Examine the options under and see the way you went!

## Part 1: A number of Selection

Query | Reply | Resolution |

1 | D | At max peak the vertical element of velocity is zero, so the projectile travels horizontally. |

2 | B | Spectral strains at totally different wavelengths point out a distinct Doppler shift, and therefore totally different velocity relative to the Earth. The identical set of spectral strains signifies the presence of the identical component, and therefore the identical composition. |

3 | C | They used alpha particles, and in accordance with Thomson’s mannequin initially anticipated them to go by way of atoms. |

4 | C | B class stars have the next temperature than M class stars. |

5 | D | An induced present will solely stream by way of the coil on the fitting when it experiences a change in flux, i.e. when the present within the coil on the left decreases attributable to opening the change. Therefore the galvanometer will learn zero to start with, then deflect, the return to zero. |

6 | A | That is dictated by Wien’s Displacement Regulation: (λ_{textrm{max}}=frac{b}{T}), giving (λ_{textrm{max}}∝ frac{1}{T}). |

7 | D | There might be an opposing pressure on the magnet in accordance with Lenz’s Regulation, requiring a North pole on the solenoid near the South pole of the magnet. The appropriate hand coil rule offers the route of the present. |

8 | A | Hubble discovered that recessional velocity was proportional to distance, therefore a straight line graph. The distances in choice A are in keeping with Hubble’s measurements. |

9 | D | Orbital interval and whole power each enhance with the radius of the orbit, therefore are each larger for GEO satellites that are additional away from Earth than LEO satellites. |

10 | A | Malus’s Regulation offers (frac{I_B}{I_0} =cos^2 30° =0.75). This helps the wave mannequin of sunshine. |

11 | C | Kepler’s Third Regulation offers (r= sqrt[3]{40,000^2}=1170 textrm{ AU}) |

12 | A | This describes beta decay the place a neutron (udd) is remodeled right into a proton (uud), an electron, and an antineutrino. Alternatively, conservation of cost can be utilized. |

13 | B | The photon power is (E=frac{hc}{λ}=3.1×10^{-19} textrm{J}). The variety of photons per second is the facility divided by the photon power: (frac{30×10^{-3}}{3.1×10^{-19}} = 9.8×10^{16}) |

14 | B | (v= sqrt{frac{GM}{r}}), giving (v∝frac{1}{sqrt{r}}). Doubling (r) would change (v) by an element of (frac{1}{sqrt{2}}≈0.7). |

15 | C | (d) is the space between the slits, given within the query. (theta) is the angle at which the sunshine propagates from the slits to a most. The trail the sunshine takes kinds the hypotenuse of a triangle shaped by becoming a member of the centre of the slits to the central most, to the higher most. The ratio (frac{0.08}{0.3}) offers the tan of (θ). |

16 | D | Within the second state of affairs the online pressure on the particle is upwards and equal in magnitude to its weight (in order to supply the identical movement upwards). The particle has each its weight and the electrical pressure (qE) appearing on it: (F=mg=qE-mg) (E=frac{2mg}{q}) |

17 | D | The present by way of (QR) would create a discipline to the fitting above (QR) in accordance with the fitting hand grip rule. The cathode rays would expertise a pressure upwards in accordance with the fitting hand palm rule (NB cathode rays are electrons, and therefore are unfavorable). Growing the resistance would lower the present in (QR) and would scale back this impact, ensuing within the cathode rays transferring downwards in the direction of (Z). |

18 | B | In accordance with Faraday’s Regulation and Lenz’s Regulation, the coil experiences an anticlockwise EMF when it enters the magnetic discipline, and a clockwise EMF when it exits the sector. For the reason that battery is producing a clockwise present, the EMF will initially oppose the prevailing present, dimming the lamp on the way in which in to the sector, after which elevated the present on the way in which out, making the lamp brighter. |

19 | D | Power needed to be launched within the response whatever the properties of (Z). The binding power of (Y) is bigger than the sum of the binding energies of the reactants (W) and (Z). Breaking (W) and (Z) into constituent particles would eat (2.22 + 8.48 = 10.7 textrm{MeV}) of power, however forming (Y) will launch (28.3 textrm{MeV}), that means (28.3 – 10.7 = 17.6 textrm{MeV}) is launched with out contemplating (Z). The binding power of (Z) might be zero (if it’s a single particle) or optimistic (if it’s a nucleus consisting of not less than two particles). Its binding power may also be launched, that means the power launched is (17.6 textrm{MeV}) + the binding power of (Z). |

20 | B | This can be a assertion of Newton’s Third Regulation. The pressure of the blackboard on the dice supplies the centripetal pressure required to maintain the dice in uniform round movement. This pressure additionally determines the friction by way of (f=μN). |

## Part 2: Brief Response

### Query 21 (2 marks)

De Broglie recommended that electrons and different matter particles would have a wave nature along with their particle nature.

