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**Question 1**
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I. Increase the melting point of the liquid.

II. Increase the boiling point of the liquid.

III. Decrease the melting point of the liquid.

IV. Decrease the boiling point of the liquid.

Which of the statement above about the effect on increase in pressure in a liquid are correct?

**Answer Details**

Increase in pressure generally raises the boiling point but but lowers the melting point. II and III

**Question 2**
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The figure above shows 4 forces 3N, 10N, 3√3N and 6N acting on a particle P. The resultant of he four forces is

**Answer Details**

__Resolution of force:__

Total vertical upward force:

= 3√3 + 10 cos 60o

= 3√3 + 5.

Total downward force

= 6 cos 30o

= 6 x √(3/2) = 3√3

Total horizontal forces towards right;

= 3N + cos 30o

= 3N + 10 x √(3/2)

= 3N + 5√3

Total horizontal forces towards left:

= 6 cos 60o

= 6 x 0.5

= 3.0N
__Summary__:

Vertical resultant forces upward:

= 3√3 + 5 - 3√3

= 5N

Horizontal resultant forces towards right:

= 3N + 5√3 - 3√3

= 5N

Horizontal resultant force towards right

= 3N + 5√3 - 3N

= 5√3N

∴ R2 = 52 + (5√3)2

= 25 + 25 x 3

= 100

∴ R = √100

= 10N

**Question 3**
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The blade of a hoe feel colder to touch in the morning than the wooden handle because the

**Answer Details**

The blade of a hoe feels colder to touch in the morning than the wooden handle because the blade is a better conductor of heat than the handle. In the morning, the temperature of the environment is usually lower than the temperature of the objects in it. The blade being a better conductor of heat than the handle conducts the heat from our hand more efficiently, making it feel colder. The handle, on the other hand, being a poor conductor of heat, does not conduct the heat from our hand very well, so it feels warmer.

**Question 4**
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From the diagram above, the bob of the pendulum has the fastest speed at

**Answer Details**

The ball has its fastest speed at its equilibrium position when its displacement X = O; at X

**Question 5**
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In the diagram above , if the magnetic field points into the paper and the force on a current - carrying conductor points upwards, what is the direction of the current?

**Answer Details**

According to Fleming's left-hand rule, which relates the direction of the magnetic field, current, and force in a conductor, the direction of the current in the conductor will be perpendicular to both the direction of the magnetic field and the direction of the force acting on it. In this case, the magnetic field points into the paper (i.e., downwards, from our perspective), and the force on the conductor points upwards. Therefore, the direction of the current in the conductor must be perpendicular to both, which is to the left. Hence, the answer is "Left".

**Question 6**
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If two charged plates are maintained at a potential difference of 3 kv, the work done in taking a charge of 600 µC across the field is?

**Answer Details**

The work done in taking a charge across a potential difference is given by the formula: work = charge x potential difference In this case, the charge taken is 600 µC (which is 600 x 10^-6 C) and the potential difference between the plates is 3 kV (which is 3 x 10^3 V). Therefore, substituting these values into the above formula, we get: work = 600 x 10^-6 C x 3 x 10^3 V = 1.8 J Hence, the work done in taking a charge of 600 µC across the field is 1.8 J. Therefore, the answer is 1.8 J.

**Question 7**
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I. They should be identical

II. They should originate from the same source.

III. They should be coherent.

IV. They should be monochromatic.

From the statements above, the conditions for two waves to interfere are?

**Answer Details**

The statements above describe the conditions for two waves to interfere. Interference occurs when two waves meet and overlap. For interference to occur, the waves should be identical (statement I), which means they have the same amplitude, frequency, and wavelength. The waves should also originate from the same source (statement II), which means they should have the same phase. In addition, the waves should be coherent (statement III), which means they have a constant phase difference. Finally, the waves should be monochromatic (statement IV), which means they have a single frequency. Therefore, the correct answer is option II, which includes statements I, III, and IV.

**Question 8**
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The instantaneous value of the induced e.m.f as a function of time is ε = εo sin ωt where εo is the peak value of the e.m.f. The instantaneous value of the e.m.f., one quarter of the period is

**Answer Details**

The question is asking for the instantaneous value of the induced e.m.f at one quarter of the period. The given equation for the e.m.f as a function of time is ε = εo sin ωt, where εo is the peak value of the e.m.f. One quarter of the period corresponds to a time t = T/4, where T is the period of the waveform. At this time, the argument of the sine function is ωt = π/2. Therefore, the instantaneous value of the e.m.f. at one quarter of the period is ε = εo sin(π/2) = εo * 1 = εo. The answer is therefore (c) εo/2.

