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Tambaya 1 Rahoto
Three resistors with resistance 200Ω, 500Ω and 1kΩ are connected in series. A 6v battery is connected to either end of the combination. Calculate the potential difference between the ends of 200Ω resistance.
Bayanin Amsa
To calculate the potential difference between the ends of the 200Ω resistance, we need to use Ohm's Law, which states that the potential difference (V) across a resistor is equal to the current (I) flowing through the resistor multiplied by the resistance (R) of the resistor. First, we need to find the total resistance of the series combination of resistors. We add up the individual resistances: Total resistance = 200Ω + 500Ω + 1kΩ = 1.7kΩ Next, we can use Ohm's Law to find the current flowing through the circuit. We know that the battery voltage is 6V, and the total resistance is 1.7kΩ: I = V / R = 6V / 1.7kΩ = 0.0035A Now we can use Ohm's Law again to find the potential difference across the 200Ω resistor: V = IR = 0.0035A * 200Ω = 0.7V Therefore, the potential difference between the ends of the 200Ω resistance is 0.7V. The correct answer is option B.
Tambaya 2 Rahoto
A train has an initial velocity of 44m/s and an acceleration of -4m/s2 . Calculate its velocity after 10 seconds
Bayanin Amsa
The velocity of the train after 10 seconds can be calculated using the formula: v = u + at where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. Substituting the given values, we get: v = 44 m/s + (-4 m/s^2) x 10 s v = 44 m/s - 40 m/s v = 4 m/s Therefore, the velocity of the train after 10 seconds is 4m/s. Answer option D is correct. Explanation: The train has an initial velocity of 44 m/s and an acceleration of -4 m/s^2. The negative sign indicates that the acceleration is in the opposite direction to the initial velocity, which means that the train is slowing down. After 10 seconds, the train's velocity decreases by 40 m/s (4 m/s^2 x 10 s) to reach a final velocity of 4 m/s.
Tambaya 3 Rahoto
Which of the following statement about the electromagnet shown above is correct?
Bayanin Amsa
A - B = S - N.
Also, starting end of the current is south while terminating end is North.
Tambaya 4 Rahoto
The pitch of a screw jack is 0.45cm and the arm is 60cm long. If the efficiency of the Jack is 75/π %, calculate the mechanical advantage.
Bayanin Amsa
P = 0.45cm, L = 60cm, Eff = 75/π%
VR | (Screw | system) | = | 2πrP | = | 2πLP |
M.A | = | Eff% × VR100 | = | 75π | × | 1100 | × | 2π × 600.45 | = | 75 × 800300 | = | 200 |
Tambaya 5 Rahoto
Which of the following is consistent with Charles' law?
I
II
III
IV.
Bayanin Amsa
This is the correct graph. The graph is volume against 1/ temperature where temperature is in Celsius.
Tambaya 7 Rahoto
A body was slightly displaced from its equilibrium position. Which one of the following is a condition for its stable equilibrium
Bayanin Amsa
The condition for stable equilibrium of a body that has been slightly displaced from its equilibrium position is "an increase in the potential energy of the body." When an object is at its equilibrium position, it has a minimum potential energy. When the object is displaced from its equilibrium position, it has a higher potential energy. For the object to be in stable equilibrium, it must be able to return to its equilibrium position after it has been displaced. If the potential energy of the object increases as it is displaced, it means that the equilibrium position is a point of stable equilibrium. This is because the object will experience a restoring force that will push it back towards its equilibrium position, as the potential energy decreases. Therefore, an increase in potential energy is a condition for a body to be in stable equilibrium after it has been slightly displaced from its equilibrium position. An increase in kinetic energy or height does not necessarily indicate stability, as it depends on the specific situation and other factors at play.
Tambaya 8 Rahoto
A well 1km deep is filled with a liquid of density 950kg/m3 and g = 10m/s2 , the pressure at the bottom of the well is
Bayanin Amsa
P = Pa + ρgh = (1.00 × 105
) + (950 × 10 × 1000)
P = 105
+ (95 × 105
) = 105
(1 + 95) = 96 × 105
P = 9.6 × 106
N/m2
Tambaya 9 Rahoto
When blue and green colours of light are mixed, the resultant colour is
Tambaya 10 Rahoto
Which of the following media allow the transmission of sound waves through them?
