9702_m23_qp_12.pdf_

Question

What represents a physical quantity?

A 3.0

B kilogram

C $7.0 \mathrm{~N}$

D $40 \%$

▶️Answer/Explanation

Ans:C

A physical quantity is typically represented by a numerical value accompanied by a unit that indicates what is being measured. In the options you’ve provided:

A) “3.0” is just a numerical value and doesn’t have a unit, so it doesn’t fully represent a physical quantity.

B) “kilogram” is a unit of mass, but it needs to be associated with a numerical value to fully represent a physical quantity.

C) $7.0 \, \mathrm{N}$ represents a physical quantity. It’s a force of 7.0 Newtons (N).

D) “40%” is a relative value, representing 40% of something, but it doesn’t fully represent a specific physical quantity without additional context.

So, option C ($7.0 \, \mathrm{N}$) is the one that best represents a physical quantity on its own.

Question

The relationship between the variables $D$ and $T$ is given by the equation
$$
\frac{1}{T}=\frac{b}{\sqrt{D}}+c
$$
where $b$ and $c$ are constants. The unit of $D$ is $\mathrm{m}^2$ and the unit of $T$ is s.

What are the units of $b$ and $c$ ?

▶️Answer/Explanation

Ans:B

Let’s analyze the given equation and determine the units of the constants $b$ and $c$:

The equation is:
\[
\frac{1}{T} = \frac{b}{\sqrt{D}} + c
\]

Given that the unit of $D$ is $\mathrm{m}^2$ and the unit of $T$ is seconds ($s$), we can substitute the units into the equation to find the units of $b$ and $c$:

For the left side of the equation:
\[
\text{Unit of }\frac{1}{T} = \frac{1}{s}
\]

For the right side of the equation:
\[
\text{Unit of }\frac{b}{\sqrt{D}} = \frac{\text{Unit of } b}{\sqrt{\text{Unit of } D}} = \frac{\text{Unit of } b}{\sqrt{\mathrm{m}^2}} = \frac{\text{Unit of } b}{\mathrm{m}}
\]

So, the right side of the equation has units of $b$ divided by meters ($m$).

\[
\text{Unit of } c = \text{Unit of constant } c
\]

Since both sides of the equation must have the same units for the equation to be consistent, we can equate the units of the left and right sides:

\[
\frac{1}{s} = \frac{\text{Unit of } b}{\mathrm{m}} + \text{Unit of constant } c
\]

This means that the units of $b$ must be $\mathrm{m/s}$ to cancel out the meters ($m$) in the denominator of the right side, and the units of $c$ must be $\mathrm{1/s}$ to match the units on the left side.

In conclusion:
$\bullet$ The unit of $b$ is $\mathrm{m/s}$.
$\bullet$ The unit of $c$ is $\mathrm{1/s}$.

Question

A hollow cylinder, which is open at both ends, has a radius of $(3.0 \pm 0.1) \mathrm{cm}$ and a length of $(15.0 \pm 0.1) \mathrm{cm}$. What is the value, with its absolute uncertainty, of the surface area of the cylinder?

A $(280 \pm 10) \mathrm{cm}^2$

B $(282.7 \pm 0.2) \mathrm{cm}^2$

C $(420 \pm 30) \mathrm{cm}^2$

D $(424.1 \pm 0.3) \mathrm{cm}^2$

▶️Answer/Explanation

Ans:A

Question

A snooker ball of mass $0.20 \mathrm{~kg}$ has a collision so that its direction of movement changes by an angle of $90^{\circ}$, as shown.

The ball has a speed of $0.40 \mathrm{~m} \mathrm{~s}^{-1}$ before the collision and a speed of $0.30 \mathrm{~m} \mathrm{~s}^{-1}$ after the collision.

What is the magnitude of the change in momentum of the snooker ball?

A $0.020 \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}$

B $0.10 \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}$

C $0.14 \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}$

D $0.50 \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}$

▶️Answer/Explanation

Ans:B

Given:
Initial velocity ($v_{\text{initial}}$) = $0.40 \, \mathrm{m/s}$
Final velocity ($v_{\text{final}}$) = $0.30 \, \mathrm{m/s}$

The change in velocity ($\Delta \mathbf{v}$) is the vector difference between the final velocity and the initial velocity. Since they are perpendicular and at $90^\circ$ angle, we can treat this as a right-angled triangle:

\[|\Delta \mathbf{v}| = \sqrt{v_{\text{final}}^2 + v_{\text{initial}}^2}\]

Substituting the values:
\[|\Delta \mathbf{v}| = \sqrt{(0.30 \, \mathrm{m/s})^2 + (0.40 \, \mathrm{m/s})^2} \approx 0.5 \, \mathrm{m/s}\]

Now, we can calculate the change in momentum ($\Delta \mathbf{p}$) using the formula:
\[\Delta \mathbf{p} = m \cdot \Delta \mathbf{v} = (0.20 \, \mathrm{kg}) \cdot (0.5 \, \mathrm{m/s}) = 0.10 \, \mathrm{kg \cdot m/s}\]

The magnitude of the change in momentum is $0.10 \, \mathrm{kg \cdot m/s}$, which corresponds to option B: $0.10 \, \mathrm{kg \cdot m/s}$.

Question

A ball is kicked upwards at an angle of $45^{\circ}$ to horizontal ground. After a short flight, the ball returns to the ground. It may be assumed that air resistance is negligible. What is never zero during the flight of the ball?

A the horizontal component of the ball’s acceleration

B the horizontal component of the ball’s velocity

C the vertical component of the ball’s momentum

D the vertical component of the ball’s velocity

▶️Answer/Explanation

Ans:B

During the flight of the ball, the only force acting on it is gravity. Since air resistance is negligible, the only force is the weight of the ball, which acts vertically downward. This force causes changes in the ball’s velocity and momentum. However, it doesn’t directly influence the horizontal motion of the ball.

In projectile motion, the horizontal and vertical components of motion are independent of each other. The horizontal velocity remains constant throughout the motion (assuming no external horizontal forces like air resistance).

