Loading....
|
Press & Hold to Drag Around |
|||
|
Click Here to Close |
|||
Question 1 Report
A body of mass 5kg movimg with a velocity of 30ms-1 due east is suddenly hit by another body and changes its velocity to 50ms-1 in the same direction. Calculate the magnitude of the impulse received.
Answer Details
To calculate the magnitude of the impulse received by the body, we need to use the impulse-momentum theorem, which states that the impulse applied to a body is equal to the change in momentum of the body. The momentum of an object is defined as the product of its mass and velocity, i.e., P = mv, where P is momentum, m is mass, and v is velocity. The change in momentum (ΔP) of the body can be calculated as the difference between the final momentum (Pf) and the initial momentum (Pi), i.e., ΔP = Pf - Pi. In this case, the initial momentum of the body is given by Pi = mv = 5 kg x 30 m/s = 150 kg m/s (due east), and the final momentum is given by Pf = mv = 5 kg x 50 m/s = 250 kg m/s (due east). Therefore, the change in momentum (ΔP) of the body is: ΔP = Pf - Pi = 250 kg m/s - 150 kg m/s = 100 kg m/s (due east). The magnitude of the impulse (J) received by the body is equal to the change in momentum (ΔP), i.e., J = ΔP = 100 kg m/s. Therefore, the answer is option A: 100 Ns.
Question 2 Report
An inductor of inductance 1.0H is connected in series with a capacitor of capacitance 2.0\(\mu F\) in an a.c. circuit. to resonate
Answer Details
Question 3 Report
A body of volume 0.046m3 is immersed in a liquid of density 980kgm-3 with \(\frac{3}{4}\) of its volume submerged. Calculate the upthrust on the body. [g = 10ms-2]
Answer Details
When an object is immersed in a fluid, it experiences an upward force called buoyant force or upthrust. This force is equal to the weight of the fluid displaced by the object. In this case, the volume of the object submerged in the fluid is \(\frac{3}{4}\) of its total volume. So, the volume of fluid displaced by the object is \(\frac{3}{4}\) x 0.046m3 = 0.0345m3 The weight of this displaced fluid can be calculated as: mass = density x volume = 980kgm-3 x 0.0345m3 = 33.81kg weight = mass x g = 33.81kg x 10ms-2 = 338.1N Therefore, the upthrust on the body is 338.1N. So, the correct answer is (C) 338.10N.
Question 4 Report
A resistor of resistance R is connected to a battery of negligible internal resistance. If a siilar resistor is connected in series with it the
Answer Details
Question 5 Report
In a hydraulic press, a force of 40N is applied to the smaller priston of area 10cm2. If the area of the larger piston is 200cm2, calculate the force obtained
Answer Details
In a hydraulic press, the pressure exerted is constant throughout the system. The pressure exerted is given by the formula: Pressure = Force / Area The force is directly proportional to the product of the pressure and area. F ∝ P.A If F1 is the force applied on the smaller piston of area A1, and F2 is the force on the larger piston of area A2, then: F1 / A1 = F2 / A2 Rearranging the formula above to solve for F2: F2 = F1 x (A2 / A1) Substituting the values given in the question, we get: F2 = 40N x (200cm2 / 10cm2) = 800N Therefore, the force obtained is 800N. The answer is (a) 800N.
Question 6 Report
An isolated metal sphere of radius R, carrying an electric charge Q, is situated in the medium of relative permitivity, Er. A test charge is placed at a point p, distance r from the surface of the sphere. Let Eo represent the permitivity of free space. The electric potential at p is given by the expression
Answer Details
Question 7 Report
The period of a simple pendulum of length 80.0cm was found to have doubled when the length of the pendulum was increased by X. Calculate X.
Answer Details
The period of a simple pendulum depends on its length. The relationship between the period (T) of a simple pendulum and its length (L) is given by the formula: T = 2π√(L/g), where g is the acceleration due to gravity. If the period of the pendulum doubles, then the new period (T') is twice the original period (T): T' = 2T = 4π√(L/g). If the length of the pendulum is increased by X, then the new length (L') is: L' = L + X. Substituting this into the formula for the new period gives: T' = 2π√((L+X)/g) Since T' = 2T, we can equate the two expressions for T' and simplify: 4π√(L/g) = 2π√((L+X)/g) Squaring both sides and canceling the common factor of 4π2/g, we get: L = (L+X)/2 Solving for X gives: X = L Therefore, the length of the pendulum needs to be increased by 80.0 cm to double its period, so X = 80.0 cm. The answer is option D: 240.0 cm.
Question 8 Report
Which of the following factors affect the rate of evaporation of a liquid? i. Exposed surface area ii Temperature of the surroundings iii. Thermal capacity of the liquid iv. Relative humidity of the surroundings
Answer Details
Question 9 Report
It takes 4 minutes to boil a quantity of water using an electrical heating coil. How long will it take to boil the same quantity of water using the same coil if the current is doubled? [Neglect any external heat losses]
Answer Details
If the current is doubled, the heating power will also double, which means that the coil will generate twice as much heat per unit of time. Since the amount of water to be boiled is the same, the amount of heat needed to raise its temperature to boiling point will also be the same. However, with twice the heating power, the time required to deliver that amount of heat will be halved. Therefore, the time it will take to boil the water using the doubled current will be half of the time it took with the original current, which is 2 minutes. So the answer is 2 minutes.
Question 10 Report
When table salt is added to ice, the melting point of the ice
Answer Details
When table salt (sodium chloride) is added to ice, it lowers the melting point of the ice. This is because salt dissolves in the thin layer of water on the surface of the ice, forming a saltwater solution. The saltwater solution has a lower freezing point than pure water, so it requires a lower temperature to freeze. As a result, the ice melts at a lower temperature when salt is added, leading to a lowering of the melting point. This phenomenon is called freezing point depression.
Question 11 Report
Which of the following is used to measure relative humidity?
Answer Details
The instrument used to measure relative humidity is called a hygrometer. A hygrometer is a device that measures the amount of water vapor in the air or the humidity level. It measures the amount of moisture in the air by using two thermometers. One of the thermometers is dry while the other is kept wet. The difference in temperature between the two thermometers is then measured, and this difference is used to calculate the relative humidity of the air. Therefore, the correct option is (d) hygrometer.
Question 12 Report
An ammeter connected to an a.c circuit records 5.5. The peak current in the circuit is
Answer Details
In an AC circuit, the current is constantly changing direction and amplitude, and we use the root mean square (rms) value to measure the effective current flowing in the circuit. The peak value is the maximum value the current can reach, and it is equal to the square root of 2 times the rms value. So, in this case, we can use the formula: Peak value = sqrt(2) x rms value Given that the ammeter records 5.5, which is the rms value, we can calculate the peak value as follows: Peak value = sqrt(2) x 5.5 Peak value = 7.77 Rounded to one decimal place, the answer is 7.8A. Therefore, option A is the correct answer.
