ESAT Physics Mock 5

1 is correct: For a floating object in vertical equilibrium, the upward buoyant force equals the weight of the block: $\displaystyle U = ho g V_{\text{sub}} = ho g A h = mg$ . Rearranging gives $\displaystyle h = \dfrac{m}{\rho A}$ . Because the liquid expands upon heating, its density $\displaystyle ho$ decreases while $\displaystyle m$ and $\displaystyle A$ remain constant, so the submerged depth $\displaystyle h$ increases.

2 is incorrect: Since the block is floating in equilibrium both before and after heating, the buoyant force must balance the downward gravitational force ( $\displaystyle U = mg$ ). The mass of the block does not change, so the buoyant force remains constant.

3 is correct: The hydrostatic pressure due to the liquid at the bottom surface of the block is $\displaystyle p = ho g h$ . Substituting $\displaystyle h = \dfrac{m}{\rho A}$ gives $\displaystyle p = \rho g \left(\dfrac{m}{\rho A}\right) = \dfrac{mg}{A}$ . Since $\displaystyle m$ , $\displaystyle g$ , and $\displaystyle A$ are all constant, this pressure remains unchanged.

The problem asks for the ratio of the number of atoms of Q ( $\displaystyle N_Q$ ) to the number of atoms of P ( $\displaystyle N_P$ ) after a certain time.

1. First, determine the number of half-lives that have elapsed. The time elapsed is 80 years, and the half-life is 20 years.
Number of half-lives, $\displaystyle n = \dfrac{\text{time elapsed}}{\text{half-life}} = \dfrac{80}{20} = 4$ .

2. Let the initial number of atoms of P be $\displaystyle N_0$ . After $\displaystyle n$ half-lives, the number of remaining atoms of P is given by:For $\displaystyle n=4$ , the number of remaining P atoms is:3. Since the sample was initially pure P, every atom of P that decays becomes an atom of the stable isotope Q. The number of atoms of Q is the initial number of P atoms minus the number of P atoms remaining.4. The required ratio is $\displaystyle \dfrac{N_Q}{N_P}$ .Therefore, the ratio of the number of atoms of Q to the number of atoms of P is 15.

From the displacement–distance graph:
1. Amplitude $\displaystyle A = 3.0\text{ cm} = 0.030\text{ m}$ .
2. Wavelength $\displaystyle \lambda = 0.40\text{ m}$ .

The frequency of the wave is: $\displaystyle f = \dfrac{v}{\lambda} = \dfrac{12\text{ m s}^{-1}}{0.40\text{ m}} = 30\text{ Hz}$ In a time interval of $\displaystyle \Delta t = 0.50\text{ s}$ , the number of complete oscillations made by each particle is: $\displaystyle N = f \Delta t = 30\text{ s}^{-1} \times 0.50\text{ s} = 15$ During one complete oscillation, a particle moves from equilibrium to maximum displacement, through equilibrium to minimum displacement, and back to equilibrium, covering a total distance of: $\displaystyle d_{\text{cycle}} = 4A = 4 \times 3.0\text{ cm} = 12\text{ cm} = 0.12\text{ m}$ Thus, the total distance travelled by a particle in $\displaystyle 0.50\text{ s}$ is: $\displaystyle d_{\text{particle}} = 15 \times 0.12\text{ m} = 1.8\text{ m}$ The distance travelled by the wave in the same time interval is: $\displaystyle d_{\text{wave}} = v \Delta t = 12\text{ m s}^{-1} \times 0.50\text{ s} = 6.0\text{ m}$ Therefore, row E is correct.

We first calculate the total thermal energy required to raise the temperature of the block by $\displaystyle 20 \text{ K}$ :The effective heating power is the input power minus the rate of heat loss to the surroundings:The time taken is the total energy required divided by this net power:

At terminal velocity, the downward gravitational force ( $\displaystyle W$ ) is balanced by the upward drag force ( $\displaystyle F$ ).

The gravitational force is proportional to the mass, which scales with the volume of the sphere:The drag force is given as:Equating the proportionalities ( $\displaystyle W \propto F$ ):Dividing both sides by $\displaystyle r^2$ :Taking the square root gives the final relationship:

The principle of conservation of linear momentum applies. The system consists of the astronaut and the tool.