The vertical component of the ball’s velocity changes due to the acceleration caused by gravity. Initially, the ball’s vertical velocity is upward, but as it rises, its velocity decreases until it reaches its maximum height. Then, the ball’s vertical velocity becomes downward, and it increases as the ball falls back to the ground.

The horizontal component of the ball’s velocity remains constant, unaffected by the vertical motion.

Therefore, the correct answer is:

B) the horizontal component of the ball’s velocity

Question

The graph shows the variation with time t of the displacement s of an object.

Which graph represents the variation with time t of the acceleration a of the object?

▶️Answer/Explanation

Ans:D

Question

Which expression defines force?

A (mass $\times$ change in speed) $\times$ time taken

B mass $\times$ change in speed time taken

C (change of momentum) $\times$ time taken

D $\frac{\text { change of momentum }}{\text { time taken }}$

▶️Answer/Explanation

Ans:D

The correct option that defines force is: D)

Force is defined as the rate of change of momentum. In equation form:

\[F = \frac{\Delta p}{\Delta t}\]

where $F$ is the force, $\Delta p$ is the change in momentum, and $\Delta t$ is the time taken for that change.

Question

A ship of mass $8.4 \times 10^7 \mathrm{~kg}$ is approaching a harbour with speed $16.4 \mathrm{~m} \mathrm{~s}^{-1}$. By using reverse thrust it can maintain a constant total stopping force of $920000 \mathrm{~N}$.

How long will it take to stop?

A 15 seconds

B 150 seconds

C 25 minutes

D 250 minutes

▶️Answer/Explanation

Ans:C

Given:
Total stopping force ($F$) = $920000 \, \mathrm{N}$
Mass of the ship ($m$) = $8.4 \times 10^7 \, \mathrm{kg}$
Initial velocity ($v_{\text{initial}}$) = $16.4 \, \mathrm{m/s}$
Final velocity ($v_{\text{final}}$) = $0 \, \mathrm{m/s}$

The change in momentum ($\Delta p$) is given by:
\[\Delta p = m \cdot (v_{\text{final}} – v_{\text{initial}})\]

Substituting the values:
\[\Delta p = (8.4 \times 10^7 \, \mathrm{kg}) \cdot (0 \, \mathrm{m/s} – 16.4 \, \mathrm{m/s}) = -1377600000 \, \mathrm{kg \cdot m/s}\]

Now, we can use the equation $F = \frac{\Delta p}{\Delta t}$ to find the time ($\Delta t$) it takes for the ship to stop:
\[920000 \, \mathrm{N} = \frac{-1377600000 \, \mathrm{kg \cdot m/s}}{\Delta t}\]

Solving for $\Delta t$:
\[\Delta t = \frac{-1377600000 \, \mathrm{kg \cdot m/s}}{920000 \, \mathrm{N}}\]

Calculating the value of $\Delta t$:
\[\Delta t \approx \frac{1497.3}{60} \, \mathrm{min}\]

Among the given options, the correct answer is:

C) 25 seconds

So, it will take approximately 25 seconds for the ship to stop.

Question

The velocity – time graphs of four different objects are shown.

Which graph represents an object falling from rest through a long distance in the Earth’s atmosphere?

▶️Answer/Explanation

Ans:D

When an object falls from rest through a long distance in Earth’s atmosphere while experiencing air resistance, the velocity – time graph will exhibit certain characteristics that distinguish it from the graph of free fall in a vacuum. Air resistance opposes the motion of the falling object and affects its acceleration.

Here’s how the velocity – time graph might look for an object falling from rest through a long distance in the presence of air resistance:

1. Initial Phase (Acceleration Phase):
$\bullet$ The object starts from rest, so its initial velocity is $0 \, \mathrm{m/s}$.
$\bullet$ Initially, as the object begins to fall, it accelerates downwards due to gravity.
$\bullet$ The graph will show a steep curve with the velocity increasing over time.

2. Terminal Velocity Phase:
$\bullet$ As the object gains velocity, air resistance also increases.
$\bullet$ Eventually, the force of air resistance becomes equal in magnitude to the force of gravity, resulting in a net force of zero.
$\bullet$ At this point, the object reaches its terminal velocity, where the acceleration due to gravity is balanced by the decelerating effect of air resistance.
$\bullet$ The graph levels off and becomes nearly horizontal, indicating a constant velocity.

3. Equilibrium Phase:
$\bullet$ During this phase, the object continues to fall, but the velocity remains constant (terminal velocity).
$\bullet$ The graph remains horizontal.

4. Impact Phase (Upon Landing):
$\bullet$ As the object approaches the ground, air resistance continues to oppose its motion.
$\bullet$ The object’s velocity remains constant at terminal velocity until it reaches the ground.
$\bullet$ At impact, the velocity abruptly drops to $0 \, \mathrm{m/s}$.

In summary, the velocity – time graph for an object falling from rest through a long distance in Earth’s atmosphere with air resistance will show an initial steep curve representing acceleration, a horizontal plateau indicating constant terminal velocity, and a sudden drop to $0 \, \mathrm{m/s}$ upon impact. The key features include the gradual attainment of terminal velocity due to the balance between gravity and air resistance, as well as the cessation of acceleration once terminal velocity is reached.

Question

Which statement about collisions is correct?

A Kinetic energy is conserved in all collisions.

B Momentum is only conserved in perfectly elastic collisions.

C The relative speed of approach is equal to the relative speed of separation for perfectly elastic collisions.

D When two objects of different masses collide, they exert forces of different magnitudes on each other.

▶️Answer/Explanation

Ans:C

The correct statement about collisions is:

C) The relative speed of approach is equal to the relative speed of separation for perfectly elastic collisions.

In a perfectly elastic collision, both momentum and kinetic energy are conserved. Additionally, the relative speed of approach (before collision) is equal to the relative speed of separation (after collision). This means that the objects that collide will have the same speed when moving apart as they did when moving toward each other before the collision. This behavior is unique to perfectly elastic collisions and doesn’t apply to other types of collisions.

Question

A satellite uses two thrusters to adjust its motion in space. Each thruster exerts a force of 40N on the satellite. The line of action of each force is a perpendicular distance of 2.3 m from the centre of gravity of the satellite. These two parallel forces act in opposite directions.