**Question 9**
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The unit of moment of a couple can be expressed in?

**Answer Details**

The unit of moment of a couple can be expressed in "N m", which stands for Newton-meters. This is because moment is defined as the product of the magnitude of the force and the perpendicular distance between the line of action of the force and the point about which the moment is being calculated. In the case of a couple, there are two equal and opposite forces acting at a perpendicular distance from each other, and the moment of the couple is the product of one of the forces and the distance between the forces. The unit of force is Newtons (N) and the unit of distance is meters (m), so the unit of moment is Newton-meters (N m).

**Question 10**
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A reservoir 500 m deep is filled with a fluid of density 850 kgm-3. If the atmospheric pressure is 1.05 x 105 Nm-2, the pressure at the bottom of the reservoir is.

[g = 10ms-2]

**Answer Details**

In general, pressure varies with depth in the relation p = lgh

= 850 x 10 x 500

= 4250000

= 4.25 x 106 NM-2

**Question 11**
**Report**

.

Three cells each of e.m.f 1.5 V and internal resistance 2.5Ω are connected as shown in the diagram above. Find the net e.m.f and the internal resistance

**Answer Details**

Since the cells are connected in parallel the net e.m.f is that of one cell, E.

However the effective internal resistance should be added as parallel resistance.

=>1/v + 1/25 + 1/25 + 1/25 = 1.2

∴ v = 0.83Ω

∴ Effective e.m.f and internal resistance are respectively = 1.5V and 0.83Ω

**Question 12**
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In a large telecommunications auditorium, perforated absorbent materials are used to line the ceiling so as to

**Answer Details**

The purpose of perforated absorbent materials in a telecommunications auditorium is to reduce the reverberation of sound in the hall. Reverberation is the persistence of sound in an enclosed space after the sound source has stopped. The use of absorbent materials helps to reduce this effect, creating a clearer and more intelligible sound. Therefore, option D is the correct answer.

**Question 13**
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A 40 kW electric cable is used to transmit electricity through a resistor of resistance 2.0Ω at 800 V. The power loss as internal energy is?

**Answer Details**

In general, power = IV; 40kw = IV.

∴ 40,000 = 1 x 800

1 = (40.000)/800 = 50A. tune the current through resistor = 50A.

∴ power loss = 12R = 502 X 2 = 2500 X 2 = 5.0 X 103W

**Question 14**
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A gramophone record takes 5s to reach its constant angular velocity of 4π $\pi $ rads-1 from rest. Find its constant angular acceleration.

**Answer Details**

The gramophone record has an initial angular velocity of 0 and reaches a final angular velocity of 4π π rads-1 in 5 seconds. The constant angular acceleration can be found using the following formula: angular acceleration = (final angular velocity - initial angular velocity) / time Substituting the given values, we get: angular acceleration = (4π π rads-1 - 0) / 5s = 4π π rads-1 / 5s = 0.8π π rads-2 Therefore, the constant angular acceleration is 0.8π π rads-2.

**Question 15**
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The ratio of the coefficient of linear expansion of two metals ∝1/∝2 is 3:4. If, when heated through the same temperature change, the ratio of the increase in length of the two metals, e1/e2 is 1:2, the ratio of the original lengths l1/l2 is

**Answer Details**

Ratio of their linear expansion = ∝1/∝2= 3:4.

When heated to the same temperature range, the ratio of their increase in length e1/e2 = 1:2

But the increase in length of

1 = e1 = ∝1l1Δθ and increase i length of

2 = e2 = ∝2l2Δθ

=> | e1 | = | ∝1l1Δθ | = | ∝1l1 | = | 1 |

e2 | ∝2l2Δθ | ∝2l2 | 2 |

But | ∝1 | = | 3 | , ∴ | 3 | x | l1 | = | 1 |

∝2 | 4 | 4 | l2 | 2 |

=> 6l1 = 4l2

∴ | l1 | = | 4 | = | 2 | OR 2 : 3 |

l2 | 6 | 3 |

**Question 16**
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When an object is placed between the principal focus and the optical center of a convex lens, it could be used as a?

