250 m s⁻¹).
4. 4(c) Reframe: the mass of Jupiter is very large compared with the mass of the Earth (about 318 times).
5. 5(d) Reframe: the air inside this room contains a large number of molecules compared with, say, the number of people in the room (of order 10²⁷).
6. 6(e) Statement already relative: a proton is much more massive than an electron (about 1836 times).
7. 7(f) Statement already relative: the speed of sound is much smaller than the speed of light.

Step-by-step solution

1. 1Speed of light in the new unit is 1 (new unit of length) per (unit of time), by definition.
2. 2Time taken = 8 min 20 s = 8 × 60 + 20 = 500 s.
3. 3Distance = speed × time = 1 × 500 = 500 new units of length.

Step-by-step solution

1. 1Least count of the vernier callipers = 1 main-scale division / 20 = 1/20 mm = 0.05 mm = 5 × 10⁻⁵ m.
2. 2Least count of the screw gauge = pitch/divisions = 1 mm/100 = 0.01 mm = 1 × 10⁻⁵ m.
3. 3The optical instrument resolves down to one wavelength of light, λ ≈ 5 × 10⁻⁷ m.
4. 4Smallest least count ⇒ highest precision: 5 × 10⁻⁷ m < 1 × 10⁻⁵ m < 5 × 10⁻⁵ m.

Step-by-step solution

1. 1The observed width 3.5 mm is the magnified image width.
2. 2Actual thickness = observed width / magnification = 3.5 mm / 100 = 0.035 mm.
3. 30.035 mm = 3.5 × 10⁻² × 10⁻³ m = 3.5 × 10⁻⁵ m = 35 µm.

Step-by-step solution

1. 1(a) Wind the thread closely around a pencil or cylinder for N turns, measure the total length L of the N turns with the metre scale, then diameter = L/N. The more turns, the smaller the relative error in the single-turn width.
2. 2(b) No. Least count = pitch/divisions; increasing divisions reduces the least count, but accuracy is ultimately limited by the backlash of the screw, the zero error and the measurer’s ability to judge coincidence. Beyond a point, extra divisions only give false precision.
3. 3(c) Random errors average out over many readings; the random error of the mean falls roughly as 1/√N. With 100 readings the random scatter is strongly suppressed, so the mean is closer to the true value than with 5 readings. Systematic zero error, however, is not removed by either set.

Step-by-step solution

1. 1Area magnification = screen area / slide area = 1.55 m² / 1.75 cm².
2. 2Convert: 1.55 m² = 1.55 × 10⁴ cm², so area magnification = 1.55 × 10⁴/1.75 ≈ 8857.
3. 3Linear magnification is the square root of the area magnification: m = √8857 ≈ 94.

Step-by-step solution

1. 1(a) Leading zeros are not significant: 0.007 has 1 significant figure.
2. 2(b) 2.64 × 10²⁴ has 3 significant figures.
3. 3(c) 0.2370 has 4 (the trailing zero after the decimal is significant).
4. 4(d) 6.320 has 4 (trailing zero is significant).
5. 5(e) 6.032 has 4 (the embedded zero counts).
6. 6(f) 0.0006032 has 4 (only leading zeros are not significant).

Step-by-step solution

1. 1Convert the thickness: 2.01 cm = 0.0201 m (3 significant figures).
2. 2Area = length × breadth = 4.234 × 1.005 = 4.25517 m².
3. 3The two factors have 4 and 4 significant figures, so the product is reported to 4: A = 4.255 m².
4. 4Volume = area × thickness = 4.25517 × 0.0201 = 0.0855289 m³.
5. 5The thickness has only 3 significant figures, so the volume is reported to 3: V = 0.0855 m³.

Step-by-step solution

1. 1Convert the gold masses: 20.15 g = 0.02015 kg and 20.17 g = 0.02017 kg.
2. 2(a) Total = 2.30 + 0.02015 + 0.02017 = 2.34032 kg.
3. 3In addition, the result has the same number of decimal places as the least precise term: 2.30 kg is precise only to the first decimal, so total = 2.3 kg.
4. 4(b) Difference = 20.17 − 20.15 = 0.02 g, keeping the two decimal places of the least precise piece.

Step-by-step solution

1. 1The argument of a square root must be dimensionless, since m₀ and m have the same dimensions [M].
2. 21 − v² is not dimensionless: v carries dimensions [LT⁻¹] and 1 is dimensionless, so the terms cannot be added.
3. 3For dimensional consistency the subtracted quantity must be (v²/c²), which is dimensionless because v and c both have speed dimensions.
4. 4Correct form: m₀ = m(1 − v²/c²)^(1/2), i.e. put c under v inside the bracket.

Step-by-step solution

1. 1Size of the atom ≈ 0.5 Å = 0.5 × 10⁻¹⁰ m = 5 × 10⁻¹¹ m; taking this as the radius r.
2. 2Volume of one atom = (4/3)πr³ = (4/3)π(5 × 10⁻¹¹)³.
3. 3
   $$V_1=\frac{4}{3}\pi (1.25\times 10^{-31})\approx 5.24\times 10^{-31}{\text{m}}^3$$
4. 4A mole has NA = 6.023 × 10²³ atoms, so total volume = NA V₁.
5. 5
   $$V=6.023\times 10^{23}\times 5.24\times 10^{-31}\approx 3.15\times 10^{-7}{\text{m}}^3$$

Step-by-step solution

1. 1Molar volume of the gas = 22.4 L = 22.4 × 10⁻³ m³ = 2.24 × 10⁻² m³.
2. 2Size of a hydrogen molecule ≈ 1 Å, so take radius ≈ 0.5 × 10⁻¹⁰ m (as in Exercise 1.14).
3. 3Atomic volume of a mole of hydrogen ≈ 3.15 × 10⁻⁷ m³ (computed in Exercise 1.14).
4. 4
   $$\text{Ratio}=\frac{2.24\times 10^{-2}}{3.15\times 10^{-7}}\approx 7.1\times 10^4$$
5. 5The ratio is so large because gas molecules are widely separated — the gas itself is mostly empty space, so its molar volume is tens of thousands of times the volume actually filled by the molecules.

Step-by-step solution

1. 1When you look at nearby objects, the line of sight sweeps through a large angle in a short time as the train moves past, so the apparent angular speed is large — they seem to rush backwards.
2. 2For distant objects the line of sight swings through a tiny angle for the same displacement of the train, so their apparent angular motion is negligible — they seem fixed.
3. 3The same parallax idea applies to observing stars from two positions six months apart in the Earth’s orbit (Exercises 1.5 on light travel time illustrates the same large distance).
4. 4The apparent speed of an object therefore depends on its distance: apparent angular speed ∝ (train speed)/(distance).