What are the magnitudes of the torque and the resultant force acting on the satellite due to the two thrust forces?

▶️Answer/Explanation

Ans:C

1. Torque (Moment of Force):
The torque ($\tau$) due to each force can be calculated using the formula:
\[\tau = F \cdot r \cdot \sin(\theta)\]

Where:
$F$ = Force applied (40 N)
$r$ = Distance from the point to the line of action of the force (2.3 m)
$\theta$ = Angle between the force and the lever arm (90 degrees since they are perpendicular)

Substituting the values:
\[\tau = 40 \, \mathrm{N} \times 2.3 \, \mathrm{m} \times \sin(90^\circ) = 92 \, \mathrm{N \cdot m}\]

Since there are two thrusters, the total torque due to both thrust forces will be:
\[\text{Total Torque} = 2 \times 92 \, \mathrm{N \cdot m} = 184 \, \mathrm{N \cdot m}\]

2. Resultant Force:
The resultant force due to the two thrust forces will be the algebraic sum of the forces:
\[F_{\text{net}} = F_1 – F_2\]

Where $F_1$ and $F_2$ are the magnitudes of the thrust forces (40 N each).

Substituting the values:
\[F_{\text{net}} = 40 \, \mathrm{N} – 40 \, \mathrm{N} = 0 \, \mathrm{N}\]

In summary:
$\bullet$ The magnitude of the total torque due to both thrust forces is $184 \, \mathrm{N \cdot m}$.
$\bullet$ The magnitude of the resultant force due to the two thrust forces is $0 \, \mathrm{N}$.

Question

The diagrams show two ways of hanging the same picture.

In both cases, a string is attached to the same points on the picture and looped symmetrically over a nail in a wall. The forces shown are those that act on the nail.

In diagram 1, the string loop is shorter than in diagram 2. Which information about the magnitude of the forces is correct?

A $R_1=R_2 \quad T_1=T_2$

B $R_1=R_2 \quad T_1>T_2$

C $R_1>R_2 \quad T_1<T_2$

D $R_1<R_2 \quad T_1=T_2$

▶️Answer/Explanation

Ans:B

$\begin{aligned} & \quad R_1=2 T_1 \cos \theta_1 \\ & R_2=2 T_2 \cos \theta_2 \\ & \theta_1>\theta_2 \\ & \cos \theta_1<\cos \theta_2 \\ & If ~~ R_1=R_2 \\ & T_1>T_2\end{aligned}$

Question

An object shaped as a hemisphere rests with its flat surface on a table. The object has radius $r$ and density $\rho$.

The volume of a sphere is $\frac{4}{3} \pi r^3$.

Which average pressure does the object exert on the table?

A $\frac{1}{3} \rho r^2$

B $\frac{1}{3} \rho r^2 g$

C $\frac{2}{3} \rho r$

D $\frac{2}{3} \rho r g$

▶️Answer/Explanation

Ans:D

When an object is placed on a surface, the pressure it exerts on the surface is given by the force it applies per unit area. In this case, the hemisphere-shaped object is resting with its flat surface on the table, so the average pressure it exerts on the table will be the weight of the object divided by the area of the flat surface in contact with the table.

Given the density ($\rho$) of the object and its radius ($r$), the volume of the hemisphere is $\frac{1}{2} \cdot \frac{4}{3} \pi r^3 = \frac{2}{3} \pi r^3$.

The weight ($W$) of the hemisphere can be calculated using its volume and density:
\[W = \rho \cdot V \cdot g = \rho \cdot \frac{2}{3} \pi r^3 \cdot g\]

The area ($A$) of the flat surface in contact with the table is the circular base of the hemisphere:
\[A = \pi r^2\]

The average pressure ($P$) exerted by the object on the table is given by:
\[P = \frac{W}{A} = \frac{\rho \cdot \frac{2}{3} \pi r^3 \cdot g}{\pi r^2} = \frac{2}{3} \rho r g\]

Among the given options, the correct answer is:

D) $\frac{2}{3} \rho r g$

So, the average pressure that the object exerts on the table is $\frac{2}{3} \rho r g$.

Question

A probe is used to monitor the quality of the water in the sea. The probe is suspended by a vertical string which is attached to a sphere. The stationary sphere floats in equilibrium on the surface of the sea, as shown.

The sphere has a weight of $5.00 \mathrm{~N}$. The probe and string have a combined weight of $2.00 \mathrm{~N}$.
The density of the seawater is $1.03 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}$. The upthrust acting on the probe and thread is negligible.

What is the volume of the sphere below the surface of the sea?

A $1.98 \times 10^{-4} \mathrm{~m}^3$

B $2.97 \times 10^{-4} \mathrm{~m}^3$

C $4.95 \times 10^{-4} \mathrm{~m}^3$

D $6.93 \times 10^{-4} \mathrm{~m}^3$

▶️Answer/Explanation

Ans:D

Given:
Weight of the sphere ($W_{\text{sphere}}$) = $5.00 \, \mathrm{N}$
Weight of the probe and string ($W_{\text{probe}}$) = $2.00 \, \mathrm{N}$
Density of seawater ($\rho_{\text{water}}$) = $1.03 \times 10^3 \, \mathrm{kg/m^3}$
Buoyant force ($F_{\text{buoyant}}$) = $W_{\text{sphere}} + W_{\text{probe}}$

Using the buoyant force equation:
\[F_{\text{buoyant}} = \rho_{\text{water}} \cdot V_{\text{submerged}} \cdot g\]

Where:
$V_{\text{submerged}}$ = Volume of the sphere submerged below the surface
$g$ = Acceleration due to gravity

Rearranging the equation to solve for $V_{\text{submerged}}$:
\[V_{\text{submerged}} = \frac{F_{\text{buoyant}}}{\rho_{\text{water}} \cdot g}\]

Substituting the values:
\[V_{\text{submerged}} = \frac{(5.00 \, \mathrm{N} + 2.00 \, \mathrm{N})}{(1.03 \times 10^3 \, \mathrm{kg/m^3}) \cdot (9.81 \, \mathrm{m/s^2})}\]

Calculating the value of $V_{\text{submerged}}$:
\[V_{\text{submerged}} \approx 4.95 \times 10^{-4} \, \mathrm{m^3}\]

Among the given options, the closest match is:

D) $4.95 \times 10^{-4} \, \mathrm{m^3}$

Question

What is the centre of gravity of an object?