**Answer Details**

When an object is placed between the principal focus and the optical center of a convex lens, it forms a virtual and magnified image. This image can be seen through the lens and can be used as a simple microscope to observe small objects or details. Therefore, the correct answer is a "Simple microscope".

**Question 17**
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A radioactive substance has a half-life of 20 days what fraction the original radiation nuclei will remain after 80 days?

**Answer Details**

The half-life of a radioactive substance is the time it takes for half of the original radioactive nuclei to decay. If the half-life of a substance is 20 days, this means that after 20 days, half of the original nuclei would have decayed, and half would remain. After another 20 days (a total of 40 days), half of the remaining nuclei would have decayed, leaving a quarter of the original nuclei. Similarly, after 60 days, half of the remaining quarter of nuclei would have decayed, leaving one-eighth of the original nuclei. Finally, after 80 days, half of the remaining one-eighth of nuclei would have decayed, leaving one-sixteenth of the original nuclei. Therefore, the fraction of the original radiation nuclei that will remain after 80 days is 1/16. So, the correct option is: (1)/16.

**Question 18**
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The instrument that measures both a.c and d.c is?

**Answer Details**

The instrument that measures both AC and DC is the "moving iron ammeter." This type of ammeter measures the average value of the current, and its design allows it to work for both AC and DC. The other options listed are not suitable for measuring both AC and DC.

**Question 19**
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A boy drags a bag of rice along a smooth horizontal floor with a force of 2N applied at the angle of 60o to the floor. The work done after a distance of 3m is

**Answer Details**

Effective component of the force along the horizontal = F cos 60o

Work done by this force : 60

F cos 60 x X

= 2 x 0.5 x 3

= 3J

**Question 20**
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I. Use a liquid with a high melting point.

II. use a liquid of high volume expansivity.

III. use a capillary tube of large diameter.

Which of the above best describes how the sensitivity of a liquid-in-glass thermometer can be enhanced?

**Answer Details**

The question is asking how the sensitivity of a liquid-in-glass thermometer can be enhanced. The three options provided are: (I) use a liquid with a high melting point, (II) use a liquid of high volume expansivity, and (III) use a capillary tube of large diameter. Option I alone cannot enhance the sensitivity of the thermometer. The melting point of a liquid only affects the upper range of temperatures that the thermometer can measure, but it has no effect on the sensitivity or the precision of the measurements. Therefore, option I is not the correct answer. Option II can enhance the sensitivity of the thermometer. The volume expansivity of a liquid refers to how much the volume of the liquid changes with temperature. A liquid with high volume expansivity will change its volume more for a given change in temperature, resulting in a larger change in the length of the column in the thermometer. This means that the thermometer will be more sensitive to temperature changes. Therefore, option II is a correct answer. Option III alone cannot enhance the sensitivity of the thermometer. The diameter of the capillary tube only affects the rate at which the thermometer responds to temperature changes, but it has no effect on the sensitivity or the precision of the measurements. Therefore, option III is not the correct answer. Therefore, the correct answer is option II only, which is "II. use a liquid of high volume expansivity."

**Question 21**
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A 120 V, 60 W lamp is to be operated on 220 V a.c. the supply mains. calculate the values of non-inductive resistance that would be required to ensure that the lamp is run on current value.

**Answer Details**

Power of the lamp = 60W = (V2)/R and the voltage rate = 120v.

R = (V2)/60 = (120 X 120)/60 = 240 Ω, the fuse wire that should ensure current does not exceed this value should be rated 200 Ω

**Question 22**
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A machine has a velocity ratio of 4. if it requires 800 N to overcome a load of 1600 N,what is the efficiency of the machine?

**Answer Details**

The efficiency of a machine is the ratio of output work to input work. In this case, the load is 1600 N and it requires 800 N to overcome it. So, the input work is 800 N × distance moved by effort. The output work is 1600 N × distance moved by load. Velocity ratio is the ratio of the distance moved by effort to the distance moved by load. Velocity Ratio = distance moved by effort / distance moved by load Velocity Ratio = 4 distance moved by effort = 4 × distance moved by load Input work = force × distance moved by effort Input work = 800 N × 4 × distance moved by load Output work = force × distance moved by load Output work = 1600 N × distance moved by load Efficiency = Output work / Input work Efficiency = (1600 N × distance moved by load) / (800 N × 4 × distance moved by load) Efficiency = 0.5 = 50% Therefore, the efficiency of the machine is 50%. Option (B) is the correct answer.

