ESAT Paper 4 Physics Worked Solutions

Correct Answer: B. 4320 cm

Fastest Approach (🚀):
Distance per cycle $= 4A = 4(0.6) = 2.4\,\text{cm}$.
Cycles in 60 s $= f \times t = 30.0 \times 60 = 1800$.
Total distance $= 1800 \times 2.4 = 4320\,\text{cm}$.

Matches Option B.

Step-by-Step Breakdown:

• We are given a transverse wave with frequency $f = 30.0\text{ Hz}$, amplitude $A = 0.6\text{ cm}$, and time period $t = 60\text{ s}$.

Wave Speed vs. Particle Speed: It is crucial to differentiate between the speed at which the wave energy travels horizontally across the medium, and the speed at which individual particles oscillate up and down. We are asked for the distance traveled by a particle.
In one complete wave cycle, a single particle travels from equilibrium up to the crest, back down to equilibrium, down to the trough, and back up to equilibrium. This total vertical distance is exactly $4$ times the amplitude: $d_{\text{cycle}} = 4A = 4 \times 0.6\text{ cm} = 2.4\text{ cm}$.

• The total number of cycles the particle undergoes in 60 seconds is $N = f \times t = 30.0\text{ Hz} \times 60\text{ s} = 1800\text{ cycles}$.
• The total vertical distance travelled by the particle is $N \times d_{\text{cycle}} = 1800 \times 2.4\text{ cm} = 4320\text{ cm}$.
• Correct Answer: Option B.

Correct Answer: D. 2AS/F

Fastest Approach (🚀):
$v^2 - u^2 = 2AS$ factors as $(v-u)(v+u) = 2AS$.
Since $v - u = F$, substitute: $F(v+u) = 2AS \implies v+u = \frac{2AS}{F}$.

Matches Option D.

Step-by-Step Breakdown:

• We have constant acceleration $A$, distance $S$, and final/initial velocity difference $F = v - u$.
• Using the standard kinematic equation: $v^2 - u^2 = 2AS$.
• Calculus Derivation Bonus: For advanced students, recall that $A = \frac{dv}{dt}$. By the chain rule, $A = \frac{dv}{dS} \cdot \frac{dS}{dt} = v \frac{dv}{dS}$. Integrating $\int_u^v v\,dv = \int_0^S A\,dS$ yields $\frac{1}{2}v^2 - \frac{1}{2}u^2 = AS$, which rearranges to $v^2 - u^2 = 2AS$.
• We can factor the left side as a difference of squares: $(v - u)(v + u) = 2AS$.
• Substituting $F$ into the equation gives: $F(v + u) = 2AS$.
• Solving for the sum of the velocities ($v + u$) yields: $v + u = \frac{2AS}{F}$.
• Correct Answer: Option D.

Correct Answer: B. Kinetic energy is always conserved during a particle collision

Fastest Approach (🚀):
Check each statement: orbital work on Moon is zero (force is always perpendicular to velocity, TRUE); KE is conserved only in elastic collisions, not in general (FALSE — this is the incorrect one); pendulum tension is greatest at the lowest point (TRUE); velocity is perpendicular to centripetal acceleration in circular motion (TRUE); a standing wave is the sum of two counter-propagating travelling waves (TRUE).

Matches Option B.

Step-by-Step Breakdown:

• Let's evaluate each statement to find the incorrect ones.
• A. True. The Zero Dot Product: Work is the dot product of force and displacement ($W = \mathbf{F} \cdot \mathbf{d}$). In a circular orbit, gravity acts radially inward, while displacement is tangential. Because they are strictly perpendicular, their dot product is zero ($F d \cos(90^\circ) = 0$). Hence, no net work is done.

B. False. Kinetic energy is only conserved in perfectly elastic collisions. In inelastic collisions, kinetic energy is lost to heat, sound, or deformation.
C. True. The tension in a pendulum string is highest at the lowest vertical point because it must support the full weight of the bob plus provide the maximum centripetal force ($T = mg + mv^2/r$).
D. True. As established in A, velocity is tangential and acceleration (centripetal) is radial. They are perpendicular.
E. True. A standing wave is mathematically the superposition (sum) of two identical traveling waves moving in opposite directions.
Incorrect statement: Option B.
Correct Answer: Option B.

Correct Answer: A. 5

Fastest Approach (🚀):
Power in the fixed $20.0\,\Omega$ resistor is $P=I^2R$ with $R$ constant, so maximise current by minimising total resistance: set the variable resistor to its minimum, $4.0\,\Omega$.
$I = \frac{12}{20.0+4.0} = 0.5\,\text{A} \implies P = (0.5)^2 \times 20.0 = 5\,\text{W}$.

Matches Option A.