A the geometrical centre of the object

B the point at which the weight of the object may be considered to act

C the point on the object about which there is a zero net torque

D the point where gravity acts on the object

▶️Answer/Explanation

Ans:B

The correct answer is:

B) the point at which the weight of the object may be considered to act

The center of gravity of an object is the point at which the entire weight of the object appears to act when considering gravitational forces. It’s the point through which the force of gravity on the object can be considered to act, causing the object to behave as if all its mass were concentrated at that point.

Question

A system with an efficiency of $74 \%$ wastes $230 \mathrm{~W}$ of power. What is the useful output power of the system?

A $170 \mathrm{~W}$

B $310 \mathrm{~W}$

C $650 \mathrm{~W}$

D $880 \mathrm{~W}$

▶️Answer/Explanation

Ans:C

The efficiency ($\eta$) of a system is defined as the ratio of useful output power ($P_{\text{out}}$) to the input power ($P_{\text{in}}$), expressed as a percentage:

\[\eta = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\]

Given the efficiency ($\eta$) as $74\%$ and the wasted power ($P_{\text{waste}}$) as $230 \, \mathrm{W}$, we can rearrange the efficiency formula to solve for the useful output power ($P_{\text{out}}$):

\[\eta = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100 \implies P_{\text{out}} = \frac{\eta}{100} \cdot P_{\text{in}}\]

We want to find $P_{\text{out}}$, so let’s solve for it by substituting the given values and solving for $P_{\text{in}}$:

\[\frac{74}{100} = \frac{P_{\text{out}}}{P_{\text{in}}} \implies P_{\text{in}} = \frac{P_{\text{out}}}{0.74}\]

Given that $P_{\text{waste}} = 230 \, \mathrm{W}$, the total input power is the sum of the useful output power and the wasted power:

\[P_{\text{in}} = P_{\text{out}} + P_{\text{waste}}\]

Substitute the expression for $P_{\text{in}}$ from above:

\[\frac{P_{\text{out}}}{0.74} = P_{\text{out}} + 230\]

Now, solve for $P_{\text{out}}$:

\[P_{\text{out}} = 0.74 \cdot P_{\text{out}} + 230\]
\[0.26 \cdot P_{\text{out}} = 230\times 0.74\]
\[P_{\text{out}} = \frac{170}{0.26} \approx 650 \, \mathrm{W}\]

Among the given options, the closest match is:

C) $650 \, \mathrm{W}$

So, the useful output power of the system is approximately $650 \, \mathrm{W}$.

Question

A projectile of mass $0.25 \mathrm{~kg}$ is at a height of $30 \mathrm{~m}$ above horizontal ground and travelling at a speed of $15 \mathrm{~m} \mathrm{~s}^{-1}$. A short time later, it is at a height of $35 \mathrm{~m}$ above the horizontal ground and travelling at a speed of $5.0 \mathrm{~m} \mathrm{~s}^{-1}$. How much work is done against air resistance during this time?

A $0 \mathrm{~J}$

B $13 \mathrm{~J}$

C $25 \mathrm{~J}$

D $37 \mathrm{~J}$

▶️Answer/Explanation

Ans:B

To find the work done against air resistance, we can use the concept of mechanical energy conservation. The initial total mechanical energy of the projectile (potential energy + kinetic energy) is equal to the final total mechanical energy, assuming there are no non-conservative forces like air resistance acting on the projectile.

The initial mechanical energy is the sum of the initial potential energy ($PE_i$) and initial kinetic energy ($KE_i$):
\[E_{\text{initial}} = PE_i + KE_i\]

The final mechanical energy is the sum of the final potential energy ($PE_f$) and final kinetic energy ($KE_f$):
\[E_{\text{final}} = PE_f + KE_f\]

Since there’s no change in mechanical energy ($E_{\text{initial}} = E_{\text{final}}$) due to the absence of non-conservative forces like air resistance, we can equate the initial and final energies:

\[PE_i + KE_i = PE_f + KE_f\]

Let’s calculate the potential and kinetic energies for both the initial and final states:

1. Initial State:
– Potential Energy ($PE_i$) at $30 \, \mathrm{m}$ height = $mgh$ = $0.25 \, \mathrm{kg} \times 9.81 \, \mathrm{m/s^2} \times 30 \, \mathrm{m}$
– Kinetic Energy ($KE_i$) at $15 \, \mathrm{m/s}$ = $0.5 \times 0.25 \, \mathrm{kg} \times (15 \, \mathrm{m/s})^2$

2. Final State:
– Potential Energy ($PE_f$) at $35 \, \mathrm{m}$ height = $mgh$ = $0.25 \, \mathrm{kg} \times 9.81 \, \mathrm{m/s^2} \times 35 \, \mathrm{m}$
– Kinetic Energy ($KE_f$) at $5 \, \mathrm{m/s}$ = $0.5 \times 0.25 \, \mathrm{kg} \times (5 \, \mathrm{m/s})^2$

Now, substitute the values into the energy conservation equation and solve for the work done against air resistance ($W$):

\[PE_i + KE_i = PE_f + KE_f\]
\[\text{Air Resistance Work } W = PE_i + KE_i – (PE_f + KE_f)\]

$W=73.5+28.12-(85.75+3.12)$.

Among the given options, the closest match is:

B) $13 \, \mathrm{J}$

So, the work done against air resistance during this time is approximately $13 \, \mathrm{J}$.

Question

A spring has an unstretched length of $4.50 \mathrm{~cm}$. The spring is fixed at one end and a force of $35.0 \mathrm{~N}$ is applied to the other end so that the spring extends.

The spring obeys Hooke’s law and has a spring constant of $420 \mathrm{~N} \mathrm{~m}^{-1}$. What is the strain of the extended spring?