Question 13 Report
A boat is rocked by waves of speed 30ms-1 whose successive crest are 10m apart. Calculate the rate at which the boat recieves the waves
Answer Details
The rate at which the boat receives the waves is equal to the frequency of the waves. The distance between successive crests of the waves is 10m, which is also the wavelength of the wave. Therefore, the formula for the speed of the wave can be written as: speed = wavelength x frequency We are given the speed of the wave as 30m/s and the wavelength as 10m. Rearranging the formula above to solve for frequency gives: frequency = speed / wavelength Substituting the values we have: frequency = 30m/s / 10m = 3s-1 Therefore, the rate at which the boat receives the waves is 3s-1, which is option (C).
Question 14 Report
In which of the following states of matter do the molecules vibrate about their mean positions?
Answer Details
Question 15 Report
Electric energy is measured in
Answer Details
Electric energy is measured in kilowatt-hour (kWh). A kilowatt-hour is a unit of energy equivalent to the energy used by a load of one kilowatt in one hour. This unit is commonly used by electric utility companies to measure the amount of energy consumed by households or businesses. An ampere is a unit of electric current, while a coulomb is a unit of electric charge. A kilowatt is a unit of power, which is the rate at which energy is consumed or produced. Therefore, none of these units are suitable for measuring electric energy.
Question 16 Report
The following properties of a solid are measured on earth i. weight ii. specific heat capacity iii. density. Which of the properties will hav a different value if measured in space?
Answer Details
The property that will have a different value if measured in space is (i) weight. Weight is the force with which an object is attracted towards the center of the earth due to gravity. In space, there is no gravity or the gravitational force is very weak compared to that on earth. Therefore, the weight of an object measured on earth will be different from the weight of the same object measured in space. On the other hand, specific heat capacity and density are intrinsic properties of a substance that do not depend on the gravitational force, so they will have the same value whether measured on earth or in space. Thus, the answer is (i) only.
Question 17 Report
Which of the following is not a part of the electro-magnetic spectrum?
Answer Details
Alpha rays are not part of the electromagnetic spectrum. The electromagnetic spectrum includes all forms of electromagnetic radiation that differ in wavelength and frequency, including radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays, and gamma rays. Alpha rays are actually alpha particles, which consist of two protons and two neutrons and are emitted by some types of radioactive materials. Alpha particles have a positive charge and are relatively large compared to other types of radiation, so they are not part of the electromagnetic spectrum.
Question 19 Report
A stone of mass 0.5kg is dropped from a height of 12m. Calculate its maximum kinetic energy. [g = 10ms-1]
Answer Details
When the stone is dropped, it gains potential energy due to its position above the ground. This potential energy is converted into kinetic energy as the stone falls towards the ground. The maximum kinetic energy of the stone will occur just before it hits the ground when all its potential energy has been converted to kinetic energy. The potential energy of the stone is given by the formula P.E. = mgh, where m is the mass of the stone, g is the acceleration due to gravity and h is the height from which it is dropped. Substituting the given values into the formula, we get: P.E. = (0.5 kg)(10 ms^-2)(12 m) = 60 J Since the maximum kinetic energy is equal to the potential energy at the maximum height, the maximum kinetic energy of the stone is also 60 J. Therefore, the correct option is (d) 60.0 J.
Question 20 Report
A thin lens is placed 50cm from an illustrated object. The image produced has linear magnification of \(\frac{1}{4}\). Calculate the power of the lens in dioptres
Answer Details
Question 21 Report
A rod of initial lenght 2m at a temperature of 25oC is heated to 80oC. Calculate the increase in length of the rod if its linear expansivity is 4.0 x 10-3K-1
Answer Details
The increase in length of the rod can be calculated using the formula: ΔL = αLΔT where: ΔL = change in length α = linear expansivity coefficient L = initial length ΔT = change in temperature Substituting the given values, we have: ΔL = (4.0 x 10^-3 K^-1) x (2m) x (80°C - 25°C) ΔL = 0.44m Therefore, the increase in length of the rod is 0.44m. Note: When a material is heated, its length increases due to the expansion of its particles. The linear expansivity coefficient (α) represents the fractional change in length per unit change in temperature.
Question 22 Report
An isolated metal sphere of radius R, carrying an electric charge Q, is situated in the medium of relative permitivity, Er. A test charge is placed at a point p, distance r from the surface of the sphere. Let Eo represent the permitivity of free space. The magnitude of the electric field intensity at P is given by the expression
Answer Details
Question 23 Report
A machine of efficiency 80% is used to lift a box. If the effort applied by the machine is twice the weight of the box. Calculate the velocity ratio of the machine
Answer Details
Question 24 Report
Light travelling through a small pinhole usually does not make a shadow with a distinct sharp edge because of
Answer Details
The phenomenon that causes light travelling through a small pinhole to not make a shadow with a distinct sharp edge is called diffraction. Diffraction is the bending of waves around an obstacle or through an opening. When light waves pass through a small pinhole, they bend and spread out after passing through the hole. As a result, the edges of the shadow become blurred, and a distinct sharp edge cannot be seen. This effect is also observed when sound waves pass through small openings, such as in the case of musical instruments. Therefore, the correct option is "diffraction".
Question 25 Report
An electron of mass 9.1 x 10-31 kg moves with a speed of 107ms-1. Calculate the wavelength of the associates wave[h = 6.6 x 10-34Js]
Answer Details
The wavelength of the associated wave of a moving particle can be determined using the de Broglie equation, which relates the particle's momentum to its wavelength: λ = h / p where λ is the wavelength, h is Planck's constant, and p is the momentum of the particle. We can determine the momentum of the electron using the formula: p = m * v where p is the momentum, m is the mass, and v is the velocity. Substituting the given values, we get: p = (9.1 x 10^-31 kg) * (10^7 ms^-1) = 9.1 x 10^-24 kgms^-1 Substituting p into the de Broglie equation, we get: λ = h / p = (6.6 x 10^-34 Js) / (9.1 x 10^-24 kgms^-1) = 7.25 x 10^-11 m Therefore, the wavelength of the associated wave is 7.25 x 10^-11 m. The correct option is (c) 7.25 x 10^-11 m.
Question 26 Report
A concave mirror forms a magnified and errect image only when the object is placed
Answer Details
A concave mirror forms an image of an object when light rays from the object are reflected off the mirror. The image formed can either be magnified or reduced in size, and it can be erect or inverted depending on where the object is placed in relation to the mirror. In this question, we are told that the concave mirror forms a magnified and erect image. For this to happen, the object must be placed between the focus and the pole of the mirror. This is because, in a concave mirror, light rays coming parallel to the principal axis are reflected and converge at the focus. When an object is placed between the focus and the pole of the mirror, the reflected rays will still converge and form a real image that is magnified and erect. If the object is placed at the focus, the reflected rays will be parallel, and no image will be formed. If the object is placed beyond the radius of curvature, the image formed will be real, but it will be inverted. If the object is placed at the center of curvature, the image formed will be real, inverted, and the same size as the object.
