ESAT Physics Mock 3

27 questions · 40 minutes

Original full-length ESAT-style mock paper (Physics Mock C).

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Questions

Questions & worked solutions — spoilers below

Question 1
A magnet rests on a top-pan balance. A straight horizontal wire of length 4.0 cm\displaystyle 4.0\,cm is held between the magnet's poles and carries current I\displaystyle I . The balance reading increases linearly from 500.0 g\displaystyle 500.0\,g at I=0\displaystyle I=0 to 504.0 g\displaystyle 504.0\,g at I=2.0 A\displaystyle I=2.0\,A . What is the magnitude of the magnetic field strength between the poles? Take g=10 N kg−1\displaystyle g=10\,N\,kg^{-1} .
  1. A.
    0.0050 T\displaystyle 0.0050\,\text{T}
  2. B.
    0.050 T\displaystyle 0.050\,\text{T}
  3. C.
    0.50 T\displaystyle 0.50\,\text{T}
  4. D.
    2.0 T\displaystyle 2.0\,\text{T}
  5. E.
    5.0 T\displaystyle 5.0\,\text{T}
  6. F.
    50 T\displaystyle 50\,\text{T}
Answer and solution

Answer: C

A 2.0 A increase changes the balance reading by 4.0 g = 0.004 kg, so the magnetic force changes by 0.004×10=0.040 N\displaystyle 0.004\times10=0.040\,N . Using F=BIL\displaystyle F=BIL with L=0.040 m\displaystyle L=0.040\,m , B=0.040/(2.0×0.040)=0.50 T\displaystyle B=0.040/(2.0\times0.040)=0.50\,T . The correct option is C.
Question 2
A block of mass 3.0 kg\displaystyle 3.0\text{ kg} rests on a smooth, horizontal table. It is connected by two light, inextensible strings passing over smooth, frictionless pulleys at opposite ends of the table to two hanging masses of 5.0 kg\displaystyle 5.0\text{ kg} and 2.0 kg\displaystyle 2.0\text{ kg} , as shown in the diagram.

The system is released from rest.

What is the tension T1\displaystyle T_1 in the string attached to the 5.0 kg\displaystyle 5.0\text{ kg} mass and the tension T2\displaystyle T_2 in the string attached to the 2.0 kg\displaystyle 2.0\text{ kg} mass while the blocks are moving?

[Take the acceleration due to gravity to be g=10 m s−2\displaystyle g = 10\text{ m s}^{-2} .]

| | T1 / N\displaystyle T_1\text{ / N} | T2 / N\displaystyle T_2\text{ / N} |
| --- | --- | --- |
| A | 15 | 6 |
| B | 15 | 20 |
| C | 35 | 20 |
| D | 35 | 26 |
| E | 50 | 20 |
| F | 50 | 26 |
| G | 65 | 14 |
| H | 65 | 20 |

<figure class="qg-diagram" style="margin:1em 0;text-align:center;"><img src="https://bcbttpsokwoapjypwwwq.supabase.co/storage/v1/object/public/question-images/nsaa_review/nsaa-2512-far/rendered.png" alt="Exam diagram" style="max-width:100%;height:auto;" /></figure>
  1. A.
    T1=15 N, T2=6 N\displaystyle T_1 = 15\text{ N},\ T_2 = 6\text{ N}
  2. B.
    T1=15 N, T2=20 N\displaystyle T_1 = 15\text{ N},\ T_2 = 20\text{ N}
  3. C.
    T1=35 N, T2=20 N\displaystyle T_1 = 35\text{ N},\ T_2 = 20\text{ N}
  4. D.
    T1=35 N, T2=26 N\displaystyle T_1 = 35\text{ N},\ T_2 = 26\text{ N}
  5. E.
    T1=50 N, T2=20 N\displaystyle T_1 = 50\text{ N},\ T_2 = 20\text{ N}
  6. F.
    T1=50 N, T2=26 N\displaystyle T_1 = 50\text{ N},\ T_2 = 26\text{ N}
  7. G.
    T1=65 N, T2=14 N\displaystyle T_1 = 65\text{ N},\ T_2 = 14\text{ N}
  8. H.
    T1=65 N, T2=20 N\displaystyle T_1 = 65\text{ N},\ T_2 = 20\text{ N}
Answer and solution

Answer: D

1. Find the acceleration of the system:
Consider the whole connected system along the direction of motion (leftwards/downwards for the 5.0 kg\displaystyle 5.0\text{ kg} mass): Total mass Mtotal=5.0 kg+3.0 kg+2.0 kg=10.0 kg\displaystyle \text{Total mass } M_{\text{total}} = 5.0\text{ kg} + 3.0\text{ kg} + 2.0\text{ kg} = 10.0\text{ kg} Net accelerating force Fnet=m1g−m2g=(5.0−2.0)×10=30 N\displaystyle \text{Net accelerating force } F_{\text{net}} = m_1 g - m_2 g = (5.0 - 2.0) \times 10 = 30\text{ N} a=FnetMtotal=30 N10.0 kg=3.0 m s−2\displaystyle a = \dfrac{F_{\text{net}}}{M_{\text{total}}} = \dfrac{30\text{ N}}{10.0\text{ kg}} = 3.0\text{ m s}^{-2} 2. Find tension T1\displaystyle T_1 :
For the 5.0 kg\displaystyle 5.0\text{ kg} mass moving downwards with acceleration a=3.0 m s−2\displaystyle a = 3.0\text{ m s}^{-2} : m1g−T1=m1a  ⟹  T1=m1(g−a)=5.0×(10−3.0)=35 N\displaystyle m_1 g - T_1 = m_1 a \implies T_1 = m_1(g - a) = 5.0 \times (10 - 3.0) = 35\text{ N} 3. Find tension T2\displaystyle T_2 :
For the 2.0 kg\displaystyle 2.0\text{ kg} mass moving upwards with acceleration a=3.0 m s−2\displaystyle a = 3.0\text{ m s}^{-2} : T2−m2g=m2a  ⟹  T2=m2(g+a)=2.0×(10+3.0)=26 N\displaystyle T_2 - m_2 g = m_2 a \implies T_2 = m_2(g + a) = 2.0 \times (10 + 3.0) = 26\text{ N} Hence, T1=35 N\displaystyle T_1 = 35\text{ N} and T2=26 N\displaystyle T_2 = 26\text{ N} , which corresponds to option D.
Question 3
A laser beam reflects off a plane mirror M1\displaystyle M_1 and then off a second plane mirror M2\displaystyle M_2 . The direction of the beam incident on M1\displaystyle M_1 is fixed. M1\displaystyle M_1 is rotated by an angle α\displaystyle \alpha in a clockwise direction.

To ensure the final beam reflected from M2\displaystyle M_2 maintains a constant direction, M2\displaystyle M_2 must be rotated by:
  1. A.
    α\displaystyle \alpha in the clockwise direction
  2. B.
    α\displaystyle \alpha in the anticlockwise direction
  3. C.
    2α\displaystyle 2\alpha in the clockwise direction
  4. D.
    2α\displaystyle 2\alpha in the anticlockwise direction
  5. E.
    α2\displaystyle \dfrac{\alpha}{2} in the clockwise direction
Answer and solution

Answer: A

When a mirror rotates by an angle θ\displaystyle \theta , the reflected beam rotates by 2θ\displaystyle 2\theta in the same direction.