A $0.019$

B $0.083$

C $1.85$

D $2.67$

▶️Answer/Explanation

Ans:C

Hooke’s law relates the force applied to a spring ($F$) to the displacement ($x$) it undergoes from its equilibrium position and the spring constant ($k$):

\[F = kx\]

Given the force ($F = 35.0 \, \mathrm{N}$) and the spring constant ($k = 420 \, \mathrm{N/m}$), we can rearrange Hooke’s law to solve for the displacement ($x$) of the spring:

\[x = \frac{F}{k}\]

Substitute the given values:

\[x = \frac{35.0 \, \mathrm{N}}{420 \, \mathrm{N/m}} = 0.083 \, \mathrm{m}\]

The strain ($\varepsilon$) of an extended spring is defined as the ratio of the change in length ($\Delta L$) to the original unstretched length ($L_0$):

\[\varepsilon = \frac{\Delta L}{L_0}\]

Since the displacement $x$ of the spring is the change in length and the original unstretched length is $L_0 = 4.50 \, \mathrm{cm} = 0.045 \, \mathrm{m}$, we can calculate the strain:

\[\varepsilon = \frac{x}{L_0} = \frac{0.083 \, \mathrm{m}}{0.045 \, \mathrm{m}} = 1.84\]

Among the given options, the closest match is:

C) $1.85$

So, the strain of the extended spring is approximately $1.85$.

Question

A wire is fixed at one end and extended by a force that is applied to the other end. The force is slowly increased from zero and then slowly decreased back to zero.

The force-extension graph for the wire is shown.

The graph line for the wire being loaded is the same as the graph line for the wire being unloaded. Which statement describes the deformation of the wire?

A It is both elastic and plastic.

B It is elastic only.

C It is neither elastic nor plastic.

D It is plastic only.

▶️Answer/Explanation

Ans:B

If the force-extension graph for the wire being loaded and unloaded is the same, it indicates that the deformation of the wire is elastic. In an elastic deformation, the material returns to its original shape and size when the applied force is removed. The fact that the graph lines coincide for loading and unloading suggests that the wire is following Hooke’s law and that it is behaving elastically.

Question

In a progressive water wave, two particles, $\mathrm{P}$ and $\mathrm{Q}$, on the surface of the water, are a fixed horizontal distance apart. P and Q oscillate vertically.
At time $t=0$, the wave is as shown.

Which graph best represents the variation with time $t$ of the phase difference $\phi$ between the oscillation of the water particle $\mathrm{P}$ and the oscillation of the water particle $\mathrm{Q}$ ?

▶️Answer/Explanation

Ans:D

Phase difference will be constant.

Question

Which statement about longitudinal waves and transverse waves is not correct?

A Both waves can be polarised.

B Both waves can form stationary waves.

C Both waves can transfer energy as progressive waves.

D Both waves obey the equation $v=f \lambda$.

▶️Answer/Explanation

Ans:A

Polarization is a phenomenon that occurs in transverse waves, not in longitudinal waves. Polarization refers to the alignment of the oscillations of a wave in a specific direction perpendicular to the direction of wave propagation. Transverse waves, like light and electromagnetic waves, can be polarized by filtering out oscillations in certain directions, which results in only allowing waves with specific orientations to pass through.

Longitudinal waves, on the other hand, have oscillations that occur along the direction of wave propagation. Since the oscillations are parallel to the direction of motion, they cannot be polarized in the same manner as transverse waves.

So, the statement that is not correct is: “Both waves can be polarised.”

Question

An observer hears a sound wave emitted from a moving source.

The observed frequency is less than the frequency of sound emitted from the source. What could be the reason for this?

A The source is moving away from the observer.

B The source is moving towards the observer.

C The speed of the sound wave in air decreases due to the movement of the source.

D The speed of the sound wave in air increases due to the movement of the source.

▶️Answer/Explanation

Ans:A

When a sound source is moving away from the observer, the observed frequency of the sound wave is lower than the frequency of the sound emitted from the source. This phenomenon is known as the Doppler effect. As the source moves away, the waves get “stretched out,” causing a decrease in the perceived frequency. This is why the observed frequency is less than the emitted frequency.

Question

What is the approximate range of frequencies of electromagnetic radiation visible to the human eye?

A $(430-750) \mathrm{kHz}$

B $(430-750) \mathrm{MHz}$

C $(430-750) \mathrm{GHz}$

D $(430-750) \mathrm{THz}$

▶️Answer/Explanation

Ans:D

The correct answer is:

D) $(430-750) \mathrm{THz}$

The approximate range of frequencies of electromagnetic radiation visible to the human eye is indeed in the terahertz (THz) range. This corresponds to the visible light spectrum, with violet light having the highest frequency (around 750 THz) and red light having the lowest frequency (around 430 THz).

Question

A beam of vertically polarised light is incident normally on a polarising filter. The filter can be rotated so that it is always in a plane perpendicular to the beam. The transmission axis of the filter is initially vertical.

The filter is first rotated clockwise by an angle of $30^{\circ}$ so that the transmitted light waves have intensity $I_{30}$. The filter is then rotated clockwise by a further angle of $30^{\circ}$.
What is the new intensity of the transmitted light waves?
A $ 0.25 I_{30}$

B $0.33 I_{30}$

C $0.75 I_{30}$

D $0.87 I_{30}$

▶️Answer/Explanation

Ans:B

When vertically polarized light is incident normally on a polarizing filter, the intensity of the transmitted light is given by Malus’s law:

\[I = I_0 \cos^2 \theta\]

Where:
$\bullet$ $I_0$ is the initial intensity of the light incident on the filter.
$\bullet$ $\theta$ is the angle between the polarization axis of the filter and the direction of polarization of the incident light.

Given that the filter is rotated by $30^\circ$ initially, the intensity after the first rotation ($I_{30}$) is given by $I_{30} = I_0 \cos^2 30^\circ$.

Now, when the filter is rotated by another $30^\circ$, the new angle between the polarization axis of the filter and the direction of polarization of the transmitted light becomes $60^\circ$. Therefore, the new intensity ($I_{60}$) is given by $I_{60} = I_0 \cos^2 60^\circ$.