**Question 23**
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Two long parallel wires X and Y carry currents 3A and 5A each. if the force experienced by unit length is 5 x 10-5N. the force per unit length experienced by wire Y is

**Answer Details**

since the wires are parallel to each other, the forces which they exert on each other are equal and opposite, since they must agree with newton's law of action an reaction. Hence the force exerted by the wire is equally

5 x 10-5

**Question 25**
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A glass plate 0.9cm thick has a reflective index of 1.50. how long does it take for a pulse of light to pass through the plate?

[c = 3.0 x 8 ms-1]

**Answer Details**

N = 1.5 = (speed of light in glass)/speed of light in air

(1.5)/1 = (3.0 x 108)/V

V = (3.0 x 108)/1.5

V = 2.0 x 108 ms-1

But x = Vt

t = (x)/Vt

t = (x)/v = (9 x 10)4/2.0x 108 = 4.5 x 10-2S

∴ Time = 4.5 x 10-12S

**Question 26**
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Which of the following gas laws is equivalent to the work done?

**Answer Details**

Mathematically Boyle's law = PV = K

Where P = Pressure

and V = Volume

By definition Pressure = FA=FM×M $\frac{F}{A}=\frac{F}{M\times M}$

Again Workdone = force x displacement

thereforePV=FA×V=FM2×M31PV=Fx×x×x×x×x1=F×x $PV=\frac{F}{A}\times V=\frac{F}{{M}^{2}}\times \frac{{M}^{3}}{1}\phantom{\rule{0ex}{0ex}}PV=\frac{F}{x\times x}\times \frac{x\times x\times x}{1}=F\times x$

= force x displacement

**Question 27**
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The difference observed in solids, liquids and gases can be accounted for by

**Answer Details**

The question is asking what accounts for the observed differences in solids, liquids, and gases. The options provide four possible answers to choose from. The correct answer is: "The spaces and forces acting between the molecules." Solids, liquids, and gases have different physical properties because of the way their molecules are arranged and how they interact with each other. The molecules in solids are tightly packed together and vibrate in place, while the molecules in liquids and gases have more space between them and can move around more freely. The spaces and forces between molecules determine how they behave under different conditions, such as temperature and pressure. This is why solids, liquids, and gases have different melting points and boiling points, and why they can have different densities and viscosities.

**Question 28**
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I. Mass

II. Density

III. Temperature

IV. Nature of substance

Which of the above affect diffusion?

**Answer Details**

Rates of diffusion of a substance generally depend on density temp. and nature of substance II,III and IV

**Question 29**
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Lenz's law is a law of conservation of

**Answer Details**

Lenz's law is a fundamental law in physics that states that the direction of an induced current in a conductor will be such that it opposes the change that produced it. In simpler terms, Lenz's law is a law of conservation of energy that describes the direction of induced current in relation to a changing magnetic field. When a magnetic field changes, an electric current is induced in a conductor, and the direction of the induced current is such that it opposes the change in the magnetic field. Therefore, the correct answer is Energy.

**Question 30**
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A piece of iron weights 250 N in air and 200 N in a liquid of density 1000 kgm-3. the volume of the iron is?

**Answer Details**

With in the air = 250N

wt. in the liquid = 200N

loss in wt. in the liquid = 250 - 200 = 50N

∴ Wt. the liquid of equal vol. of the iron = 50N

∴If the density of the liquid is 1000 lgm-3

then the volume of the liquid = (N)/D = V = (25kg)/1000 =0.025m3 = 2.5 x 10-2 m8.

**Question 31**
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The fundamental frequency of a plucked wire under a tension of 400N is 250Hz. When the frequency is changed to 500 Hz at constant length, the tension is

**Answer Details**

The tension in a wire affects its fundamental frequency. As tension increases, so does the frequency. The relationship between tension and frequency is given by the following formula: f = (1/2L)√(T/μ) where f is the frequency, L is the length of the wire, T is the tension, and μ is the linear mass density of the wire. We are given that the wire has a tension of 400N and a frequency of 250Hz. If we keep the length of the wire constant and increase the frequency to 500Hz, we can use the formula to find the new tension: f = (1/2L)√(T/μ) Rearranging the formula, we get: T = (4L^2μ)(f^2) Since we are keeping the length of the wire constant, we can simplify the formula to: T ∝ f^2 This means that the tension is proportional to the square of the frequency. If we increase the frequency by a factor of 2 (from 250Hz to 500Hz), the tension will increase by a factor of 2^2 = 4. Therefore, the new tension is: T = 4(400N) = 1600N So the answer is 1600N.