1. M1\displaystyle M_1 rotates by α\displaystyle \alpha (clockwise). The beam reflected from M1\displaystyle M_1 therefore rotates by 2α\displaystyle 2\alpha (clockwise).
2. This reflected beam is the incident beam for M2\displaystyle M_2 . So, the incidence angle on M2\displaystyle M_2 changes such that the incoming beam has rotated by 2α\displaystyle 2\alpha clockwise.
3. We want the final reflected beam from M2\displaystyle M_2 to have zero net rotation.

Let the rotation of M2\displaystyle M_2 be β\displaystyle \beta . The change in the angle of the final beam is given by:
Δθfinal=2Δθmirror−Δθincident \Delta \theta_{\text{final}} = 2\Delta \theta_{\text{mirror}} - \Delta \theta_{\text{incident}}
Substituting the values (taking clockwise as positive):
0=2β−2α 0 = 2\beta - 2\alpha
2β=2α  ⟹  β=α 2\beta = 2\alpha \implies \beta = \alpha
Thus, M2\displaystyle M_2 must rotate by α\displaystyle \alpha in the same direction (clockwise).
Question 4
A block of mass 2.0 kg\displaystyle 2.0\text{ kg} is initially moving at a speed of 4.0 m s−1\displaystyle 4.0\text{ m s}^{-1} along a smooth horizontal surface in the positive x\displaystyle x -direction.

A single horizontal force F\displaystyle F acts on the block in the direction of its motion. The graph shows how F\displaystyle F varies with the displacement x\displaystyle x of the block.

What is the speed of the block when it reaches x=6.0 m\displaystyle x = 6.0\text{ m} ?
Exam diagram
  1. A.
    6.0 m s−1\displaystyle 6.0\text{ m s}^{-1}
  2. B.
    6.9 m s−1\displaystyle 6.9\text{ m s}^{-1}
  3. C.
    7.2 m s−1\displaystyle 7.2\text{ m s}^{-1}
  4. D.
    8.0 m s−1\displaystyle 8.0\text{ m s}^{-1}
  5. E.
    8.4 m s−1\displaystyle 8.4\text{ m s}^{-1}
  6. F.
    9.4 m s−1\displaystyle 9.4\text{ m s}^{-1}
Answer and solution

Answer: D

1. From the graph, the force F\displaystyle F decreases linearly with displacement x\displaystyle x , starting at F=12 N\displaystyle F = 12\text{ N} at x=0 m\displaystyle x = 0\text{ m} and reaching F=0 N\displaystyle F = 0\text{ N} at x=9.0 m\displaystyle x = 9.0\text{ m} .

2. The equation for the line is: F(x)=12−129x=12−43x\displaystyle F(x) = 12 - \dfrac{12}{9}x = 12 - \dfrac{4}{3}x At x=6.0 m\displaystyle x = 6.0\text{ m} : F(6.0)=12−43(6.0)=4.0 N\displaystyle F(6.0) = 12 - \dfrac{4}{3}(6.0) = 4.0\text{ N} 3. The work done W\displaystyle W by the force between x=0 m\displaystyle x = 0\text{ m} and x=6.0 m\displaystyle x = 6.0\text{ m} is the area under the F–x\displaystyle F\text{--}x graph (a trapezium): W=12+4.02×6.0=48 J\displaystyle W = \dfrac{12 + 4.0}{2} \times 6.0 = 48\text{ J} 4. By the work–energy theorem: W=ΔEk=12mv2−12mu2\displaystyle W = \Delta E_\text{k} = \dfrac{1}{2}mv^2 - \dfrac{1}{2}mu^2 48=12(2.0)v2−12(2.0)(4.0)2\displaystyle 48 = \dfrac{1}{2}(2.0)v^2 - \dfrac{1}{2}(2.0)(4.0)^2 48=v2−16\displaystyle 48 = v^2 - 16 v2=64  ⟹  v=8.0 m s−1\displaystyle v^2 = 64 \implies v = 8.0\text{ m s}^{-1}
Question 5
A spring obeys Hooke's law. A force of 6.0 N\displaystyle 6.0\,\mathrm{N} produces an extension of 3.0 cm\displaystyle 3.0\,\mathrm{cm} .

The spring is already extended by 2.0 cm\displaystyle 2.0\,\mathrm{cm} . How much additional energy must be transferred to the spring to increase its extension to 5.0 cm\displaystyle 5.0\,\mathrm{cm} ?
  1. A.
    0.09 J\displaystyle 0.09\,\mathrm{J}
  2. B.
    0.15 J\displaystyle 0.15\,\mathrm{J}
  3. C.
    0.21 J\displaystyle 0.21\,\mathrm{J}
  4. D.
    0.25 J\displaystyle 0.25\,\mathrm{J}
  5. E.
    0.29 J\displaystyle 0.29\,\mathrm{J}
Answer and solution

Answer: C

The spring constant is k=6.0/0.030=200 N m−1\displaystyle k=6.0/0.030=200\,\mathrm{N\,m^{-1}} . Subtract the initial stored energy from the final stored energy: ΔE=12(200)(0.0502−0.0202)=0.21 J.\displaystyle \Delta E=\tfrac12(200)(0.050^2-0.020^2)=0.21\,\mathrm{J}. Using only the change in extension in 12kx2\displaystyle \tfrac12kx^2 would be incorrect.
Question 6
A hot solid block is placed into a cold liquid in an insulated container.

The mass of the liquid is twice the mass of the block. The specific heat capacity of the liquid is three times the specific heat capacity of the block.

The temperature of the block decreases by ΔT\displaystyle \Delta T .

Assuming no thermal energy is transferred to the surroundings, what is the temperature increase of the liquid?
  1. A.
    ΔT6\displaystyle \dfrac{\Delta T}{6}
  2. B.
    ΔT5\displaystyle \dfrac{\Delta T}{5}
  3. C.
    ΔT3\displaystyle \dfrac{\Delta T}{3}
  4. D.
    ΔT2\displaystyle \dfrac{\Delta T}{2}
  5. E.
    5ΔT\displaystyle 5\Delta T
  6. F.
    6ΔT\displaystyle 6\Delta T
Answer and solution

Answer: A

By conservation of energy, the thermal energy lost by the block equals the thermal energy gained by the liquid: Qlost=Qgained\displaystyle Q_{\text{lost}} = Q_{\text{gained}} .
Using Q=mcΔT\displaystyle Q = mc\Delta T , we have mbcbΔT=mlclΔTl\displaystyle m_b c_b \Delta T = m_l c_l \Delta T_l .
Substituting the given ratios ml=2mb\displaystyle m_l = 2m_b and cl=3cb\displaystyle c_l = 3c_b gives mbcbΔT=(2mb)(3cb)ΔTl\displaystyle m_b c_b \Delta T = (2m_b)(3c_b)\Delta T_l .
Simplifying this yields mbcbΔT=6mbcbΔTl\displaystyle m_b c_b \Delta T = 6 m_b c_b \Delta T_l , which rearranges to ΔTl=ΔT6\displaystyle \Delta T_l = \dfrac{\Delta T}{6} .
Question 7
The activity of a radioactive source is measured. Its activity after 1 hour is 8 times its activity after 4 hours.
Its activity after 2 hours is 120 Bq.