Initially, both are at rest, so the total initial momentum of the system is zero.After the tool is thrown, the total final momentum must also be zero. Let $\displaystyle m_a$ and $\displaystyle v_a$ be the mass and velocity of the astronaut, and $\displaystyle m_t$ and $\displaystyle v_t$ be the mass and velocity of the tool.By conservation of momentum, $\displaystyle p_{\text{initial}} = p_{\text{final}}$ :We are looking for the recoil speed of the astronaut, which is the magnitude of $\displaystyle v_a$ . Rearranging the equation for $\displaystyle v_a$ :Now, we substitute the given values: $\displaystyle m_a = 100\ \text{kg}$ , $\displaystyle m_t = 4.0\ \text{kg}$ , and $\displaystyle v_t = 5.0\ \text{m/s}$ . We can take the direction of the tool as positive.The negative sign indicates that the astronaut moves in the opposite direction to the tool. The question asks for the recoil speed, which is the magnitude of the velocity.

The speed is $\displaystyle |-0.20\ \text{m/s}| = 0.20\ \text{m/s}$ .

1. Calculate the current in the wire: $\displaystyle \Delta t = 2.5\text{ minutes} = 2.5 \times 60\text{ s} = 150\text{ s}$ $\displaystyle I = \dfrac{\Delta Q}{\Delta t} = \dfrac{45\text{ C}}{150\text{ s}} = 0.30\text{ A}$ 2. Calculate the magnetic force on the wire ( $\displaystyle L = 15\text{ cm} = 0.15\text{ m}$ ): $\displaystyle F = B I L = 0.40\text{ T} \times 0.30\text{ A} \times 0.15\text{ m} = 0.018\text{ N}$ 3. Determine the force on the balance using Newton's third law:
The magnetic force on the wire acts vertically upwards. By Newton's third law, the wire exerts an equal and opposite downward force of $\displaystyle 0.018\text{ N}$ on the magnet.

4. Calculate the change in the balance reading: $\displaystyle \Delta m = \dfrac{F}{g} = \dfrac{0.018\text{ N}}{10\text{ m s}^{-2}} = 0.0018\text{ kg} = 1.8\text{ g}$ Since the additional force on the magnet is downwards, the balance reading increases by $\displaystyle 1.8\text{ g}$ .

1. When the switch is open:
- The lower branch is disconnected, so the circuit consists of the main-line resistor in series with the top-branch resistor and the ammeter.
- Total resistance: $\displaystyle R_{\text{open}} = R + R = 2R$ .
- Using the power formula $\displaystyle P = V I$ : $\displaystyle V = \dfrac{P_{\text{open}}}{I_{\text{open}}} = \dfrac{18\text{ W}}{3.0\text{ A}} = 6.0\text{ V}$ - Using Ohm's law: $\displaystyle 2R = \dfrac{V}{I_{\text{open}}} = \dfrac{6.0\text{ V}}{3.0\text{ A}} = 2.0\ \Omega \implies R = 1.0\ \Omega$ 2. When the switch is closed:
- The two parallel branches each have a resistance of $\displaystyle R = 1.0\ \Omega$ .
- Their combined equivalent resistance is: $\displaystyle R_p = \dfrac{1.0 \times 1.0}{1.0 + 1.0} = 0.5\ \Omega$ - The total resistance of the circuit becomes: $\displaystyle R_{\text{closed}} = R + R_p = 1.0 + 0.5 = 1.5\ \Omega$ - The total current drawn from the battery is: $\displaystyle I_{\text{total}} = \dfrac{V}{R_{\text{closed}}} = \dfrac{6.0\text{ V}}{1.5\ \Omega} = 4.0\text{ A}$ - The new power transferred from the battery is: $\displaystyle P_{\text{new}} = V I_{\text{total}} = 6.0\text{ V} \times 4.0\text{ A} = 24\text{ W}$ - The total current of $\displaystyle 4.0\text{ A}$ splits equally between the two identical parallel branches, so the current through the ammeter is: $\displaystyle I_{\text{ammeter}} = \dfrac{4.0\text{ A}}{2} = 2.0\text{ A}$ Therefore, the new ammeter reading is $\displaystyle 2.0\text{ A}$ and the new power transferred is $\displaystyle 24\text{ W}$ (Option D).