Step-by-step solution

1. 1Guess: even though the Sun is a plasma, its density is expected to lie in the range of solids and liquids (roughly 10³ kg m⁻³), not of gases at ordinary pressure.
2. 2Volume of the Sun (sphere) = (4/3)πR³ = (4/3)π(7.0 × 10⁸)³.
3. 3
   $$V=\frac{4}{3}\pi (3.43\times 10^{26})\approx 1.437\times 10^{27}{\text{m}}^3$$
4. 4Density = mass/volume = 2.0 × 10³⁰/1.437 × 10²⁷ ≈ 1.4 × 10³ kg m⁻³.
5. 51.4 × 10³ kg m⁻³ is indeed in the range of densities of solids and liquids (about 10³ kg m⁻³), confirming the guess — the plasma has ionic densities far above ordinary gas.

Step-by-step solution

1. 1A body is treated as a point object when its size is negligible compared with the distance it moves or the size of its path.
2. 2(a) The size of the carriage is very small compared with the distance between two stations, so it can be a point object.
3. 3(b) The size of the monkey is very small compared with the size of the circular track, so it too can be a point object.
4. 4(c) The size of the spinning cricket ball is comparable to the distance through which it turns sharply on hitting the ground — it cannot be a point object.
5. 5(d) The size of the tumbling beaker is comparable to the height of the table from which it slips — it cannot be a point object.

Step-by-step solution

1. 1(a) From the graph, the distance OP is less than OQ, so A lives closer to the school than B.
2. 2(b) At x = 0, t = 0 for A, but for B at x = 0, t has some finite positive value — so A starts from the school earlier than B.
3. 3(c) The slope of the x-t graph gives speed; the slope of B’s graph is greater than A’s, so B walks faster than A.
4. 4(d) Both graphs end at the same time on the t-axis, so A and B reach home at the same time.
5. 5(e) B, who starts later but walks faster, meets A exactly once — so B overtakes A once on the road.

Step-by-step solution

1. 1Time for the outward walk: t = distance/speed = 2.5/5 = 0.5 h = 30 min. She reaches the office at 9.30 am.
2. 2Return trip by auto: t = 2.5/25 = 0.1 h = 6 min. She leaves the office at 5.00 pm and reaches home at 5.06 pm.
3. 3On the x-t graph: a straight line of slope 5 km h⁻¹ from t = 0 to t = 30 min (position rising to 2.5 km), a horizontal line at x = 2.5 km from 30 min to 480 min (9.30 am to 5.00 pm), and a steeper falling straight line from x = 2.5 km to x = 0 between 480 min and 486 min, slope 25 km h⁻¹.
4. 4Suitable scales: 1 cm = 30 min on the time axis and 1 cm = 0.5 km on the position axis make the three legs of the graph clearly visible.

Step-by-step solution

1. 1Each step is 1 m long and takes 1 s, so 5 steps forward take 5 s and 3 steps backward take 3 s.
2. 2Net distance per cycle = 5 − 3 = 2 m, covered in 5 + 3 = 8 s.
3. 3After one cycle (8 s) he is at 2 m; after two (16 s) at 4 m; after three (24 s) at 6 m; after four (32 s) at 8 m.
4. 4From 8 m, he takes 5 more forward steps covering 5 m in 5 s, reaching 8 + 5 = 13 m at t = 32 + 5 = 37 s, and falls into the pit.
5. 5The x-t graph is a staircase: rising 1 m per second for 5 s, falling 1 m per second for 3 s, repeating until the pit is reached at 13 m.

Step-by-step solution

1. 1Initial speed u = 126 km h⁻¹ = 126 × (5/18) = 35 m s⁻¹; final speed v = 0; stopping distance s = 200 m.
2. 2Use v² − u² = 2as: 0 − (35)² = 2 × a × 200.
3. 3
   $$a=-\frac{35\times 35}{2\times 200}=-3.06{\text{m s}}^{-2}$$
4. 4So the retardation (magnitude of the deceleration) is 3.06 m s⁻².
5. 5Use v = u + at: t = (v − u)/a = (0 − 35)/(−3.06) ≈ 11.44 s.

Step-by-step solution

1. 1(a) Irrespective of direction of motion, acceleration (due to gravity) always acts downward, towards the centre of the Earth.
2. 2(b) At the highest point the velocity is zero, while the acceleration is still g = 9.8 m s⁻² acting downward.
3. 3(c) With x = 0 at the highest point and downward positive: position x > 0 for both motions; velocity v < 0 during upward motion and v > 0 during downward motion; acceleration a > 0 throughout.
4. 4(d) For the rise: v² − u² = 2(−g)(s), with v = 0, u = 29.4 m s⁻¹: s = (0² − 29.4²)/(2 × (−9.8)) = 44.1 m.
5. 5Time of ascent using v = u + at: t = (0 − 29.4)/(−9.8) = 3 s.
6. 6Time of ascent equals time of descent, so total time = 3 + 3 = 6 s.

Step-by-step solution

1. 1(a) True. When an object is thrown vertically up, its speed becomes zero at the maximum height, yet its acceleration is g = 9.8 m s⁻² downward at that instant.
2. 2(b) False. Speed is the magnitude of velocity; if speed is zero, the magnitude of velocity is zero, so the velocity itself is zero.
3. 3(c) True. A car moving on a straight highway with constant speed has constant velocity, and acceleration — the rate of change of velocity — is zero.
4. 4(d) False in general. If acceleration is positive while velocity is negative (e.g., a particle projected upwards after the origin is chosen), the particle slows down until velocity becomes zero. The statement is true only when both velocity and acceleration are positive (e.g., falling vertically downward).

Step-by-step solution

1. 1First fall: s = ut + ½at² with u = 0, a = g: 90 = ½ × 9.8 × t², so t = √18.38 ≈ 4.29 s.
2. 2Impact speed: v = u + at = 0 + 9.8 × 4.29 ≈ 42.04 m s⁻¹.
3. 3After the collision the ball keeps 9/10 of its speed: rebound speed uᵣ = (9/10)(42.04) ≈ 37.84 m s⁻¹.
4. 4Time to reach maximum height on rebound: 0 = uᵣ + (−g)t′, so t′ = 37.84/9.8 ≈ 3.86 s; total time so far = 4.29 + 3.86 = 8.15 s.
5. 5Time of ascent = time of descent, so the second impact is at t = 8.15 + 3.86 ≈ 12.01 s, with rebound speed (9/10)(37.84) ≈ 34.05 m s⁻¹.
6. 6Speed-time graph: for each fall the speed rises linearly from 0 to the impact value; at each bounce it drops discontinuously to 9/10 of that value and falls again linearly.

Step-by-step solution

1. 1(a) The magnitude of displacement is the shortest distance (a straight line) between the initial and final positions; the path length is the actual distance traversed. E.g., moving from A to B and back to C: magnitude of displacement = AC, while path length = AB + BC.
2. 2The path length can equal the displacement only when there is no reversal of direction — AB + BC ≥ AC, so path length ≥ magnitude of displacement.
3. 3(b) Magnitude of average velocity = |displacement| / Δt; average speed = total path length / Δt.
4. 4In the example above, average speed = (AB + BC)/t while |average velocity| = AC/t; since AB + BC > AC, average speed is greater than magnitude of average velocity.
5. 5Equality holds exactly when the motion never reverses direction, i.e., the particle moves along a straight line without turning back.