What was the initial activity of the source in Bq?
  1. A.
    30\displaystyle 30
  2. B.
    120\displaystyle 120
  3. C.
    240\displaystyle 240
  4. D.
    480\displaystyle 480
  5. E.
    960\displaystyle 960
  6. F.
    1920\displaystyle 1920
Answer and solution

Answer: D

Let A(t)\displaystyle A(t) be the activity at time t\displaystyle t in hours, and let A0\displaystyle A_0 be the initial activity at t=0\displaystyle t=0 . Radioactive decay is an exponential process, so we can write the activity as A(t)=A0rt\displaystyle A(t) = A_0 r^t , where r\displaystyle r is the constant decay factor per hour.

From the first piece of information:
A(1)=8×A(4) A(1) = 8 \times A(4)
Substituting our expression for activity:
A0r1=8×(A0r4) A_0 r^1 = 8 \times (A_0 r^4)
Since A0\displaystyle A_0 and r\displaystyle r are non-zero, we can divide both sides by A0r\displaystyle A_0 r :
1=8r3 1 = 8r^3
r3=18 r^3 = \frac{1}{8}
Taking the cube root gives the decay factor per hour:
r=12 r = \frac{1}{2}
This means the half-life of the source is 1 hour.

Now we use the second piece of information: A(2)=120\displaystyle A(2) = 120 Bq.
We relate this to the initial activity A0\displaystyle A_0 using our formula:
A(2)=A0r2 A(2) = A_0 r^2
Substituting the known values:
120=A0(12)2=A0(14) 120 = A_0 \left(\frac{1}{2}\right)^2 = A_0 \left(\frac{1}{4}\right)
Solving for A0\displaystyle A_0 :
A0=120×4=480 Bq A_0 = 120 \times 4 = 480\ \text{Bq}
Question 8
A sealed syringe with a freely moving, frictionless piston contains a fixed mass of an ideal gas.

At a depth of 30 m\displaystyle 30\text{ m} below the surface of a lake, the volume of the gas in the syringe is 80 cm3\displaystyle 80\text{ cm}^3 .

When the syringe is moved to a depth of 10 m\displaystyle 10\text{ m} below the surface, the volume of the gas increases to 120 cm3\displaystyle 120\text{ cm}^3 .

The temperature of the water is uniform throughout the lake, and the gas remains in thermal equilibrium with the water.

What is the volume of the gas in the syringe at the surface of the lake?
  1. A.
    135 cm3\displaystyle 135\text{ cm}^3
  2. B.
    140 cm3\displaystyle 140\text{ cm}^3
  3. C.
    150 cm3\displaystyle 150\text{ cm}^3
  4. D.
    160 cm3\displaystyle 160\text{ cm}^3
  5. E.
    180 cm3\displaystyle 180\text{ cm}^3
  6. F.
    200 cm3\displaystyle 200\text{ cm}^3
  7. G.
    240 cm3\displaystyle 240\text{ cm}^3
  8. H.
    360 cm3\displaystyle 360\text{ cm}^3
Answer and solution

Answer: D

Let p0\displaystyle p_0 be the atmospheric pressure at the surface of the lake, ρ\displaystyle \rho be the density of the water, and g\displaystyle g be the acceleration due to gravity.

The absolute pressure at a depth h\displaystyle h is given by: p(h)=p0+ρgh\displaystyle p(h) = p_0 + \rho g h Since the temperature of the ideal gas remains constant, Boyle's law applies: p(h)⋅V(h)=constant\displaystyle p(h) \cdot V(h) = \text{constant} Using the given data at depths h1=30 m\displaystyle h_1 = 30\text{ m} and h2=10 m\displaystyle h_2 = 10\text{ m} : (p0+30ρg)×80=(p0+10ρg)×120\displaystyle (p_0 + 30\rho g) \times 80 = (p_0 + 10\rho g) \times 120 Dividing both sides by 40\displaystyle 40 : 2(p0+30ρg)=3(p0+10ρg)\displaystyle 2(p_0 + 30\rho g) = 3(p_0 + 10\rho g) 2p0+60ρg=3p0+30ρg\displaystyle 2p_0 + 60\rho g = 3p_0 + 30\rho g p0=30ρg\displaystyle p_0 = 30\rho g Thus, at a depth of 30 m\displaystyle 30\text{ m} , the absolute pressure is: p(30)=p0+30ρg=p0+p0=2p0\displaystyle p(30) = p_0 + 30\rho g = p_0 + p_0 = 2p_0 At the surface of the lake ( h=0\displaystyle h = 0 ), the pressure is simply p0\displaystyle p_0 . Applying Boyle's law between depth 30 m\displaystyle 30\text{ m} and the surface: p0×Vsurface=p(30)×V(30)\displaystyle p_0 \times V_{\text{surface}} = p(30) \times V(30) p0×Vsurface=2p0×80 cm3\displaystyle p_0 \times V_{\text{surface}} = 2p_0 \times 80\text{ cm}^3 Vsurface=160 cm3\displaystyle V_{\text{surface}} = 160\text{ cm}^3
Question 9
A detector records the mean count rates shown for a source emitting two types of radiation. Its background count rate is 20 counts min−1\displaystyle 20\,\mathrm{counts\,min^{-1}} .

The paper stops all alpha particles. The aluminium stops all alpha and beta particles. Neither absorber significantly affects gamma radiation.

Which two types are detected from the source, and what is the count rate due to the more penetrating type alone?

<div class="md-table-wrap"><table class="md-table"><thead><tr><th>Absorber</th><th>Mean count rate / counts min⁻¹</th></tr></thead><tbody><tr><td>none</td><td>240</td></tr><tr><td>paper</td><td>240</td></tr><tr><td>aluminium</td><td>90</td></tr></tbody></table></div>
  1. A.
    alpha and beta; 70 counts min−1\displaystyle 70\,\mathrm{counts\,min^{-1}}
  2. B.
    alpha and gamma; 90 counts min−1\displaystyle 90\,\mathrm{counts\,min^{-1}}
  3. C.
    beta and gamma; 70 counts min−1\displaystyle 70\,\mathrm{counts\,min^{-1}}
  4. D.
    beta and gamma; 90 counts min−1\displaystyle 90\,\mathrm{counts\,min^{-1}}
  5. E.
    beta and gamma; 150 counts min−1\displaystyle 150\,\mathrm{counts\,min^{-1}}
Answer and solution

Answer: C

Paper causes no reduction, so no detected alpha component is present. Aluminium removes a beta component but leaves gamma plus background. The gamma contribution is 90−20=70 counts min−1\displaystyle 90-20=70\,\mathrm{counts\,min^{-1}} .
Question 10
A 0.050 kg\displaystyle 0.050\,\mathrm{kg} block is launched along a horizontal track by a spring of spring constant 800 N m−1\displaystyle 800\,\mathrm{N\,m^{-1}} , initially compressed by 5.0 cm\displaystyle 5.0\,\mathrm{cm} . The spring obeys Hooke's law and is not attached to the block.