From the graph, $\displaystyle y$ is directly proportional to $\displaystyle x$ (a straight line passing through the origin $\displaystyle (0, 0)$ ). The gradient of the line is: $\displaystyle \text{gradient} = \dfrac{12.0 - 0}{2.0 - 0} = 6.0\text{ SI units}$ Let us analyse each statement:

1. Momentum $\displaystyle p = m v$ . If $\displaystyle y = p$ and $\displaystyle x = v$ , then $\displaystyle y = 6.0 x$ . This is a straight line through the origin with gradient $\displaystyle 6.0\text{ kg}$ . This matches the graph. (Statement 1 is correct)

2. Kinetic energy $\displaystyle E_k = \dfrac{1}{2} m v^2 = \dfrac{1}{2}(12) v^2 = 6.0 v^2$ . A plot of $\displaystyle E_k$ against $\displaystyle v$ is a parabola ( $\displaystyle y = 6.0 x^2$ ), not a straight line. (Statement 2 is incorrect)

3. Electrical energy transferred $\displaystyle \Delta E = V \Delta Q$ . If $\displaystyle y = \Delta E$ and $\displaystyle x = \Delta Q$ , then $\displaystyle y = 6.0 x$ . This is a straight line through the origin with gradient $\displaystyle 6.0\text{ V} = 6.0\text{ J C}^{-1}$ . This matches the graph. (Statement 3 is correct)

4. For two identical springs of spring constant $\displaystyle k = 6.0\text{ N m}^{-1}$ connected in series, the effective spring constant is: $\displaystyle \dfrac{1}{k_{\text{eff}}} = \dfrac{1}{6.0} + \dfrac{1}{6.0} = \dfrac{2}{6.0} \implies k_{\text{eff}} = 3.0\text{ N m}^{-1}$ Tension is $\displaystyle T = k_{\text{eff}} x = 3.0 x$ . The gradient would be $\displaystyle 3.0\text{ N m}^{-1}$ , not $\displaystyle 6.0\text{ N m}^{-1}$ . (Statement 4 is incorrect)

Therefore, the graph can represent relationships 1 and 3.

Average speed is given by: $\displaystyle \text{average speed} = \dfrac{\text{total distance}}{\text{total time}}$ The total time for the journey is: $\displaystyle T = \dfrac{800\text{ m}}{16\text{ m s}^{-1}} = 50\text{ s}$ The time taken for the first segment of distance $\displaystyle d$ is: $\displaystyle t_1 = \dfrac{d}{10}$ The time taken for the remainder of the journey of distance $\displaystyle (800 - d)$ is: $\displaystyle t_2 = \dfrac{800 - d}{25}$ Since $\displaystyle t_1 + t_2 = T$ : $\displaystyle \dfrac{d}{10} + \dfrac{800 - d}{25} = 50$ Multiplying the entire equation by $\displaystyle 50$ : $\displaystyle 5d + 2(800 - d) = 2500$ $\displaystyle 5d + 1600 - 2d = 2500$ $\displaystyle 3d = 900$ $\displaystyle d = 300\text{ m}$

At constant temperature, Boyle's Law states that for a fixed mass of gas, pressure $\displaystyle P$ is inversely proportional to volume $\displaystyle V$ , so $\displaystyle PV = k$ (a constant). This can be written as $\displaystyle 1/V = P/k$ .

The total pressure $\displaystyle P$ on the bubble at a depth $\displaystyle h$ is the sum of atmospheric pressure $\displaystyle P_\text{atm}$ and the hydrostatic pressure $\displaystyle \rho g h$ . So, $\displaystyle P = P_\text{atm} + \rho g h$ .

Substituting this into the Boyle's Law relation gives:This is an equation of a straight line $\displaystyle y = mx + c$ , where $\displaystyle y = 1/V$ and $\displaystyle x = h$ . The graph confirms this linear relationship.