Question 27 Report
Two forces 6N and 8N, act eastwards and northwards respectively on a body. Calculate the magnitude of their equilibrant
Answer Details
When two forces act on a body, their resultant force can be calculated using vector addition. The equilibrant is the force required to balance the two forces and bring the body into a state of equilibrium. It has the same magnitude as the resultant force but acts in the opposite direction. To calculate the equilibrant, we need to find the resultant force of the two given forces. We can do this by drawing a vector diagram or using trigonometry. Using trigonometry, we can find the magnitude of the resultant force as follows: \begin{align*} F_{\text{resultant}} &= \sqrt{(6\text{N})^2 + (8\text{N})^2} \\ &= \sqrt{36\text{N}^2 + 64\text{N}^2} \\ &= \sqrt{100\text{N}^2} \\ &= 10\text{N} \end{align*} The equilibrant has the same magnitude as the resultant force but acts in the opposite direction. Therefore, the magnitude of the equilibrant is also 10N. So, the correct answer is 10N.
Question 28 Report
When a conductor mounted on an insulating stand is charged and left for sometime, the conductor eventually loses all its charges. This is because the
Answer Details
When a conductor mounted on an insulating stand is charged and left for some time, the conductor eventually loses all its charges. This happens because the charges on the conductor attract opposite charges from the surrounding air, which neutralize the charge on the conductor. This process is called electrostatic discharge or simply discharge. As a result, the charges are dissipated, and the conductor becomes neutral again. Therefore, the correct option is "charges on the conductor are neutralized by opposite charges from the surrounding air".
Question 29 Report
A body moving in a circle at constant speed has i. a velocity tangent to the circle ii. a constant kinetic energy ii. an acceleration directed towards the circmference of the circle ii. an acceleration directed towards the circumference of the circle. Which of the statements above are correct?
Question 30 Report
If the direction of the curent in a straight wire is reversed, the magnetic field
Answer Details
When an electric current flows through a straight wire, it generates a magnetic field around the wire. The direction of the magnetic field is dependent on the direction of the current flow. If the direction of the current flow is reversed, then the direction of the magnetic field will also be reversed. Therefore, the correct answer is: the magnetic field is oppositely directed.
Question 31 Report
Which of the following types of types of motion will be produced when a pair of equal and opposite non-collinear parallel forces acts on a body?
Answer Details
When a pair of equal and opposite non-collinear parallel forces act on a body, a rotational motion will be produced. This is because the two forces will cause a torque or moment about the center of mass of the body, which will result in the body rotating around that point. The magnitude of the torque is proportional to the product of the forces and the perpendicular distance between them, which is also known as the lever arm. Therefore, the greater the force or the longer the lever arm, the greater the torque and the faster the body will rotate. The other options, such as vibrational motion, random motion, and translational motion, do not apply in this scenario because the forces are not acting in such a way as to cause those types of motion.
Question 32 Report
A faulty thermometer registers 102.5oC t 100oC. If the thermometer has no zero hour, what will it register at 55oC?
Answer Details
Question 34 Report
A satellite in circular motion around the earth does not have
Answer Details
A satellite in circular motion around the earth does not have a uniform velocity. Even though its speed is constant, the direction of its velocity is constantly changing due to the centripetal force provided by the gravitational force of the earth. Therefore, it has an acceleration towards the center of the circle, which is equal in magnitude to the centripetal force acting on it.
Question 35 Report
A battery of e.m.f. 12.0V and internal resistance0.5\(\Omega\) is connected to 1.5\(\Omega\) and 4.0\(\Omega\) series resistor. Calculate the terminal voltage of the battery.
Question 36 Report
The angle of incidence of a ray of light on a plane mirror is 55o. Determine the angle between the reflected ray and the mirror
Question 37 Report
A cell of e.m.f. 1.5V is connected in series with a resistor of resistance 3.0\(\Omega\). A voltmeter connected across the cell registers 0.9V. Calculate the internal resistance of the cell
Answer Details
The internal resistance of a cell can be calculated using the equation: r = (E - V)/I where r is the internal resistance of the cell, E is the e.m.f. of the cell, V is the voltage across the cell, and I is the current passing through the circuit. In this case, the e.m.f. of the cell is given as 1.5V, and the resistance of the resistor is 3.0\(\Omega\). Since the cell and the resistor are connected in series, the current passing through them will be the same. We can calculate the current using Ohm's law: I = V/R = 0.9V/3.0\(\Omega\) = 0.3A Now we can use the formula for the internal resistance: r = (E - V)/I = (1.5V - 0.9V)/0.3A = 2.0\(\Omega\) Therefore, the internal resistance of the cell is 2.0\(\Omega\). Answer option A is correct.
Question 38 Report
as a bicycle Tyre was being pumped up, it was noticed that contrary to Boyle's law the volume increased as the pressure increased. The best explanation of this is that Boyle's law is only true for
Answer Details
Question 39 Report
A wire of resistivity 4.40 x 10-5\(\Omega\) cm has a cross sectional area of 7.50 x 10-4 cm2. Calculate the length of the wire that will be required to make a 4.0\(\Omega\) resistor
Answer Details
The resistance of a wire can be calculated using the formula R = \(\frac{\rho L}{A}\), where R is the resistance, \(\rho\) is the resistivity of the wire, L is the length of the wire, and A is the cross-sectional area of the wire. In this problem, we are given the resistivity of the wire and the cross-sectional area of the wire. We are also told that the wire needs to have a resistance of 4.0\(\Omega\). We can rearrange the formula to solve for L: L = \(\frac{AR}{\rho}\) Substituting the values given in the question, we have: L = \(\frac{(7.50 \times 10^{-4} cm^{2}) (4.0 \Omega)}{4.40 \times 10^{-5} \Omega cm}\) = 68.18 cm Therefore, the length of the wire required to make a 4.0\(\Omega\) resistor is 68.18 cm. The correct option is: 68.18cm.
Question 40 Report
The main difference between between x-rays and \(\gamma-rays\) lies in their
Answer Details
Question 41 Report
Which of the following characteristics of light determines its colour?
Answer Details
The characteristic of light that determines its color is wavelength. Wavelength is the distance between consecutive peaks or troughs of a wave. Different colors of light have different wavelengths, with longer wavelengths corresponding to colors like red and orange, and shorter wavelengths corresponding to colors like blue and violet. So, when light interacts with an object and some wavelengths are absorbed while others are reflected or transmitted, the color that we perceive is determined by the wavelengths that are reflected or transmitted to our eyes. Therefore, the wavelength is responsible for the color of light we see.
Question 42 Report
Which of the following properties of a steel bar can be measured in terms of the dimension of length only?