He proposed their wavelength would depend upon Planck’s fixed and their momentum, (λ=frac{h}{mv}=frac{h}{p}).

### Query 22 (3 marks)

Linear velocity: The Doppler shift of spectral strains noticed within the star’s mild will point out i’s linear velocity alongside the road of sight between the Earth and the star. A pink shift signifies the star is transferring away from Earth and a blue shift signifies the star is transferring in the direction of Earth. The quantity of shift signifies the velocity.

Rotational velocity: The broadening of spectral strains (attributable to Doppler broadening) signifies the rotational velocity of the star. Extra broadening signifies the next rotational velocity.

Stress: The broadening of the sides of spectral strains (attributable to stress broadening) signifies the stress within the star’s ambiance. Extra broadening signifies the next stress.

### Query 23 (3 marks)

The information given needs to be plotted on the graph supplied. The road of finest match needs to be drawn.

The road of finest match needs to be extrapolated to search out the y-intercept which is the unfavorable of the work operate.

The work operate is round (4 textrm{eV}).

### Query 24a (3 marks)

The enter energy is given within the query (V_P I_P=500 textual content{W}). The output energy may be calculated from (V_S I_S= 50×9=450 textrm{W}). Though there’s a discrepancy, with the output energy being (50 textrm{W}) decrease than the enter energy, conservation of power is just not violated.

The cast-iron core used will expertise a change in flux from the coils, which can end in an induced EMF and induced eddy currents. These eddy currents will generate warmth.

Warmth can also be produced within the wires of the coils attributable to their resistance. This warmth and the power of the eddy currents will correspond to the (50 textrm{W}) famous above.

### Query 24b (4 marks)

Laminating the iron core will stop eddy currents from forming giant loops because of the insulating materials. Because of this solely a lot smaller eddy currents will type inside every lamination. This reduces the power consumed by eddy currents, growing the effectivity of the transformer.

Growing the thickness of the wires within the coils will scale back their resistance and the power misplaced as warmth, once more growing the effectivity of the transformer.

### Query 25 (4 marks)

Maxwell, by way of the equations named after him, confirmed that electrical and magnetic fields are associated and components of the one phenomenon – electromagnetism.

He argued that simply as a altering magnetic discipline produces an electrical discipline (by way of induction and Faraday’s Regulation), a altering electrical discipline would additionally produce a magnetic discipline. Therefore he proposed that electromagnetic waves would type whereby altering electrical and magnetic fields will produce one another leading to a propagating wave.

Maxwell calculated the velocity of electromagnetic waves and confirmed that they might journey on the velocity of sunshine. Therefore he proposed that mild is an electromagnetic wave, a part of a large spectrum of such waves.

As a way to create an electromagnetic wave the electrical discipline and magnetic discipline are each required to alter. This may be achieved by an accelerating electrical cost. An oscillating cost will produce electromagnetic waves because the cost repeatedly accelerates throughout the oscillation. The change in place of the cost produces a altering electrical discipline, and the change in velocity of the cost produces a altering present and therefore a altering magnetic discipline. The fields then proceed to mutually generate one another.

### Query 26a (3 marks)

The factors seem to lie on a straight line, thus the coed concludes that that the torque is proportional to the angle.

The gradient is roughly (frac{42-8}{25-5}=1.7)

Therefore the equation of the road of finest match could be (τ=1.7×θ).

### Query 26b (3 marks)

We all know that (tau = rF sin theta) , and therefore (tau propto sin theta) .

For small angles, so the coed’s mannequin might be legitimate as (sin theta approx theta).

For bigger angles the mannequin will turn out to be more and more inaccurate because it doesn’t have the proper dependence on angle.

### Query 27a (3 marks)

Have a laser on a prepare transferring at excessive velocity. A pulse of sunshine leaves the laser at one finish of the carriage, displays off a mirror on the different finish of the carriage, after which returns to the laser.

Two observers measure the time it takes for the sunshine to return to the laser. One observer is within the prepare carriage with the laser. A second observer is stationary on the bottom outdoors the prepare.

The 2 observers will calculate the identical velocity for the sunshine, however will measure totally different instances for the sunshine’s journey attributable to time dilation. The observer on the prepare will measure correct time for the sunshine’s journey, and the observer on the bottom will measure the dilated time which might be longer and can depend upon the velocity of the prepare relative to the bottom.

[There are other acceptable answers.]

### Query 27b (3 marks)

Within the Hafele-Keating experiment three excessive precision atomic clocks have been used. One remained stationary on the bottom, one was flown world wide eastwards and one was flown world wide westwards.

Every clock, when thought of within the reference body of the Earth’s axis, is travelling at a distinct velocity, and therefore experiences a distinct period of time dilation. The eastwards transferring clock experiences probably the most, and the westwards transferring clock the least.