The block loses 0.20 J\displaystyle 0.20\,\mathrm{J} of energy on a rough section, then rises up a smooth slope. All other energy losses are negligible.

What maximum vertical height does it reach above its starting level? Take g=10 N kg−1\displaystyle g=10\,\mathrm{N\,kg^{-1}} .
  1. A.
    0.40 m\displaystyle 0.40\,\mathrm{m}
  2. B.
    0.80 m\displaystyle 0.80\,\mathrm{m}
  3. C.
    1.0 m\displaystyle 1.0\,\mathrm{m}
  4. D.
    1.6 m\displaystyle 1.6\,\mathrm{m}
  5. E.
    2.0 m\displaystyle 2.0\,\mathrm{m}
Answer and solution

Answer: D

The initial spring energy is 12(800)(0.050)2=1.0 J\displaystyle \tfrac12(800)(0.050)^2=1.0\,\mathrm{J} . After the rough section, 0.80 J\displaystyle 0.80\,\mathrm{J} remains. At the highest point this is gravitational potential energy: h=0.800.050×10=1.6 m.\displaystyle h=\dfrac{0.80}{0.050\times10}=1.6\,\mathrm{m}.
Question 11
A U-tube of uniform cross-sectional area, open to the atmosphere at both ends, initially contains a liquid of density ρ1\displaystyle \rho_1 .

A second immiscible liquid of density ρ2\displaystyle \rho_2 is poured into the left arm, forming a column of height h\displaystyle h .

A third immiscible liquid of density ρ3\displaystyle \rho_3 is then poured into the right arm until the top liquid surfaces in both arms are at the exact same horizontal level, as shown in the diagram.

Given that ρ1>ρ2>ρ3\displaystyle \rho_1 > \rho_2 > \rho_3 , which expression gives the height h3\displaystyle h_3 of the column of the third liquid?
Exam diagram
  1. A.
    (ρ2ρ3)h\displaystyle \left(\dfrac{\rho_2}{\rho_3}\right) h
  2. B.
    (ρ3ρ2)h\displaystyle \left(\dfrac{\rho_3}{\rho_2}\right) h
  3. C.
    (ρ1−ρ2ρ1−ρ3)h\displaystyle \left(\dfrac{\rho_1 - \rho_2}{\rho_1 - \rho_3}\right) h
  4. D.
    (ρ1−ρ3ρ1−ρ2)h\displaystyle \left(\dfrac{\rho_1 - \rho_3}{\rho_1 - \rho_2}\right) h
  5. E.
    (ρ2−ρ3ρ1−ρ3)h\displaystyle \left(\dfrac{\rho_2 - \rho_3}{\rho_1 - \rho_3}\right) h
  6. F.
    (ρ1−ρ2ρ2−ρ3)h\displaystyle \left(\dfrac{\rho_1 - \rho_2}{\rho_2 - \rho_3}\right) h
  7. G.
    (ρ1+ρ2ρ1+ρ3)h\displaystyle \left(\dfrac{\rho_1 + \rho_2}{\rho_1 + \rho_3}\right) h
  8. H.
    (ρ1−ρ2ρ1+ρ3)h\displaystyle \left(\dfrac{\rho_1 - \rho_2}{\rho_1 + \rho_3}\right) h
Answer and solution

Answer: C

Let atmospheric pressure be P0\displaystyle P_0 . Choose a horizontal reference level at the interface between liquid 1 and liquid 2 in the left arm (let this be height y=0\displaystyle y = 0 ).

In the left arm, the column of liquid 2 has height h\displaystyle h , so the absolute pressure at the reference level is: Pleft=P0+ρ2gh\displaystyle P_{\text{left}} = P_0 + \rho_2 g h In the right arm, the top surface is at the same horizontal level as the top surface in the left arm (at height y=h\displaystyle y = h ). Liquid 3 extends downwards by height h3\displaystyle h_3 , occupying the region from y=h−h3\displaystyle y = h - h_3 to y=h\displaystyle y = h . Liquid 1 fills the remainder of the right arm below this, from y=0\displaystyle y = 0 to y=h−h3\displaystyle y = h - h_3 .

The absolute pressure in the right arm at the reference level y=0\displaystyle y = 0 is: Pright=P0+ρ3gh3+ρ1g(h−h3)\displaystyle P_{\text{right}} = P_0 + \rho_3 g h_3 + \rho_1 g (h - h_3) Since liquid 1 is continuous across the bottom of the tube at height y=0\displaystyle y = 0 , Pleft=Pright\displaystyle P_{\text{left}} = P_{\text{right}} : ρ2gh=ρ3gh3+ρ1g(h−h3)\displaystyle \rho_2 g h = \rho_3 g h_3 + \rho_1 g (h - h_3) ρ2h=ρ3h3+ρ1h−ρ1h3\displaystyle \rho_2 h = \rho_3 h_3 + \rho_1 h - \rho_1 h_3 (ρ1−ρ3)h3=(ρ1−ρ2)h\displaystyle (\rho_1 - \rho_3) h_3 = (\rho_1 - \rho_2) h h3=(ρ1−ρ2ρ1−ρ3)h\displaystyle h_3 = \left(\dfrac{\rho_1 - \rho_2}{\rho_1 - \rho_3}\right) h
Question 12
Plane water waves travel along a straight canal. At a stationary post, wave crests are observed to pass every 2.0 s\displaystyle 2.0 \text{ s} . A boat travels along the canal in the same direction as the waves with a constant speed of 4.0 m/s\displaystyle 4.0 \text{ m/s} . An observer on the boat records that wave crests overtake the boat every 6.0 s\displaystyle 6.0 \text{ s} .

What is the speed of the waves?
  1. A.
    2.0 m/s\displaystyle 2.0 \text{ m/s}
  2. B.
    3.0 m/s\displaystyle 3.0 \text{ m/s}
  3. C.
    4.0 m/s\displaystyle 4.0 \text{ m/s}
  4. D.
    6.0 m/s\displaystyle 6.0 \text{ m/s}
  5. E.
    12.0 m/s\displaystyle 12.0 \text{ m/s}
Answer and solution

Answer: D

Let the speed of the waves be v\displaystyle v .