Question 43 Report
Two isotopes of uranium are designed as \(^{238}U\) and \(^{235}U\). The numbers 238 and 235 represent their
Answer Details
The numbers 238 and 235 represent the nucleon numbers of the two isotopes of uranium, namely \(^{238}U\) and \(^{235}U\). The nucleon number represents the total number of protons and neutrons in the nucleus of an atom. Therefore, \(^{238}U\) has a nucleus consisting of 238 protons and neutrons, and \(^{235}U\) has a nucleus consisting of 235 protons and neutrons. Atomic number, on the other hand, represents the number of protons in the nucleus of an atom, and proton number and neutron number are not commonly used scientific terms.
Question 44 Report
Sound interference is necessary to produce the phenomenon of
Answer Details
Sound interference is necessary to produce the phenomenon of beats. Beats are produced when two sound waves of slightly different frequencies are superimposed on each other. The interference of these sound waves causes the sound intensity to oscillate, resulting in a periodic variation of loudness known as beats. These beats can be heard as a throbbing sound. The number of beats produced per second is equal to the difference in frequency between the two sound waves. The phenomenon of beats is used in tuning musical instruments to ensure they are in tune with each other.
Question 45 Report
The maximum displacement of a vibrating tunning fork is its
Answer Details
The maximum displacement of a vibrating tuning fork is its amplitude. In a vibrating system, amplitude refers to the maximum displacement of the vibrating object from its equilibrium or rest position. For a tuning fork, the amplitude refers to the maximum distance that the prongs of the fork move away from their resting position when they are vibrating. The amplitude of a tuning fork determines the intensity or loudness of the sound it produces. A larger amplitude produces a louder sound, while a smaller amplitude produces a softer sound. Therefore, the correct option is "amplitude".
Question 46 Report
(a) Explain why it is not advisable to sterilize a clinical thermometer in boiling water at normal atmospheric pressure.
(b) State the effect of an increase in pressure on the (i) boiling point; and (ii) melting point of water.
(c) The graph below shows the saturated vapour pressure (s.v.p.) of water plotted against temperature.
Pure water is known to boil at 100°C and at an atmospheric pressure of 760 mmHg. What general conclusion can be drawn from the information given above?
(d) A thread of mercury of length 20 cm is used to trap some air in a capillary tube with uniform cross-sectional area and closed at one end. With the tube vertical and the open end uppermost, the length of the trapped air column is 15cm. Calculate the length of the air column when the tube is held: (i) horizontally; (ii) vertically with the open end underneath. [Atmospheric pressure = 76 cmHg]
(a) A clinical thermometer has a small temperature range, approximately \(35\,^{\circ}\mathrm{C}\) to \(43\,^{\circ}\mathrm{C}\). In boiling water, the mercury expands excessively and may produce enough pressure to crack or burst the glass thermometer.
(b)
(c) A liquid boils when its saturated vapour pressure is equal to the external atmospheric pressure. Thus, at \(100\,^{\circ}\mathrm{C}\), the saturated vapour pressure of pure water is \(760\ \mathrm{mmHg}\).
(d) The arrangements of the trapped air and mercury thread are represented below.
Since the capillary tube has uniform cross-sectional area, volume is proportional to length. Hence, Boyle's law may be written as:
\[P_1L_1=P_2L_2\]
Initially, with the open end uppermost, the pressure of the trapped air is
\[P_1=76+20=96\ \mathrm{cmHg}\]
and \(L_1=15\ \mathrm{cm}\). Therefore,
\[P_1L_1=96\times15=1440\ \mathrm{cmHg\,cm}.\]
(i) Tube held horizontally
The pressure of the trapped air is atmospheric pressure:
\[P_2=76\ \mathrm{cmHg}\]
\[96\times15=76L_2\]
\[L_2=\frac{1440}{76}=18.95\ \mathrm{cm}\approx\boxed{18.9\ \mathrm{cm}}\]
(ii) Tube held vertically with the open end underneath
The pressure of the trapped air is
\[P_3=76-20=56\ \mathrm{cmHg}\]
\[96\times15=56L_3\]
\[L_3=\frac{1440}{56}=25.7\ \mathrm{cm}\]
\[\boxed{L_3=25.7\ \mathrm{cm}}\]
Answer Details
(a) A clinical thermometer has a small temperature range, approximately \(35\,^{\circ}\mathrm{C}\) to \(43\,^{\circ}\mathrm{C}\). In boiling water, the mercury expands excessively and may produce enough pressure to crack or burst the glass thermometer.
(b)
(c) A liquid boils when its saturated vapour pressure is equal to the external atmospheric pressure. Thus, at \(100\,^{\circ}\mathrm{C}\), the saturated vapour pressure of pure water is \(760\ \mathrm{mmHg}\).
(d) The arrangements of the trapped air and mercury thread are represented below.
Since the capillary tube has uniform cross-sectional area, volume is proportional to length. Hence, Boyle's law may be written as:
\[P_1L_1=P_2L_2\]
Initially, with the open end uppermost, the pressure of the trapped air is
\[P_1=76+20=96\ \mathrm{cmHg}\]
and \(L_1=15\ \mathrm{cm}\). Therefore,
\[P_1L_1=96\times15=1440\ \mathrm{cmHg\,cm}.\]
(i) Tube held horizontally
The pressure of the trapped air is atmospheric pressure:
\[P_2=76\ \mathrm{cmHg}\]
\[96\times15=76L_2\]
\[L_2=\frac{1440}{76}=18.95\ \mathrm{cm}\approx\boxed{18.9\ \mathrm{cm}}\]
(ii) Tube held vertically with the open end underneath
The pressure of the trapped air is
\[P_3=76-20=56\ \mathrm{cmHg}\]
\[96\times15=56L_3\]
\[L_3=\frac{1440}{56}=25.7\ \mathrm{cm}\]
\[\boxed{L_3=25.7\ \mathrm{cm}}\]
Question 47 Report
A particle is projected horizontally at 15ms\(^{-1}\) from a height of 20m.
Calculate the horizontal distance covered by the particle just before hitting the ground.
[g = 10 ms\(^{-2}\)]
The particle is projected horizontally, so its vertical motion is free fall from rest and its horizontal velocity stays constant.
Time to reach the ground (vertical fall of \( h = 20\,\text{m} \)): \[ h = \tfrac{1}{2} g t^2 \Rightarrow t = \sqrt{\dfrac{2h}{g}} = \sqrt{\dfrac{2 \times 20}{10}} = \sqrt{4} = 2\,\text{s}. \]
Horizontal distance (range): \[ x = u \times t = 15 \times 2 = 30\,\text{m}. \]
The horizontal distance covered before hitting the ground is 30 m.
Answer Details
The particle is projected horizontally, so its vertical motion is free fall from rest and its horizontal velocity stays constant.