From the graph, we can find the equation of the line. It passes through $\displaystyle (0, 2.0)$ and $\displaystyle (10, 4.0)$ .
The $\displaystyle y$ -intercept is $\displaystyle c = 2.0\,\text{units}$ .
The gradient is $\displaystyle m = \dfrac{\Delta(1/V)}{\Delta h} = \dfrac{4.0 - 2.0}{10 - 0} = \dfrac{2.0}{10} = 0.2\,\text{units/m}$ .

The equation of the line is $\displaystyle 1/V = 0.2h + 2.0$ .

To find the atmospheric pressure, we can find the theoretical depth $\displaystyle h_\text{int}$ at which the total pressure would be zero. This corresponds to the $\displaystyle h$ -intercept of the graph (where $\displaystyle 1/V = 0$ ).

Setting $\displaystyle 1/V = 0$ : $\displaystyle 0 = 0.2h_\text{int} + 2.0$ $\displaystyle 0.2h_\text{int} = -2.0$ $\displaystyle h_\text{int} = -10\,\text{m}$ From the physical relationship $\displaystyle P = P_\text{atm} + \rho g h$ , if the total pressure $\displaystyle P$ were zero, we would have: $\displaystyle 0 = P_\text{atm} + \rho g h_\text{int}$ $\displaystyle P_\text{atm} = -\rho g h_\text{int}$ Substituting the values: $\displaystyle P_\text{atm} = -(1000\,\text{kg}\,\text{m}^{-3}) \times (10\,\text{N}\,\text{kg}^{-1}) \times (-10\,\text{m})$ $\displaystyle P_\text{atm} = 1000 \times 10 \times 10 \,\text{Pa} = 100\,000\,\text{Pa} = 1.0 \times 10^5\,\text{Pa}$ .

The fixed resistor has $\displaystyle 8.0\,\mathrm{V}$ , leaving $\displaystyle 4.0\,\mathrm{V}$ across the LDR. Their current is the same, so $\displaystyle R_{\rm LDR}/3.0=4.0/8.0$ , giving $\displaystyle 1.5\,\mathrm{k\Omega}$ . More light reduces the LDR resistance, increasing current and the fixed-resistor voltage.

The rate of heat loss depends on three mechanisms: conduction, convection, and radiation.

1. Conduction and Convection: Wrapping the beaker in mineral wool (E), placing it on a cork mat (D), and adding a lid (C) all serve to insulate the beaker or prevent air currents/evaporation, thereby reducing heat loss.
2. Radiation: The rate of thermal radiation emission depends on the emissivity of the surface. Shiny metallic surfaces have very low emissivity (they are poor emitters). Dull, black surfaces have high emissivity (they are good emitters).

Therefore, coating the shiny copper beaker with dull black paint (A) increases its emissivity, which increases the rate of heat loss by radiation. Polishing the beaker (B) would decrease emissivity and help retain heat. Thus, A is the counter-productive modification.

At $\displaystyle 2.0\,\mathrm{N}$ , the graph gives an extension of $\displaystyle 3.0\,\mathrm{cm}$ relative to the original length, so the length is $\displaystyle 13.0\,\mathrm{cm}$ . When the force reaches zero, $\displaystyle 2.0\,\mathrm{cm}$ of extension remains permanently.

The constant temperature means that average particle kinetic energy is unchanged. Energy is still supplied and is used to change the arrangement and separation of particles during melting, increasing their potential energy.

The solution is a chain of five logical steps:

1. Equilibrium Condition: For an object to be neutrally buoyant, its weight must be equal to the buoyant force acting on it. Let $\displaystyle W$ be weight and $\displaystyle F_B$ be buoyant force. So, $\displaystyle W_X = F_{BX}$ and $\displaystyle W_Y = F_{BY}$ .

2. Mass and Buoyant Force: The weight of each tube is $\displaystyle W = mg$ . Since we are given $\displaystyle m_X > m_Y$ , it follows that $\displaystyle W_X > W_Y$ . Therefore, for equilibrium, the buoyant force on X must be greater than the buoyant force on Y: $\displaystyle F_{BX} > F_{BY}$ .