We know that $\cos^2 60^\circ = \frac{1}{4}$, so the new intensity is $I_{60} = \frac{1}{4} I_0$.

Now, we have $I_{60}$ in terms of $I_0$, and we also know that $I_{30} = I_0 \cos^2 30^\circ$, which simplifies to $\frac{3}{4} I_0$.

Comparing $I_{60}$ and $I_{30}$, we have:

\[I_{60} = \frac{1}{4} I_0\]
\[I_{30} = \frac{3}{4} I_0\]

Dividing $I_{60}$ by $I_{30}$, we get:

\[\frac{I_{60}}{I_{30}} = \frac{\frac{1}{4} I_0}{\frac{3}{4} I_0} = \frac{1}{3}\]

Therefore, the new intensity of the transmitted light waves after the additional $30^\circ$ rotation is $0.33 I_{30}$, which corresponds to option B.

Question

A musical instrument is made using a long tube with a mouthpiece at one end. The other end is open and flared, as shown.

A musician maintains stationary sound waves with a node at the mouthpiece and an antinode at the other end. The lowest frequency of sound that the instrument can produce is $92 \mathrm{~Hz}$.

Which different frequencies of sound can be produced by the instrument?

A $92 \mathrm{~Hz}, 138 \mathrm{~Hz}, 184 \mathrm{~Hz}, 230 \mathrm{~Hz}$

B $92 \mathrm{~Hz}, 184 \mathrm{~Hz}, 276 \mathrm{~Hz}, 368 \mathrm{~Hz}$

C $92 \mathrm{~Hz}, 276 \mathrm{~Hz}, 460 \mathrm{~Hz}, 644 \mathrm{~Hz}$

D $92 \mathrm{~Hz}, 276 \mathrm{~Hz}, 828 \mathrm{~Hz}, 1288 \mathrm{~Hz}$

▶️Answer/Explanation

Ans:C

We are given that the lowest frequency of sound produced by the instrument is $92 \, \mathrm{Hz}$, which corresponds to the fundamental frequency.

The frequencies of different harmonics (overtones) in a closed-open pipe are given by the formula:

\[f_n = \frac{(2n – 1)v}{4L}\]

Where:
$\bullet$ $f_n$ is the frequency of the $n$th harmonic,
$\bullet$ $n$ is the harmonic number,
$\bullet$ $v$ is the speed of sound,
$\bullet$ $L$ is the length of the pipe.

Since we are dealing with a closed-open pipe, the first harmonic is the fundamental frequency, and the formula holds for the higher harmonics as well.

Using the information provided and substituting the values:
$\bullet$ $n = 1$,
$\bullet$ $v$ is the speed of sound in air (approximately $343 \, \mathrm{m/s}$),
$\bullet$ $L$ is the length of the pipe.

For the fundamental frequency ($n = 1$):
\[92 \, \mathrm{Hz} = \frac{(2 \times 1 – 1) \times 343}{4L}\]
Solving for $L$:
\[L = \frac{(2 \times 1 – 1) \times 343}{4 \times 92} \approx 1.875 \, \mathrm{m}\]

For the third harmonic ($n = 3$):
\[f_3 = \frac{(2 \times 3 – 1) \times 343}{4 \times 1.875} \approx 276 \, \mathrm{Hz}\]

Similarly, for the fifth harmonic ($n = 5$):
\[f_5 = \frac{(2 \times 5 – 1) \times 343}{4 \times 1.875} \approx 460 \, \mathrm{Hz}\]

And for the seventh harmonic ($n = 7$):
\[f_7 = \frac{(2 \times 7 – 1) \times 343}{4 \times 1.875} \approx 644 \, \mathrm{Hz}\]

Among the given options, the closest match is:

C) $92 \, \mathrm{Hz}, 276 \, \mathrm{Hz}, 460 \, \mathrm{Hz}, 644 \, \mathrm{Hz}$

Question

Two waves of equal frequency and amplitude are travelling in opposite directions along a stretched string. When they meet, they form a stationary wave with three nodes and two antinodes.
The frequency of both waves is doubled and a new stationary wave is formed. How many antinodes are there in the new stationary wave?

A 1

B 2

C 3

D 4

▶️Answer/Explanation

Ans:D

The frequency ($f$) of a stationary wave is directly proportional to the number of antinodes ($n$) present in the wave. As the number of antinodes increases, the frequency of the wave also increases. This relationship is a fundamental characteristic of stationary waves, and it’s based on the fact that the number of antinodes affects how many complete cycles (oscillations) occur within a given time interval.

So, in the context of the question, when the frequency of the waves is doubled and a new stationary wave is formed, the number of antinodes will also change accordingly. If the original stationary wave had two antinodes, doubling the frequency will lead to a new stationary wave with four antinodes.

Question

A transmitting mast sends out microwaves of wavelength 1.5 cm and radio waves of wavelength 1.5 km.

A receiving aerial behind a mountain can detect the radio waves but not the microwaves. What is the reason for this?

A The radio waves are coherent but the microwaves are not.

B The radio waves are diffracted around the mountain but the microwaves are not.

C The radio waves are reflected by the mountain but the microwaves are not.

D The radio waves travel at the speed of light but the microwaves do not.

▶️Answer/Explanation

Ans:B

The correct answer is:

B) The radio waves are diffracted around the mountain but the microwaves are not.

Diffracted is the phenomenon of waves bending around obstacles or spreading out as they pass through openings. Radio waves have much longer wavelengths compared to microwaves, making them more capable of diffracting around obstacles like mountains. This is why the receiving aerial behind the mountain can detect the radio waves but not the microwaves. Microwaves, due to their shorter wavelengths, have less ability to diffract around obstacles and are more likely to be blocked by the mountain.

Question

Waves are emitted from two coherent sources. Which statement about the waves must be correct?

A They are in phase.

B They are transverse waves.

C They have a constant phase difference.

D They have the same amplitude.

▶️Answer/Explanation

Ans:C

The correct answer is:

C) They have a constant phase difference.