Time to reach the ground (vertical fall of \( h = 20\,\text{m} \)): \[ h = \tfrac{1}{2} g t^2 \Rightarrow t = \sqrt{\dfrac{2h}{g}} = \sqrt{\dfrac{2 \times 20}{10}} = \sqrt{4} = 2\,\text{s}. \]
Horizontal distance (range): \[ x = u \times t = 15 \times 2 = 30\,\text{m}. \]
The horizontal distance covered before hitting the ground is 30 m.
Question 48 Report
(a) Briefly explain the following terms:
(i) emission line spectra;
(ii) line absorption spectra.
(b) Draw a labeled diagram showing the structure of a simple type of photocell and explain its mode of operation.
(c) State two
(i) reasons to show that x-rays are waves;
(ii) uses of x-rays other than in medicine.
(d) An electron jumps from an energy level of \(-1.6\ \text{eV}\) to one of \(-1.4\ \text{eV}\) in an atom. Calculate the energy and wavelength of the emitted radiation. [ \(h = 6.6 \times 10^{-34}\ \text{Js}\); \(c = 3.00 \times 10^8\ \text{ms}^{-1}\); \(\text{eV} = 1.6 \times 10^{-19}\ \text{J}\) ]
(a)(i) Emission line spectrum: This consists of separate bright lines of definite wavelengths on a dark background. It is produced when excited atoms emit radiation as their electrons fall from higher to lower energy levels.
(ii) Line absorption spectrum: This consists of dark lines on a continuous bright spectrum. It is obtained when continuous light passes through a cool gas, which absorbs particular wavelengths corresponding to its energy transitions.
(b) Simple photocell
When radiation of sufficiently high frequency falls on the photosensitive cathode, photoelectrons are emitted from its surface. The anode is maintained positive relative to the cathode, so it attracts and collects the emitted electrons. The resulting movement of electrons through the external circuit constitutes a photocurrent, which is indicated by the microammeter.
(c)(i) Evidence that X-rays are waves:
(ii) Uses of X-rays other than in medicine:
(d)
For an electron falling from [?25l[?25h\,[?25l[?25h-1.6\,\text{eV}\) to \(-10.4\,\text{eV}\), the energy emitted is
Since \(1\,\text{eV}=1.6\times10^{-19}\,\text{J}\),
Using \(E=\dfrac{hc}{\lambda}\),
Energy of radiation = \(1.41\times10^{-18}\,\text{J}\) (or \(8.8\,\text{eV}\)); wavelength = \(1.41\times10^{-7}\,\text{m}\).
Answer Details
(a)(i) Emission line spectrum: This consists of separate bright lines of definite wavelengths on a dark background. It is produced when excited atoms emit radiation as their electrons fall from higher to lower energy levels.
(ii) Line absorption spectrum: This consists of dark lines on a continuous bright spectrum. It is obtained when continuous light passes through a cool gas, which absorbs particular wavelengths corresponding to its energy transitions.
(b) Simple photocell
When radiation of sufficiently high frequency falls on the photosensitive cathode, photoelectrons are emitted from its surface. The anode is maintained positive relative to the cathode, so it attracts and collects the emitted electrons. The resulting movement of electrons through the external circuit constitutes a photocurrent, which is indicated by the microammeter.
(c)(i) Evidence that X-rays are waves:
(ii) Uses of X-rays other than in medicine:
(d)
For an electron falling from [?25l[?25h\,[?25l[?25h-1.6\,\text{eV}\) to \(-10.4\,\text{eV}\), the energy emitted is
Since \(1\,\text{eV}=1.6\times10^{-19}\,\text{J}\),
Using \(E=\dfrac{hc}{\lambda}\),
Energy of radiation = \(1.41\times10^{-18}\,\text{J}\) (or \(8.8\,\text{eV}\)); wavelength = \(1.41\times10^{-7}\,\text{m}\).
Question 49 Report
Explain why mercury does not wet glass while water does.
Whether a liquid wets a solid surface depends on the relative sizes of two types of intermolecular force:
Water on glass: The adhesive force between water molecules and glass molecules is greater than the cohesive force between the water molecules themselves. Water molecules are therefore pulled towards the glass, spread over it, and the water forms a concave meniscus with an acute angle of contact. Because it clings to and spreads on the surface, water is said to wet glass.
Mercury on glass: The cohesive force between mercury molecules is greater than the adhesive force between mercury and glass. The mercury molecules are pulled more strongly towards one another than towards the glass, so the mercury draws itself together into rounded drops, forms a convex meniscus with an obtuse angle of contact, and does not spread. Hence mercury does not wet glass.
In short: a liquid wets a solid when adhesion exceeds cohesion (water and glass) and does not wet it when cohesion exceeds adhesion (mercury and glass).
Answer Details
Whether a liquid wets a solid surface depends on the relative sizes of two types of intermolecular force:
Water on glass: The adhesive force between water molecules and glass molecules is greater than the cohesive force between the water molecules themselves. Water molecules are therefore pulled towards the glass, spread over it, and the water forms a concave meniscus with an acute angle of contact. Because it clings to and spreads on the surface, water is said to wet glass.
Mercury on glass: The cohesive force between mercury molecules is greater than the adhesive force between mercury and glass. The mercury molecules are pulled more strongly towards one another than towards the glass, so the mercury draws itself together into rounded drops, forms a convex meniscus with an obtuse angle of contact, and does not spread. Hence mercury does not wet glass.
In short: a liquid wets a solid when adhesion exceeds cohesion (water and glass) and does not wet it when cohesion exceeds adhesion (mercury and glass).
Question 50 Report
a) List two properties of cathode rays.
(b) Explain how the intensity and energy of cathode rays may be increased
(a) Two properties of cathode rays
(Other valid properties: they possess kinetic energy and momentum, they can produce heat, they cause certain substances to fluoresce, and they can produce X-rays on striking a metal target.)
(b) Increasing the intensity and energy of cathode rays
Intensity: The intensity (number of electrons emitted per second) is increased by raising the temperature of the cathode/filament, that is by increasing the heating (filament) current. A hotter cathode emits more electrons per second by thermionic emission, giving a more intense beam.
Energy: The energy (speed) of the cathode rays is increased by raising the accelerating potential difference (the anode voltage) between the cathode and anode. Each electron gains energy \(eV\), so a larger \(V\) gives faster, more energetic electrons.
Answer Details
(a) Two properties of cathode rays
(Other valid properties: they possess kinetic energy and momentum, they can produce heat, they cause certain substances to fluoresce, and they can produce X-rays on striking a metal target.)
(b) Increasing the intensity and energy of cathode rays
Intensity: The intensity (number of electrons emitted per second) is increased by raising the temperature of the cathode/filament, that is by increasing the heating (filament) current. A hotter cathode emits more electrons per second by thermionic emission, giving a more intense beam.
Energy: The energy (speed) of the cathode rays is increased by raising the accelerating potential difference (the anode voltage) between the cathode and anode. Each electron gains energy \(eV\), so a larger \(V\) gives faster, more energetic electrons.
Question 51 Report
a) Given a retort stand and clamp, a stout pin, a simple pendulum and a pencil, describe how you would use these apparatus to determine the centre of gravity of an irregularly shaped piece of cardboard of a moderate size.
(b) Using a suitable diagram, explain how the following can be obtained from a velocity-time graph:
(i) acceleration;
(ii) total distance covered.
(c ) A body at rest is given an initial uniform acceleration of \(6.0\ \mathrm{ms}^{-2}\) for \(20\mathrm{s}\) after which the acceleration is reduced to \(4.0\ \mathrm{ms}^{-2}\) for the next \(10\mathrm{s}\).
The body maintains the speed attained for \(30\mathrm{s}\).
Draw the velocity-time graph of the motion using the information given above. From the graph, calculate the:
(a) Determination of the centre of gravity of the cardboard
Precautions: The pin must be firmly clamped; the cardboard must swing freely; and the pendulum must be at rest before each line is traced.
(b) Velocity-time graph
(i) Acceleration
Acceleration is the gradient of the velocity-time graph. For the straight-line section joining points A and B,
\[a=\frac{\text{change in velocity}}{\text{time taken}}=\frac{v_B-v_A}{t_B-t_A}.\]
(ii) Total distance covered
The total distance covered is the area between the velocity-time graph and the time axis. Thus, for the graph shown, the distance is the area under the graph from the starting time to the final time.
(c) Velocity-time graph of the motion
For the first 20 s:
\[v_1=0+(6.0\times20)=120\ \text{m s}^{-1}.\]
For the next 10 s:
\[v_2=120+(4.0\times10)=160\ \text{m s}^{-1}.\]
The body then continues at \(160\ \text{m s}^{-1}\) for a further 30 s.
The graph consists of straight-line sections through \((0,0)\), \((20,120)\), \((30,160)\), and \((60,160)\), where time is in seconds and velocity is in \(\text{m s}^{-1}\).
(i) Maximum speed
\[\boxed{160\ \text{m s}^{-1}}\]
(ii) Total distance travelled during the first 30 s
This is the area under the graph from 0 s to 30 s:
\[\begin{aligned}s_1&=\frac12\times20\times120=1200\ \text{m},\\s_2&=\frac12(120+160)\times10=1400\ \text{m}.\end{aligned}\]
\[s=1200+1400=\boxed{2600\ \text{m}}.\]
(iii) Average speed during the first 30 s
\[\text{Average speed}=\frac{\text{total distance}}{\text{total time}}=\frac{2600}{30}=86.7\ \text{m s}^{-1}.\]
\[\boxed{86.7\ \text{m s}^{-1}}\]
Answer Details
(a) Determination of the centre of gravity of the cardboard
Precautions: The pin must be firmly clamped; the cardboard must swing freely; and the pendulum must be at rest before each line is traced.
(b) Velocity-time graph
(i) Acceleration
Acceleration is the gradient of the velocity-time graph. For the straight-line section joining points A and B,
\[a=\frac{\text{change in velocity}}{\text{time taken}}=\frac{v_B-v_A}{t_B-t_A}.\]
(ii) Total distance covered
The total distance covered is the area between the velocity-time graph and the time axis. Thus, for the graph shown, the distance is the area under the graph from the starting time to the final time.
(c) Velocity-time graph of the motion
For the first 20 s:
\[v_1=0+(6.0\times20)=120\ \text{m s}^{-1}.\]
For the next 10 s:
\[v_2=120+(4.0\times10)=160\ \text{m s}^{-1}.\]
The body then continues at \(160\ \text{m s}^{-1}\) for a further 30 s.
The graph consists of straight-line sections through \((0,0)\), \((20,120)\), \((30,160)\), and \((60,160)\), where time is in seconds and velocity is in \(\text{m s}^{-1}\).
(i) Maximum speed
\[\boxed{160\ \text{m s}^{-1}}\]
(ii) Total distance travelled during the first 30 s
This is the area under the graph from 0 s to 30 s:
\[\begin{aligned}s_1&=\frac12\times20\times120=1200\ \text{m},\\s_2&=\frac12(120+160)\times10=1400\ \text{m}.\end{aligned}\]
\[s=1200+1400=\boxed{2600\ \text{m}}.\]
(iii) Average speed during the first 30 s
\[\text{Average speed}=\frac{\text{total distance}}{\text{total time}}=\frac{2600}{30}=86.7\ \text{m s}^{-1}.\]
\[\boxed{86.7\ \text{m s}^{-1}}\]
Question 52 Report
State three methods of polirizing an unpolarized light
Unpolarized light (in which the vibrations occur in all planes perpendicular to the direction of travel) can be polarized by the following methods:
(Scattering, for example by dust or air molecules, also produces partially polarized light.)
Answer Details
Unpolarized light (in which the vibrations occur in all planes perpendicular to the direction of travel) can be polarized by the following methods:
(Scattering, for example by dust or air molecules, also produces partially polarized light.)
Question 53 Report
(a) Explain what is meant by cations
(b) Draw and label an electrolytic cell
(a) Cations are positively charged ions. During electrolysis, they are attracted to and discharged at the cathode.
(b) A labelled electrolytic cell is shown below.
The anode is connected to the positive terminal of the d.c. supply, while the cathode is connected to the negative terminal. Cations move towards the cathode and anions move towards the anode through the electrolyte.
Answer Details
(a) Cations are positively charged ions. During electrolysis, they are attracted to and discharged at the cathode.
(b) A labelled electrolytic cell is shown below.
The anode is connected to the positive terminal of the d.c. supply, while the cathode is connected to the negative terminal. Cations move towards the cathode and anions move towards the anode through the electrolyte.
Question 54 Report
Give three observations in support of de Broglie's assumption that moving particles behave like waves
De Broglie proposed that a moving particle of momentum \(p = mv\) has an associated wavelength \(\lambda = \dfrac{h}{mv}\). The following observations support this wave nature of moving particles:
(A further supporting observation is the diffraction of other particles such as neutrons and protons by crystals.)
Answer Details
De Broglie proposed that a moving particle of momentum \(p = mv\) has an associated wavelength \(\lambda = \dfrac{h}{mv}\). The following observations support this wave nature of moving particles:
(A further supporting observation is the diffraction of other particles such as neutrons and protons by crystals.)
Question 55 Report
Explain the following terms:
(a) tensile stress;
(b)Young’s modulus
(a) Tensile stress
Tensile stress is the stretching force acting per unit cross-sectional area of a material when it is subjected to a pull along its length.
\[ \text{Tensile stress} = \frac{\text{Force (tension)}}{\text{Cross-sectional area}} = \frac{F}{A} \]Its S.I. unit is the pascal (\(\text{Pa}\)) or \(\text{N m}^{-2}\).
(b) Young's modulus
Young's modulus (the modulus of elasticity) is the ratio of tensile stress to tensile strain for a material, within the limit of proportionality (the region where Hooke's law holds).
\[ E = \frac{\text{tensile stress}}{\text{tensile strain}} = \frac{F/A}{e/l} = \frac{Fl}{Ae} \]where \(F\) is the stretching force, \(A\) the cross-sectional area, \(l\) the original length and \(e\) the extension. Its S.I. unit is the pascal (\(\text{Pa}\)) or \(\text{N m}^{-2}\). A large value of \(E\) indicates a stiff material that is difficult to stretch.
Answer Details
(a) Tensile stress
Tensile stress is the stretching force acting per unit cross-sectional area of a material when it is subjected to a pull along its length.
\[ \text{Tensile stress} = \frac{\text{Force (tension)}}{\text{Cross-sectional area}} = \frac{F}{A} \]Its S.I. unit is the pascal (\(\text{Pa}\)) or \(\text{N m}^{-2}\).
(b) Young's modulus
Young's modulus (the modulus of elasticity) is the ratio of tensile stress to tensile strain for a material, within the limit of proportionality (the region where Hooke's law holds).
\[ E = \frac{\text{tensile stress}}{\text{tensile strain}} = \frac{F/A}{e/l} = \frac{Fl}{Ae} \]where \(F\) is the stretching force, \(A\) the cross-sectional area, \(l\) the original length and \(e\) the extension. Its S.I. unit is the pascal (\(\text{Pa}\)) or \(\text{N m}^{-2}\). A large value of \(E\) indicates a stiff material that is difficult to stretch.
Question 56 Report
(a) State two essential differences between a moving coil galvanometer and a d.c. generator.
(b) Explain the term eddy currents and state two devices in which the currents are applied.
(c) State the principle on which the potentiometer is based when it is functioning.
(d) A source of e.m.f. 110 V and frequency 60Hz is connected to a resistor, an inductor and a capacitor in series. When the current in the capacitor is 2A, the potential differences across the resistor is 80 V and that across the inductor is 40 V. Draw the vector diagram of the potential differences across the inductor, the capacitor and the resistor.
Calculate the:
(i) potential difference across the capacitor;
(ii) capacitance of the capacitor;
(iii) inductance of the inductor. [π = 3.14]
(a) Differences between a moving-coil galvanometer and a d.c. generator
| Moving-coil galvanometer | D.C. generator |
|---|---|
| It detects or measures small electric currents. | It generates electrical energy. |
| It converts electrical energy to mechanical deflection; its coil turns through a limited angle and is controlled by hair springs. | It converts mechanical energy to electrical energy; its coil rotates continuously and uses a split-ring commutator and carbon brushes. |
(b) Eddy currents
Eddy currents are circulating currents induced in the body of a conductor when the magnetic flux through it changes. They flow in closed paths and oppose the change producing them. They are applied in an induction furnace and an eddy-current brake.
(c) Principle of a potentiometer
When a steady current flows through a uniform wire of constant cross-sectional area, the potential difference across the wire is directly proportional to its length:
\[V \propto l.\]
(d) Vector diagram
Taking the current, and hence \(V_R\), as the horizontal reference, \(V_L\) leads the current by \(90^\circ\), while \(V_C\) lags the current by \(90^\circ\).
\[V^2=V_R^2+(V_L-V_C)^2\]
\[110^2=80^2+(40-V_C)^2\]
\[(40-V_C)^2=12100-6400=5700\]
Since the circuit is net capacitive, \(V_C>V_L\):
\[V_C=40+\sqrt{5700}=40+75.5=115.5\text{ V}\]
(i) Potential difference across the capacitor:
\[\boxed{V_C=115.5\text{ V}}\]
(ii) Capacitive reactance:
\[X_C=\frac{V_C}{I}=\frac{115.5}{2}=57.75\ \Omega\]
\[C=\frac{1}{2\pi fX_C}=\frac{1}{2\times3.14\times60\times57.75}=4.59\times10^{-5}\text{ F}\]
\[\boxed{C\approx46\ \mu\text{F}}\]
(iii) Inductive reactance:
\[X_L=\frac{V_L}{I}=\frac{40}{2}=20\ \Omega\]
\[L=\frac{X_L}{2\pi f}=\frac{20}{2\times3.14\times60}=0.0531\text{ H}\]
\[\boxed{L\approx0.053\text{ H}}\]
Answer Details
(a) Differences between a moving-coil galvanometer and a d.c. generator
| Moving-coil galvanometer | D.C. generator |
|---|---|
| It detects or measures small electric currents. | It generates electrical energy. |
| It converts electrical energy to mechanical deflection; its coil turns through a limited angle and is controlled by hair springs. | It converts mechanical energy to electrical energy; its coil rotates continuously and uses a split-ring commutator and carbon brushes. |
(b) Eddy currents
Eddy currents are circulating currents induced in the body of a conductor when the magnetic flux through it changes. They flow in closed paths and oppose the change producing them. They are applied in an induction furnace and an eddy-current brake.
(c) Principle of a potentiometer
When a steady current flows through a uniform wire of constant cross-sectional area, the potential difference across the wire is directly proportional to its length:
\[V \propto l.\]
(d) Vector diagram
Taking the current, and hence \(V_R\), as the horizontal reference, \(V_L\) leads the current by \(90^\circ\), while \(V_C\) lags the current by \(90^\circ\).
\[V^2=V_R^2+(V_L-V_C)^2\]
\[110^2=80^2+(40-V_C)^2\]
\[(40-V_C)^2=12100-6400=5700\]
Since the circuit is net capacitive, \(V_C>V_L\):
\[V_C=40+\sqrt{5700}=40+75.5=115.5\text{ V}\]
(i) Potential difference across the capacitor:
\[\boxed{V_C=115.5\text{ V}}\]
(ii) Capacitive reactance:
\[X_C=\frac{V_C}{I}=\frac{115.5}{2}=57.75\ \Omega\]
\[C=\frac{1}{2\pi fX_C}=\frac{1}{2\times3.14\times60\times57.75}=4.59\times10^{-5}\text{ F}\]
\[\boxed{C\approx46\ \mu\text{F}}\]
(iii) Inductive reactance:
\[X_L=\frac{V_L}{I}=\frac{40}{2}=20\ \Omega\]
\[L=\frac{X_L}{2\pi f}=\frac{20}{2\times3.14\times60}=0.0531\text{ H}\]
\[\boxed{L\approx0.053\text{ H}}\]
Question 57 Report
A particle is dropped from a vertical height \(h\) and falls freely for a time \(t\). With the aid of a sketch, explain how \(h\) varies with
(a) \(t\);
(b) \(t^2\).
For a particle dropped from rest, the initial velocity, \(u=0\). Hence,
\[h=ut+\frac{1}{2}gt^2=\frac{1}{2}gt^2.\]
Taking \(g=9.8\ \text{m s}^{-2}\),
\[h=4.9t^2.\]
(a) Graph of \(h\) against \(t\)
The graph is a parabola passing through the origin. Thus, \(h\) varies parabolically with \(t\), since \(h\propto t^2\). Its gradient increases as time increases.
(b) Graph of \(h\) against \(t^2\)
This is a straight line through the origin. Therefore,
\[h\propto t^2.\]
The gradient of the graph is
\[\frac{h}{t^2}=4.9=\frac{g}{2}.\]
Hence, \(g=2\times4.9=9.8\ \text{m s}^{-2}\).
Answer Details
For a particle dropped from rest, the initial velocity, \(u=0\). Hence,
\[h=ut+\frac{1}{2}gt^2=\frac{1}{2}gt^2.\]
Taking \(g=9.8\ \text{m s}^{-2}\),
\[h=4.9t^2.\]
(a) Graph of \(h\) against \(t\)
The graph is a parabola passing through the origin. Thus, \(h\) varies parabolically with \(t\), since \(h\propto t^2\). Its gradient increases as time increases.
(b) Graph of \(h\) against \(t^2\)
This is a straight line through the origin. Therefore,
\[h\propto t^2.\]
The gradient of the graph is
\[\frac{h}{t^2}=4.9=\frac{g}{2}.\]
Hence, \(g=2\times4.9=9.8\ \text{m s}^{-2}\).
Question 58 Report
a) Define diffusion.
(b) State two applications of electrical conduction through gases.
(a) Diffusion
Diffusion is the gradual movement or spreading of the molecules of a substance from a region of higher concentration to a region of lower concentration until the molecules are uniformly distributed. It occurs because the molecules are in continuous random motion, and it is faster in gases than in liquids because gas molecules move faster and are farther apart.
(b) Two applications of electrical conduction through gases
(Other acceptable examples: the cathode-ray tube / X-ray tube, and the lightning discharge, which is conduction through air.)
Answer Details
(a) Diffusion
Diffusion is the gradual movement or spreading of the molecules of a substance from a region of higher concentration to a region of lower concentration until the molecules are uniformly distributed. It occurs because the molecules are in continuous random motion, and it is faster in gases than in liquids because gas molecules move faster and are farther apart.
(b) Two applications of electrical conduction through gases
(Other acceptable examples: the cathode-ray tube / X-ray tube, and the lightning discharge, which is conduction through air.)
Question 59 Report
(a) State two differences between a sound wave and a radio wave.
(b) Explain why a vibrating tuning fork sounds louder when its stem is pressed against a table top than when held in air.
(c)State two conditions necessary for the:
(d) A ray of light is incident on one face of an equilateral glass prism.
(a)
| Sound wave | Radio wave |
|---|---|
| It is a mechanical wave and requires a material medium for propagation. | It is an electromagnetic wave and can travel through a vacuum. |
| It is longitudinal in air. | It is transverse. |
(b)
The vibrating tuning fork forces the table top to vibrate. The table top has a much larger vibrating surface than the prongs of the fork and therefore sets a larger volume of air into vibration. More sound energy is transmitted to the air per second, so the sound is louder.
(c)
(d)(i) At minimum deviation, the path through the equilateral prism is symmetrical and the refracted ray is parallel to the base.
(d)(ii)
For an equilateral prism,
\[A=60^\circ, \qquad D_m=41^\circ\]
\[n=\frac{\sin\left(\frac{A+D_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}\]
\[n=\frac{\sin\left(\frac{60^\circ+41^\circ}{2}\right)}{\sin\left(\frac{60^\circ}{2}\right)}=\frac{\sin 50.5^\circ}{\sin 30^\circ}\]
\[n=\frac{0.7716}{0.5000}=1.54\]
Therefore, the refractive index of the glass is \(1.54\).
Answer Details
(a)
| Sound wave | Radio wave |
|---|---|
| It is a mechanical wave and requires a material medium for propagation. | It is an electromagnetic wave and can travel through a vacuum. |
| It is longitudinal in air. | It is transverse. |
(b)
The vibrating tuning fork forces the table top to vibrate. The table top has a much larger vibrating surface than the prongs of the fork and therefore sets a larger volume of air into vibration. More sound energy is transmitted to the air per second, so the sound is louder.
(c)
(d)(i) At minimum deviation, the path through the equilateral prism is symmetrical and the refracted ray is parallel to the base.
(d)(ii)
For an equilateral prism,
\[A=60^\circ, \qquad D_m=41^\circ\]
\[n=\frac{\sin\left(\frac{A+D_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}\]
\[n=\frac{\sin\left(\frac{60^\circ+41^\circ}{2}\right)}{\sin\left(\frac{60^\circ}{2}\right)}=\frac{\sin 50.5^\circ}{\sin 30^\circ}\]
\[n=\frac{0.7716}{0.5000}=1.54\]
Therefore, the refractive index of the glass is \(1.54\).
Question 60 Report
(a) State Faraday’s second law of electrolysis.
(b) An electric charge of 9.6 x 10\(^4\) C liberates 1 mole of substance containing 6.0 x 10\(^{23}\) atoms. Determine the value of the electronic charge
(a) Faraday's second law of electrolysis
When the same quantity of electricity is passed through different electrolytes, the masses of the different substances liberated (or deposited) are directly proportional to their chemical equivalents (that is, to the ratio of their relative atomic mass to valency).
(b) Value of the electronic charge
The substance is monovalent, so each atom is discharged by one electron. One mole contains \(N = 6.0\times10^{23}\) atoms and is liberated by a charge \(Q = 9.6\times10^{4}\ \text{C}\).
The charge carried by one electron (one atom) is:
\[ e = \frac{Q}{N} = \frac{9.6\times10^{4}}{6.0\times10^{23}} \] \[ e = 1.6\times10^{-19}\ \text{C} \]The value of the electronic charge is \(1.6\times10^{-19}\ \text{C}\).
Answer Details
(a) Faraday's second law of electrolysis
When the same quantity of electricity is passed through different electrolytes, the masses of the different substances liberated (or deposited) are directly proportional to their chemical equivalents (that is, to the ratio of their relative atomic mass to valency).
(b) Value of the electronic charge
The substance is monovalent, so each atom is discharged by one electron. One mole contains \(N = 6.0\times10^{23}\) atoms and is liberated by a charge \(Q = 9.6\times10^{4}\ \text{C}\).
The charge carried by one electron (one atom) is:
\[ e = \frac{Q}{N} = \frac{9.6\times10^{4}}{6.0\times10^{23}} \] \[ e = 1.6\times10^{-19}\ \text{C} \]The value of the electronic charge is \(1.6\times10^{-19}\ \text{C}\).
Would you like to proceed with this action?