When the clocks arrived again on the identical place, their time readings have been in contrast and so they have been in keeping with the theories of relativity, together with time dilation.

### Query 28 (3 marks)

Earlier than the rotation, the present by way of (XY) is parallel to the magnetic discipline, so (XY) experiences no pressure, (F=ILB sinθ=ILB sin0°=0).

After the rotation, the angle between (XY) and the magnetic discipline will increase to 90°, so the pressure will increase to (F=ILB sin90°=ILB)

The present by way of (WX) is at all times perpendicular to the magnetic discipline, and the wire experiences a continuing pressure of (F = ILB) each earlier than and after the rotation.

### Query 29 (3 marks)

The work finished is (W=qV.)

The work finished will enhance the cost’s kinetic power such that (W=ΔK=frac{1}{2} mv^2-frac{1}{2} mu^2).

Since (u=0), we get:

(qV=frac{1}{2} mv^2) which supplies (v= sqrt{frac{2qV}{m}}).

### Query 30a (3 marks)

The potential power at P is transformed to kinetic power at (Q):

(frac{1}{2} mv^2=mgh),

the place ( v ) is the velocity because it leaves the monitor, and ( h ) is the preliminary peak above level (Q). Therefore:

(h=frac{v^2}{2g}=frac{1.5^2}{2×9.8}=0.115 m)

### Query 30b (3 marks)

The ball’s preliminary vertical velocity is (u_y= 1.5 sin50°=1.149 textrm{ms}^{-1}) downwards.

Take all the way down to be the optimistic route. Utilizing the equation (s=ut+frac{1}{2}at^2) we resolve for (s) as follows:

(s=1.149×0.5+frac{1}{2×9.8}×0.5^2)

Giving a peak of (s = 1.8 textual content{ m}) above the ground.

### Query 31a (4 marks)

Because the fan’s rotation velocity will increase, it should apply an more and more robust pressure on the air molecules to push them downwards.

In accordance with Newton’s Third Regulation, the air will apply an equal and reverse pressure on the fan.

The fan will therefore expertise two forces: its weight downwards which is fixed, and the pressure from the air upwards which will increase with the fan’s velocity.

The spring stability will initially learn the fan’s weight. When the fan is switched on the studying will lower over the primary 10 seconds because the fan hurries up and the upwards pressure will increase. It’ll then stay fixed.

### Query 31b (4 marks)

The scholar’s prediction is inaccurate.

The present by way of the fan’s motor might be affected by the again EMF which depends upon rotation velocity. Initially the coil of the motor is stationary, and the coil experiences no change in flux. Therefore there isn’t a induced (again) EMF within the coil. The ability provide to the coil will then end in most present, which leads to most torque and the fan’s rotation velocity growing.

Because the velocity will increase, the induced (again) EMF will increase attributable to flux altering at a sooner fee (Faraday’s Regulation). The again EMF opposes the facility provide (Lenz’s Regulation) and reduces the present, and in flip, the torque.

Therefore the present will begin at most and can scale back over the primary 10 s because the fan hurries up. As soon as the fan rotates at fixed velocity, there might be a continuing present that produces simply sufficient torque to beat any friction and end in a internet zero torque on the fan.

### Query 32 (5 marks)

Thomson’s experiments with cathode rays allowed the charge-to-mass ratio of electrons to be measured. Thomson used an electrical discipline to speed up the electrons and used a magnetic discipline to deflect them. He then used an electrical discipline to cancel the deflection attributable to the magnetic discipline. By contemplating the movement of the electrons beneath the affect of the magnetic discipline and the pressure stability between the electrical and magnetic forces, Thomson was capable of calculate the charge-to-mass ratio.

Millikan’s oil drop experiments measured the cost of the electron. Millikan used charged drops of oil between two steel plates. By observing the drops utilizing a microscope and measuring their terminal velocity when falling beneath the affect of gravity, Millikan calculated the scale and mass of the drops.

By making use of an electrical discipline between the plates, Millikan balanced the electrical and gravitational forces and calculated the cost of the oil drops. He calculated the cost of an electron from the smallest distinction in cost between two oil drops. For the reason that charge-to-mass was already recognized, Millikan was capable of additionally calculate the electron’s mass.

Firing electrons at protons at excessive velocity, within the Stanford Linear Accelerator, in an experiment known as Deep Inelastic Scattering, confirmed that protons weren’t elementary particles.

When the scattering was inelastic, a few of the kinetic power of the collision was used to take away a quark from the proton. This led to the invention of quarks.

[There are other acceptable answers for the second part, however they need to relate to a fundamental particle, like a quark or a lepton.]