From the stationary observer, the period is T=2.0 s\displaystyle T = 2.0 \text{ s} . The wavelength is therefore:
λ=vT=2v \lambda = vT = 2v
The boat travels at u=4.0 m/s\displaystyle u = 4.0 \text{ m/s} in the same direction. Since the waves overtake the boat, the relative speed of the waves with respect to the boat is:
vrel=v−u=v−4 v_{\text{rel}} = v - u = v - 4
The observer on the boat measures the time for one wavelength to pass as Tobs=6.0 s\displaystyle T_{\text{obs}} = 6.0 \text{ s} . We can write:
Tobs=λvrel  ⟹  6=2vv−4 T_{\text{obs}} = \frac{\lambda}{v_{\text{rel}}} \implies 6 = \frac{2v}{v - 4}
Solving for v\displaystyle v :
3=vv−43(v−4)=v3v−12=v2v=12  ⟹  v=6.0 m/s 3 = \frac{v}{v - 4} \\ 3(v - 4) = v \\ 3v - 12 = v \\ 2v = 12 \implies v = 6.0 \text{ m/s}
Alternatively, using proportional reasoning: the observed period is 3\displaystyle 3 times the stationary period ( 6s\displaystyle 6\text{s} vs 2s\displaystyle 2\text{s} ), so the relative speed must be 13\displaystyle \dfrac{1}{3} of the wave speed. Thus v−4=v3\displaystyle v - 4 = \dfrac{v}{3} , which yields v=6\displaystyle v=6 .
Question 13
A small block of mass 2.0 kg\displaystyle 2.0\text{ kg} is released from rest at the top of a smooth, curved ramp at a height H=20 m\displaystyle H = 20\text{ m} above horizontal ground. The end of the ramp is horizontal and is at a height h=5.0 m\displaystyle h = 5.0\text{ m} above the ground, where the block leaves the ramp.

Air resistance is negligible.
The gravitational field strength is g=10 N kg−1\displaystyle g = 10\text{ N kg}^{-1} .

What is the horizontal distance d\displaystyle d travelled by the block between leaving the ramp and landing on the ground?
Exam diagram
  1. A.
    53 m\displaystyle 5\sqrt{3}\text{ m}
  2. B.
    10 m\displaystyle 10\text{ m}
  3. C.
    103 m\displaystyle 10\sqrt{3}\text{ m}
  4. D.
    15 m\displaystyle 15\text{ m}
  5. E.
    20 m\displaystyle 20\text{ m}
  6. F.
    203 m\displaystyle 20\sqrt{3}\text{ m}
Answer and solution

Answer: C

1. Conservation of energy as the block slides down the smooth track: ΔEp=Ek  ⟹  mg(H−h)=12mv2\displaystyle \Delta E_p = E_k \implies mg(H - h) = \dfrac{1}{2}mv^2 v=2g(H−h)=2×10×(20−5.0)=300=103 m s−1\displaystyle v = \sqrt{2g(H - h)} = \sqrt{2 \times 10 \times (20 - 5.0)} = \sqrt{300} = 10\sqrt{3}\text{ m s}^{-1} 2. Vertical motion after leaving the ramp horizontally: h=12gt2  ⟹  5.0=12(10)t2  ⟹  t2=1.0  ⟹  t=1.0 s\displaystyle h = \dfrac{1}{2}gt^2 \implies 5.0 = \dfrac{1}{2}(10)t^2 \implies t^2 = 1.0 \implies t = 1.0\text{ s} 3. Horizontal distance travelled: d=vt=103 m s−1×1.0 s=103 m\displaystyle d = v t = 10\sqrt{3}\text{ m s}^{-1} \times 1.0\text{ s} = 10\sqrt{3}\text{ m}
Question 14
Three identical resistors are connected in a circuit with a DC power supply. One resistor is connected in series with a parallel combination of the other two.

What fraction of the total power supplied by the source is dissipated by the parallel combination?
  1. A.
    19\displaystyle \dfrac{1}{9}
  2. B.
    16\displaystyle \dfrac{1}{6}
  3. C.
    14\displaystyle \dfrac{1}{4}
  4. D.
    13\displaystyle \dfrac{1}{3}
  5. E.
    12\displaystyle \dfrac{1}{2}
  6. F.
    23\displaystyle \dfrac{2}{3}
Answer and solution

Answer: D

Let the resistance of each identical resistor be R\displaystyle R .

The parallel combination of two such resistors has an equivalent resistance, Rp\displaystyle R_p , given by:
1Rp=1R+1R=2R  ⟹  Rp=R2 \frac{1}{R_p} = \frac{1}{R} + \frac{1}{R} = \frac{2}{R} \implies R_p = \frac{R}{2}
This combination is in series with the third resistor, so the total equivalent resistance of the circuit is:
Rtotal=R+Rp=R+R2=3R2 R_{\text{total}} = R + R_p = R + \frac{R}{2} = \frac{3R}{2}
The total current I\displaystyle I from the supply flows through both the single series resistor and the parallel combination block. Since power is given by P=I2Req\displaystyle P = I^2 R_{\text{eq}} , the power dissipated by each part is directly proportional to its equivalent resistance.

The fraction of the total power dissipated by the parallel combination is therefore the ratio of its resistance to the total resistance:
PparallelPtotal=RpRtotal=R/23R/2=13 \frac{P_{\text{parallel}}}{P_{\text{total}}} = \frac{R_p}{R_{\text{total}}} = \frac{R/2}{3R/2} = \frac{1}{3}
Question 15
An electric pump lifts oil of density 800 kg m−3\displaystyle 800 \text{ kg m}^{-3} through a vertical height of 5.0 m\displaystyle 5.0 \text{ m} at a constant volumetric flow rate of 2.5×10−3 m3 s−1\displaystyle 2.5 \times 10^{-3} \text{ m}^3 \text{ s}^{-1} . The pump system has an overall efficiency of 50%\displaystyle 50\% and is connected to a 50 V\displaystyle 50 \text{ V} DC power supply.

Assuming g=10 m s−2\displaystyle g = 10 \text{ m s}^{-2} , what is the current drawn from the supply?
  1. A.
    1.0 A\displaystyle 1.0 \text{ A}
  2. B.
    2.0 A\displaystyle 2.0 \text{ A}
  3. C.
    2.5 A\displaystyle 2.5 \text{ A}
  4. D.
    4.0 A\displaystyle 4.0 \text{ A}
  5. E.
    5.0 A\displaystyle 5.0 \text{ A}
Answer and solution

Answer: D

First, we determine the mass flow rate m˙\displaystyle \dot{m} from the volumetric flow rate Q\displaystyle Q and density ρ\displaystyle \rho :
m˙=ρQ=800×(2.5×10−3)=2.0 kg s−1 \dot{m} = \rho Q = 800 \times (2.5 \times 10^{-3}) = 2.0 \text{ kg s}^{-1}
Next, we calculate the useful mechanical power output Pout\displaystyle P_{\text{out}} required to lift this mass against gravity:
Pout=m˙gh=2.0×10×5.0=100 W P_{\text{out}} = \dot{m} g h = 2.0 \times 10 \times 5.0 = 100 \text{ W}
The efficiency β=0.50\displaystyle \beta = 0.50 relates the output mechanical power to the input electrical power ( Pin\displaystyle P_{\text{in}} ). Since β=PoutPin\displaystyle \beta = \dfrac{P_{\text{out}}}{P_{\text{in}}} :
Pin=Poutβ=1000.50=200 W P_{\text{in}} = \frac{P_{\text{out}}}{\beta} = \frac{100}{0.50} = 200 \text{ W}
Finally, we find the current I\displaystyle I using the electrical power formula Pin=IV\displaystyle P_{\text{in}} = IV :
I=PinV=20050=4.0 A I = \frac{P_{\text{in}}}{V} = \frac{200}{50} = 4.0 \text{ A}
Question 16
A continuous wave travels from medium P, where its speed is v\displaystyle v , into medium Q, where its speed is 0.8v\displaystyle 0.8v .
Which of the following gives the fractional change in the wavelength of the wave?
  1. A.
    −14\displaystyle -\dfrac{1}{4}
  2. B.
    −15\displaystyle -\dfrac{1}{5}
  3. C.
    0\displaystyle 0
  4. D.
    15\displaystyle \dfrac{1}{5}
  5. E.
    14\displaystyle \dfrac{1}{4}
Answer and solution

Answer: B

When a wave passes from one medium to another, its frequency f\displaystyle f remains constant (determined by the source).

