ESAT Physics Mock 1

First, determine the total number of oscillations ( $\displaystyle N$ ) in the given time interval.In one complete oscillation, the point moves from equilibrium to the peak ( $\displaystyle A$ ), back to equilibrium ( $\displaystyle A$ ), to the trough ( $\displaystyle A$ ), and back to equilibrium ( $\displaystyle A$ ). The total distance travelled in one cycle is therefore $\displaystyle 4A$ .The total distance travelled is:

The question asks for the change in the reading on the force sensor. This change is caused by the magnetic force, $\displaystyle F_m$ , that acts on the wire when current flows. The weight of the wire is constant and is part of the initial reading, so it does not contribute to the change.

1. First, calculate the current $\displaystyle I$ flowing through the wire using the definition of current: $\displaystyle I = \dfrac{Q}{t}$ . $\displaystyle I = \dfrac{12 \text{ C}}{2.0 \text{ s}} = 6.0 \text{ A}$ .

2. Next, calculate the magnetic force $\displaystyle F_m$ on the current-carrying wire. The formula is $\displaystyle F_m = BIL\sin(\theta)$ . Since the field is perpendicular to the wire, $\displaystyle \theta = 90^\circ$ and $\displaystyle \sin(90^\circ) = 1$ . So, $\displaystyle F_m = BIL$ .

3. Convert the length to SI units: $\displaystyle L = 20 \text{ cm} = 0.20 \text{ m}$ .

4. Substitute the values into the force equation: $\displaystyle F_m = (0.50 \text{ T}) \times (6.0 \text{ A}) \times (0.20 \text{ m})$ . $\displaystyle F_m = (0.50 \times 0.20) \times 6.0 = 0.10 \times 6.0 = 0.60 \text{ N}$ .

This magnetic force is the change in the reading on the sensor.

1. Rearrange the relationship $\displaystyle E_k = \dfrac{p^2}{2m}$ to make mass the subject: $\displaystyle m = \dfrac{p^2}{2E_k}$ .
2. This reveals the proportional relationship $\displaystyle m \propto \dfrac{p^2}{E_k}$ .
3. We are given the scaling factors for momentum ( $\displaystyle p_{\text{X}} = 3p_{\text{Y}}$ ) and kinetic energy ( $\displaystyle E_{k\text{X}} = 4E_{k\text{Y}}$ ).
4. Substitute these factors into the proportional relationship: $\displaystyle \dfrac{m_{\text{X}}}{m_{\text{Y}}} = \dfrac{(3)^2}{4} = \dfrac{9}{4}$ .

First, determine the power law exponent $\displaystyle n$ .

Comparing the two data points:Since the intensity decreases by a factor of 4 when the distance doubles ( $\displaystyle 2^2 = 4$ ), the relationship is an inverse square law ( $\displaystyle I \propto r^{-2}$ or $\displaystyle n = -2$ ).

We can find the constant $\displaystyle k$ using the first point (or the second):Now calculate $\displaystyle I$ for $\displaystyle r = 3.0 \text{ m}$ :

Heating increases average particle kinetic energy and speed. More frequent and more forceful wall collisions increase momentum transferred per unit time, hence force and pressure. The fixed particle number and container volume rule out an increase in particle number or average spacing.

Initially $\displaystyle I=20000/2000=10\,\mathrm{A}$ , so the loss is $\displaystyle I^2R=500\,\mathrm{W}$ . Afterwards $\displaystyle I=20000/10000=2\,\mathrm{A}$ , so the loss is $\displaystyle 20\,\mathrm{W}$ . The reduction is $\displaystyle 500-20=480\,\mathrm{W}$ .

The circuit described is a Wheatstone bridge with five identical resistors, each of resistance $\displaystyle R$ . Let the two series resistors in the first branch be $\displaystyle R_1$ and $\displaystyle R_2$ , and in the second branch be $\displaystyle R_3$ and $\displaystyle R_4$ . The fifth resistor, $\displaystyle R_5$ , connects the junction between $\displaystyle R_1$ and $\displaystyle R_2$ to the junction between $\displaystyle R_3$ and $\displaystyle R_4$ .

Since all resistors are identical ( $\displaystyle R_1 = R_2 = R_3 = R_4 = R_5 = R$ ), we can check the balance condition for the Wheatstone bridge: $\displaystyle \dfrac{R_1}{R_2} = \dfrac{R}{R} = 1$ $\displaystyle \dfrac{R_3}{R_4} = \dfrac{R}{R} = 1$ Since the ratios are equal ( $\displaystyle 1=1$ ), the bridge is balanced. This means that the electric potential at the midpoint of the first branch is identical to the potential at the midpoint of the second branch. Consequently, there is no potential difference across the central connecting resistor ( $\displaystyle R_5$ ).