3. Buoyant Force and Volume: The buoyant force is given by Archimedes' principle: $\displaystyle F_B = \rho V g$ , where $\displaystyle \rho$ is the density of the water and $\displaystyle V$ is the volume of displaced water. In this case, the displaced volume is the volume of the trapped air ( $\displaystyle V_{gas}$ ). Since $\displaystyle F_{BX} > F_{BY}$ , it must be that the volume of trapped air in X is greater than in Y: $\displaystyle V_X > V_Y$ .

4. Volume and Pressure (Boyle's Law): The temperature is constant, so Boyle's Law ( $\displaystyle P \propto 1/V$ or $\displaystyle PV = \text{constant}$ ) applies to the trapped air. Since both tubes initially trapped the same amount of air, the constant is the same for both. As $\displaystyle V_X > V_Y$ , the pressure of the gas in tube X must be lower than the pressure in tube Y: $\displaystyle P_X < P_Y$ .

5. Pressure and Depth: The total pressure on the trapped gas at depth $\displaystyle h$ is the sum of atmospheric pressure $\displaystyle P_{atm}$ and the hydrostatic pressure $\displaystyle \rho g h$ . So, $\displaystyle P = P_{atm} + \rho g h$ . Since $\displaystyle P_X < P_Y$ , we have $\displaystyle P_{atm} + \rho g h_X < P_{atm} + \rho g h_Y$ . This simplifies to $\displaystyle \rho g h_X < \rho g h_Y$ , which means $\displaystyle h_X < h_Y$ .

Therefore, the heavier tube X comes to rest at a shallower depth than the lighter tube Y.

The power dissipated by a resistor is given by $\displaystyle P = \dfrac{V^2}{R}$ .

Since the resistance $\displaystyle R$ is constant, the power is proportional to the square of the potential difference:The ratio of the new voltage to the old voltage is:Therefore, the new power $\displaystyle P_{\text{new}}$ is:

In motion 1 the rod cuts field lines so that opposite charges are separated towards its two ends. Moving along its own length does not produce end-to-end charge separation. Moving parallel to the field does not cut the field lines. Only motion 1 gives an end-to-end induced voltage.

The relationship between kinetic energy ( $\displaystyle KE$ ), momentum ( $\displaystyle p$ ), and mass ( $\displaystyle m$ ) is given by the formula:This equation can be rearranged to highlight the relationship between $\displaystyle KE$ and $\displaystyle p^2$ :The question states that a graph of $\displaystyle KE$ (on the y-axis) against $\displaystyle p^2$ (on the x-axis) is a straight line with gradient $\displaystyle G$ . The equation of a straight line through the origin is $\displaystyle y = Gx$ .

By comparing the two equations, we can identify the corresponding terms:
- $\displaystyle y$ corresponds to $\displaystyle KE$ .
- $\displaystyle x$ corresponds to $\displaystyle p^2$ .
- The gradient $\displaystyle G$ corresponds to the coefficient of $\displaystyle p^2$ .

Therefore, we can equate the gradient $\displaystyle G$ with the term in the brackets:The question asks for the mass $\displaystyle m$ in terms of $\displaystyle G$ . We need to rearrange this equation to make $\displaystyle m$ the subject:Thus, the correct option is A.

Beta radiation is sensitive to the relevant thickness changes. A long half-life avoids rapid changes in source output. A thinner sheet absorbs fewer beta particles, so the detected count rate increases.

The frequency $\displaystyle f$ of a wave is determined by the source and remains constant when the wave crosses a boundary between two media.

The relationship between wave speed $\displaystyle v$ , frequency $\displaystyle f$ , and wavelength $\displaystyle \lambda$ is given by:Since $\displaystyle f$ is constant, the wavelength is directly proportional to the speed ( $\displaystyle \lambda \propto v$ ).

We calculate the ratio of the speeds:Therefore, the wavelength in medium Q is:

Let the mass be $\displaystyle m$ and the initial speed be $\displaystyle u$ .