Using the wave equation v=fλ\displaystyle v = f\lambda , we have the proportionality:
λ∝v \lambda \propto v
The speed changes from v\displaystyle v to 0.8v\displaystyle 0.8v .
Therefore, the new wavelength is:
λnew=0.8λold=45λold \lambda_{\text{new}} = 0.8 \lambda_{\text{old}} = \frac{4}{5} \lambda_{\text{old}}
The fractional change is:
Δλλold=λnew−λoldλold=0.8λold−λoldλold=0.8−1=−0.2=−15 \frac{\Delta \lambda}{\lambda_{\text{old}}} = \frac{\lambda_{\text{new}} - \lambda_{\text{old}}}{\lambda_{\text{old}}} = \frac{0.8\lambda_{\text{old}} - \lambda_{\text{old}}}{\lambda_{\text{old}}} = 0.8 - 1 = -0.2 = -\frac{1}{5}
Question 17
A solid cuboid is made from a material of density 2000 kg m−3\displaystyle 2000\, \text{kg m}^{-3} . Its side lengths are in the ratio 1:2:4\displaystyle 1:2:4 . The cuboid is placed on a horizontal surface to exert the maximum possible pressure. This pressure is 8.0 kPa\displaystyle 8.0\, \text{kPa} .

What is the mass of the cuboid?

(The gravitational field strength g\displaystyle g is 10 N kg−1\displaystyle 10\, \text{N kg}^{-1} )
  1. A.
    2.0 kg\displaystyle 2.0\, \text{kg}
  2. B.
    16 kg\displaystyle 16\, \text{kg}
  3. C.
    128 kg\displaystyle 128\, \text{kg}
  4. D.
    160 kg\displaystyle 160\, \text{kg}
  5. E.
    1024 kg\displaystyle 1024\, \text{kg}
  6. F.
    16000 kg\displaystyle 16000\, \text{kg}
Answer and solution

Answer: B

Let the side lengths of the cuboid be k\displaystyle k , 2k\displaystyle 2k , and 4k\displaystyle 4k for some scaling factor k\displaystyle k .

The volume of the cuboid is V=(k)(2k)(4k)=8k3\displaystyle V = (k)(2k)(4k) = 8k^3 .

The mass of the cuboid is m=density×volume=ρV=2000×8k3=16000k3\displaystyle m = \text{density} \times \text{volume} = \rho V = 2000 \times 8k^3 = 16000k^3 .

The weight of the cuboid is F=mg=(16000k3)(10)=160000k3\displaystyle F = mg = (16000k^3)(10) = 160000k^3 .

Pressure is given by P=F/A\displaystyle P = F/A . To exert the maximum pressure, the cuboid must rest on its smallest face area.
The three possible face areas are: A1=(k)(2k)=2k2\displaystyle A_1 = (k)(2k) = 2k^2 (smallest) A2=(k)(4k)=4k2\displaystyle A_2 = (k)(4k) = 4k^2 A3=(2k)(4k)=8k2\displaystyle A_3 = (2k)(4k) = 8k^2 (largest)

The minimum area is Amin=2k2\displaystyle A_{min} = 2k^2 .

The maximum pressure is Pmax=FAmin=160000k32k2=80000k\displaystyle P_{max} = \dfrac{F}{A_{min}} = \dfrac{160000k^3}{2k^2} = 80000k .

We are given that Pmax=8.0 kPa=8000 Pa\displaystyle P_{max} = 8.0\, \text{kPa} = 8000\, \text{Pa} .
Setting the two expressions for Pmax\displaystyle P_{max} equal: 8000=80000k\displaystyle 8000 = 80000k This gives k=800080000=0.1 m\displaystyle k = \dfrac{8000}{80000} = 0.1\, \text{m} .

Now we can find the mass using our expression for m\displaystyle m : m=16000k3=16000×(0.1)3=16000×0.001=16 kg\displaystyle m = 16000k^3 = 16000 \times (0.1)^3 = 16000 \times 0.001 = 16\, \text{kg} .
Question 18
An NTC thermistor is connected in series with a 2.0 kΩ\displaystyle 2.0\,\mathrm{k\Omega} resistor across a constant 12 V\displaystyle 12\,\mathrm{V} supply. At the initial temperature the thermistor resistance is 4.0 kΩ\displaystyle 4.0\,\mathrm{k\Omega} .

The thermistor is warmed until its resistance becomes 1.0 kΩ\displaystyle 1.0\,\mathrm{k\Omega} . What happens to the electrical power transferred to the thermistor?
  1. A.
    It decreases to one quarter of its initial value.
  2. B.
    It decreases to one half of its initial value.
  3. C.
    It remains unchanged.
  4. D.
    It doubles.
  5. E.
    It quadruples.
Answer and solution

Answer: C

Initially I=12/6000=2.0 mA\displaystyle I=12/6000=2.0\,\mathrm{mA} , giving PT=I2RT=16 mW\displaystyle P_T=I^2R_T=16\,\mathrm{mW} . Afterwards I=12/3000=4.0 mA\displaystyle I=12/3000=4.0\,\mathrm{mA} , giving PT=(0.004)2(1000)=16 mW\displaystyle P_T=(0.004)^2(1000)=16\,\mathrm{mW} . The current doubles while the thermistor resistance falls by four.
Question 19
An object P collides with a stationary object Q on a straight, frictionless track. The graph shows the momentum of object P as a function of time.
0 4.0 2.0 time / s momentum of P / kg m s⁻¹
What is the momentum of object Q after the collision?
  1. A.
    −4.0 kg m s−1\displaystyle -4.0\,\text{kg m s}^{-1}
  2. B.
    −2.0 kg m s−1\displaystyle -2.0\,\text{kg m s}^{-1}
  3. C.
    0 kg m s−1\displaystyle 0\,\text{kg m s}^{-1}
  4. D.
    2.0 kg m s−1\displaystyle 2.0\,\text{kg m s}^{-1}
  5. E.
    4.0 kg m s−1\displaystyle 4.0\,\text{kg m s}^{-1}
  6. F.
    6.0 kg m s−1\displaystyle 6.0\,\text{kg m s}^{-1}
Answer and solution

Answer: D

The problem must be solved by applying the principle of conservation of linear momentum to the system of two objects, P and Q.

Let pP\displaystyle p_P and pQ\displaystyle p_Q be the momenta of objects P and Q, respectively.