Step-by-step solution

1. 1Time outward = 2.5/5 = 0.5 h = 30 min; time return = 2.5/7.5 = 1/3 h = 20 min; total = 50 min.
2. 2(i) 0 to 30 min: displacement = 2.5 km, time = 0.5 h. Average velocity magnitude = 2.5/0.5 = 5 km h⁻¹; average speed = 5 km h⁻¹ too (no reversal yet).
3. 3(ii) 0 to 50 min: net displacement = 0, time = 5/6 h. Average velocity = 0; total distance = 5 km, so average speed = 5/(5/6) = 6 km h⁻¹.
4. 4(iii) 0 to 40 min: he has been returning for 10 min, covering 7.5 × (10/60) = 1.25 km. Net displacement = 2.5 − 1.25 = 1.25 km; total distance = 2.5 + 1.25 = 3.75 km.
5. 5Average velocity magnitude = 1.25/(40/60) = 1.875 ≈ 1.9 km h⁻¹; average speed = 3.75/(40/60) = 5.625 ≈ 5.6 km h⁻¹.

Step-by-step solution

1. 1Instantaneous velocity is the first derivative of position with respect to time.
2. 2In the infinitesimal interval dt, the displacement is so small that the particle does not change the direction of its motion within it.
3. 3Because there is no reversal of direction in that interval, the path length travelled equals the magnitude of displacement in dt.
4. 4Hence instantaneous speed (ds/dt) equals the magnitude of instantaneous velocity (|dx/dt|) — the distinction vanishes at the instantaneous limit.

Step-by-step solution

1. 1Graph (a) is an x-t graph that is not one-valued in position: it cannot represent one-dimensional motion because a particle cannot have two positions at the same instant of time.
2. 2Graph (b) is a v-t graph that is not one-valued in velocity: a particle can never have two values of velocity at the same instant of time.
3. 3Graph (c) is a v-t graph taking negative speed values: speed, being a scalar magnitude, can never be negative.
4. 4Graph (d) shows the x-t curve turning back so that the position decreases: the total path length travelled can never decrease with time.

Step-by-step solution

1. 1No — the x-t graph does not give the shape of the particle’s path; t and x are both coordinates, so the ‘path’ picture is a misreading of the graph.
2. 2The claim that the particle ‘moves in a straight line for t < 0 and on a parabolic path for t > 0’ is therefore not correct.
3. 3A suitable physical situation matching the plot: a freely falling body that is held at a height for some time (so x stays constant), then released from rest at t = 0 and falls under gravity, giving the parabolic x-t section beyond t = 0.

Step-by-step solution

1. 1Convert speeds: police van vₚ = 30 km h⁻¹ = 8.33 m s⁻¹; thief’s car vₜ = 192 km h⁻¹ = 53.33 m s⁻¹; muzzle speed of bullet in the van’s frame v_b = 150 m s⁻¹.
2. 2Since the bullet is fired from the moving van, its resultant speed in the ground frame = 150 + 8.33 = 158.33 m s⁻¹.
3. 3Both vehicles move in the same direction, so the speed of the bullet relative to the thief’s car = 158.33 − 53.33 = 105 m s⁻¹.

Step-by-step solution

1. 1Graph (a), an x-t graph: the body is initially at rest, its speed builds up to a constant value, drops to zero, then increases in the opposite direction and levels off. A suitable situation: a football (initially at rest) kicked towards a rigid wall, rebounding with reduced speed, passing the kicker, and finally coming to rest.
2. 2Graph (b), a v-t graph: the sign of velocity changes and its magnitude decreases with time. A suitable situation: a ball dropped on a hard floor from a height — it strikes with some velocity, rebounds with reduced velocity, and this repeats until the ball comes to rest.
3. 3Graph (c), an a-t graph: the body moves with a uniform velocity, its acceleration rises for a short interval and drops back to zero, and the uniform velocity resumes. A suitable situation: a hammer moving with uniform velocity striking a nail.

Step-by-step solution

1. 1For SHM, acceleration a = −ω²x, so the acceleration always points opposite to the displacement; the velocity sign comes from the slope of the x-t graph.
2. 2At t = 0.3 s: x is negative, and the slope of the x-t graph is negative, so v is negative; from a = −ω²x, a is positive. Answer: (−, −, +).
3. 3At t = 1.2 s: x is positive and the slope is positive; hence v is positive and a is negative. Answer: (+, +, −).
4. 4At t = −1.2 s: x is negative, and with t negative the slope gives a positive velocity; a is positive. Answer: (−, +, +).

Step-by-step solution

1. 1Average speed over an interval is given by the magnitude of the slope of the x-t graph in that interval.
2. 2From the graph, the slope is maximum in interval 3 and minimum in interval 2: average speed is greatest in interval 3 and least in interval 2.
3. 3The sign of the average velocity follows the sign of the slope: positive in intervals 1 and 2 (slope positive), negative in interval 3 (slope negative).

Step-by-step solution

1. 1Acceleration is the slope of the speed-time graph; the slope is greatest (in magnitude) in interval 2, so the average acceleration is greatest in interval 2.
2. 2The height of the curve above the time-axis gives the speed; it is greatest in interval 3, so the average speed is greatest in interval 3.
3. 3Interval 1: slope positive, so a > 0; speed positive, so v > 0. Interval 2: slope negative, so a < 0; speed (a scalar) still positive. Interval 3: slope zero, so a = 0; speed is uniform and positive.
4. 4At the points A, B, C and D the v-t curve runs parallel to the time-axis, making the slope — and hence the acceleration — zero at each point.

Step-by-step solution

1. 1A scalar is specified by magnitude alone and has no direction; a vector has both magnitude and direction.
2. 2Scalars: volume, mass, speed, density, number of moles, angular frequency.
3. 3Vectors: acceleration, velocity, displacement, angular velocity.

Step-by-step solution

1. 1Work is the dot product of force and displacement; a dot product is always a scalar, so work is a scalar.
2. 2Current is described only by its magnitude — its direction is not taken into account, so it is a scalar.
3. 3All the remaining quantities (force, angular momentum, linear momentum, electric field, magnetic moment) have an associated direction and are vectors; average velocity and relative velocity are velocity quantities — vectors. Angular momentum and magnetic moment are vectors too.

Step-by-step solution

1. 1Impulse is the product of force and time (J = F × Δt).
2. 2Force is a vector quantity; multiplying a vector by the scalar time gives a vector along the force direction, so impulse is a vector.
3. 3Every other item (temperature, pressure, time, power, total path length, energy, gravitational potential, coefficient of friction, charge) has magnitude only and is a scalar.

Step-by-step solution

1. 1(a) Meaningful, but only if the two scalars represent the same physical quantity.
2. 2(b) Not meaningful — a scalar quantity cannot be added to a vector quantity.
3. 3(c) Meaningful — e.g., multiplying the force vector by the scalar time gives the vector impulse.
4. 4(d) Meaningful — a scalar can be multiplied with another scalar of the same or different dimensions.
5. 5(e) Meaningful, but only if the two vectors represent the same physical quantity.
6. 6(f) Meaningful — a component of a vector has the same dimensions as the vector itself, so they may be added.