**Question 32**
**Report**

118.8cm2 surface of the copper cathode of a voltameter is to be coated with 10-6m thick copper of density 9 x 103kgm-3. How long will the process run with 10A constant current?

[3.3 x 10-7kgC-1]

**Answer Details**

The problem is asking for the time it will take to deposit a certain thickness of copper on a copper cathode in a voltameter given the surface area of the cathode, the desired thickness of the deposit, the density of copper and a constant current. The formula to use is: time = (mass of copper deposited) / (current x electrochemical equivalent of copper) The mass of copper deposited can be calculated as: mass = volume x density = (surface area x thickness) x density The electrochemical equivalent of copper is the amount of copper deposited by one coulomb of charge passing through a copper ion in the electrolyte, and its value is 0.000329 gC^-1. Substituting the given values in the above formulas, we get: mass = (118.8 cm^2 x 10^-6 m) x (9 x 10^3 kgm^-3) = 1.0692 x 10^-3 kg electrochemical equivalent of copper = 0.000329 kgC^-1 time = (1.0692 x 10^-3 kg) / (10 A x 0.000329 kgC^-1) = 3.2548 s Converting the time to minutes, we get: time = 3.2548 s x (1 min / 60 s) = 0.0542 min ≈ 5.4 min Therefore, the answer is 5.4 min.

**Question 33**
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A 40KW electric cable was uses to transmit electricity through a resistor of resistance 2.00Ω at 800V. The power loss as internal energy is

**Answer Details**

In general, Power = IV;⟹40KW=IVTherefore 40000=1×800⟹I=40000800=50A,tune the current through $\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}40KW=IV\phantom{\rule{0ex}{0ex}}\text{Therefore}40000=1\times 800\phantom{\rule{0ex}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}I=\frac{40000}{800}=50A,\text{tune the current through}$

Resistor = 50A

power loss= I2R = 502 x 2

= 2500 x 2 = 5.0 x 103W

**Question 34**
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A conductor has a diameter of 1.00 mm and length 2.00m. if the resistance of the material is 0.1Ω, its resistivity is?

**Answer Details**

R = (1)/a, P = (R.a)/1 = (0.1 x 3.14)/2 (10-3/2)3

∴ P = (0.1 x 3.14 x 10-6)/2 x 4

= 0.3925 x 10-7 OR -3.930 X 10-8 Ωm

**Question 35**
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A car of mass 1500 kg goes round a circular curve of radius 50m at a speed of 40 ms-1. The magnitude of the centripetal force on the car is ?

**Answer Details**

Centipetal force of the car = ma = (MV)2/r

= (1500 x 40 x40)/50

= 48000 = 4.8 x 104 N

**Question 36**
**Report**

A string of length 4m is extended by 0.02m when a load of 0.4kg is suspended at its end. What will be the length of the string when the applied force is 15N?

**Answer Details**

The problem involves finding the final length of a string when a certain force is applied to it, given its initial length and how much it extends under a certain load. To solve the problem, we can use Hooke's law, which states that the extension of an elastic object is directly proportional to the applied force, provided the elastic limit is not exceeded. In this case, we know that the string of length 4m extends by 0.02m when a load of 0.4kg is suspended at its end. This means that the spring constant k is: k = F/x = (0.4 kg * 9.81 m/s^2)/0.02 m = 196.2 N/m where F is the force, x is the extension, and 9.81 m/s^2 is the acceleration due to gravity. To find the length of the string when a force of 15N is applied, we can use Hooke's law again: F = kΔL where ΔL is the change in length of the string. Rearranging the equation to solve for ΔL, we get: ΔL = F/k = 15 N/196.2 N/m = 0.0764 m Therefore, the final length of the string is: L = 4m + ΔL = 4m + 0.0764m = 4.0764m So the answer is 4.08m.

**Question 37**
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The energy E of a photon and it's wavelength are related by Eλ = X. the numerical value of X is?

[h = 6.63 x 10-34js, c = 3 x 108 ms-1]

**Answer Details**

The equation Eλ = X relates the energy E of a photon to its wavelength λ. The value of X can be found by substituting the given values for Planck's constant (h) and the speed of light (c) in the equation. Using the given values, X = hc = (6.63 x 10^-34 J s) x (3 x 10^8 m/s) = 1.99 x 10^-25 J m. Therefore, the correct option is 1.99 x 10-25.