Before the collision, we can read the momentum of P from the graph, and we are told Q is stationary.
pP,initial=4.0 kg m s−1 p_{P,\text{initial}} = 4.0\,\text{kg m s}^{-1}
pQ,initial=0 kg m s−1 p_{Q,\text{initial}} = 0\,\text{kg m s}^{-1}
So, the total initial momentum of the system is:
ptotal, initial=pP,initial+pQ,initial=4.0+0=4.0 kg m s−1 p_{\text{total, initial}} = p_{P,\text{initial}} + p_{Q,\text{initial}} = 4.0 + 0 = 4.0\,\text{kg m s}^{-1}
After the collision, we can read the new momentum of P from the graph.
pP,final=2.0 kg m s−1 p_{P,\text{final}} = 2.0\,\text{kg m s}^{-1}
Let the final momentum of Q be pQ,final\displaystyle p_{Q,\text{final}} . The total final momentum of the system is:
ptotal, final=pP,final+pQ,final=2.0+pQ,final p_{\text{total, final}} = p_{P,\text{final}} + p_{Q,\text{final}} = 2.0 + p_{Q,\text{final}}
By the conservation of momentum, the total initial momentum equals the total final momentum:
ptotal, initial=ptotal, final p_{\text{total, initial}} = p_{\text{total, final}}
4.0=2.0+pQ,final 4.0 = 2.0 + p_{Q,\text{final}}
Solving for the final momentum of Q gives:
pQ,final=4.0−2.0=2.0 kg m s−1 p_{Q,\text{final}} = 4.0 - 2.0 = 2.0\,\text{kg m s}^{-1}
Question 20
The graph shows the voltage produced by an electromagnetic generator during one complete cycle. It is connected in series with a 2.0 kΩ\displaystyle 2.0\,\mathrm{k\Omega} resistor and an ideal diode. The diode allows current only when the plotted voltage is positive. The voltage is unchanged by the connection.

How much charge passes through the resistor during one complete cycle?
Electromagnetic induction diagram
  1. A.
    2.0 μC\displaystyle 2.0\,\mu\mathrm{C}
  2. B.
    4.0 μC\displaystyle 4.0\,\mu\mathrm{C}
  3. C.
    8.0 μC\displaystyle 8.0\,\mu\mathrm{C}
  4. D.
    16 μC\displaystyle 16\,\mu\mathrm{C}
  5. E.
    zero
Answer and solution

Answer: B

The peak current in the positive half-cycle is 4.0/2000=2.0 mA\displaystyle 4.0/2000=2.0\,\mathrm{mA} . Its current-time graph is a triangle of base 4.0 ms\displaystyle 4.0\,\mathrm{ms} . The charge is its area: Q=12(4.0×10−3)(2.0×10−3)=4.0 μC.\displaystyle Q=\tfrac12(4.0\times10^{-3})(2.0\times10^{-3})=4.0\,\mu\mathrm{C}. The diode blocks the negative half-cycle, so it does not cancel this charge.
Question 21
An LDR and a fixed 1.0 kΩ\displaystyle 1.0\,\mathrm{k\Omega} resistor are connected in series across an ideal 9.0 V\displaystyle 9.0\,\mathrm{V} supply. At the fixed light intensity, the working LDR has resistance 2.0 kΩ\displaystyle 2.0\,\mathrm{k\Omega} .

A fault develops in exactly one component. An ideal voltmeter across the LDR now reads 9.0 V\displaystyle 9.0\,\mathrm{V} , and an ideal ammeter in series with the supply reads 4.5 mA\displaystyle 4.5\,\mathrm{mA} . All connecting wires remain intact.

Which fault is consistent with both readings?
  1. A.
    The LDR has become an open circuit.
  2. B.
    The LDR has become a short circuit.
  3. C.
    The fixed resistor has become an open circuit.
  4. D.
    The fixed resistor has become a short circuit.
  5. E.
    The LDR resistance has increased to 3.0 kΩ\displaystyle 3.0\,\mathrm{k\Omega} .
Answer and solution

Answer: D

The observed total resistance is 9.0/0.0045=2000 Ω\displaystyle 9.0/0.0045=2000\,\Omega , exactly the working LDR resistance. The LDR has the whole supply voltage, so the fixed resistor must have zero voltage across it while carrying current: it is short-circuited. An open LDR could give the voltage reading but would give zero current.
Question 22
An ideal gas bubble of volume V\displaystyle V is released at a depth 3H\displaystyle 3H below the surface of a liquid. It rises to a depth H\displaystyle H .
The atmospheric pressure at the surface is equal to the pressure exerted by a column of the liquid of height H\displaystyle H .
Assuming the temperature of the gas remains constant, what is the volume of the bubble at depth H\displaystyle H ?
  1. A.
    V3\displaystyle \dfrac{V}{3}
  2. B.
    V2\displaystyle \dfrac{V}{2}
  3. C.
    2V\displaystyle 2V
  4. D.
    3V\displaystyle 3V
  5. E.
    4V\displaystyle 4V
Answer and solution

Answer: C

Since the temperature of the gas is constant, the bubble obeys Boyle's Law, which states that pressure times volume is constant ( PV=k\displaystyle PV = k ).

The total pressure at a depth h\displaystyle h below the surface is the sum of the atmospheric pressure and the hydrostatic pressure. We are given that the atmospheric pressure is equivalent to the pressure exerted by a column of the liquid of height H\displaystyle H .

Therefore, the total pressure at a depth h\displaystyle h is proportional to (H+h)\displaystyle (H+h) .

At the initial depth of 3H\displaystyle 3H , the pressure P1\displaystyle P_1 is proportional to H+3H=4H\displaystyle H+3H = 4H .
At the final depth of H\displaystyle H , the pressure P2\displaystyle P_2 is proportional to H+H=2H\displaystyle H+H = 2H .

Let the initial volume be V\displaystyle V . By Boyle's Law, P1V=P2Vfinal\displaystyle P_1 V = P_2 V_{\text{final}} . We can find the final volume:
Vfinal=V×P1P2=V×4H2H=2V V_{\text{final}} = V \times \frac{P_1}{P_2} = V \times \frac{4H}{2H} = 2V
The new volume of the bubble is 2V\displaystyle 2V .
Question 23
A laser beam from a fixed source is incident on a plane mirror M1\displaystyle M_1 . The reflected beam then strikes a second plane mirror M2\displaystyle M_2 . M1\displaystyle M_1 rotates with angular speed ω\displaystyle \omega and M2\displaystyle M_2 rotates with angular speed 4ω\displaystyle 4\omega . Both mirrors rotate in the same direction about parallel axes perpendicular to the plane of incidence.

What is the angular speed of the final beam reflected from M2\displaystyle M_2 ?
  1. A.
    3ω\displaystyle 3\omega
  2. B.
    5ω\displaystyle 5\omega
  3. C.
    6ω\displaystyle 6\omega
  4. D.
    8ω\displaystyle 8\omega
  5. E.
    10ω\displaystyle 10\omega
Answer and solution

Answer: C

Let θ\displaystyle \theta denote the angle of a beam and ϕ\displaystyle \phi the angle of a mirror normal. The law of reflection θout−ϕ=−(θin−ϕ)\displaystyle \theta_{out} - \phi = -(\theta_{in} - \phi) leads to the angular velocity relation:
θ˙out=2ϕ˙−θ˙in \dot{\theta}_{out} = 2\dot{\phi} - \dot{\theta}_{in}
**First Reflection ( M1\displaystyle M_1 ):**
The source is fixed, so θ˙in,1=0\displaystyle \dot{\theta}_{in,1} = 0 . The mirror rotates at ω\displaystyle \omega .
θ˙out,1=2(ω)−0=2ω \dot{\theta}_{out,1} = 2(\omega) - 0 = 2\omega
The beam incident on M2\displaystyle M_2 rotates at 2ω\displaystyle 2\omega .