I. air
II. liquid
III. solids
Bayanin Amsa
Sound waves are disturbances in a medium that propagate through the medium and transfer energy from one point to another. The transmission of sound waves depends on the physical properties of the medium, including its elasticity and density. Air (Option I) is a gas that is compressible and has a relatively low density, which makes it an excellent medium for transmitting sound waves. Liquids (Option II) are also able to transmit sound waves, although the speed of sound in liquids is slower than in gases because liquids are more dense and less compressible. Solids (Option III) are able to transmit sound waves as well, but their density and elasticity make them more rigid, which means that sound waves in solids tend to be transmitted as elastic waves or mechanical waves, rather than as acoustic waves. Therefore, the correct answer is "I, II, and III".
Tambaya 11 Rahoto
In the molecular explanation of conduction, heat is transferred by the
Bayanin Amsa
In the molecular explanation of conduction, heat is transferred by the Free electrons. In metals, free electrons move randomly and collide with other particles as they gain kinetic energy. These free electrons transfer the energy to the adjacent particles, which in turn gain kinetic energy and transmit it to other adjacent particles, thus transferring heat energy from one part of the material to another. This process of heat transfer by free electrons is called conduction. Therefore, the correct option is "Free electrons."
Tambaya 12 Rahoto
In semi-conductor, the carriers of current at room temperature are
Bayanin Amsa
In a semiconductor, the carriers of current at room temperature are both electrons and holes. Semiconductors are materials with properties that are in between those of conductors (e.g. metals) and insulators (e.g. rubber). At room temperature, a semiconductor crystal contains both free electrons and positively charged vacancies called holes. When a voltage is applied across the semiconductor, the electrons move towards the positive end of the circuit and the holes move towards the negative end. This movement of charge carriers constitutes an electric current. In summary, both electrons and holes can carry current in a semiconductor at room temperature, making the correct answer.
Tambaya 14 Rahoto
Efficiency of conduction in liquids and gases compared to solids is
Bayanin Amsa
The efficiency of conduction in liquids and gases compared to solids is generally less efficient. This means that solids are better conductors of heat and electricity than liquids and gases. This is because the particles in solids are closely packed and are tightly bound to one another, allowing heat and electricity to flow easily through the material. On the other hand, the particles in liquids and gases are more spread out and less tightly bound, making it more difficult for heat and electricity to flow through these materials. However, it is important to note that the efficiency of conduction can vary depending on the specific liquid or gas and the specific solid being compared. Some liquids and gases may have properties that make them better conductors than certain solids, but this is not a general rule.
Tambaya 15 Rahoto
Gases conduct electricity under
Bayanin Amsa
Gases conduct electricity under low pressure and high voltage
Tambaya 16 Rahoto
A coil X is moved quickly away from the end Y of a stationary metal bar and a current then flows in X as shown above.
Then
Bayanin Amsa
N - S magnet is moved towards a coil production clockwise direction of current in the coil.
- This is the same as a coil moved away from S-N (Y - North pole)
Tambaya 17 Rahoto
In which of the points labelled A, B, C, D and E on the conductor shown would electric charge tend to concentrate most
Bayanin Amsa
- Charge are mostly concentrated at the outermost part of a hollow conductor
- Charge are also mostly concentrated at the pointed ends or places with high density point.
Tambaya 18 Rahoto
When two objects A and B are supplied with the same quantity of heat, the temperature change in A is obtained to be twice that of B. The mass of P is half that of Q. The ratio of the specific heat capacity of A to B is
Bayanin Amsa
θA = 2θB ,
mA | = | 12 | mB |
H = MCθ
mA
cA
θA
= mB
cB
θB
( 1/2 mB
)CA
(2θB
) = mB
cB
θB
CA CB | = | 11 |
⇒ 1 : 1
Tambaya 19 Rahoto
Which of the following characteristics of a wave is used in the measurement of the depth of the Sea?
Bayanin Amsa
Depth of sea can be measured by echo, a reflected sound waves.
Tambaya 20 Rahoto
A mixture of blue and red pigment when illuminated by white light will appear
Bayanin Amsa
A mixture of blue and red pigment when illuminated by white light will appear purple. This is because when white light shines on a surface, it contains all the colors of the visible spectrum. When blue and red pigments are mixed together, they absorb all the other colors in the spectrum except for blue and red. Therefore, when white light shines on this mixture, the blue pigment absorbs all the colors except blue, while the red pigment absorbs all the colors except red. The result of this is that the blue and red pigments reflect only blue and red light, which then combines to form purple. Therefore, the mixture of blue and red pigments appears purple when illuminated by white light.