**Question 38**
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A clinical thermometer is different from other mercury in glass thermometers owing to

**Answer Details**

A clinical thermometer is different from other mercury in glass thermometers because of the constriction on its stem. The constriction prevents the mercury from falling back when the thermometer is removed from the patient's body. This allows the temperature reading to remain constant until it is reset. The other options are not correct because the grade of mercury used in a thermometer is usually the same, the length of the stem can vary depending on the thermometer's purpose, and the wide range of temperatures is not necessarily unique to clinical thermometers.

**Question 39**
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A particle of weight 120N is placed on a plane inclined at 300 to the horizontal. If the plane has an efficiency of 60% to the floor what is the force required to push the weight uniformly up the plane?

**Answer Details**

Efficiency = W(120N)INPUTWORKOUTPUT×1001 $\frac{W(120N)INPUT}{WORKOUTPUT}\times \frac{100}{1}$

And for incline plane

V.R =1Sinθ=V.R=1Sin30=10.5=2therefore 60120=M.A2=M.A=120100=1.2but M.A =loadeffort=120Etherefore120E=1.2=E=1201.2=100N $\frac{1}{Sin\theta}\phantom{\rule{0ex}{0ex}}=V.R=\frac{1}{Sin30}=\frac{1}{0.5}=2\phantom{\rule{0ex}{0ex}}\text{therefore}\frac{60}{120}=\frac{M.A}{2}=M.A=\frac{120}{100}=1.2\phantom{\rule{0ex}{0ex}}\text{but M.A}=\frac{load}{effort}=\frac{120}{E}\phantom{\rule{0ex}{0ex}}\text{therefore}\frac{120}{E}=1.2\phantom{\rule{0ex}{0ex}}=E=\frac{120}{1.2}=100N\phantom{\rule{0ex}{0ex}}$

Therefore Effort up the plane = 100N

**Question 40**
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The bond that forms a semiconductor is?

**Answer Details**

The bond that forms a semiconductor is a covalent bond. In a covalent bond, two or more atoms share electrons in order to achieve a stable configuration. In semiconductors, such as silicon and germanium, each atom has four valence electrons which form covalent bonds with neighboring atoms, resulting in a crystal lattice structure. This lattice structure creates a small energy gap between the valence and conduction bands, which allows some electrons to move from the valence band to the conduction band when energy is added to the system, resulting in electrical conductivity.

**Question 41**
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Silicon doped with aluminum and germanium doped with arsenic become?

**Answer Details**

When silicon, a four valency element, is doped with aluminum, a three valency element, a p-type semiconductor is produced where the major carriers are the electrons.

Thus, the correct option is p-type and n-type respectively

**Question 42**
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Two spheres of masses 5.0 kg and 10.0 kg are 0.3 m apart. Calculate the force of attraction between them?

[G = 6.67 X 10-11 Nm2 Kg2]

**Answer Details**

The force of attraction between the two spheres is given by F (GM1 M2)/r2 = (6.67X 10-11 X5X10)/(0.3)2

= 3.71 X 10-8N.

**Question 43**
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A blacksmith heated a metal whose cubic expansivity is 6.3 x 10-6K-1. The areal expansivity is

**Answer Details**

If the lineal expansivity = ∝

=> area expansivity β= 2∝.

And cubic expansivity δ = 3∝.

=> 6.3 x 10-6 = 3∝

∴ ∝ =

∴ ∝ = | 6.3 x 10-6 | = 2.1x 10-6 |

__3 |

∴ Area = 2∝ = 2 x 2.1x10-6

= 4.2 x 10-6K-1

**Question 44**
**Report**

From the diagram above, the inductive reactance and the resistance R are respectively

**Answer Details**

The inductive reactance is given as

Xl = 2πfl

= 2 x 50/π x 0.1

= 10Ω

Again from ohms law,

R = V/I = 75/1.5 = 50Ω

∴ Thus the inductive reactance and the resistance are respectively 10Ω, and 50Ω

**Question 45**
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In the diagram above, a rod 50 cm long of uniform cross-section is suspended horizontally on a fulcrum, F, by the action of two forces. What is the weight of the rod?