Because there is no potential difference across $\displaystyle R_5$ , no current flows through it. Therefore, $\displaystyle R_5$ dissipates no power and does not contribute to the overall equivalent resistance of the circuit. The total power dissipated by the circuit, $\displaystyle P$ , is determined by the equivalent resistance of the four outer resistors.

When the central resistor ( $\displaystyle R_5$ ) is removed, the circuit's equivalent resistance remains unchanged because it was not carrying any current. The circuit effectively consists of two parallel branches, each with a total resistance of $\displaystyle R+R=2R$ . The equivalent resistance of these two parallel branches is: $\displaystyle R_{eq} = \dfrac{(2R)(2R)}{2R+2R} = \dfrac{4R^2}{4R} = R$ The initial power dissipated was $\displaystyle P = \dfrac{V^2}{R_{eq}} = \dfrac{V^2}{R}$ .
After removing the central resistor, the equivalent resistance remains $\displaystyle R$ . Thus, the new total power dissipated will be $\displaystyle P' = \dfrac{V^2}{R} = P$ .

Therefore, the total power dissipated by the circuit remains the same.

First, we extract the amplitude ( $\displaystyle A$ ) and period ( $\displaystyle T$ ) from the graph.

The maximum displacement is $\displaystyle 0.5\ \text{m}$ , so the amplitude is $\displaystyle A = 0.5\ \text{m}$ .

The motion starts at maximum positive displacement, returns to maximum positive displacement at $\displaystyle t = 2.0\ \text{s}$ . So the period is $\displaystyle T = 2.0\ \text{s}$ .

Average speed is defined as total distance divided by total time. In one period ( $\displaystyle T$ ), the particle moves from its positive amplitude position to its negative amplitude position and back again. The distance covered is:
- from $\displaystyle x = +A$ to $\displaystyle x = 0$ (distance $\displaystyle A$ )
- from $\displaystyle x = 0$ to $\displaystyle x = -A$ (distance $\displaystyle A$ )
- from $\displaystyle x = -A$ to $\displaystyle x = 0$ (distance $\displaystyle A$ )
- from $\displaystyle x = 0$ to $\displaystyle x = +A$ (distance $\displaystyle A$ )

The total distance in one period is $\displaystyle 4A$ .

So, the distance travelled in one period is $\displaystyle 4 \times 0.5\ \text{m} = 2.0\ \text{m}$ .

The time interval is $\displaystyle 4.0\ \text{s}$ , which is exactly two periods ( $\displaystyle 2T$ ).

Total distance travelled in $\displaystyle 4.0\ \text{s}$ is $\displaystyle 2 \times (4A) = 2 \times 2.0\ \text{m} = 4.0\ \text{m}$ .

Total time taken is $\displaystyle 4.0\ \text{s}$ .

The thermal camera detects infrared. The frequency order is infrared, ultraviolet, X-rays. All three are electromagnetic waves and travel at the same speed in a vacuum, despite their different frequencies.

The ray must bend towards the normal on entering glass and away from the normal on leaving. For parallel faces with air on both sides, the emergent ray is parallel to the incident ray. Only R satisfies all these conditions.

Five crests span four wavelengths: $\displaystyle \lambda_d=24/4=6\,\mathrm{cm}$ . Thus $\displaystyle f=30/6=5\,\mathrm{Hz}$ . Frequency is unchanged by refraction, and $\displaystyle \lambda_s=16/4=4\,\mathrm{cm}$ , so $\displaystyle v_s=5\times4=20\,\mathrm{cm\,s^{-1}}$ .

The steam production rate is $\displaystyle 1.0\,g\,s^{-1}=1.0\times10^{-3}\,kg\,s^{-1}$ . The useful power used for vaporisation is $\displaystyle P=\dot m L_v=(1.0\times10^{-3})(2.0\times10^6)=2000\,W$ . The input is 2500 W, so the loss rate is $\displaystyle 2500-2000=500\,W$ . Therefore the correct option is B.

Let $\displaystyle N_\alpha$ be the number of alpha particles ( $\displaystyle ^{4}_{2}\text{He}$ ) and $\displaystyle N_\beta$ be the number of beta-minus particles ( $\displaystyle ^{0}_{-1}\text{e}$ ) emitted.

We can determine $\displaystyle N_\alpha$ and $\displaystyle N_\beta$ by conserving the mass number (superscript) and the atomic number (subscript).

First, for the mass number:This simplifies to $\displaystyle 24 = 4N_\alpha$ , which gives $\displaystyle N_\alpha = 6$ .

Next, for the atomic number:Substituting our value for $\displaystyle N_\alpha$ :Solving for $\displaystyle N_\beta$ gives $\displaystyle N_\beta = 4$ .

Therefore, 6 $\displaystyle \alpha$ particles and 4 $\displaystyle \beta^-$ particles are emitted.