Tambaya 21 Rahoto
The mass of water vapour in a given volume of air is 0.05g at 20°C, while the mass of water vapour required to saturate it at the same temperature is 0.15g. Calculate the relative humidity of the air.
Bayanin Amsa
Relative humidity is a measure of how much water vapor the air is holding compared to the maximum amount it could hold at a given temperature. It is expressed as a percentage. To calculate the relative humidity of the air in this problem, we need to use the formula: Relative humidity = (mass of water vapor in air / mass of water vapor required for saturation) x 100% We are given that the mass of water vapor in the air is 0.05g and the mass of water vapor required for saturation at the same temperature is 0.15g. Plugging these values into the formula, we get: Relative humidity = (0.05 / 0.15) x 100% = 33.33% Therefore, the relative humidity of the air is 33.33%. So the answer is 33.33%.
Tambaya 22 Rahoto
A metal rod has a length of 100cm at 200oC . At what temperature will its length be 99.4cm. If the linear expansivity of the material of the rod is 2 × 10−5C−1
Bayanin Amsa
The linear expansivity of a material describes how its length changes with temperature. If the linear expansivity is given as 2 × 10^-5/°C, this means that for every 1°C change in temperature, the length of the material will change by 2 × 10^-5 times its original length. Given that the rod has a length of 100 cm at 200°C, we can use this information to find its length at a different temperature. If we let L be the length of the rod at temperature T, we can write the relationship as follows: L = 100 cm * (1 + 2 × 10^-5 * (T - 200°C)) To find the temperature at which the rod will have a length of 99.4 cm, we can set L equal to 99.4 cm and solve for T: 99.4 cm = 100 cm * (1 + 2 × 10^-5 * (T - 200°C)) 99.4 cm / 100 cm = 1 + 2 × 10^-5 * (T - 200°C) 0.994 = 1 + 2 × 10^-5 * (T - 200°C) -0.006 = 2 × 10^-5 * (T - 200°C) -0.006 / 2 × 10^-5 = T - 200°C -0.006 / (2 × 10^-5) = T - 200°C -0.006 / (2 × 10^-5) + 200°C = T So the temperature at which the rod will have a length of 99.4 cm is approximately equal to -0.006 / (2 × 10^-5) + 200°C, or -100°C. Therefore, the answer is -100°C.
Tambaya 23 Rahoto
When water is boiling, it
Bayanin Amsa
When water is boiling, it changes from a liquid state to a gaseous state called steam. This happens when the water is heated to its boiling point, which is when it reaches a temperature of 100 degrees Celsius (212 degrees Fahrenheit) at sea level. As the water is heated, it absorbs energy and the molecules start to move faster and faster, eventually reaching a point where they escape into the air as steam. The temperature of the water during boiling does not change, as all the energy is being used to break the bonds between the water molecules rather than increasing the temperature. Therefore, the options "gets hotter," "increase in mass," and "decreases in mass" are not correct when describing what happens when water is boiling.
Tambaya 24 Rahoto
When the temperature of a liquid is increased, its surface tension
Bayanin Amsa
Surface tension or elasticity of a fluid decreases with increased in temperature
Tambaya 25 Rahoto
A single force which produces the same effect as a set of forces acting together at a point is known as the
Bayanin Amsa
The single force which produces the same effect as a set of forces acting together at a point is known as the "resultant". In other words, the resultant is the net force that results from combining all the individual forces acting on an object. It represents the combined effect of all the forces acting on the object and is the force that would produce the same motion as the original set of forces acting together. Therefore, when solving problems in physics, it is often useful to find the resultant force in order to determine the overall effect of multiple forces on an object.
Tambaya 26 Rahoto
A rectangular solid black has length 10cm, breadth 5cm and height 2cm. If it lies on a horizontal surface, and has density 100kg/m3 , calculate the pressure it exerts on the surface.