Step-by-Step Breakdown:

• A circuit has a $20.0\Omega$ fixed resistor and a variable resistor ($4.0\Omega$ to $35\Omega$) in series with a 12 V battery. We want the maximum power dissipated in the $20.0\Omega$ resistor.
• Internal Resistance & Maximum Power Transfer: Normally, to maximize power in a variable load, we match it to the fixed internal resistance (Maximum Power Transfer Theorem). However, here we are asked to maximize power in the fixed $20.0\Omega$ resistor, not the variable one!
• Power in the fixed resistor is $P = I^2 R$. Since $R = 20.0\Omega$ is constant, we must maximize the current $I$.
• Current $I = \frac{V}{R_{\text{total}}} = \frac{12}{20.0 + R_{\text{variable}}}$.
• To maximize $I$, we must minimize the denominator by setting $R_{\text{variable}}$ to its minimum value, which is $4.0\Omega$.
• Max Current $I = \frac{12}{20.0 + 4.0} = \frac{12}{24} = 0.5\text{ A}$.
• Maximum Power $P = (0.5)^2 \times 20.0 = 0.25 \times 20.0 = 5\text{ W}$.
• Correct Answer: Option A.

Correct Answer: C. R1=10 Ω, R2=50 Ω, R3=50 Ω, R4=10 Ω

Step-by-Step Breakdown:
We want the bulb to shine brightest, which means we must maximize the current flowing through it. The bulb is in series with the main battery, so we need the lowest possible total circuit resistance.
Mathematical Limit for Parallel Resistors: The parallel block consists of three branches. $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2+R_3} + \frac{1}{R_4}$. A fundamental property of parallel circuits is that the equivalent resistance $R_p$ is strictly less than the smallest individual branch resistance. Therefore, to drive $R_p$ down, we must assign our lowest available resistance ($10\Omega$) to the single-resistor branches ($R_1$ and $R_4$).

• We assign the largest resistance ($50\Omega$ or $100\Omega$) to the series branch ($R_2+R_3$) to isolate it.
• Looking at Option C ($R_1=10$, $R_2=50$, $R_3=50$, $R_4=10$): $\frac{1}{R_p} = \frac{1}{10} + \frac{1}{100} + \frac{1}{10} = 0.21\Omega^{-1} \implies R_p \approx 4.76\Omega$.
• Compare with Option B ($R_1=10$, $R_2=100$, $R_3=10$, $R_4=10$): $\frac{1}{R_p} = \frac{1}{10} + \frac{1}{110} + \frac{1}{10} \implies R_p \approx 4.78\Omega$ — very close, but not quite as low.

Option C yields the lowest total resistance, maximizing the current and bulb brightness.
Correct Answer: Option C.

Correct Answer: B. a+c < e+g

Fastest Approach (🚀):
The middle step ($^{c}_{d}\text{Y} \to {}^{e}_{f}\text{Y}$) keeps the same element, so it must be gamma decay: $c=e$ and $d=f$. So Option A ($d=f$) is certainly TRUE, not the answer.
Substituting $c=e$ into Option B: $a+c<e+g \implies a<g$. But mass number never increases (alpha: $-4$; beta/gamma: unchanged), so $g \le a$ always. Therefore $a<g$ is NEVER true — Option B is certainly INCORRECT.

Matches Option B.

Step-by-Step Breakdown:

• We have the decay chain: $^{a}_{b}\text{X} \to ^{c}_{d}\text{Y} \to ^{e}_{f}\text{Y} \to ^{g}_{h}\text{Z}$.
• Decay rules:
• Alpha decay ($\alpha$): mass number $-4$, atomic number $-2$.
• Beta-minus decay ($\beta^-$): mass number unchanged, atomic number $+1$.
• Gamma decay ($\gamma$): mass number and atomic number both unchanged.
• Step 1 — the middle transition is forced to be gamma. The element in the middle step goes from Y to Y. The only decay type that leaves the atomic number (and therefore the element) unchanged is gamma decay. So $^{c}_{d}\text{Y} \to {}^{e}_{f}\text{Y}$ must be pure gamma: $c=e$ and $d=f$.
• Step 2 — evaluate Option A. Since $d=f$ is forced by the argument above, Option A ('$d=f$') is CERTAINLY TRUE. It cannot be the 'certainly incorrect' statement.
• Step 3 — evaluate Option B. $a+c<e+g$. Substituting $c=e$ (from Step 1) gives $a<g$. But across the whole chain, mass number only ever decreases (via alpha emissions) or stays the same (via beta or gamma) — it can never increase. So $g \le a$ is always true, which means $a<g$ is NEVER true, for any valid decay sequence. Option B is CERTAINLY INCORRECT — regardless of how the beta decay (required by the question, since electrons must be emitted at least once) is distributed between the first and third steps.
• Step 4 — check the remaining options aren't also 'certainly incorrect'. Option C ($g=a$) is true only if no alpha decay occurs anywhere in the chain — possible but not forced, so it's not 'certainly' either way. Options D and E depend on how the alpha and beta decays split between the first and third steps, so they are also not certain in either direction.
• Conclusion: only Option B is guaranteed false for every decay sequence consistent with the question. Correct Answer: Option B.

Why the Other Options Are Wrong (❌):

• A. d=f — Sign/Direction Error
  d=f is actually forced to be TRUE by the requirement that the middle Y-to-Y step must be gamma decay (the only decay type preserving atomic number). Since it's certainly correct, it cannot be the 'certainly incorrect' statement.
• C. g=a — Overgeneralization
  g=a is true only in the special case where no alpha decay occurs anywhere in the chain. Since that's a possible (not forced) scenario, this statement isn't certainly true OR certainly false — so it doesn't fit 'certainly incorrect' either.
• D. h+f < b+d — Insufficient Constraint Check
  After substituting f=d, this reduces to h<b, whose truth depends on how the alpha/beta decays are split between the first and third steps — it isn't fixed by the given information.
• E. h<d — Insufficient Constraint Check
  h<d depends only on the alpha/beta split within the third step alone, which also isn't fixed by the given information.