1. The heated parcel consists of a fixed number of particles, so its mass is unchanged.
2. Since mass is unchanged and the gravitational field strength is constant, its weight ( $\displaystyle W = mg$ ) is also unchanged.
3. Heating causes the parcel to expand, meaning its volume increases.
4. Upthrust is equal to the weight of the surrounding (unheated) fluid displaced by the parcel. Because the parcel's volume has increased, it displaces a larger volume of the surrounding fluid, so the upthrust acting on it increases.
5. The increased upthrust now exceeds the unchanged weight, resulting in a net upward force that causes the parcel to rise.

In a vacuum, heat loss occurs only by radiation. A matte black surface is a better emitter than a polished silver surface, so $\displaystyle R_{\text{X}} > R_{\text{Y}}$ .

When argon is introduced, heat loss occurs by radiation and convection.

The rate of convection depends on the geometry of the object and the temperature difference, not the surface colour (emissivity). Since the spheres are identical in shape and temperature, the convective heat loss term ( $\displaystyle C$ ) is the same for both.

The new rates are:Since $\displaystyle R_{\text{X}} > R_{\text{Y}}$ and $\displaystyle C > 0$ , adding the same positive constant to the numerator and denominator of the ratio $\displaystyle \dfrac{R_{\text{X}}}{R_{\text{Y}}}$ brings the value closer to $\displaystyle 1$ . Therefore, the ratio decreases.

Example: If $\displaystyle R_{\text{X}}=10$ , $\displaystyle R_{\text{Y}}=2$ , and $\displaystyle C=6$ :

Original ratio $\displaystyle = 5$ .

New ratio $\displaystyle = \dfrac{16}{8} = 2$ .

The speed $\displaystyle v$ , frequency $\displaystyle f$ , and wavelength $\displaystyle \lambda$ of a wave are related by $\displaystyle v = f\lambda$ .

When a wave passes from one medium to another, its frequency $\displaystyle f$ remains constant (determined by the source). Therefore, the wavelength is directly proportional to the speed:Given that the new speed $\displaystyle v_Y$ is $\displaystyle \dfrac{2}{3}$ of the original speed $\displaystyle v_X$ :The fractional change is:

An $\displaystyle \alpha$ particle ( $\displaystyle ^{4}_{2}\text{He}$ ) reduces the mass number, $\displaystyle A$ , by 4 and the atomic number, $\displaystyle Z$ , by 2. A $\displaystyle \beta^-$ particle ( $\displaystyle ^{0}_{-1}\text{e}$ ) leaves $\displaystyle A$ unchanged and increases $\displaystyle Z$ by 1.

The change in mass number is solely due to $\displaystyle \alpha$ decay. The total change is $\displaystyle 235 - 207 = 28$ .
Since each $\displaystyle \alpha$ particle reduces $\displaystyle A$ by 4, the number of $\displaystyle \alpha$ particles is:Now we consider the atomic number, $\displaystyle Z$ . The overall change is from 92 to 82, a net decrease of 10.
The 7 $\displaystyle \alpha$ decays would cause a decrease in $\displaystyle Z$ of $\displaystyle 7 \times 2 = 14$ .
Let $\displaystyle n_\beta$ be the number of $\displaystyle \beta^-$ decays. The total change in $\displaystyle Z$ is the sum of the changes from both decay types:This gives $\displaystyle n_\beta = 4$ .
So, the decay involves 7 $\displaystyle \alpha$ particles and 4 $\displaystyle \beta^-$ particles.

The rate of temperature increase is determined by the net power supplied to the block. This is the constant power from the heater minus the power lost to the surroundings.
The rate of heat loss is proportional to the temperature difference, $\displaystyle \theta - 20^\circ\text{C}$ . As the block heats up, this difference increases, so the rate of heat loss increases.
This means the net power supplied to the block, and therefore its rate of temperature rise, decreases over time.

In the first 10 minutes, the temperature increased by $\displaystyle 50^\circ\text{C} - 20^\circ\text{C} = 30^\circ\text{C}$ .
Over the next 10 minutes, the average temperature is higher, so the average rate of heat loss is greater. The temperature increase in this second interval must therefore be less than the $\displaystyle 30^\circ\text{C}$ increase seen in the first.
The final temperature $\displaystyle \theta$ must be greater than $\displaystyle 50^\circ\text{C}$ but less than $\displaystyle 50^\circ\text{C} + 30^\circ\text{C} = 80^\circ\text{C}$ .

The relationship between net force $\displaystyle F$ , mass $\displaystyle m$ , and acceleration $\displaystyle a$ is given by Newton's Second Law, $\displaystyle F = ma$ .

The acceleration is constant because the velocity-time graph is a straight line. The value of the acceleration is the gradient of the graph.