Case 1: No atmosphere (Vacuum)
Conservation of mechanical energy applies. The initial kinetic energy is converted entirely to gravitational potential energy:Case 2: With atmosphere
The initial kinetic energy is converted into gravitational potential energy and work done against the resistive force $\displaystyle F_R$ .
The resistive force is $\displaystyle F_R = \dfrac{1}{4}mg$ .
The work done against resistance acts over the *actual* distance travelled, $\displaystyle H$ :The energy balance equation is:Substituting $\displaystyle \dfrac{1}{2}mu^2 = mgH_0$ :

The negatively charged droplets repel one another. They deliver excess electrons to the panel, reducing its positive charge. Electrons must therefore leave through the connection to maintain the panel's positive charge.

Changing the approaching pole reverses the voltage, so the first stage is negative. A stationary magnet gives no changing field and hence zero voltage. Reversing its motion reverses the voltage again, giving positive.

The angles to the normal are $\displaystyle 90^\circ-35^\circ=55^\circ$ in X and $\displaystyle 90^\circ-60^\circ=30^\circ$ in Y. The ray bends towards the normal, so its speed decreases. Its frequency is unchanged at the boundary.

Let $\displaystyle V_X$ be the potential difference across the parallel combination of the two identical components, and let $\displaystyle I_X$ be the current passing through each component.

Because the two components are identical and connected in parallel, each has potential difference $\displaystyle V_X$ across it, and each carries current $\displaystyle I_X$ . The total current delivered by the power supply to the circuit is therefore: $\displaystyle I_{\text{total}} = 2 I_X$ By Kirchhoff's voltage law around the circuit loop: $\displaystyle E = V_X + V_R$ $\displaystyle 9.0 = V_X + I_{\text{total}} R$ $\displaystyle 9.0 = V_X + (2 I_X)(3.0)$ $\displaystyle V_X = 9.0 - 6.0 I_X$ We find the operating point $\displaystyle (V_X, I_X)$ that satisfies both this linear equation and the given $\displaystyle I-V$ characteristic curve:
- If $\displaystyle I_X = 1.0\text{ A}$ , then $\displaystyle V_X = 9.0 - 6.0(1.0) = 3.0\text{ V}$ .
- Reading the graph at $\displaystyle V = 3.0\text{ V}$ , the current is indeed $\displaystyle I = 1.0\text{ A}$ .

Thus, under operating conditions:
- Current through each component: $\displaystyle I_X = 1.0\text{ A}$ - Total circuit current: $\displaystyle I_{\text{total}} = 2.0\text{ A}$ The total power dissipated in the entire circuit is equal to the total power supplied by the battery: $\displaystyle P_{\text{total}} = E \times I_{\text{total}} = 9.0\text{ V} \times 2.0\text{ A} = 18\text{ W}$ Alternatively, sum the powers in each component:
- Power in resistor: $\displaystyle P_R = I_{\text{total}}^2 R = (2.0)^2 \times 3.0 = 12\text{ W}$ - Power in each component: $\displaystyle P_X = V_X I_X = 3.0\text{ V} \times 1.0\text{ A} = 3.0\text{ W}$ - Total power: $\displaystyle P_{\text{total}} = 12\text{ W} + 3.0\text{ W} + 3.0\text{ W} = 18\text{ W}$

The question requires careful application of the wave equation $\displaystyle v = f\lambda$ in the correct frame of reference.

1. The source emits sound at a frequency $\displaystyle f_s = 3300 \, \text{Hz}$ . This is the rate at which wave crests are produced, and it does not change.

2. The sound propagates through the air. The speed of sound relative to the air is $\displaystyle v_{air} = 330 \, \text{m/s}$ . The wavelength of the sound waves in the air is determined by the source frequency and this speed:3. The air itself (the medium) is moving from the source to the observer at $\displaystyle v_{wind} = 30 \, \text{m/s}$ . For the stationary observer on the ground, the effective speed of the sound waves is the sum of the speed of sound in air and the speed of the wind.4. The observer detects waves with wavelength $\displaystyle \lambda = 0.1 \, \text{m}$ arriving at this effective speed $\displaystyle v_{eff}$ . The observed frequency $\displaystyle f_{obs}$ is therefore: