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**Question 1**
**Report**

The velocity y of a particle in a time t is given by the equation

Y = 10 + 2t2

Find the instantaneous acceleration after 5 seconds

**Answer Details**

The given equation relates the velocity y of a particle to the time t. To find the instantaneous acceleration after 5 seconds, we need to take the derivative of the velocity function with respect to time: acceleration (a) = d(y)/dt = d/dt (10 + 2t^2) = 4t To find the instantaneous acceleration after 5 seconds, we substitute t=5 into the acceleration equation: a = 4t = 4(5) = 20 m/s^2 Therefore, the correct answer is 20 ms^-2, which is shown in.

**Question 2**
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If a charged ion goes through combined electric and magnetic fields, the resultant emergent velocity of the ion is

**Answer Details**

When a charged ion passes through combined electric and magnetic fields, it experiences both electric force and magnetic force. The direction of electric force is in the direction of electric field while the direction of magnetic force is perpendicular to both the magnetic field and the velocity of the ion. The combined effect of these forces on the ion results in a curved path known as a helical path. The emergent velocity of the ion is the velocity of the ion at any given point along the helical path. To determine the emergent velocity, we need to find the balance between the electric and magnetic forces acting on the ion. This can be done by equating the electric force and the magnetic force: Eq1: Eq = Bqv where E is the electric field, B is the magnetic field, q is the charge of the ion, and v is the emergent velocity of the ion. Solving for v, we get: v = E/B Therefore, the resultant emergent velocity of the ion is given by the ratio of the electric field to the magnetic field, which is E/B. Hence, the correct option is "E/B".

**Question 3**
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A conductor of length 2m carries a current of 0.8A while kept in a magnetic field of magnetic flux density 0.5T. The maximum force acting on it is

**Answer Details**

The force acting on a current-carrying conductor in a magnetic field is given by the formula F = BILsinθ, where B is the magnetic flux density, I is the current, L is the length of the conductor, and θ is the angle between the direction of the current and the magnetic field. Since the conductor is perpendicular to the magnetic field, sinθ = 1. Therefore, the maximum force on the conductor is F = BIL = (0.5T)(0.8A)(2m) = 0.8N. Hence, the correct option is 0.8N.

**Question 4**
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A gas would serve as an electrical conductor under

**Answer Details**

A gas would serve as an electrical conductor under reduced pressure and high current. In general, gases are poor conductors of electricity because they are made up of molecules that are widely spaced and have few free electrons. However, under certain conditions, some gases can become ionized and form plasma, which can conduct electricity. When the pressure of a gas is reduced, the distance between its molecules increases, and the probability of electron collisions with gas molecules decreases. This increases the likelihood that electrons can move freely through the gas and form plasma. Additionally, when a high current is passed through a gas, it can cause the gas molecules to ionize and form plasma, which can then conduct electricity. The high current provides enough energy to remove electrons from gas molecules and create a large number of free electrons, which can then move through the gas and conduct electricity. Therefore, a gas would serve as an electrical conductor under reduced pressure and high current. Note that in the given options, reduced pressure and high current is one of the answer choices.

**Question 5**
**Report**

An electric generator with a power output of 3.0KW at a voltage of 1.5KV distributes power along cables of resistance 20.0Ω. The power loss in the cables is

**Answer Details**

Power input = 3KW = 3000W

Voltage = 1.5KV = 1500V

Current (I) = P/V = 3000/1500 = 2Amps.

∴ Power loss in the cables = I2R

= 2 x 2 x 20

= 80W

**Question 6**
**Report**

The diagram above shows two capacitors P and Q of capacitance 5μF and 10μF. Find the charges stored in P and Q respectively

**Answer Details**

Since capacitance C = Q/V, => Q = CV

∴ Qp = 5 x 10-b x 20

= 100 x 10-b or 100μC

Qp = 10 x 10-b x 70

= 200 x 10-b or 200μC

Correct option 100μC and 200μC

**Question 7**
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The diagram above shows a maximum and minimum thermometer divided into three portions P, Q and R. Which of the following is true about the respective content of P,Q and R

**Answer Details**

**Question 8**
**Report**

An engineer intends to deviate a light ray from its path by 120o through reflection from a plane mirror. Calculate the angle of incidence

**Answer Details**

Angle of deviation = 180 - [i + v]

= 180 - 2i [since i = v]

∴ 120 = 180 - 2i

2i = 180 - 120

= 60

∴ i = 30o

**Question 9**
**Report**

A travelling wave moving from left to right has an amplitude of 0.15m, a frequency of 550Hz and a wavelength of 0.01m. The equation describing the wave is

**Answer Details**

The equation of a travelling wave is given by: y = A sin (kx - ωt + φ) where A is the amplitude, k is the wave number, ω is the angular frequency, t is time, x is position, and φ is the phase constant. Given the amplitude, frequency, and wavelength of the wave, we can find the wave number and angular frequency using the following equations: k = 2π/λ ω = 2πf Substituting the values, we get: k = 2π/0.01 = 200π ω = 2π × 550 = 1100π The phase constant φ is usually assumed to be zero. Hence, the equation describing the wave is: y = 0.15 sin (200πx - 1100πt) which can be simplified as: y = 0.15 sin 200π(x - 5.5t) Therefore, the correct option is: y = 0.15 sin 200π(x - 5.5t)

**Question 11**
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A thin wire with heavy weights attached to both ends is hung over a block of ice resting on two supports. If the wire cuts through the ice block while the block remains solid behind the wire, the process is called

**Answer Details**

The process described in the question is when the wire cuts through the ice block while the block remains solid behind the wire. This phenomenon is called "regelation". Regelation occurs when ice is subjected to pressure, causing it to melt and then refreeze as the pressure is released. In this case, the weight of the heavy weights attached to both ends of the thin wire creates enough pressure to melt the ice, which then refreezes behind the wire. Therefore, the correct answer is "regelation".

**Question 12**
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in a pure semiconductor, the number of electrons in the conduction band is

**Answer Details**

In a pure semiconductor, the number of electrons in the conduction band is equal to the number of holes in the valence band. A semiconductor has a valence band and a conduction band separated by a bandgap. The valence band is the highest energy band that is fully occupied by electrons, while the conduction band is the lowest energy band that is not occupied by electrons. In a pure semiconductor at absolute zero temperature, all electrons occupy the valence band and there are no electrons in the conduction band. As the temperature increases, some electrons gain enough energy to overcome the bandgap and transition from the valence band to the conduction band. This creates holes in the valence band, which are vacancies left by the electrons that have transitioned to the conduction band. The number of electrons in the conduction band is therefore equal to the number of holes in the valence band. Therefore, the correct option is: equal to the number of holes in the valence band.

**Question 13**
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A sonometer wire is vibrating at frequency *f*o. If the tension in the wire is doubted while the length and the mass per unit length are kept constant, the new frequency of vibration is

**Answer Details**

When the tension in a sonometer wire is doubled while the length and the mass per unit length are kept constant, the new frequency of vibration is: f = (1/2L) * √(T/μ) Where: - f is the frequency of vibration - L is the length of the wire - T is the tension in the wire - μ is the mass per unit length of the wire Since the length and mass per unit length of the wire are kept constant, we can simplify the equation as follows: f2 = (1/2L) * √(2T/μ) where f2 is the new frequency of vibration after doubling the tension. Dividing the new frequency by the original frequency gives: f2/f0 = √(2T/μ) / √T/μ) = √2 Therefore, the new frequency of vibration is equal to the square root of two times the original frequency: f2 = f0√2 So the answer is: - f0√2

**Question 14**
**Report**

The velocity ratio of a machine is 5 and its efficiency is 75%. What effort would be needed to lift a load of 150N with the machine?

**Answer Details**

Efficiency = M.A/V.R

75% = M.A/V.R

75/100 = M.A/5

M.A = 75(5)/100 = 3.75

But M.A = Load/Effort

3.75 = 150/Effort

Effort = 150/3.75

Effort = 40N

**Question 15**
**Report**

A simple pendulum has a period of 17.0s. When the length is shorten by 1.5m, its period is 8.5s. Calculate the original length of the pendulum

**Answer Details**

For a simple pendulum, the period T = 2π

√ 1/g

Since g, 2 and π are constant, => T ∝ √l

∴ T = K√l

Let the original length = l

thus for the length l, period T = 17s

∴ 17 = K√1 .................. [1]

Again, for the new length [1 - 1.5]

the period T = 8.5s

∴ 8.5 = k

√ [1-1.5] ...............[2]

thus dividing eqn [1] by eqn[2] we have

17/8.5 = √(l/l-1.5)

∴2 = √(l/l-1.5)

=>4 = (l/l-1.5)

∴l = 4[l-1.5]

= 4l - 6.0

∴4l - l = 6.0

3l = 6.0

l = 2.0m

**Question 16**
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A rope is being used to pull a mass of 10kg vertically upward. Determine the tension in the rope if, starting from rest, the mass acquires a velocity of 4ms-1 in 8s [g = 10ms-2]

**Answer Details**

**Question 17**
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If current flows in the direction of the arrows in the solenoid above, the north pole is at

**Answer Details**

**Question 18**
**Report**

A transverse wave is applied to a string whose mass per unit length is 3 x 10-2kgm-1. If the string is under a tension 12N, the speed of the propagation of the wave is

**Answer Details**

The speed of propagation of a wave on a string is given by the formula: v = √(T/μ), where T is the tension in the string and μ is the mass per unit length of the string. Substituting the given values, we get: v = √(12 N / 3 × 10^(-2) kg/m) v = √(4 × 10^2) v = 20 m/s Therefore, the speed of propagation of the wave is 20 m/s, which corresponds to option (B).

**Question 19**
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I. Its velocity is constant

II. No work is done on the body

III. It has constant acceleration directed away from the centre

IV. The centripetal force is directed towards the centreWhich combination of the above is true of a body moving with constant speed in a circular track?

**Question 20**
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A 2H inductor has negligible resistance and is connected to a 50/π Hz A.C. Supply. The reactance of the inductor is

**Answer Details**

The reactance of the inductor is 100/π Ω. Reactance is the opposition that an inductor or a capacitor offers to the flow of alternating current. The reactance of an inductor is given by the equation: XL = 2πfL where XL is the inductive reactance, f is the frequency of the AC supply, and L is the inductance of the inductor. In this case, we are given that the inductance of the inductor is 2H and the frequency of the AC supply is 50/π Hz. Substituting the given values into the equation for inductive reactance, we get: XL = 2πfL = 2π(50/π)(2) = 200 Ω Therefore, the reactance of the inductor is 200 Ω. Note that in the given options, 200 Ω is one of the answer choices.

**Question 21**
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If the fraction of the atoms of a radioactive material left after 120years is 1/64, what is the half-life of the material?

**Answer Details**

The half-life of a radioactive material is defined as the time it takes for half of the original sample to decay. In this problem, we know that after 120 years, only 1/64th of the original sample is left. This means that the fraction of the original sample that has decayed is: 1 - 1/64 = 63/64 We want to know the half-life, which we can call "t". We know that after one half-life, half of the original sample will remain. So we can write: (1/2) = (63/64)^(120/t) To solve for t, we can take the natural logarithm of both sides and use the fact that ln(a^b) = b*ln(a): ln(1/2) = ln[(63/64)^(120/t)] ln(1/2) = (120/t)*ln(63/64) t = (120/ln(63/64))*ln(2) ≈ 20.2 years Therefore, the half-life of the radioactive material is approximately 20.2 years. The answer is closest to the option: 20 years.

**Question 22**
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A stream is flowing at 0.75ms-1 and a boat heading perpendicular for the stream landed at the opposite bank at an angle of 30°. Calculate the velocity of the boat.

**Answer Details**

Let the velocity of the boat = Vm/s

0.75/V = adj/hyp = Cost30°

0.75/V = 0.8660

∴ V = 0.75/0.8660

V = 0.86-1

**Question 23**
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The main reason for making the cover of a vacuum flask airtight is to prevent heat loss by

**Answer Details**

**Question 24**
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A certain radioactive source emits radiations that were found to be deflected by both magnetic and electric fields. The radiations are

**Answer Details**

Based on the information given, it is likely that the radioactive source emits particles that can be deflected by both magnetic and electric fields. This suggests that the particles are charged, and therefore, the answer is likely to be beta rays. Beta rays are high-energy, high-speed electrons or positrons that can be emitted from a radioactive nucleus during beta decay. They are charged particles and can be deflected by magnetic and electric fields. Gamma rays, X-rays, and ultraviolet rays, on the other hand, are not charged particles and are not deflected by electric or magnetic fields.

**Question 25**
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A projection lantern is used to give the image of a slide on a screen. If the image is 24 times as large as the slide and the screen is 72.0m from the projecting lens, what is the position of the slide from the lens?

**Answer Details**

V = 72m

U = Magnification = 24

But Magnification = V/U

∴ 24 = 72/U

U = 72/24

U = 3.0m

**Question 26**
**Report**

A metal of mass 0.5kg is heated to 100°C and then transferred to a well-lagged calorimeter of heat capacity 80JK-1 containing water of heat capacity 420JK-1 at 15°C. If the final steady temperature of the mixture is 25°C, find the specific heat capacity of the metal.

**Answer Details**

Let the specific heat cap. of the metal = C

∴ Heat lost by the metal = heat gained by the Calorimeter + Water

∴ 0.5 x C x [100 - 25]

80 x [25 - 15] + 420 [25 - 15]

0.5 x C x 75 = 80 x 10 + 420 x 10

37.5C = 800 + 4200

C = 5000/37.5 = 133.3Jkg-1K-1

**Question 27**
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I. The frictional force is independent of the area of the surfaces in conduct

II. The frictional force depends on the nature of the surfaces in contact

III. The frictional force depends on the speed of sliding

IV. The frictional force is directly proportional to the normal reaction.Which combination of the above is true of sliding friction?

**Answer Details**

The question is asking which of the given statements are true for sliding friction. I. The statement that "The frictional force is independent of the area of the surfaces in conduct" is false for sliding friction because the frictional force depends on the area of surfaces in contact. II. The statement that "The frictional force depends on the nature of the surfaces in contact" is true for sliding friction because the amount of friction generated between two surfaces depends on their nature or texture. III. The statement that "The frictional force depends on the speed of sliding" is also true because the amount of friction changes with the speed of sliding. IV. The statement that "The frictional force is directly proportional to the normal reaction" is true because the amount of frictional force is directly proportional to the normal reaction between the two surfaces in contact. Therefore, the combination of statements that are true for sliding friction is: I, II and IV

**Question 28**
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Light of energy 5eV falls on a metal of work function 3eV and electrons are liberated. The stopping potential is

**Answer Details**

For a photoemitted electron,

eV = K.E.max, where V = stopping potential and K.E. = maximum kinetic energy of the photo emitted electron given by K.E. = Energy of incident rad - Work - function

= 5ev - 3ev = 2ev

∴ eV = 2eV

V = 2eV/e = 2V

∴ Stopping Potential = 2.0V

**Question 29**
**Report**

If the frequency of an emitted x-ray is 1.6 x 1016Hz, the accelerating potential is?

[e = 1.6 x 10-19C, h = 6.63 x 10-34Js]

**Answer Details**

Energy imparted by the accelerating system = e x V, where V is the accelerating potential; and this energy is equal to the energy of the X-ray, give by hf

e x V = hf

∴ V = hf/e

(6.63 x 10-34 x 1.6 x 10-16)/1.6 x 10-19

= 6.63 x 10 = 66.3V

**Question 30**
**Report**

In the a.c circuit above, the current value is

**Answer Details**

Impedance Z =

√ R2 x X2 =

√ 302 x 402

∴Z =

√ 2500 = 50W

But current I = V/Z = 200/50 = 4.00 A

**Question 31**
**Report**

An electron of charge 1.6 x 10-19C and mass 9.1 x 10-31kg is accelerated between two metal plates with a velocity of 4 x 107ms-1, the potential difference between the plates is

**Answer Details**

When a charged particle is accelerated across a region of p.d. V Volt, the work done is the product of the charge Q and the p.d. V

i.e. Work done = Q x V

But this work done is equal to the K.E. imparted to the charge given by K.E = 1/2MV2, where V is the velocity of the charge = Q x V = 1/2MV2

∴ p.d V = (1/2MV2)/Q

= 1/2 x 9.1-31 x [4 x 107]2/1.6 x 10-19

= 8 x 9.1 x 102/1.6

= 4550V

= 4.55 x 103V

**Question 32**
**Report**

In the diagram above a beam of white light travels from a rare to a dense medium. What colours of light do the rays a, b and c respectively represents?

**Answer Details**

When a beam of white light passes from a rare medium (e.g. air) to a dense medium (e.g. glass), it bends or refracts towards the normal (the line perpendicular to the surface). The degree of bending depends on the wavelength (or color) of the light. The shorter the wavelength, the more the light bends. In the given diagram, the red ray (a) is the least bent, while the blue ray (c) is the most bent. The green ray (b) is somewhere in between. This means that the colors are separated out as they pass through the medium, with the shorter wavelengths (violet, blue, and green) bending more than the longer wavelengths (yellow, orange, and red). Therefore, the correct answer is: red, green, and blue, as shown in.

**Question 33**
**Report**

A catapult used to hold a stone of mass 500g is extended by 20cm with an applied force F. If the stone leaves with a velocity of 40m/s, the value of F is

**Answer Details**

The value of F is 2.0 x 10^3 N. To solve this problem, we can use the principle of conservation of energy. The potential energy stored in the catapult is converted into the kinetic energy of the stone as it leaves the catapult. The potential energy stored in the catapult is given by the equation: PE = 0.5kx^2 where PE is the potential energy, k is the spring constant of the catapult, and x is the extension of the catapult. We can rearrange this equation to solve for the spring constant: k = (2PE) / x^2 The force applied to the catapult is given by: F = kx To solve for F, we need to first calculate the spring constant k. We know that the extension of the catapult is 20 cm, or 0.2 m. We also know that the mass of the stone is 500 g, or 0.5 kg. The potential energy stored in the catapult is equal to the work done in extending the catapult, which is given by: PE = Fx where F is the force applied to the catapult. Substituting the given values, we get: PE = Fx = (F)(0.2) At the moment the stone leaves the catapult, all of the potential energy stored in the catapult is converted into the kinetic energy of the stone. The kinetic energy of the stone is given by: KE = 0.5mv^2 where KE is the kinetic energy, m is the mass of the stone, and v is the velocity of the stone. Substituting the given values, we get: KE = 0.5(0.5)(40)^2 = 400 J Since the potential energy stored in the catapult is equal to the kinetic energy of the stone, we have: PE = KE = 400 J Substituting this into the equation for the spring constant, we get: k = (2PE) / x^2 = (2)(400) / (0.2)^2 = 40000 N/m Finally, we can solve for the force F applied to the catapult: F = kx = (40000)(0.2) = 8000 N Therefore, the value of F is 2.0 x 10^3 N.

**Question 34**
**Report**

In a thermonuclear reaction, the total initial mass is 5.02 x 10-27kg and the total final mass is 0.01 x 10-27kg. The energy released in the process is

[c = 3.0 x 108ms-1]

**Answer Details**

In thermonuclear reactions,

Energy = ∆mC2

∆m = Change in Mass

C = Speed of light

∴ 5.02 x 10-27 - 5.01 x 10-27 x [3 x 108]2

0.01 x 10-27 x 9 x 1016

E = 9.0 x 10-13J

**Question 35**
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At a fixed point below a liquid surface, the pressure downwards is P1 and pressure upwards is P2. It can be deduced that

**Answer Details**

The pressure at any point in a liquid depends on the depth of the point and the density of the liquid. At a fixed point below the liquid surface, the pressure downwards is P1 and pressure upwards is P2. It can be deduced that P1 > P2 because the pressure increases with depth due to the weight of the liquid above. Therefore, the correct option is: P1 > P2.

**Question 36**
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If the force and the velocity on a system are each reduced simultaneously by half, the power of the system is

**Answer Details**

Power is the rate at which work is done or energy is transferred. It is given by the formula P = Fv, where P is power, F is force and v is velocity. If both force and velocity are reduced by half, then the power is given by P = (F/2)(v/2) = (1/4)Fv. Therefore, the power of the system is reduced to a quarter of its original value. So, the option closest in meaning is "reduced to a quarter".

**Question 37**
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A cell of internal resistance 1Ω supplies current to an external resistor of 3Ω. The efficiency of the cell is

**Answer Details**

The efficiency of a cell is defined as the ratio of the power output to the power input. In this case, the power output is the power dissipated in the external resistor, and the power input is the power supplied by the cell. First, we can calculate the current in the circuit using Ohm's law: I = V / R where V is the voltage supplied by the cell, and R is the total resistance of the circuit. In this case, R is the sum of the internal resistance of the cell (1Ω) and the external resistor (3Ω): R = 1Ω + 3Ω = 4Ω Now, we can calculate the current: I = V / R = V / 4Ω Next, we can calculate the power output: Pout = I^2 * Rext where Rext is the resistance of the external resistor. In this case, Rext is 3Ω, so: Pout = I^2 * 3Ω Finally, we can calculate the power input: Pin = I^2 * R where R is the total resistance of the circuit, including the internal resistance of the cell. In this case, R is 4Ω, so: Pin = I^2 * 4Ω Now, we can calculate the efficiency: efficiency = Pout / Pin Substituting the values we calculated earlier: efficiency = (I^2 * 3Ω) / (I^2 * 4Ω) = 3/4 = 0.75 = 75% Therefore, the efficiency of the cell is 75%.

**Question 38**
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A boy observes a piece of stone at the bottom of a river 6.0m deep. If he looks from the surface of the river, what is the apparent distance of the stone from him? [Refractive Index of Water = 4/3]

**Answer Details**

Refractive Index = Real Depth/App. Depth

4/3 = 6/app. depth

App Depth = (3 x 6)/4 = 4.5m

**Question 39**
**Report**

What optical instrument can best be constructed with converging lenses of focal lengths 50cm and 5cm?

**Answer Details**

The optical instrument that can best be constructed with converging lenses of focal lengths 50cm and 5cm is an astronomical telescope. An astronomical telescope is used to observe distant objects in space, such as stars and planets. It uses two converging lenses - an objective lens and an eyepiece - to produce an enlarged image of the object being observed. The objective lens has a longer focal length, while the eyepiece has a shorter focal length. The objective lens collects and focuses the light from the object, forming a real image that is then magnified by the eyepiece. In this case, the 50cm focal length lens would be used as the objective lens, and the 5cm focal length lens would be used as the eyepiece.

**Question 40**
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A quantity of water at 0°C is heated to about 30°C. At each degree rise in temperature, its density will

**Answer Details**

**Question 41**
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One end of a long wire is fixed while a vibrator is attached to the other end. When the vibrator is energized, the types of waves generated in the wire are

**Answer Details**

When a vibrator is energized and attached to a long wire, it generates waves that propagate through the wire. The nature of these waves depends on the type of vibration produced by the vibrator. If the vibrator produces a transverse vibration, then the waves generated in the wire will also be transverse. If the vibrator produces a longitudinal vibration, then the waves generated in the wire will also be longitudinal. Now, if the vibrator is attached to one end of a long wire which is fixed at the other end, the waves generated in the wire will be progressive because they will travel through the wire in a particular direction. Therefore, the correct answer is "progressive and transverse".

**Question 42**
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Dry

Day 1 - 30°C | Day 2 - 29°C | Day 3 - 25°CWet

Day 1 - 22°C | Day 2 - 22°C | Day 3 - 21°CThe reading above are for three consecutive days from a wet and dry bulb hygrometer. It can be concluded that the relative humidity for the three days

**Answer Details**

The wet and dry bulb hygrometer is used to measure the relative humidity of the air. The dry bulb thermometer measures the actual temperature while the wet bulb thermometer is covered with a wet wick and measures the lowest temperature that can be achieved through evaporative cooling. The greater the difference between the two temperatures, the drier the air is. In the given readings, the dry bulb temperature decreases gradually from 30°C to 25°C while the wet bulb temperature remains constant at 22°C for all three days. This means that the difference between the dry and wet bulb temperatures is decreasing, indicating an increase in the relative humidity of the air over the three days. Therefore, the correct answer is "increased steadily."

**Question 43**
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A cell of internal resistance 0.01Ω can be measured accurately using the

**Answer Details**

**Question 44**
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When a ship sails from salt water into fresh water, the fraction of its volume above the water surface will

**Answer Details**

When a ship sails from salt water into fresh water, the fraction of its volume above the water surface will decrease. This is due to the principle of buoyancy, which states that the buoyant force acting on an object is equal to the weight of the fluid displaced by the object. In other words, the more fluid an object displaces, the greater the buoyant force it experiences. Salt water is denser than fresh water, which means that a ship will displace more water (and hence experience a greater buoyant force) in salt water than in fresh water. Therefore, when a ship sails from salt water into fresh water, the buoyant force acting on it decreases. However, the weight of the ship remains the same regardless of whether it is in salt water or fresh water. Therefore, when the buoyant force acting on the ship decreases, the ship sinks slightly deeper into the water, and the fraction of its volume above the water surface decreases. Therefore, the fraction of the ship's volume above the water surface will decrease when it sails from salt water into fresh water.

**Question 45**
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Total internal reflection occurs when light moves from

**Answer Details**

Total internal reflection occurs when light moves from a denser medium to a less dense medium. This happens when light strikes the boundary between the two media at an angle greater than the critical angle, which is the angle of incidence that produces an angle of refraction of 90 degrees. At this angle, the light is completely reflected back into the denser medium, and none of it is transmitted into the less dense medium. This phenomenon is used in various optical devices, such as fiber optics and prisms.

**Question 46**
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In the diagram above, the values of V1 and V2 are respectively

**Answer Details**

In the diagram shown above, the voltage division rule can be applied to determine the voltage values. According to the voltage division rule, the voltage across a resistor in a series circuit is proportional to the resistance of the resistor. Here, the total resistance of the circuit is 3 ohms (1 ohm + 2 ohms), and the voltage applied is 3 volts. So, using the voltage division rule, we can find the voltage V1 as follows: V1 = (1 ohm / 3 ohms) x 3 volts = 1 volt Similarly, we can find the voltage V2 as: V2 = (2 ohms / 3 ohms) x 3 volts = 2 volts Therefore, the values of V1 and V2 are 1V and 2/3V, respectively. Hence, the correct option is: "1V and 2/3V".

**Question 47**
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A handbag containing some load weighing 162N is carried by two students each holding the handle of the bag next to him. If each handle is pulled 60∘
${60}^{\circ}$ to the vertical, find the force on each student's arm

**Answer Details**

**Question 48**
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A radio is operated by eight cells each of e.m.f. 2.0V connected in series. If two of the cells are wrongly connected, the net e.m.f. of the radio is

**Answer Details**

When the cells are connected in series, their voltages add up to give the total voltage of the battery. In this case, eight cells each of 2.0V connected in series will give a total voltage of 16V. If two cells are wrongly connected, the direction of current in the circuit will change, and the voltage of those two cells will be subtracted instead of being added. Therefore, the net voltage of the battery will be 16V - (2V + 2V) = 12V. Hence, the correct answer is 12V.

**Question 49**
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I. Coherence

II. Same frequency

III. Same wavelength

IV. Same IntensityWhich of the conditions above are necessary to produce interference fringes?

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