Using the points $\displaystyle (0\,\text{s}, 2\,\text{m s}^{-1})$ and $\displaystyle (4\,\text{s}, 10\,\text{m s}^{-1})$ from the graph:Now, we can calculate the net force:

Reversal blocks the diode branch completely. The remaining conducting path contains $\displaystyle 4.0\,\Omega$ and $\displaystyle 12\,\Omega$ in series, so $\displaystyle R=16\,\Omega$ . Hence $\displaystyle I=12/16=0.75\,\mathrm{A}$ and $\displaystyle P=VI=9.0\,\mathrm{W}$ . The blocked branch is an open circuit, not a short circuit.

Anticlockwise current makes the viewed left end a north pole, so the right end is a south pole. Just to the right of a south pole, the field points left towards it. Reversing the current reverses the poles and the field, so the compass then points right.

Let the total current entering the cube be $\displaystyle I$ . By symmetry, this current must split equally among the three resistors connected to the input vertex. The current in one of these 'input' resistors is therefore $\displaystyle I_{\text{in}} = \dfrac{I}{3}$ .

Each of these currents then reaches a vertex (e.g., from input vertex A to vertex B). From such a vertex (B), the current $\displaystyle I/3$ splits into two equivalent paths leading to the next layer of vertices (e.g., to C and F). By symmetry, the current divides equally between them.

The current in a 'middle' resistor (e.g., BC or BF), which is not connected to the input or output vertex, is therefore half of the current in an 'input' resistor: $\displaystyle I_{\text{mid}} = \dfrac{1}{2} I_{\text{in}} = \dfrac{1}{2} \left(\dfrac{I}{3}\right) = \dfrac{I}{6}$ .

The power dissipated in a resistor with resistance $\displaystyle R$ is given by $\displaystyle P = I^2 R$ . We want to find the ratio $\displaystyle \dfrac{P_{\text{in}}}{P_{\text{mid}}}$ .

Let $\displaystyle \rho_F$ be the density of the fluid. For the object to be neutrally buoyant, its average density, $\displaystyle \rho_{\text{avg}}$ , must be equal to the fluid density. So, $\displaystyle \rho_{\text{avg}} = \rho_F = 1100 \text{ kg m}^{-3}$ .

The total mass of the object is $\displaystyle M_{\text{total}} = \rho_{\text{avg}} \times V_{\text{total}} = 1100 \times (4.0 \times 10^{-3}) = 4.4 \text{ kg}$ .

Let $\displaystyle V_X$ and $\displaystyle V_Y$ be the volumes of materials X and Y, and $\displaystyle m_X$ and $\displaystyle m_Y$ be their masses. We have two simultaneous equations:
1. For total volume: $\displaystyle V_X + V_Y = V_{\text{total}} = 4.0 \times 10^{-3} \text{ m}^3$ .
2. For total mass: $\displaystyle m_X + m_Y = M_{\text{total}} = 4.4 \text{ kg}$ .

We can write the masses in terms of volumes and densities ( $\displaystyle m_X = \rho_X V_X$ and $\displaystyle m_Y = \rho_Y V_Y$ ):From the volume equation, $\displaystyle V_Y = 4.0 \times 10^{-3} - V_X$ . Substituting this into the mass equation:The question asks for the mass of material X, which is $\displaystyle m_X = \rho_X V_X$ .

The secondary voltages are $\displaystyle 12\,\mathrm{V}$ and $\displaystyle 24\,\mathrm{V}$ . The load powers are $\displaystyle 12^2/6=24\,\mathrm{W}$ and $\displaystyle 24^2/24=24\,\mathrm{W}$ . The input supplies both: $\displaystyle I_p=(24+24)/120=0.40\,\mathrm{A}$ .

For a fixed mass of an ideal gas at constant temperature, the product of pressure and volume is constant (Boyle's Law). We write this relationship as:Letting the final pressure be $\displaystyle P_{\text{final}}$ , we substitute the initial values $\displaystyle P, V$ and the final volume $\displaystyle V/3$ :Rearranging to solve for $\displaystyle P_{\text{final}}$ :The pressure increases by a factor of 3.

The original transit time is $\displaystyle 0.12/(2.0\times10^8)=0.60\,\mathrm{ns}$ . The replacement gives $\displaystyle 0.12/(1.5\times10^8)=0.80\,\mathrm{ns}$ . Subtracting gives an extra delay of $\displaystyle 0.20\,\mathrm{ns}$ . No bending occurs at normal incidence, but speed still changes.

In a series circuit, the current is the same through all components. To find this current, we must determine the equivalent resistance of the entire circuit.

The total resistance $\displaystyle R_{\text{total}}$ is the sum of the individual resistances:Applying Ohm's Law to the whole circuit:This is the current flowing through the source and both resistors.