Common Mistake (⚠️):
Concluding that 'd=f' (Option A) must be the incorrect statement simply because the question asks for an incorrect statement and A is the first option discussing the forced gamma step. In fact the gamma-decay argument PROVES d=f is true, which should rule A out rather than select it. The genuinely unprovable-as-true statement (B, since mass number can never increase) is easy to miss because it requires substituting c=e first.

Takeaway (📌):
In a nuclide decay chain, mass number is non-increasing under alpha, beta, and gamma decay — it only ever goes down (via alpha) or stays flat. Any statement implying a mass number increase across a decay chain is automatically false, regardless of the specific unknowns involved.

Correct Answer: B. 80/9

Fastest Approach (🚀):
Constant braking power $\implies$ kinetic energy $E_k$ falls linearly with time (not velocity).
$E_0 = \frac12 m(15)^2 = 112.5m$, reaching 0 at $t=10\,\text{s} \implies P = 11.25m\,\text{W}$.
At $v=5$: $E_k = \frac12 m(5)^2 = 12.5m$. Solve $12.5m = 112.5m - 11.25m\,t \implies t = \frac{80}{9}\,\text{s}$.

Matches Option B.

Step-by-Step Breakdown:

• A car moving at 15 m/s brakes to a stop over 10 seconds, dissipating energy at a constant rate. We want to know when velocity $< 5$ m/s.
• Velocity-Time Graph Alternative: Instead of purely algebraic energy formulas, let's visualize this. The car's kinetic energy is $E_k = \frac{1}{2}mv^2$. The rate of energy dissipation (power $P$) is constant.
• Therefore, the kinetic energy decreases linearly with time: $E_k(t) = E_{\text{initial}} - P \cdot t$.
• Because $E_k \propto v^2$, this means $v^2$ decreases linearly with time. The velocity-time graph is a curve (a square root function), not a straight line!
• Initial energy $E_0 = \frac{1}{2}m(15)^2 = 112.5m$. Final energy is $0$ at $t=10$.
• The constant power $P = \frac{112.5m}{10} = 11.25m$ Joules per second.
• We want the time $t$ when $v = 5$. At $v=5$, $E_k = \frac{1}{2}m(5)^2 = 12.5m$.
• Set up the linear energy equation: $12.5m = 112.5m - 11.25m \cdot t$.
• Cancel $m$: $12.5 = 112.5 - 11.25t \implies 11.25t = 100 \implies t = \frac{100}{11.25} = \frac{400}{45} = \frac{80}{9}$ seconds.
• Correct Answer: Option B.

Correct Answer: D. 1, 2, and 3

Step-by-Step Breakdown:
A ball freefalls and bounces. We must identify what kinematics variables the three graphs could represent.
Defining Axis Variables:

• The ball is under constant acceleration $g$ (downwards) while in the air. Upon impact, it experiences massive, brief upward acceleration.
• Graph 1 (Square Wave): This represents constant values that abruptly flip. This perfectly describes Acceleration. It is $-g$ while falling, and spikes to a huge positive value during the bounce.
• Graph 3 (Sawtooth Wave): This represents a value that changes linearly over time, then abruptly resets. This perfectly describes Velocity. It linearly decreases (becomes more negative) at rate $-g$, then instantly jumps to a positive value upon bouncing.
• Graph 2 (Parabolic Humps): This represents a quadratic relationship with time. This perfectly describes Position (Displacement/Height). The height follows $y = y_0 + v_0t - \frac{1}{2}gt^2$, forming downward parabolas.

Because the question asks which graphs could* describe the behaviour, and we've successfully mapped all three to standard kinematic variables, all three are valid.

• Correct Answer: Option D.

Correct Answer: D. 2.5 × 10²⁹ W

Fastest Approach (🚀):
$P \propto R^2T^4$. Radius $\times 100$, temperature $\times \frac12$: multiplier $= 100^2 \times (\frac12)^4 = \frac{10000}{16} = 625$.
$P_{\text{new}} = 625 \times 4.0\times10^{26} = 2.5\times10^{29}\,\text{W}$.

Matches Option D.

Step-by-Step Breakdown:

• The power radiated by a star is $P = kR^2T^4$.
• Stefan-Boltzmann Law: This equation is an explicit formulation of the Stefan-Boltzmann Law, which states that the luminosity (power) of a black body is $L = 4\pi R^2 \sigma T^4$. The constant $k$ encapsulates $4\pi\sigma$.
• Initial power $P_0 = 4.0 \times 10^{26}$ W.
• The radius increases by a factor of $100$ ($R_{\text{new}} = 100R$). The temperature decreases by a factor of 2 ($T_{\text{new}} = T/2$).
• Substitute these into the power equation: $P_{\text{new}} = k(100R)^2(T/2)^4$.
• Expanding the terms: $P_{\text{new}} = k(10000 R^2)(\frac{T^4}{16}) = \frac{10000}{16} (kR^2T^4)$.
• Therefore, the new power is $\frac{10000}{16}$ times the original power.
• $P_{\text{new}} = 625 \times (4.0 \times 10^{26}) = 2500 \times 10^{26} = 2.5 \times 10^{29}$ W.
• Correct Answer: Option D.

Correct Answer: A. The change in momentum of a molecule over the collision is zero

Fastest Approach (🚀):
Elastic collision reverses the molecule's velocity: $u_i \to -u_i$.
Momentum is a vector: change $= -mu_i - mu_i = -2mu_i \ne 0$, so 'change in momentum is zero' is FALSE.
Kinetic energy is a scalar depending on speed$^2$, so it's unchanged (TRUE), and the other statements about momentum's magnitude are also TRUE.

Matches Option A (the false statement).

Step-by-Step Breakdown:

• Gas molecules of mass $m$ and velocity $u_i$ collide elastically at right angles with a container wall.
• Kinetic Theory of Gases: This scenario is the foundational derivation of pressure in the kinetic theory of gases. The particle strikes the wall at $u_i$ and rebounds at $-u_i$ (elastic collision means no kinetic energy is lost).

A. The change in momentum is not* zero. Momentum is a vector. Initial momentum is $mu_i$, final is $-mu_i$. Change = $-mu_i - mu_i = -2mu_i$. Statement A is incorrect.

• B. True by Newton's Third Law (action-reaction).
• C. True. Kinetic energy is a scalar: $\frac{1}{2}m(u_i)^2$ before and $\frac{1}{2}m(-u_i)^2 = \frac{1}{2}m(u_i)^2$ after. Change is zero.
• D. True. As calculated in A, the magnitude of the change is $|-2mu_i| = 2mu_i$.
• Correct Answer: Option A.

Correct Answer: H. None of the above

Fastest Approach (🚀):
$g$ has units $\text{N/kg} = \text{m/s}^2$. Test each option by substituting base units:
(1) $\text{J}^2/(\text{N}^2\text{s})$ does not reduce to $\text{m/s}^2$.
(2) $\text{Nm/kg} = \frac{(\text{kg m/s}^2)\text{m}}{\text{kg}} = \text{m}^2/\text{s}^2$ (velocity$^2$, not acceleration).
(3) $\text{W}/(\text{kg m}) = \frac{\text{kg m}^2/\text{s}^3}{\text{kg m}} = \text{m/s}^3$ (jerk, not acceleration).

None match $g$'s units. Matches Option H.

Step-by-Step Breakdown:

• We need to identify units equivalent to gravitational field strength $g$.
• Step-by-Step Dimensional Analysis: The standard SI unit for $g$ is Newtons per kilogram (N/kg) or meters per second squared (m/s$^2$).
• 1. J$^2$/(n$^2$s): Joules are $kg \cdot m^2/s^2$. This becomes incredibly complex and does not reduce to m/s$^2$.
• 2. Nm/kg: A Newton is $kg \cdot m/s^2$. Substituting this in: $\frac{(kg \cdot m/s^2) \cdot m}{kg} = m^2/s^2$. This is velocity squared, not acceleration!
• 3. W/(kg·m): A Watt is Joules per second ($kg \cdot m^2/s^3$). Substituting this in: $\frac{kg \cdot m^2/s^3}{kg \cdot m} = m/s^3$. This is jerk (rate of change of acceleration), not acceleration!
• Therefore, absolutely none of the provided options reduce to m/s$^2$ or N/kg.
• Correct Answer: Option H.

Correct Answer: D. 175/√24

Step-by-Step Breakdown:
A 2.8 m string connects A and B via a smooth pulley C, holding a 50 N weight.
Free Body Diagram at the Pulley: Consider the knot exactly at pulley C. Because the pulley is smooth, the tension $T$ is uniform throughout the entire string. The knot experiences three forces: Tension $T$ pulling diagonally up toward A, Tension $T$ pulling diagonally up toward B, and the 50 N weight pulling straight down.

• For vertical equilibrium, the upward components of tension must balance the 50 N weight: $T\cos(\theta) + T\cos(\phi) = 50$.
• For horizontal equilibrium, the horizontal pulls must cancel exactly: $T\sin(\theta) = T\sin(\phi)$, which implies $\theta = \phi$. The setup is perfectly symmetrical!
• Because it's symmetrical, the pulley C hangs exactly in the middle. The horizontal distance from A to the centerline is 1 m (half of 2 m). The string length on each side is 1.4 m (half of 2.8 m).
• We have a right triangle with hypotenuse 1.4 and horizontal leg 1.0. We need $\cos(\theta)$, which is adjacent (vertical) over hypotenuse.
• Vertical depth = $\sqrt{1.4^2 - 1.0^2} = \sqrt{1.96 - 1.0} = \sqrt{0.96}$.
• $\cos(\theta) = \frac{\sqrt{0.96}}{1.4}$.
• Substitute into vertical balance: $2T\cos(\theta) = 50 \implies 2T \left(\frac{\sqrt{0.96}}{1.4}\right) = 50 \implies T = \frac{50 \times 1.4}{2\sqrt{0.96}} = \frac{35}{\sqrt{0.96}}$.
• We can multiply numerator and denominator by 5 to match the options: $\frac{175}{5\sqrt{0.96}} = \frac{175}{\sqrt{25 \times 0.96}} = \frac{175}{\sqrt{24}}$.
• Correct Answer: Option D.