### Query 33 (4 marks)

**Similarity:** Each particles will bear uniform round movement. They expertise a continuing pressure (F=qvB) perpendicular to their velocity. Because the pressure is perpendicular to their movement it does no work, so their velocity is fixed, that means their route modifications at a continuing fee and so they comply with a round path at fixed velocity. The magnetic pressure supplies the centripetal pressure: (qvB=frac{mv^2}{r}) .

**Distinction:** The radius of the round path might be totally different. Given the mass of the alpha is roughly 4 instances the mass of the proton, and its cost is double the cost of the proton, and we get (r=frac{mv}{qB}) from above:

The radius of the proton’s path might be (r_{textrm{proton}}=frac{m_{textrm{proton}} v}{eB})

The radius of the alpha’s path might be (r_{textrm{alpha}}=frac{4m_{textrm{proton} }v}{2eB}=2r_{textrm{proton}})

### Query 34 (9 marks)

Manufacturing of power is thru nuclear fusion, particularly the fusion of hydrogen (protons) into helium within the core of the star. Within the core the place temperature may be very excessive, protons journey at excessive velocity and may overcome the electrostatic repulsion between them. They can get sufficiently shut for the nuclear robust pressure to bind them into a bigger nucleus.

The mass of helium is lower than the mass of 4 protons, and the distinction in mass is transformed to power by way of Einstein’s power mass equivalence (E=mc^2).

The fusion response can proceed immediately by way of the proton-proton chain to supply helium, or it could be catalysed by a big nucleus by way of the CNO cycle.

The power launched from fusion heats up the core of the Solar, and this power propagates by way of the Solar’s floor. Approximating the Solar as a black physique, the black physique spectrum and Wien’s Regulation permits the floor temperature to be estimated:

(λ_{textrm{max}}=510 textrm{nm})

(T=frac{b}{λ_{textrm{max}}} =frac{2.898×10^{-3}}{510×10^{-9}}=5682 textual content{Ok})

The Solar would radiate power within the type of black physique radiation, which can unfold out over a sphere because it travels away from the Solar. The depth of the radiated power follows an inverse sq. regulation. From the worth of the depth at Earth, the entire energy emitted by the Solar may be calculated:

(I=frac{P}{A}=frac{P}{4πr^2 }) (P=I×A=1360×4π(1.5×10^11 )^2=3.85×10^{26} textual content{W})

The Solar is in thermal equilibrium and maintains a continuing temperature, that means that the facility radiated is the same as the facility generated by fusion within the core.

### Query 35 (4 marks)

The equipment is an accelerometer that may measure the automobile’s acceleration, within the route perpendicular to the perimeters of the automobile (i.e. left and proper).

If the string is hanging at an angle to the fitting, then there’s a internet pressure to the left attributable to pressure. Contemplate the forces on the mass hanging from the string. The vertical forces on the mass would be the vertical element of pressure ((T cos θ)) and it’s weight. These forces cancel since there isn’t a vertical internet pressure. The horizontal pressure on the mass is the horizontal element of the strain ((T sin θ)) which would be the internet pressure on the mass.

This offers:

(T cosθ=mg), based mostly on the vertical forces, therefore (T=frac{mg}{cosθ})

Substituting into an expression for the online pressure:

(T sinθ =ma), therefore (a=g tan θ).

The angle at which the mass hangs may give the magnitude of acceleration (perpendicular to the perimeters of the automobile) and the aspect it leans in the direction of may give the route (turning left or proper). Within the instance described, the automobile is in uniform round movement, turning left, because it accelerates the left while travelling at fixed velocity, therefore (T sinθ =frac{mv^2}{r}).

### Query 36 (7 marks)

Calculate the mass defect:

(Δm= m_R- m_p=197.999-(193.988+4.00260)=0.0084 textual content{ u})

Calculate the power launched in (textrm{MeV}), then joules:

(E= 0.0084×931.5=7.8246 textual content{ MeV}

=7.8246×10^6 textrm{eV}

=(7.8246×10^6×1.6×10^{-19}) textrm{J}

=1.25×10^{-12} textual content{ J})

Conservation of power signifies that the response merchandise will need to have a complete kinetic power equal to this. Subsequently the kinetic power of the alpha particle is (1.25×10^{-12}-2.55×10^{-14}=1.23×10^{-12} textual content{ J}) which is round 50 instances bigger than the power of the polonium.

For the reason that radon was initially at relaxation the preliminary momentum of this response was 0. To preserve momentum, the ultimate momentum of the polonium and alpha should add to zero. This implies they must be equal in magnitude and reverse in route.

For the reason that mass of polonium is round 50 instances bigger than the mass of helium, the velocity of the helium atom have to be 50 instances bigger than the polonium to present the identical momentum, (p=mv).

(Ok=frac{1}{2} mv^2=frac{1}{2}mv×v).

Since their momenta (mv) are equal, and the helium has a velocity that’s 50 instances bigger, it should even have a kinetic power that’s 50 instances bigger.