Coherent sources are sources of waves that maintain a constant phase relationship with each other. This means that the waves emitted from coherent sources have a consistent phase difference that does not change over time. The term “coherent” implies that the waves maintain a predictable relationship as they propagate, ensuring that they interfere constructively or destructively in a controlled manner.

Question

The diagram shows a screen that is a distance $L$ from a diffraction grating. The grating has a total number of $N$ lines. Any two adjacent lines are a distance $d$ apart. A beam of parallel light of wavelength $\lambda$ is incident normally on the grating.

Which quantities affect the distance between the first-order diffraction maxima on the screen?

▶️Answer/Explanation

Ans:A

The quantities that affect the distance between the first-order diffraction maxima on the screen are $d$, $\lambda$, and $L$. The total number of lines ($N$) on the grating does not directly affect this distance.

The first-order diffraction maxima are given by the formula:

\[d \sin \theta = m \lambda\]

Where:
$d$ is the spacing between adjacent lines on the grating.
$\theta$ is the angle of diffraction for the first-order maximum.
$m$ is the order of diffraction (in this case, $m = 1$ for the first-order maximum).
$\lambda$ is the wavelength of the incident light.

Rearranging the formula to solve for $\theta$:

\[\theta = \arcsin \left( \frac{m \lambda}{d} \right)\]

Since $L$ is the distance between the grating and the screen, it affects the angle $\theta$ and, consequently, the position of the diffraction maxima on the screen.

So, the correct answer is:

A) $d$, $\lambda$, $L$

Question

A wire carries a current of $0.10 \mu \mathrm{A}$. The potential difference across the wire is $10 \mathrm{mV}$. How much energy is dissipated by the wire in a time of $10 \mathrm{~s}$ ?

A $1.0 \mathrm{pJ}$

B $10 \mathrm{pJ}$

C $1.0 \mathrm{~nJ}$

D $10 \mathrm{~nJ}$

▶️Answer/Explanation

Ans:D

The power ($P$) dissipated by an electrical component can be calculated using the formula:

\[P = IV\]

Where:
– $I$ is the current flowing through the component,
– $V$ is the potential difference (voltage) across the component.

Given that the current $I = 0.10 \, \mu \mathrm{A} = 0.10 \times 10^{-6} \, \mathrm{A}$ and the potential difference $V = 10 \, \mathrm{mV} = 10 \times 10^{-3} \, \mathrm{V}$, we can calculate the power $P$ dissipated by the wire.

\[P = (0.10 \times 10^{-6} \, \mathrm{A}) \times (10 \times 10^{-3} \, \mathrm{V}) = 1.0 \times 10^{-9} \, \mathrm{W}\]

The energy ($E$) dissipated by the wire over a time period $t$ can be calculated using the formula:

\[E = Pt\]

Given that the time $t = 10 \, \mathrm{s}$, we can calculate the energy $E$ dissipated by the wire.

\[E = (1.0 \times 10^{-9} \, \mathrm{W}) \times (10 \, \mathrm{s}) = 10.0 \times 10^{-9} \, \mathrm{J}\]

Since $1 \, \mathrm{J} = 1 \, \mathrm{Ws}$, the energy can also be expressed in joules (J) as $10 \times 10^{-9} \, \mathrm{J}$.

Among the given options, the closest match is:

D) $10 \, \mathrm{~nJ}$

So, the energy dissipated by the wire in a time of $10 \, \mathrm{~s}$ is approximately $10 \, \mathrm{~nJ}$.

Question

What is the definition of the potential difference across an electrical component?

A energy transferred per unit charge

B energy transferred per unit current

C energy transferred per unit resistance

D energy transferred per unit time

▶️Answer/Explanation

Ans:A

The potential difference (voltage) across an electrical component is defined as the energy transferred per unit charge as a charge moves from one point to another within the component. It represents the work done or energy expended in moving a unit positive charge between the two points and is measured in volts (V).

Question

Which graph represents the way the current $I$ through a filament lamp varies with the potential difference $V$ across it?

▶️Answer/Explanation

Ans:B

The $I-V$ characteristic of a filament lamp, also known as its current-voltage characteristic, describes how the current through the lamp varies with the applied voltage (potential difference) across it. In the case of a filament lamp, the characteristic graph typically exhibits certain features:

Non-Ohmic Behavior: Unlike linear components such as resistors, filament lamps exhibit non-Ohmic behavior. This means that the relationship between current and voltage is not directly proportional, and the graph is not a straight line passing through the origin.
Exponential Increase: As the voltage increases, the current through the filament lamp increases, but this increase is often not linear. Instead, it follows an exponential-like curve.
Gradual Increase in Slope: The slope of the graph gradually increases as the voltage increases. At lower voltages, the resistance of the filament is relatively low, and the current increases more slowly. As the filament heats up due to the increased current, its resistance also increases, leading to a steeper increase in current with voltage.
Negative Temperature Coefficient: Filament lamps typically have a negative temperature coefficient, meaning that as the temperature of the filament increases due to the current passing through it, the resistance also increases. This contributes to the exponential-like behavior of the
$$I-V$$
characteristic.
Gradual Saturation: At high voltages, the graph may show a tendency to level off or saturate. This is because the filament approaches a state where its resistance remains relatively constant due to the maximum temperature it can achieve without melting.
Curve Variability: The exact shape of the
$$I-V$$
characteristic curve can vary based on factors such as the specific material of the filament, its length, diameter, and the initial temperature of the filament.

Overall, the

\(I-V\)

characteristic graph of a filament lamp illustrates the relationship between the applied voltage and the resulting current, demonstrating the lamp’s unique behavior due to the interplay between resistance, temperature, and current.

Question

The table shows the properties of two different wires, P and Q.

Wire $\mathrm{P}$ has a cross-section of diameter $d$. What is the diameter of the cross-section of wire Q?

A $0.41 d$

B $1.6 d$

C $2.7 d$

D $7.1 d$

▶️Answer/Explanation

Ans:B

Question

A cell has a constant electromotive force. A variable resistor is connected between the terminals of the cell. The resistance of the variable resistor is decreased. Which statement about the change of the cell’s terminal potential difference (p.d.) is correct?