Correct Answer: B. γ increases so L decreases

Fastest Approach (🚀):
As $v \to c$, $v^2/c^2 \to 1$, so $\sqrt{1-v^2/c^2} \to 0$ and $\gamma = 1/\sqrt{1-v^2/c^2} \to \infty$.
Since $L = L'/\gamma$, dividing by an ever-larger $\gamma$ drives $L \to 0$.

So $\gamma$ increases and $L$ decreases. Matches Option B.

Step-by-Step Breakdown:

• We have the length contraction formula $L = L'/\gamma$, where $\gamma = 1/\sqrt{1 - v^2/c^2}$.

Physical Intuition for Length Contraction: In special relativity, an object moving at relativistic speeds appears contracted (shortened) along its direction of motion to a stationary observer. The faster it goes, the shorter it looks.
Let's track the math: as velocity $v$ approaches the speed of light $c$, the fraction $v^2/c^2$ approaches 1.

• The denominator of the Lorentz factor becomes $\sqrt{1 - 1} = 0$.
• Dividing by a number approaching 0 causes the Lorentz factor $\gamma$ to increase towards infinity.
• Looking at the length formula $L = L'/\gamma$, dividing the rest length $L'$ by an infinitely large $\gamma$ causes the measured length $L$ to decrease towards zero.
• Therefore, $\gamma$ increases and $L$ decreases.
• Correct Answer: Option B.

Correct Answer: F. 3 and 4

Fastest Approach (🚀):
Gradient of the graph: $\frac{10}{2.0} = 5$. Statement 3: $a = F/m = 100/20 = 5$, matches. Statement 4: gradient is force $F = 5\text{N}$, matches directly.

Matches Option F.

Step-by-Step Breakdown:

1. Gradient of the Graph

The graph is a straight line through the origin passing through (2.0, 10). The gradient is $m = \frac{\Delta y}{\Delta x} = \frac{10}{2.0} = 5$.

2. Testing Each Statement

The physical meaning of the gradient depends on which quantities are plotted. Testing each option for a constant gradient of 5:

1. $E_k$ vs $v$: $E_k = \frac{1}{2}mv^2$. Graphing $E_k$ against $v$ gives a parabola, not a straight line — ruled out regardless of gradient.
2. $E_p$ vs $h$: $E_p = mgh$. The gradient is $mg$. With $m=20$ and $g=10$, gradient = 200, not 5.
3. $v$ vs $t$: $v = at$. The gradient is acceleration $a$. By Newton's second law, $a = F/m = 100/20 = 5$. This matches the gradient of 5.
4. Work vs distance: $W = Fd$. The gradient is the force $F$. We are told the external force is 5N — this also matches the gradient of 5.

3. Conclusion

Statements 3 and 4 accurately describe the graph.

Matches Option F.

Common Mistake (⚠️):
Checking only whether a relationship is linear (rules out statement 1) without also checking that its gradient numerically matches the graph's gradient of 5 (which also rules out statement 2).

Takeaway (📌):
A straight-line graph through the origin only tells you the two plotted quantities are directly proportional — you still need to check both the shape (linear vs. not) and the numerical gradient against each candidate relationship.

Correct Answer: C. 2 only

Step-by-Step Breakdown:

• A Uranium-235 nucleus undergoes fission into two smaller nuclei. We must evaluate three proposed decay equations.

Explicit Conservation Laws: In any nuclear reaction, two fundamental properties must be strictly conserved: Baryon number (total mass number, the top number) and Charge (atomic number, the bottom number).
A neutron ($n$) is $^{1}_{0}n$, and a proton ($p$) is $^{1}_{1}p$. Uranium-235 is $^{235}_{92}\text{U}$.

• Diagram 1: $^{1}_{1}p + ^{235}_{92}\text{U} \to ^{144}_{56}\text{Ba} + ^{89}_{36}\text{Kr} + 2^{1}_{0}n$.
• Mass check: $1 + 235 = 144 + 89 + 2(1) \implies 236 = 235$. Mass is NOT conserved. Incorrect.
• Diagram 2: $^{1}_{0}n + ^{235}_{92}\text{U} \to ^{137}_{54}\text{Xe} + ^{96}_{38}\text{Sr} + 3^{1}_{0}n$.
• Mass check: $1 + 235 = 137 + 96 + 3(1) \implies 236 = 236$. (Conserved)
• Charge check: $0 + 92 = 54 + 38 + 3(0) \implies 92 = 92$. (Conserved). This is a valid fission path.
• Diagram 3: $^{1}_{0}n + ^{235}_{92}\text{U} \to ^{87}_{35}\text{Br} + ^{145}_{57}\text{La} + 3^{1}_{0}n$.
• Mass check: $1 + 235 = 87 + 145 + 3(1) \implies 236 = 235$. Mass is NOT conserved. Incorrect.