**Answer Details**

The weight of the rod can be determined by balancing the moments on either side of the fulcrum, F. The moment of a force is its magnitude multiplied by the perpendicular distance from the fulcrum to the line of action of the force. Let the weight of the rod be W, and the distance from the left end of the rod to the fulcrum be x. Then, the distance from the right end of the rod to the fulcrum is (50 - x). Since the rod is in equilibrium, the total clockwise moment about F must be equal to the total anticlockwise moment. The weight W acts downwards at the centre of gravity of the rod, which is at its midpoint, so its moment about F is W(50/2 - x) = 25W - Wx. The force acting to the left at the left end of the rod is 40N, and its moment about F is 40x. The force acting to the right at the right end of the rod is unknown, and its moment about F is P(50 - x), where P is the magnitude of the force. Setting the clockwise moment equal to the anticlockwise moment gives: 25W - Wx = 40x + P(50 - x) Simplifying, we get: 25W = 41x + 50P We need to find W, so we need to find P. We can use the fact that the rod is in equilibrium to say that the total upward force must be equal to the total downward force. Upward force = weight of rod + 40N Downward force = P Therefore: W + 40 = P Substituting this into the previous equation, we get: 25W = 41x + 50(W + 40) Simplifying, we get: 25W = 41x + 50W + 2000 Solving for W, we get: W = 80N Therefore, the weight of the rod is 80N.

**Question 46**
**Report**

During the nuclear reaction described by

235 | W → | 235 | X→ | 231 | Y |

92 | 93 | 91 |

The particles emitted are respectively

**Answer Details**

The decay process illustrated above shows that the lost β -particle to decay to the X by increase its atomic number ( from 92 to 93) and leaving the mass no. unaffected. There after, x decayed by -α emission to give Y, by decreasing the mass no. by 2 ( from 93 to 91).

Thus , the emission is β and α particles respectively

**Question 47**
**Report**

A charge 50 µc has electric field strength of 360 NC-1 at a certain point. The electric field strength due to another charge 120 µc kept at the same distance apart and in the same medium is?

**Answer Details**

E = Fq
$\frac{\text{F}}{\text{q}}$ and Q1 = 50μF, E1 = 360μF, E2 = ?, Q2 = 120μF

Using E = Fq
$\frac{\text{F}}{\text{q}}$.

So we have F = Q ×
$\times $ E

F = 50 ×
$\times $ 10−6
$-6$ ×
$\times $ 360 = 0.018N

When E2 = F/q

0.018120×10−6
$\frac{0.018}{120\times {10}^{-6}}$ = 150NC−1

**Question 48**
**Report**

The instrument used by designers to obtain different color pattern is called?

**Answer Details**

The instrument used by designers to obtain different color patterns is called a kaleidoscope. A kaleidoscope is a tube-shaped optical instrument that reflects multiple mirrors to produce a symmetrical pattern of colors and shapes. Designers use a kaleidoscope to get inspiration for creating new patterns and designs in various fields, including textile design, graphic design, and art.

**Question 49**
**Report**

2 kg of water is heated with a heating coil which draws 3.5 A from a 200 V mains for 2 minutes. what is the increase in temperature of the water?

[specific heat capacity of water = 4200 jkg -1k-1]

**Answer Details**

The increase in temperature of the water can be calculated using the formula: Q = mcΔT Where Q is the heat energy transferred to the water, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature. First, let's calculate the heat energy transferred to the water: Power = VI = 200V × 3.5A = 700W Energy = Power × time = 700W × 2 minutes = 84000J Now, let's calculate the mass of the water: 1 kg of water has a volume of 1 liter, and the density of water is 1000 kg/m³. Therefore, 2 kg of water has a volume of 2 liters or 0.002 m³. Next, we can use the formula to calculate the change in temperature: ΔT = Q / (mc) ΔT = 84000J / (2kg × 4200J/kg.K) ΔT = 10°C Therefore, the increase in temperature of the water is 10°C. Option (d) is the correct answer.

**Question 50**
**Report**

The phenomenon of light bending round an obstacle is?

**Answer Details**

The phenomenon of light bending around an obstacle is called diffraction. When light waves encounter an obstacle or an opening in an object that is comparable in size to the wavelength of the light, the waves will bend and spread out. This bending and spreading out of waves is known as diffraction. It is a characteristic of waves that causes them to bend around obstacles or to spread out after passing through small openings. Refraction is the bending of light as it passes through a medium, such as a lens or a prism. Reflection is the bouncing of light off a surface. Polarization refers to the orientation of the electric field in a light wave, which can be either parallel or perpendicular to the direction of the wave.

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