Step-by-step solution

1. 1(a) True — the magnitude of a vector is a number (with a unit), so it is a scalar.
2. 2(b) False — each component of a vector is itself a vector (it has a direction along the axis).
3. 3(c) False — the total path length is a scalar and generally exceeds the magnitude of the displacement vector; the two are equal only for straight-line motion without reversal.
4. 4(d) True — since total path length ≥ magnitude of displacement, dividing both by the same time interval keeps the inequality: average speed ≥ magnitude of average velocity.
5. 5(e) True — three vectors not lying in one plane cannot be represented by the three sides of a triangle taken in order, so their vector sum cannot be the null vector.

Step-by-step solution

1. 1(a) Represent a and b as adjacent sides of a parallelogram; the diagonal gives |a + b|. In the triangle formed, each side is smaller than the sum of the other two: |a + b| ≤ |a| + |b|.
2. 2(b) In the same triangle, the third side exceeds the difference of the other two: |a + b| ≥ ||a| − |b||.
3. 3(c) Vector a − b is the third side of the triangle built on a and −b, so |a − b| ≤ |a| + |−b| = |a| + |b|.
4. 4(d) For a − b, the triangle inequality on the differences gives |a − b| ≥ ||a| − |b|| (taking moduli so both sides are non-negative).
5. 5Equality in (a) and (c) holds when the two vectors act along the same straight line and same sense; equality in (b) and (d) holds when they act along the same straight line in opposite directions.

Step-by-step solution

1. 1(a) Incorrect — a + b + c + d = 0 does not require each vector to be null; many non-zero combinations give a zero sum.
2. 2(b) Correct — rewrite as a + c = −(b + d); taking magnitudes, |a + c| = |−(b + d)| = |b + d|.
3. 3(c) Correct — a = −(b + c + d), so |a| = |b + c + d| ≤ |b| + |c| + |d|; hence |a| can never exceed that sum.
4. 4(d) Correct — a + (b + c) + d = 0 means (b + c) is the side that closes the triangle with a and d: (b + c) lies in the plane of a and d, or along their line if they are collinear.

Step-by-step solution

1. 1Displacement depends only on the initial and final positions: all three girls go from P to Q.
2. 2P and Q are diametrically opposite, so the displacement of each girl equals the diameter of the ground.
3. 3Diameter = 2 × radius = 2 × 200 = 400 m for every girl.
4. 4This equals the actual path length only for the girl who skates along the straight diameter path (girl B in Fig. 3.19).

Step-by-step solution

1. 1(a) The cyclist ends where he started (back at O), so the net displacement is zero.
2. 2(b) Average velocity = net displacement / total time = 0 / (10 min) = 0.
3. 3(c) Total path length = OP + arc PQ + QO = 1 + (1/4)(2π × 1) + 1 = 2 + π/2 ≈ 3.570 km.
4. 4Time = 10 min = 10/60 h = 1/6 h.
5. 5Average speed = 3.570 / (1/6) = 21.42 km h⁻¹.

Step-by-step solution

1. 1Turning left by 60° after each 500 m side traces a regular hexagon of side 500 m.
2. 2Third turn (vertex S): the two opposite vertices are two diameters apart through the centre — displacement = PS = 500 + 500 = 1000 m; total path length = 3 × 500 = 1500 m.
3. 3Sixth turn: the motorist is back at the starting point P — displacement = 0; total path length = 6 × 500 = 3000 m.
4. 4Eighth turn (vertex R): displacement PR = √(500² + 500² + 2·500·500·cos 60°) = √(500² + 500² + 500²) = 500√3 ≈ 866.03 m, directed at 30° to the first side PQ; total path length = 8 × 500 = 4000 m.
5. 5Summary — third: 1000 m vs 1500 m, sixth: 0 vs 3000 m, eighth: 866.03 m at 30° vs 4000 m.

Step-by-step solution

1. 1(a) Total distance = 23 km, time = 28 min = 28/60 h.
2. 2Average speed = 23/(28/60) = 23 × 60/28 ≈ 49.29 km h⁻¹.
3. 3(b) Displacement = straight-line distance between hotel and station = 10 km.
4. 4Magnitude of average velocity = 10/(28/60) ≈ 21.43 km h⁻¹.
5. 5The two quantities are not equal because the path taken (23 km) is not a straight line — average speed uses path length, average velocity uses displacement.

Step-by-step solution

1. 1Maximum height h = u² sin²θ / 2g. The ball must not exceed h = 25 m with u = 40 m s⁻¹.
2. 225 = (40)² sin²θ / (2 × 9.8), so sin²θ = 25 × 2 × 9.8 / 1600 = 0.30625.
3. 3sin θ ≈ 0.5534, hence θ ≈ 33.60°.
4. 4Range R = u² sin 2θ / g = 1600 × sin 67.2° / 9.8 = 1600 × 0.922 / 9.8 ≈ 150.53 m.

Step-by-step solution

1. 1Maximum horizontal range occurs at θ = 45°: R = u² sin 2θ/g = u²/g × sin 90° = u²/g.
2. 2Given R = 100 m, u²/g = 100.
3. 3The maximum height is reached when the ball is thrown vertically upward, where v = 0 at the top.
4. 4Using v² − u² = −2gH: H = u²/2g = (1/2)(u²/g) = (1/2)(100) = 50 m.

Step-by-step solution

1. 1Length of string = radius r = 80 cm = 0.8 m.
2. 2Frequency ν = 14/25 Hz; angular frequency ω = 2πν = 2 × (22/7) × (14/25) = 88/25 rad s⁻¹.
3. 3Centripetal acceleration a_c = ω²r = (88/25)² × 0.8 ≈ 9.91 m s⁻².
4. 4Direction: acceleration is always along the string, radially towards the centre of the circle.

Step-by-step solution

1. 1r = 1 km = 1000 m; v = 900 km h⁻¹ = 900 × (5/18) = 250 m s⁻¹.
2. 2Centripetal acceleration a_c = v²/r = (250)²/1000 = 62.5 m s⁻².
3. 3With g = 9.8 m s⁻², a_c/g = 62.5/9.8 ≈ 6.38.
4. 4So a_c ≈ 6.38 g — the aircraft feels a centripetal acceleration about 6.4 times the acceleration due to gravity.

Step-by-step solution

1. 1(a) False — the net acceleration points radially inward only for uniform circular motion. If the speed also changes, a tangential acceleration component exists, so the net acceleration is not purely radial.
2. 2(b) True — at any point of the path, the particle moves tangentially, so the velocity vector is always tangent to the path at that point.
3. 3(c) True — in uniform circular motion the acceleration always points toward the centre and its direction keeps changing symmetrically; the average of these vectors over one full cycle is the null vector.

Step-by-step solution

1. 1(a) Differentiate r with respect to t: v = dr/dt = 3.0 î − 4.0t ĵ (in m s⁻¹).
2. 2Differentiate again: a = dv/dt = −4.0 ĵ (in m s⁻²), a constant.
3. 3(b) At t = 2.0 s: v = 3.0 î − 4.0(2.0) ĵ = 3.0 î − 8.0 ĵ.
4. 4Magnitude |v| = √(3² + (−8)²) = √73 ≈ 8.54 m s⁻¹.
5. 5Direction: θ = tan⁻¹(v_y/v_x) = tan⁻¹(−8/3) ≈ −69.45° — the negative sign means the direction is 69.45° below the x-axis.

Step-by-step solution

1. 1Acceleration a = d v/d t = 8.0 î + 2.0 ĵ. Integrating, v(t) = 8.0t î + 2.0t ĵ + u with u = 10.0 ĵ at t = 0.
2. 2Integrating again from the origin gives r(t) = 4.0t² î + (10.0t + t²) ĵ.
3. 3(a) x = 4.0t² = 16 ⇒ t = 2 s.
4. 4At t = 2 s: y = 10(2) + (2)² = 24 m.
5. 5(b) v at t = 2 s: v = 8(2) î + 2(2) ĵ + 10 ĵ = 16 î + 14 ĵ.
6. 6Speed = √(16² + 14²) = √452 ≈ 21.26 m s⁻¹.

Step-by-step solution

1. 1For P = î + ĵ: components P_x = P_y = 1, so |P| = √(1² + 1²) = √2.
2. 2Direction of î + ĵ: tan θ = 1/1 = 1 ⇒ θ = 45° with the x-axis.
3. 3For Q = î − ĵ: |Q| = √2 and θ = tan⁻¹(−1/1) = −45° with the x-axis.
4. 4Angle between A = 2 î + 3 ĵ and î + ĵ is θ′ = 56.31° − 45° = 11.31°, where tan θ of A is 3/2 ⇒ 56.31°.
5. 5Component of A along (î + ĵ)/√2 = |A| cos θ′ = √13 × cos 11.31° ≈ 3.54 = 5/√2.
6. 6Angle between A and (î − ĵ) is θ″ = 56.31° + 45° = 101.31°; the component along (î − ĵ)/√2 = √13 cos 101.31° ≈ −0.71 = −1/√2.

Step-by-step solution

1. 1(a) False — this linear average is valid only when acceleration is constant; the motion here is arbitrary.
2. 2(b) True — by definition, average velocity is the total displacement divided by the time interval.
3. 3(c) False — requires constant acceleration, which need not hold for arbitrary motion.
4. 4(d) False — this is the constant-acceleration position formula, not valid for arbitrary, non-uniform acceleration.
5. 5(e) True — by definition, average acceleration is the change in velocity divided by the time interval.

Step-by-step solution

1. 1(a) False — energy is a scalar yet is not conserved in inelastic collisions.
2. 2(b) False — temperature is a scalar but can take negative values.
3. 3(c) False — total path length is a scalar and yet it has the dimension of length.
4. 4(d) False — gravitational potential is a scalar and yet varies from point to point in space.
5. 5(e) True — a scalar is defined so that its value is the same for observers with differently oriented axes; a vector's components change with orientation, but a scalar does not.

Step-by-step solution

1. 1The aircraft is at height OR = 3400 m, and the two positions P and Q subtend ∠POQ = 30° at the observer.
2. 2In triangle PRO, tan 15° = PR/OR, so PR = OR tan 15° = 3400 × tan 15°.
3. 3Triangle PRO is similar to RQO and PR = RQ, so PQ = PR + RQ = 2 × 3400 × tan 15°.
4. 4PQ = 6800 × 0.268 ≈ 1822.4 m covered in 10 s.
5. 5Speed = 1822.4/10 ≈ 182.24 m s⁻¹.

Step-by-step solution

1. 1(a) The raindrop falls at constant speed, so its acceleration is zero: net force = 0.
2. 2(b) The cork's weight downward is balanced by the upthrust of water: net force = 0.
3. 3(c) The kite is stationary (zero acceleration): net force = 0 by Newton's first law.
4. 4(d) Constant velocity means zero acceleration: net force = 0.
5. 5(e) Far from material objects and free of fields, nothing acts on the electron: net force = 0.

Step-by-step solution

1. 1The only force acting on the pebble is gravity, which acts vertically downward in all three cases.
2. 2F = ma = mg = 0.05 × 10 = 0.5 N, directed downward.
3. 3(a) During upward motion: 0.5 N downward.
4. 4(b) During downward motion: 0.5 N downward.
5. 5(c) At the highest point: still 0.5 N downward (acceleration due to gravity is constant).
6. 6If thrown at 45°, only the vertical component of velocity is zero at the top — gravity still acts downward with the same magnitude, so the answers do not change.

Step-by-step solution

1. 1(a) Just after release, only gravity acts: F = mg = 0.1 × 10 = 1 N, vertically downward.
2. 2(b) The train moves at constant velocity, so there is no horizontal force; once released, the stone's horizontal motion does not change. Net force = 1 N, vertically downward.
3. 3(c) The train accelerates at 1 m s⁻², but the force causing that acceleration acts between train and train — it is not transmitted to the released stone. After release, F = mg = 1 N, vertically downward.
4. 4(d) The stone at rest on the accelerating floor receives the full horizontal push: F = ma = 0.1 × 1 = 0.1 N in the direction of motion of the train (the vertical weight is balanced by the floor's normal reaction).

Step-by-step solution

1. 1On a smooth horizontal table, the weight and the normal reaction cancel vertically.
2. 2The only horizontal force on the particle is the tension T of the string.
3. 3This tension alone provides the centripetal force mv²/l towards the centre.
4. 4Hence the net force on the particle is T itself. (Option i).

Step-by-step solution

1. 1Retarding force F = −50 N (negative — opposing motion).
2. 2Using F = ma, the deceleration a = F/m = −50/20 = −2.5 m s⁻².
3. 3From v = u + at with v = 0 and u = 15 m s⁻¹: t = −u/a = −15/(−2.5) = 6 s.

Step-by-step solution

1. 1a = (v − u)/t = (3.5 − 2.0)/25 = 1.5/25 = 0.06 m s⁻².
2. 2F = ma = 3.0 × 0.06 = 0.18 N.
3. 3Since the direction of motion is unchanged, the force acts in the direction of motion.

Step-by-step solution

1. 1Resultant of the two perpendicular forces: R = √(8² + 6²) = √100 = 10 N.
2. 2Direction: θ = tan⁻¹(6/8) = 36.87° with the 8 N force.
3. 3Acceleration a = F/m = 10/5 = 2 m s⁻².
4. 4The acceleration acts along the resultant force, at 36.87° with the 8 N force.

Step-by-step solution

1. 1u = 36 km/h = 10 m s⁻¹; v = 0; total mass = 400 + 65 = 465 kg.
2. 2a = (v − u)/t = (0 − 10)/4 = −2.5 m s⁻².
3. 3F = ma = 465 × (−2.5) = −1162.5 N.
4. 4The average retarding force is 1162.5 N, opposing the motion.

Step-by-step solution

1. 1Upward equation of motion for the rocket: F − mg = ma.
2. 2F = m(g + a) = 20,000 × (10 + 5.0).
3. 3F = 20,000 × 15 = 3 × 10⁵ N.

Step-by-step solution

1. 1Acceleration due to the force: a = F/m = −8.0/0.40 = −20 m s⁻² (southward; north taken positive).
2. 2At t = −5 s (before the force is applied, a′ = 0): x = ut = 10 × (−5) = −50 m.
3. 3At t = 25 s: x = ut + ½at² = 10 × 25 + ½(−20)(25)² = 250 − 6250 = −6000 m.
4. 4From t = 0 to 30 s: x₁ = 10 × 30 + ½(−20)(30)² = 300 − 9000 = −8700 m.
5. 5Velocity at t = 30 s: v = u + at = 10 − 20 × 30 = −590 m s⁻¹.
6. 6From t = 30 s to 100 s (force ends, a′ = 0): x₂ = vt = −590 × 70 = −41300 m.
7. 7At t = 100 s: x = x₁ + x₂ = −8700 − 41300 = −50000 m.

Step-by-step solution

1. 1Velocity of truck at t = 10 s: v = u + at = 0 + 2 × 10 = 20 m s⁻¹ — this is the stone's horizontal velocity at release.
2. 2At t = 11 s the horizontal velocity is unchanged (no horizontal force): vₓ = 20 m s⁻¹.
3. 3Vertical velocity: vᵧ = uᵧ + aᵧ Δt = 0 + 10 × (1) = 10 m s⁻¹ downward.
4. 4Resultant speed: v = √(vₓ² + vᵧ²) = √(20² + 10²) = √500 ≈ 22.36 m s⁻¹.
5. 5Direction: θ = tan⁻¹(vᵧ/vₓ) = tan⁻¹(10/20) = 26.57° with the horizontal (direction of the truck's motion).
6. 6(b) After release only gravity acts: acceleration = g = 10 m s⁻² vertically downward.

Step-by-step solution

1. 1(a) At an extreme position the bob is momentarily at rest (speed = 0). If the string is cut, only gravity acts, so the bob falls vertically downward with acceleration g.
2. 2(b) At the mean position the bob moves horizontally at 1 m s⁻¹, tangential to its arc.
3. 3After the string is cut, it keeps that horizontal velocity and accelerates downward under gravity.
4. 4This combination (constant horizontal velocity + uniform vertical acceleration) gives a parabolic trajectory.

Step-by-step solution

1. 1(a) Uniform speed ⇒ a = 0: R = mg = 70 × 10 = 700 N; reading = 700/10 = 70 kg.
2. 2(b) Downward acceleration: R = m(g − a) = 70(10 − 5) = 350 N; reading = 35 kg.
3. 3(c) Upward acceleration: R = m(g + a) = 70(10 + 5) = 1050 N; reading = 105 kg.
4. 4(d) Free fall: a = g ⇒ R = m(g − g) = 0; reading = 0 (the man is weightless).

Step-by-step solution

1. 1(a) For t < 0: the displacement is zero (position coincides with the time axis), so the velocity and force are both zero.
2. 2For 0 < t < 4 s: the x-t graph is a straight line with constant slope — constant velocity, zero acceleration, hence zero force.
3. 3For t > 4 s: the position stays at 3 m (graph parallel to time axis) — the particle is at rest, so again zero force.
4. 4(b) At t = 0 the velocity jumps from 0 to the slope value 3/4 m s⁻¹: impulse = m(v − u) = 4(3/4 − 0) = 3 kg m s⁻¹.
5. 5At t = 4 s the velocity drops from 3/4 m s⁻¹ to 0: impulse = 4(0 − 3/4) = −3 kg m s⁻¹.

Step-by-step solution

1. 1Total mass m = m_A + m_B = 10 + 20 = 30 kg.
2. 2System acceleration a = F/m = 600/30 = 20 m s⁻².
3. 3(i) Force applied to A: F − T = m_A a ⇒ T = 600 − 10 × 20 = 400 N.
4. 4(ii) Force applied to B: F − T = m_B a ⇒ T = 600 − 20 × 20 = 200 N.

Step-by-step solution

1. 1For the 8 kg mass going up: T − m₁g = m₁a.
2. 2For the 12 kg mass going down: m₂g − T = m₂a.
3. 3Adding: (m₂ − m₁)g = (m₁ + m₂)a.
4. 4a = (12 − 8)/(12 + 8) × 10 = 4/20 × 10 = 2 m s⁻².
5. 5T = (2m₁m₂/(m₁ + m₂))g = (2 × 12 × 8)/20 × 10 = 96 N.

Step-by-step solution

1. 1The parent nucleus is at rest, so the initial linear momentum of the system is zero.
2. 2Let m₁, v₁ and m₂, v₂ be the masses and velocities of the two fragments.
3. 3Conservation of linear momentum: 0 = m₁v₁ + m₂v₂.
4. 4Hence v₁ = −(m₂/m₁)v₂.
5. 5The minus sign shows the two fragments move in opposite directions.

Step-by-step solution

1. 1Initial momentum of each ball: pᵢ = 0.05 × 6 = 0.3 kg m s⁻¹.
2. 2After collision, the direction reverses: p_f = −0.3 kg m s⁻¹ for each ball.
3. 3Impulse = change in momentum = p_f − pᵢ = −0.3 − 0.3 = −0.6 kg m s⁻¹.
4. 4Each ball receives an impulse of 0.6 kg m s⁻¹, directed opposite to its initial motion (by Newton's third law, equal and opposite).

Step-by-step solution

1. 1The gun and shell are initially at rest: total initial momentum = 0.
2. 2Final momentum = m_{shell}v − MV (gun recoils opposite to the shell).
3. 3Conservation of momentum: 0 = 0.020 × 80 − 100 × V.
4. 4V = (0.020 × 80)/100 = 1.6/100 = 0.016 m s⁻¹.

Step-by-step solution

1. 1v = 54 km/h = 15 m s⁻¹; mass m = 0.15 kg.
2. 2The deflection is 45°, so each velocity makes θ = 22.5° with the angle bisector.
3. 3The momentum component along the bisector reverses; the perpendicular component is unchanged.
4. 4Impulse = change in momentum = 2mv cos θ = 2 × 0.15 × 15 × cos 22.5°.
5. 5= 4.5 × 0.9239 ≈ 4.16 kg m s⁻¹.

Step-by-step solution

1. 1n = 40 rev/min = 40/60 = 2/3 rev s⁻¹; angular speed ω = 2πn.
2. 2Tension provides the centripetal force: T = m r ω² = m r (2πn)².
3. 3T = 0.25 × 1.5 × (2π × 2/3)² = 0.25 × 1.5 × (4.19)² ≈ 6.57 N.
4. 4Maximum speed: T_max = m v_max²/r ⇒ v_max = √(T_max r/m) = √(200 × 1.5/0.25).
5. 5v_max = √1200 ≈ 34.64 m s⁻¹.

Step-by-step solution

1. 1At the instant the string breaks, the centripetal force disappears and no net force acts on the stone.
2. 2By Newton's first law, the stone continues moving in the direction of its velocity at that instant.
3. 3The velocity is always tangential to the circular path at the break point.
4. 4Hence the stone flies off tangentially: option (b).

Step-by-step solution

1. 1(a) To move forward the horse pushes the ground backward; the ground's reaction throws the horse (and cart) forward. Empty space gives no such reaction force, so the horse cannot move forwards — nothing to push against.
2. 2(b) When the bus stops suddenly, the lower body (in contact with the seat) stops, but the upper body continues forward by inertia (Newton's first law) — passengers are thrown forward.
3. 3(c) Pulling at an angle θ: the vertical component of the force, F sin θ, acts upward and reduces the effective weight (mg − F sin θ). Pushing: the vertical component acts downward, increasing the effective weight (mg + F sin θ). A smaller effective weight during pulling makes it easier.
4. 4(d) By F = ma = mΔv/Δt, the stopping force is inversely proportional to impact time. Moving the hands backward increases Δt, decreasing the average force on the hands and preventing injury.

Step-by-step solution

1. 1(a) Positive: the man pulls the bucket upward and it moves upward — force and displacement are in the same direction.
2. 2(b) Negative: gravity acts downward while the bucket moves upward — force and displacement are opposite.
3. 3(c) Negative: friction opposes the motion of the body sliding down.
4. 4(d) Positive: to keep a uniform velocity on a rough plane an applied force must act forward — it acts in the direction of motion. (Friction, of course, does negative work.)
5. 5(e) Negative: the resistive force of air opposes the direction of motion of the pendulum.

Step-by-step solution

1. 1m = 2 kg, F = 7 N, μ = 0.1, u = 0, t = 10 s.
2. 2Friction: f = μmg = 0.1 × 2 × 9.8 = 1.96 N (opposing motion).
3. 3Net force F_net = 7 − 1.96 = 5.04 N; acceleration a = 5.04/2 = 2.52 m s⁻².
4. 4Distance travelled: s = ½at² = ½ × 2.52 × (10)² = 126 m.
5. 5(a) W_applied = F·s = 7 × 126 = 882 J.
6. 6(b) W_friction = −f·s = −1.96 × 126 = −247 J.
7. 7(c) W_net = F_net·s = 5.04 × 126 = 635 J.
8. 8(d) v = at = 2.52 × 10 = 25.2 m s⁻¹; ΔKE = ½mv² − 0 = ½ × 2 × (25.2)² = 635 J.
9. 9Interpretation: the net work (635 J) equals the change in kinetic energy — the work-energy theorem. The applied work minus the frictional work gives the same 635 J.

Step-by-step solution

1. 1Since E = V + K and kinetic energy cannot be negative, the particle cannot exist wherever V exceeds E.
2. 2(a) V rises steeply and crosses E at x = a: the particle cannot be found for x > a. Minimum total energy = 0.
3. 3(b) V = V₀ is greater than E everywhere: the particle cannot be found in any region. Minimum total energy = V₀.
4. 4(c) The well falls to −V₁ between x = a and x = b; outside this region V > E. Forbidden: x < a and x > b. Minimum total energy = −V₁.
5. 5(d) The double well touches −V₁ at two minima. V exceeds E in the central hump between the two wells and outside the wells. Forbidden: the middle region and x < −b/2, x > b/2. Minimum total energy = −V₁.
6. 6Physical contexts: (a) a particle near a repulsive wall, (b) motion confined by a high barrier, (c) a simple potential well (an attached mass on a spring track), (d) a double well (a diatomic molecule's asymmetric stretch about a central maximum).

Step-by-step solution

1. 1At the turning points the particle momentarily stops: K = 0, so E = V.
2. 2E = ½kx² with k = 0.5 N m⁻¹ and E = 1 J.
3. 31 = ½ × 0.5 × x² = 0.25 x².
4. 4x² = 4, so x = ±2 m.
5. 5Beyond x = ±2 m, V would exceed E, which is impossible; hence the particle must turn back at ±2 m.

Step-by-step solution

1. 1(a) From the rocket. The burning reduces the rocket's own mass, and its total energy (mgh + ½mv²) drops by the amount burned.
2. 2(b) Gravitation is a conservative force: work over any closed path (a complete orbit) is zero.
3. 3(c) As the orbit shrinks the potential energy decreases; the total energy stays essentially constant, so the lost P.E. appears as kinetic energy — the speed rises. (The thin drag only slowly drains the total energy; its dominant effect near the earth is a steeper dive, not a slowing.)
4. 4(d) In case (i) the man's supporting force is vertical while the displacement is horizontal: W = mgs cos 90° = 0.
5. 5In case (ii) the rope is pulled in its own direction: W = mgs = 15 × 9.8 × 2 = 294 J.
6. 6So the work done is greater in the second case — pulling the rope.

Step-by-step solution

1. 1(a) Decreases: positive work by a conservative force moves the body along the force, reducing separation from the centre of force.
2. 2(b) Kinetic: friction acts against motion and the work done drains kinetic energy.
3. 3(c) External force: internal forces always cancel in pairs and cannot change total momentum.
4. 4(d) Total linear momentum: momentum is conserved in every collision, elastic or inelastic (kinetic energy is not).

Step-by-step solution

1. 1(a) False: in an elastic collision the total momentum and total energy of the two bodies are conserved, not necessarily the energy or momentum of each individual body.
2. 2(b) False: external forces can do work and change the energy of a system; energy is not conserved when unbalanced external forces act.
3. 3(c) False: closed-loop work is zero only for a conservative force (e.g. gravity, spring force), not for friction or other dissipative forces.
4. 4(d) True: inelastic collisions lose energy to heat, sound, deformation, so the final kinetic energy is always less than the initial.

Step-by-step solution

1. 1(a) No: during contact the balls deform and some kinetic energy is briefly stored as elastic potential energy; only the total (KE + PE) is momentarily conserved.
2. 2(b) Yes: no external horizontal force acts during the collision, so linear momentum is conserved throughout.
3. 3(c) For an inelastic collision: (a) becomes "total kinetic energy is not conserved" — some is permanently lost; (b) stays yes — momentum is always conserved.
4. 4(d) Elastic: a force that depends only on separation is conservative, so the collision stores and returns energy — the result is an elastic collision.

Step-by-step solution

1. 1With constant acceleration a, the force F = ma is constant.
2. 2The velocity grows as v = at, so v ∝ t.
3. 3Power P = F·v, with F constant and v ∝ t: P ∝ t.
4. 4Hence option (ii).

Step-by-step solution

1. 1P = F·v = mav = mv(dv/dt) = constant.
2. 2So v dv = (P/m) dt; integrating: v²/2 = (P/m)t, i.e. v = √(2Pt/m).
3. 3v = dx/dt ∝ t¹ᐟ².
4. 4Integrating again: x ∝ t³ᐟ².
5. 5Hence option (iii).

Step-by-step solution

1. 1Displacement s = 4 m along z: s = 4k̂ m.
2. 2W = F·s = (−î + 2ĵ + 3k̂)·(4k̂).
3. 3The î and ĵ components are perpendicular to s, so they contribute nothing: W = 3 × 4 = 12 J.

Step-by-step solution

1. 1E_Ke = 10 keV = 1.60 × 10⁻¹⁵ J; E_Kp = 100 keV = 1.60 × 10⁻¹⁴ J.
2. 2v_e = √(2E_Ke/m_e) = √(2 × 1.60 × 10⁻¹⁵ / 9.11 × 10⁻³¹) ≈ 5.93 × 10⁷ m s⁻¹.
3. 3v_p = √(2E_Kp/m_p) = √(2 × 1.60 × 10⁻¹⁴ / 1.67 × 10⁻²⁷) ≈ 4.38 × 10⁶ m s⁻¹.
4. 4The electron is faster.
5. 5v_e/v_p = (5.93 × 10⁷)/(4.38 × 10⁶) = 13.54.

Step-by-step solution

1. 1r = 2 × 10⁻³ m; volume V = (4/3)πr³; density of water ρ = 10³ kg m⁻³.
2. 2m = ρV = (4/3) × 3.14 × (2 × 10⁻³)³ × 10³ kg.
3. 3First half (h = 250 m): W₁ = mgh = 0.082 J.
4. 4Second half: gravitational force is the same and the fall is again 250 m, so W₂ = 0.082 J.
5. 5Without resistance, total energy left over at the ground would be mg × 500 = 0.164 J.
6. 6With resistance the drop lands at 10 m s⁻¹, so final kinetic energy = ½mv² = 1.675 × 10⁻³ J.
7. 7Work by the resistive force = E_ground − E_top = (−0.164) + 0.001675 ≈ −0.162 J.

Step-by-step solution

1. 1Momentum is conserved in every collision — here the (very massive) wall recoils a negligible amount while the molecule's momentum change is absorbed by wall + earth.
2. 2The molecule rebounds with the same speed 200 m s⁻¹.
3. 3Speed is unchanged, so kinetic energy is conserved: the collision is elastic.

Step-by-step solution

1. 1Mass of water m = ρV = 10³ × 30 = 3 × 10⁴ kg; t = 900 s; h = 40 m.
2. 2Useful (output) power P₀ = mgh/t = (3 × 10⁴ × 9.8 × 40)/900 = 13.07 × 10³ W.
3. 3Efficiency η = P₀/Pᵢ = 30%.
4. 4Pᵢ = P₀/0.30 = (13.07 × 10³)/0.30 ≈ 4.36 × 10⁴ W = 43.6 kW.

Step-by-step solution

1. 1Check momentum and kinetic energy for each proposed result. (Mass of each ball = m.)
2. 2Before: momentum = mV; KE = ½mV².
3. 3Result (i): the striking ball stops and the two move together with V/2. Momentum mV = (2m)(V/2) ✓ but KE = ½(2m)(V/2)² = ¼mV² ≠ ½mV² ✗ — possible result.
4. 4Result (ii): the striking ball stops; the balls separate, one moves off with speed V. Momentum mV ✓ and KE = ½mV² ✓.
5. 5Result (iii): all three move together with V/3. KE = ½(3m)(V/3)² = mV²/6 ≠ ½mV² ✗.
6. 6Only result (ii) conserves both momentum and kinetic energy — it is the possible elastic outcome.

Step-by-step solution

1. 1For an elastic head-on collision of two equal masses, one at rest: the moving mass comes to rest and the stationary mass moves off with the full incoming velocity.
2. 2Bob B (initially at rest) receives the entire velocity of A.
3. 3Bob A comes to rest at the point of collision, so it rises by zero height.

Step-by-step solution

1. 1Initial energy (horizontal position): E = mgl = m × 9.8 × 1.5.
2. 25% is dissipated, so the energy at the lowermost point is 95% of E.
3. 3½mv² = 0.95 × m × 9.8 × 1.5.
4. 4v² = 2 × 0.95 × 9.8 × 1.5 = 27.93.
5. 5v ≈ 5.28 m s⁻¹.

Step-by-step solution

1. 1The system (trolley + sandbag) is on a frictionless track, so no external horizontal force acts on it.
2. 2Sand falls vertically out of the hole, carrying no horizontal momentum with it — it separates with the same forward velocity as the trolley.
3. 3By Newton's first law / conservation of momentum, the trolley's horizontal velocity is unchanged.
4. 4Hence the speed remains 27 km/h even after the bag is empty.

Step-by-step solution

1. 1At x = 0: v = 0, so u = 0.
2. 2At x = 2 m: v = a x³ᐟ² = 5 × 2³ᐟ² = 5 × 2√2 = 10√2 m s⁻¹.
3. 3By the work-energy theorem: W = ΔKE = ½m(v² − u²).
4. 4W = ½ × 0.5 × (200 − 0) = 50 J.

Step-by-step solution

1. 1(a) Volume swept in time t: A v t; mass m = ρ A v t.
2. 2(b) Kinetic energy of that air = ½ m v² = ½ ρ A v³ t.
3. 3(c) v = 36 km/h = 10 m s⁻¹; electrical energy = 25% of wind kinetic energy = (¼) × ½ ρ A v³ t = (1/8) ρ A v³ t.
4. 4Electrical power = energy/time = (1/8) ρ A v³.
5. 5P = (1/8) × 1.2 × 30 × (10)³ = 4500 W = 4.5 kW.

Step-by-step solution

1. 1(a) W = n m g h = 1000 × 10 × 9.8 × 0.5 = 49,000 J = 49 kJ.
2. 2(b) Usable energy from 1 kg of fat = 20% × 3.8 × 10⁷ = 7.6 × 10⁶ J.
3. 3Fat consumed = 49,000 / 7.6 × 10⁶ ≈ 6.45 × 10⁻³ kg.

Step-by-step solution

1. 1(a) Useful power per square metre = 20% × 200 = 40 W m⁻².
2. 2Area needed A = 8000/40 = 200 m².
3. 3(b) 200 m² is roughly a 14 m × 14 m roof — comparable to (a little larger than) the roof of a typical house.

Step-by-step solution

1. 1For any body of uniform mass density and geometric symmetry, the centre of mass lies at its centre of symmetry.
2. 2(i) Sphere: at its centre.
3. 3(ii) Cylinder: at the midpoint of its axis.
4. 4(iii) Ring: at its geometric centre — which is inside the hole, not on the material.
5. 5(iv) Cube: at its geometric centre.
6. 6No — the centre of mass need not lie inside the body, as the ring shows (its CM is at its centre, in empty space).

Step-by-step solution

1. 1Take the H nucleus at x = 0 and the Cl nucleus at x = 1.27 Å.
2. 2Let m_H = m; then m_Cl = 35.5m.
3. 3x_CM = (m_H · 0 + m_Cl × 1.27)/(m_H + m_Cl) = 35.5m × 1.27/(36.5m).
4. 4x_CM = (35.5 × 1.27)/36.5 ≈ 1.235 Å from the hydrogen nucleus.
5. 5Equivalently, it is about 0.035 Å from the chlorine nucleus, which is why we say the CM lies very close to the chlorine atom.