Only diagram 2 represents a valid process.
Correct Answer: Option C.

Correct Answer: F. 72 | 2.5

Step-by-Step Breakdown:

• A circuit has a 6.0 V battery, a resistor, and a heater in series. Total resistance is $15\Omega$. The heater transfers 180 J of energy in 3.0 minutes.
• Kirchhoff's Rules and Circuit Mapping: By Kirchhoff's Loop Rule, the sum of voltage drops equals the battery voltage, but because everything is in series, the current $I$ is constant everywhere. Let's find $I$ first.
• By Ohm's law for the total circuit, $I = V_{\text{total}}/R_{\text{total}} = 6.0 / 15 = 0.40\text{ A}$.
• We need the total charge $Q$ flowing through the heater. Charge is current times time: $Q = I \times t$.
• Convert time to seconds: $t = 3.0 \text{ minutes} = 180\text{ s}$.
• $Q = 0.40 \times 180 = 72\text{ Coulombs}$.
• We need the voltage across the heater. Electrical energy (Work) is given by $W = VQ$.
• Rearranging for voltage: $V = W/Q = 180\text{ J} / 72\text{ C} = 2.5\text{ V}$.
• Correct Answer: Option F.

Correct Answer: F. 4.0 g cm⁻³

Step-by-Step Breakdown:

• A 10 cm cubic block has a 5.0 cm square hole through it. It weighs 30 N. We need its density in $g/cm^3$.
• Standard SI Unit Conversion Strategy: To avoid order-of-magnitude errors with densities, start by calculating volume strictly in $cm^3$, then convert weight to mass in grams.
• The volume of the solid $10\ \text{cm} \times 10\ \text{cm} \times 10\ \text{cm}$ block is $1000\text{ cm}^3$.
• The volume of the hollow hole passing completely through it is $5.0\ \text{cm} \times 5.0\ \text{cm} \times 10\ \text{cm} = 250\text{ cm}^3$.
• The volume of the actual material is $1000 - 250 = 750\text{ cm}^3$.
• The weight is $W = 30\text{N}$. Using $W = mg$, the mass is $m = W/g = 30/10 = 3.0\text{ kg}$.
• Convert mass to grams: $3.0\text{ kg} = 3000\text{ g}$.
• Finally, calculate density: $\rho = \text{mass}/\text{volume} = 3000 / 750 = 4.0\text{ g/cm}^3$.
• Correct Answer: Option F.

Correct Answer: D. P and Q | S

Step-by-Step Breakdown:

• We have balls P, Q, R, S with varying surfaces (shiny/dull) and temperatures ($20^\circ$C/$40^\circ$C room, ball temps $30^\circ$C/$40^\circ$C).
• Convection Analysis: Convection is the transfer of heat through fluid (air) currents. A ball loses heat to convection if it is hotter than its surrounding fluid.
• Balls P and Q are at $30^\circ$C in a $20^\circ$C room. They are hotter than the room, so they lose energy by convection.
• Balls R and S are at $40^\circ$C in a $40^\circ$C room. They are in thermal equilibrium with the room, so net convection is zero.
• Emissivity and Absorptivity in Thermal Radiation: All objects emit thermal radiation based on their temperature and surface properties. The Stefan-Boltzmann law states radiated power $P = \epsilon \sigma A T^4$.
• "Greatest rate of emission" requires the highest temperature $T$ and the highest emissivity $\epsilon$.
• Balls R and S are the hottest ($40^\circ$C).
• A dull (matte black) surface is a much better emitter (and absorber) of radiation than a shiny (reflective) surface. Therefore, the dull ball S has a higher emissivity $\epsilon$ than the shiny ball R.
• Ball S emits at the greatest rate.
• Correct Answer: Option D.

Correct Answer: A. 90 | 70 | 3.0

Step-by-Step Breakdown:

• We have a velocity-time graph showing motion from $t=0$ to $t=30\text{ s}$: velocity starts at $8.0\text{ m/s}$, decreases linearly to $0$ at $t=20\text{ s}$, and continues decreasing linearly to $-2.0\text{ m/s}$ at $t=30\text{ s}$.

Scalar Distance vs. Vector Displacement: Distance is a scalar that sums up all ground covered, regardless of direction (area under the curve, all treated as positive). Displacement is a vector measuring net change in position from the start (area above the axis is positive, area below is negative).
The graph crosses the axis at $t=20\text{ s}$, giving two triangular regions.

• Area 1 (forward motion, $t=0$ to $20\text{ s}$): $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 20 \times 8.0 = 80\text{ m}$.
• Area 2 (backward motion, $t=20$ to $30\text{ s}$): $\frac{1}{2} \times 10 \times 2.0 = 10\text{ m}$.
• Total distance travelled (scalar sum of both areas) $= 80 + 10 = 90\text{ m}$.
• Distance from the starting position (net displacement) $= 80 - 10 = 70\text{ m}$.
• Average speed = Total Distance / Total Time $= 90 / 30 = 3.0\text{ m/s}$.
• Correct Answer: Option A.

Correct Answer: G. Y/(9X+Y) × 100

Fastest Approach (🚀):
Mass $=$ density $\times$ volume. $m_{\text{tin}} = Y(0.1V)$, $m_{\text{copper}} = X(0.9V)$.
Volume $V$ cancels in the ratio: $\%\text{tin} = \frac{0.1Y}{0.1Y+0.9X}\times100 = \frac{Y}{Y+9X}\times100$.

Matches Option G.

Step-by-Step Breakdown:

• Bronze contains 10% tin by volume ($V_{\text{tin}} = 0.1V$) and 90% copper ($V_{\text{copper}} = 0.9V$). We need the percentage of mass that is tin.
• Deriving the Mixture Formula: Density is $\rho = m/V$, so mass is $m = \rho V$.
• The mass of tin in the sample is $m_{\text{tin}} = Y \times (0.1V) = 0.1Y V$.
• The mass of copper in the sample is $m_{\text{copper}} = X \times (0.9V) = 0.9X V$.
• The total mass of the bronze sample is $m_{\text{total}} = m_{\text{tin}} + m_{\text{copper}} = 0.1Y V + 0.9X V$.

The percentage of mass that is tin is the ratio of tin mass to total mass, multiplied by 100.
Percentage = $\frac{0.1Y V}{0.1Y V + 0.9X V} \times 100$.

• Notice that the volume $V$ cancels out completely: $\frac{0.1Y}{0.1Y + 0.9X} \times 100$.
• Multiply numerator and denominator by 10 to remove decimals: $\frac{Y}{Y + 9X} \times 100$.
• Correct Answer: Option G.

Correct Answer: D. 234E/238

Step-by-Step Breakdown:

• A stationary U-238 decays into Th-234 and an alpha particle. Total kinetic energy is E. We want the kinetic energy of the alpha particle.
• Q-Value and Momentum Conservation: The total kinetic energy E is the "Q-value" of the decay. Because the initial U-238 was stationary, initial momentum was zero. By conservation of momentum, the two daughter particles must fly off in opposite directions with exactly equal and opposite momenta: $p_{Th} = p_{\alpha} = p$.
• Kinetic energy can be expressed in terms of momentum: $K = \frac{p^2}{2m}$.
• The total energy is $E = K_{Th} + K_{\alpha} = \frac{p^2}{2m_{Th}} + \frac{p^2}{2m_{\alpha}}$.
• The ratio of their kinetic energies is inversely proportional to their masses: $\frac{K_{\alpha}}{K_{Th}} = \frac{m_{Th}}{m_{\alpha}} = \frac{234}{4}$.

Therefore, the alpha particle gets a much larger share of the energy because it is much lighter.
The fraction of total energy $E$ given to the alpha particle is $\frac{m_{Th}}{m_{Th} + m_{\alpha}} = \frac{234}{234 + 4} = \frac{234}{238}$.

• $K_{\alpha} = \frac{234}{238} E$.
• Correct Answer: Option D.

Correct Answer: G. 10 Hz

Step-by-Step Breakdown:

• The loudspeaker is $48\ \text{m}$ from one building and $128 - 48 = 80\ \text{m}$ from the other. The speed of sound is $320\ \text{m/s}$.
• Echo 1: travels $2 \times 48 = 96\ \text{m}$, so $t_1 = \frac{96}{320} = 0.3\ \text{s}$.
• Echo 2: travels $2 \times 80 = 160\ \text{m}$, so $t_2 = \frac{160}{320} = 0.5\ \text{s}$.
• For each echo to return at the same instant as a later click, the click period $T$ must divide both $0.3\ \text{s}$ and $0.5\ \text{s}$ exactly.
• The largest such period is the greatest common divisor:

$$T = \gcd(0.3,\ 0.5) = 0.1\ \text{s}$$
At $T = 0.1\ \text{s}$ the first echo returns exactly 3 clicks later and the second exactly 5 clicks later, so both coincide with clicks.

• The corresponding frequency is

$$f = \frac{1}{T} = \frac{1}{0.1} = 10\ \text{Hz}$$

• Correct Answer: Option G.

Correct Answer: C. 9600 N

Step-by-Step Breakdown:

• A helicopter (4200 kg) carries a crate (800 kg) accelerating upward at 2.0 m/s$^2$. We want the tension in the rope.
• Newton's Second Law Application: To find internal forces like tension, isolate the specific object experiencing that force. We apply $F_{\text{net}} = ma$ strictly to the crate.
• Draw the free body diagram for the crate: Tension $T$ pulls up, gravity $mg$ pulls down.
• The net force equation is $F_{\text{net}} = T - mg$.
• By Newton's Second law, $F_{\text{net}} = ma$. So, $T - mg = ma$.
• Rearranging for Tension gives $T = mg + ma = m(g + a)$.
• Plug in the crate's values: $m = 800\ \text{kg}$, $g = 10\text{m/s}^2$, $a = 2.0\text{m/s}^2$.
• $T = 800(10 + 2.0) = 800(12) = 9600\text{ N}$.
• Correct Answer: Option C.

Correct Answer: A. 380 | 600

Step-by-Step Breakdown:

• A shopper pushes a 350 N trolley 15 m horizontally with a 50 N force at $37^\circ$ below horizontal.
• Trigonometric Force Breakdown: Force is a vector. We must split the 50 N diagonal push into horizontal and vertical components.
• Horizontal force: $F_x = 50 \cos(37^\circ) = 50(0.80) = 40\text{N}$.
• Vertical force (downward): $F_y = 50 \sin(37^\circ) = 50(0.60) = 30\text{N}$.

The total vertical force exerted by the surface (the Normal force) must support both the weight of the trolley and the downward push from the shopper.
Normal force = Weight + $F_y = 350 + 30 = 380\text{N}$.

• Work done by the pushing force is calculated strictly using the component of force in the direction of motion ($F_x$).
• Work $W = F_x \times d = 40\text{N} \times 15\text{m} = 600\text{J}$.
• Correct Answer: Option A.

Correct Answer: B. 2 only

Fastest Approach (🚀):
Horizontal balance: friction must equal P and act left, so friction can't be zero — Statement 2 true, 1 and 3 false.
Moment of P: height above the pivot is $d\sin(30°) = d/2$, so the moment is $Pd/2$, not $Pd$ — Statement 4 false.

Matches Option B.

Step-by-Step Breakdown:

1. Horizontal Equilibrium (Statements 1-3)

The cube is stationary, so all forces on it balance. The only horizontal forces acting are P (pushing right, per the diagram) and friction from the table. For horizontal equilibrium, friction must exactly cancel P — so friction cannot be zero (Statement 1 false), and it must act to the left, directly opposing P (Statement 2 true, Statement 3 false).

2. Moment of P About the Pivot (Statement 4)

P is applied at the vertex a distance $d$ from the pivot, along an edge tilted at 30° to the table. The moment of a force about a point is the force multiplied by the perpendicular distance from that point to the force's line of action. Since P is horizontal, that perpendicular distance is simply the vertical height of P's application point above the pivot — not the full edge length $d$, unless that edge were vertical.

That height is $d\sin(30°) = d/2$. So the moment of P about the pivot is $P \times \frac{d}{2}$ (clockwise, since P tends to rotate the cube further over the pivot) — not $P \times d$ as Statement 4 claims. Statement 4 is false.

3. Conclusion

Only Statement 2 is true.

Matches Option B.

Common Mistake (⚠️):
Treating the moment arm of P as the full edge length $d$, instead of only the vertical component of that edge above the pivot ($d\sin\theta$).

Takeaway (📌):
The moment of a horizontal force about a pivot is force × height of its application point above the pivot — never force × the length of whatever edge or rod it happens to act along, unless that edge is vertical.

Correct Answer: G. 3.0 | 1.0

Fastest Approach (🚀):
Find the height via $v^2=u^2+2as$: $2.0^2 = 8.0^2 - 2(10)s \implies s = 3.0\,\text{m}$.
By the time-symmetry of projectile motion, the object passes through this same height at the same speed (now directed downward) on the way down.
Time via $v=u+at$: $-2.0 = 8.0 - 10t \implies t = 1.0\,\text{s}$.

Matches Option G.

Step-by-Step Breakdown:

• An object is fired vertically at 8.0 m/s. We need its height when $v = 2.0$ m/s on the way up, and the time it takes to reach that height on the way down.
• Symmetry of Projectile Motion: Projectile motion under uniform gravity is perfectly symmetrical in time and velocity. The time to rise from $v=8$ to $v=0$ is exactly the time to fall from $v=0$ to $v=-8$.
• First, find the height using $v^2 = u^2 + 2as$.
• $2.0^2 = 8.0^2 + 2(-10)s \implies 4 = 64 - 20s \implies 20s = 60 \implies s = 3.0\text{m}$.
• Now, we want the time to reach $v = -2.0$ m/s (this is the velocity at 3.0 m on the way down, due to symmetry).
• Use $v = u + at$.
• $-2.0 = 8.0 - 10t \implies 10t = 10.0 \implies t = 1.0\text{s}$.
• Correct Answer: Option G.

Correct Answer: B. 0.40 m on right of pivot | 1100

Step-by-Step Breakdown:

• Take moments about the pivot at the centre of the plank. The woman sits $0.80\text{ m}$ to the left of the pivot and the boy sits $1.20\text{ m}$ to the right.
• Woman (left): $60 \times 10 \times 0.80 = 480\ \text{N m}$, tending to rotate the left side down (counter-clockwise).
• Boy (right): $35 \times 10 \times 1.20 = 420\ \text{N m}$, tending to rotate the right side down (clockwise).
• The woman's moment exceeds the boy's by $60\ \text{N m}$ counter-clockwise, so the plank's own weight ($15 \times 10 = 150\ \text{N}$) must supply $60\ \text{N m}$ clockwise to balance it — the same rotational sense as the boy, meaning the plank's centre of gravity lies on the right of the pivot.
• Distance of the plank's centre of gravity from the pivot:

$$d = \frac{60}{150} = 0.40\ \text{m (to the right)}$$

• Force on the pivot. The pivot supports the entire system, so it carries the total weight:

$$F = (15 + 35 + 60) \times 10 = 1100\ \text{N}$$

• Correct Answer: Option B ($0.40\ \text{m}$ to the right of the pivot, $1100\ \text{N}$).