**Second Reflection ( M2\displaystyle M_2 ):**
The mirror rotates at 4ω\displaystyle 4\omega in the same direction.
θ˙out,2=2(4ω)−θ˙in,2 \dot{\theta}_{out,2} = 2(4\omega) - \dot{\theta}_{in,2}
Substituting θ˙in,2=2ω\displaystyle \dot{\theta}_{in,2} = 2\omega :
θ˙out,2=8ω−2ω=6ω \dot{\theta}_{out,2} = 8\omega - 2\omega = 6\omega
Question 24
A solid cuboid of mass 1.0 kg\displaystyle 1.0\,\text{kg} is placed on a horizontal surface. Depending on which face it rests on, it can exert one of three possible pressures: 0.40 N cm−2\displaystyle 0.40\,\text{N}\,\text{cm}^{-2} , 0.50 N cm−2\displaystyle 0.50\,\text{N}\,\text{cm}^{-2} , or 2.0 N cm−2\displaystyle 2.0\,\text{N}\,\text{cm}^{-2} .

What is the density of the cuboid's material, in g cm−3\displaystyle \text{g}\,\text{cm}^{-3} ?

(Take the gravitational field strength g\displaystyle g to be 10 N kg−1\displaystyle 10\,\text{N}\,\text{kg}^{-1} .)
  1. A.
    0.05\displaystyle 0.05
  2. B.
    0.40\displaystyle 0.40
  3. C.
    2.0\displaystyle 2.0
  4. D.
    2.5\displaystyle 2.5
  5. E.
    20\displaystyle 20
  6. F.
    20000\displaystyle 20000
Answer and solution

Answer: E

The cuboid's weight is 10 N. The three possible contact areas are 10/0.40=25 cm2\displaystyle 10/0.40=25\,cm^2 , 10/0.50=20 cm2\displaystyle 10/0.50=20\,cm^2 , and 10/2.0=5 cm2\displaystyle 10/2.0=5\,cm^2 . If the side lengths are a,b,c\displaystyle a,b,c , then ab=25\displaystyle ab=25 , bc=20\displaystyle bc=20 , and ca=5\displaystyle ca=5 , so (abc)2=(25)(20)(5)=2500\displaystyle (abc)^2=(25)(20)(5)=2500 and the volume is abc=50 cm3\displaystyle abc=50\,cm^3 . The mass is 1000 g, hence the density is 1000/50=20 g cm−3\displaystyle 1000/50=20\,g\,cm^{-3} . The correct option is E.
Question 25
A resistor of resistance 4.0 Ω\displaystyle 4.0\, \Omega is connected to a power supply. The graph shows the total charge that has passed through the resistor as a function of time.
time / s total charge / C 0 10 5.0
What is the potential difference across the resistor?
  1. A.
    0.50 V\displaystyle 0.50\,\mathrm{V}
  2. B.
    2.0 V\displaystyle 2.0\,\mathrm{V}
  3. C.
    8.0 V\displaystyle 8.0\,\mathrm{V}
  4. D.
    20 V\displaystyle 20\,\mathrm{V}
  5. E.
    25 V\displaystyle 25\,\mathrm{V}
  6. F.
    40 V\displaystyle 40\,\mathrm{V}
Answer and solution

Answer: C

The electric current I\displaystyle I is defined as the rate of flow of charge Q\displaystyle Q with respect to time t\displaystyle t . For a constant current, this is given by:
I=ΔQΔt I = \frac{\Delta Q}{\Delta t}
This corresponds to the gradient of the charge-time graph.

Using the points (0,0)\displaystyle (0, 0) and (5.0 s,10 C)\displaystyle (5.0\,\mathrm{s}, 10\,\mathrm{C}) , we can calculate the gradient:
I=10 C−0 C5.0 s−0 s=105.0 A=2.0 A I = \frac{10\,\mathrm{C} - 0\,\mathrm{C}}{5.0\,\mathrm{s} - 0\,\mathrm{s}} = \frac{10}{5.0}\,\mathrm{A} = 2.0\,\mathrm{A}
Now, we use Ohm's law, V=IR\displaystyle V = IR , to find the potential difference V\displaystyle V across the resistor.
V=(2.0 A)×(4.0 Ω)=8.0 V V = (2.0\,\mathrm{A}) \times (4.0\,\Omega) = 8.0\,\mathrm{V}
Question 26
A charged particle of mass m\displaystyle m is released from rest in a uniform field. The particle experiences a constant force F\displaystyle F due to the field.

What is the kinetic energy of the particle after it has travelled a distance d\displaystyle d parallel to the force?
  1. A.
    12Fd\displaystyle \dfrac{1}{2} Fd
  2. B.
    2Fd\displaystyle 2Fd
  3. C.
    Fd\displaystyle Fd
  4. D.
    Fdm\displaystyle \dfrac{Fd}{m}
  5. E.
    Fdm\displaystyle Fdm
Answer and solution

Answer: C

According to the work-energy theorem, the net work done on an object is equal to the change in its kinetic energy. Wnet=ΔKE\displaystyle W_{net} = \Delta KE The work done, W\displaystyle W , by a constant force F\displaystyle F acting over a distance d\displaystyle d in the direction of the force is given by: W=F×d\displaystyle W = F \times d The particle starts from rest, so its initial kinetic energy, KEinitial\displaystyle KE_{initial} , is zero. The change in kinetic energy is therefore equal to its final kinetic energy, KEfinal\displaystyle KE_{final} . ΔKE=KEfinal−KEinitial=KEfinal−0=KEfinal\displaystyle \Delta KE = KE_{final} - KE_{initial} = KE_{final} - 0 = KE_{final} Equating the work done to the change in kinetic energy: KEfinal=W=Fd\displaystyle KE_{final} = W = Fd Thus, the kinetic energy of the particle is Fd\displaystyle Fd .
Question 27
An insulated metal tank acquires a positive charge as fuel flows out. It is then connected to Earth by a conducting wire and becomes uncharged.

Which row correctly describes the charging process and the particle movement through the wire during discharge?
  1. A.
    The tank gained protons; protons move from the tank to Earth.
  2. B.
    The tank lost electrons; electrons move from the tank to Earth.
  3. C.
    The tank gained electrons; electrons move from Earth to the tank.
  4. D.
    The tank lost protons; protons move from Earth to the tank.
  5. E.
    The tank lost electrons; electrons move from Earth to the tank.
Answer and solution

Answer: E

Positive charging means a loss of electrons, not a gain of protons. Earthing allows electrons to enter from Earth and replace that deficit. The conducting path prevents a large charge build-up that could cause a spark.

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