A The terminal p.d. is decreased because more work is done moving unit charge through the internal resistance of the cell.

B The terminal p.d. is decreased because the current in the variable resistor is decreased.

C The terminal p.d. is increased because more work is done moving unit charge through the variable resistor.

D The terminal p.d. is increased because the current in the variable resistor is increased.

▶️Answer/Explanation

Ans:A

When the resistance of the variable resistor is decreased, according to Ohm’s law ($V = IR$), the current ($I$) in the circuit will increase. In the context of a cell with internal resistance, as the current increases, the voltage drop across the internal resistance of the cell ($r$) also increases due to the higher current passing through it.

This means that more work is done moving unit charge through the internal resistance of the cell, leading to a greater voltage drop across the internal resistance. As a result, the effective terminal potential difference ($V$) of the cell decreases.

So, the correct answer is indeed:A)

The terminal p.d. is decreased because more work is done moving unit charge through the internal resistance of the cell.

Question

Kirchhoff’s two laws for electric circuits can be derived by using conservation laws. On which conservation laws do Kirchhoff’s laws depend?

▶️Answer/Explanation

Ans:B

Kirchhoff’s laws for electric circuits are derived based on the principles of conservation of charge and conservation of energy.

$\bullet$ Kirchhoff’s first law (junction rule) is derived from the conservation of charge. It states that the total current entering a junction in an electric circuit is equal to the total current leaving the junction. This law ensures that charge is conserved at a junction point.

$\bullet$ Kirchhoff’s second law (loop rule) is derived from the conservation of energy. It is also known as the voltage law. It states that the sum of the electromotive forces (emfs) and potential drops around any closed loop in a circuit is zero. This law ensures that energy is conserved as it accounts for the changes in potential energy of charges as they move around a closed loop.

So, the correct answers are:

A) Kirchhoff’s first law depends on conservation of charge.
D) Kirchhoff’s second law depends on conservation of energy.

Question

A battery of electromotive force (e.m.f.) $10 \mathrm{~V}$ and internal resistance $r$ is connected to three resistors of resistances $R, 2.0 \Omega$ and $15 \Omega$, as shown. A current of $0.45 \mathrm{~A}$ is in the resistor of resistance $2.0 \Omega$ and a current of $0.48 \mathrm{~A}$ is in the resistor of resistance $15 \Omega$.

What are the values of r and R?

▶️Answer/Explanation

Ans:A

$V_{AB}=10-(0.93)r$

$r=\frac{2.8}{0.93}\Rightarrow 3.01$

$V_{AB}=(R+2)0.45$

$R+2=\frac{7.2}{0.45}\Rightarrow 16$

$R=14 \Omega\).

Question

A battery of negligible internal resistance is connected in series with a thermistor and a fixed resistor of resistance $12.0 \mathrm{k} \Omega$, as shown.

The table shows the resistance of the thermistor at two different temperatures.

The potential difference $V_{\text {out }}$ across the fixed resistor is $4.50 \mathrm{~V}$ when the thermistor is a temperature of $20.0^{\circ} \mathrm{C}$.
What is $V_{\text {out }}$ when the thermistor is at a temperature of $50.0^{\circ} \mathrm{C}$ ?

A $2.65 \mathrm{~V}$

B $3.18 \mathrm{~V}$

C $6.35 \mathrm{~V}$

D $10.8 \mathrm{~V}$

▶️Answer/Explanation

Ans:C

Question

What is a conclusion from the alpha-particle scattering experiment?

A Protons and electrons have equal but opposite charges.

B Protons have a much larger mass than electrons.

C The nucleus contains most of the mass of the atom.

D The nucleus of an atom contains protons and neutrons.

▶️Answer/Explanation

Ans:C

The correct answer is:

C) The nucleus contains most of the mass of the atom.

The alpha-particle scattering experiment, also known as the Rutherford gold foil experiment, led to the conclusion that most of the mass of an atom is concentrated in a small, dense nucleus at the center of the atom. This experiment provided evidence against the “plum pudding” model of the atom and led to the development of the modern atomic model with a central nucleus containing positively charged protons and uncharged neutrons, orbited by negatively charged electrons.

Question

Americium-241 is a radioactive nuclide used in smoke detectors. It undergoes $\alpha$-decay to form nuclide $\mathrm{X}$. This decay may be represented by the equation shown.
$
{ }_{95}^{241} \mathrm{Am} \rightarrow{ }_{\mathrm{Z}}^{\mathrm{A}} \mathrm{X}+\alpha
$
What are the values of $A$ and $Z$ ?

▶️Answer/Explanation

Ans:A

In alpha decay, an alpha particle ($\alpha$) is emitted from the nucleus of an atom, resulting in the formation of a new nucleus. An alpha particle consists of 2 protons and 2 neutrons, which means the mass number ($A$) of the new nucleus will be reduced by 4, and the atomic number ($Z$) will be reduced by 2.

For the given alpha decay equation:

$_{95}^{241} \mathrm{Am} \rightarrow_{Z}^{A} \mathrm{X}+\alpha$

The starting nucleus is americium-241, with atomic number $Z = 95$ and mass number $A = 241$.

Since an alpha particle is emitted ($2$ protons and $2$ neutrons), the atomic number $Z$ of the resulting nucleus will be $95 – 2 = 93$, and the mass number $A$ will be $241 – 4 = 237$.

So, the values of $A$ and $Z$ for nuclide $\mathrm{X}$ are $A = 237$ and $Z = 93$.

Question

A top quark has a charge of $+\frac{2}{3} e$, where $e$ is the elementary charge. What is the charge of an anti top quark?

A $-\frac{2}{3} e$

B $-\frac{1}{3} e$

C $+\frac{1}{3} e$

D $+\frac{2}{3} e$

▶️Answer/Explanation

Ans:A

The correct answer is:

A) $-\frac{2}{3} e$

An anti top quark (also known as an antiquark) has a charge that is opposite in sign to that of a top quark. Since a top quark has a charge of $+\frac{2}{3} e$, its corresponding anti top quark has a charge of $-\frac{2}{3} e$.

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