Bayanin Amsa
To calculate the pressure that the rectangular solid exerts on the surface, we need to use the formula for pressure: Pressure = Force / Area In this case, the force is the weight of the rectangular solid, which we can calculate using the formula: Weight = Mass x Gravity The mass of the rectangular solid can be calculated using its density and volume: Mass = Density x Volume The volume of the rectangular solid is simply its length x breadth x height: Volume = Length x Breadth x Height = 10 cm x 5 cm x 2 cm = 100 cm3 We need to convert this volume to cubic meters to use the density given in kg/m3: Volume = 100 cm3 = 0.0001 m3 Now we can calculate the mass: Mass = Density x Volume = 100 kg/m3 x 0.0001 m3 = 0.01 kg The gravity is the acceleration due to gravity, which we can assume to be 9.81 m/s2. Therefore, the weight is: Weight = Mass x Gravity = 0.01 kg x 9.81 m/s2 = 0.0981 N Now we can use this weight to calculate the pressure on the surface. The surface area in contact with the rectangular solid is simply its length x breadth: Area = Length x Breadth = 10 cm x 5 cm = 50 cm2 We need to convert this area to square meters: Area = 50 cm2 = 0.005 m2 Therefore, the pressure is: Pressure = Force / Area = 0.0981 N / 0.005 m2 = 19.62 N/m2 We can convert this to units of N/cm2 or N/mm2 if desired. This is equivalent to: Pressure = 0.1962 N/cm2 = 0.0001962 N/mm2 So the pressure that the rectangular solid exerts on the surface is 19.62 N/m2, which is approximately 20 N/m2. Therefore, the answer is 200 N/m2.
Tambaya 27 Rahoto
Electrons were discovered by
Bayanin Amsa
Electrons were discovered by J.J. Thompson. In the late 19th century, he performed a series of experiments using cathode ray tubes, which are glass tubes containing low-pressure gas and electrodes. By applying high voltage, he observed a beam of negatively charged particles traveling from the negative electrode to the positive electrode. He concluded that these particles, which he called "corpuscles," were fundamental units of negative charge and later were renamed electrons. This discovery led to the development of the modern understanding of atomic structure and the electron's role in it.
Tambaya 28 Rahoto
The equilibrium position of objects in any field corresponds to situation of
Bayanin Amsa
The equilibrium position of an object in any field corresponds to the situation of minimum potential energy. This means that at the equilibrium position, the object has the lowest possible potential energy within the field. In other words, the forces acting on the object are balanced, and the object is not being pushed or pulled in any direction. Therefore, the object will remain at rest at the equilibrium position unless it is acted upon by an external force. Of the options given, the correct answer is "minimum potential energy".
Tambaya 29 Rahoto
The volume of 0.354g of helium at 273°C and 114cm of mercury pressure is 2667cm3 . Calculate the volume
Bayanin Amsa
m = 0.354g, T1
= 273°C = 273 + 273 = 576K
P1
= 114cmHg, V1
= 2667cm3
at STP
T2
= 273K, P2
= 76cmHg, V2
= ?
P1 V1 T1 | = | P2 V2 T1 |
V2 | = | 114 × 2667 × 27376 × 576 | = | 2000.25cm3 |
Tambaya 30 Rahoto
The lower fixed part of a faulty thermometer reads 2°C while the upper fixed point is 100°C.
What is the true temperature when the thermometer reads 51°C?
Bayanin Amsa
Since the thermometer is faulty, it is not measuring the temperature accurately. To find the true temperature, we need to determine the extent of the error in the thermometer. We can do this by comparing the difference between the lower fixed point and the reading with the difference between the upper fixed point and the true temperature. Since the lower fixed point reads 2°C and the upper fixed point reads 100°C, and the thermometer reading is 51°C, we can calculate the error as follows: True temperature = (51°C - 2°C) / (51°C - 2°C) * (100°C - 51°C) + 51°C = 50°C So, the true temperature when the thermometer reads 51°C is 50°C, which is option B.
Tambaya 31 Rahoto
A boy pushes a 500kg box along a floor with a force of 2000N. If the velocity of the box is uniform, the co-efficient of friction between the box and the floor is
Bayanin Amsa
The coefficient of friction is a measure of the amount of friction between two surfaces. It is represented by the symbol "μ" and is a dimensionless quantity. The coefficient of friction between two surfaces depends on the nature of the surfaces in contact and the force pressing them together. In this problem, the boy is pushing the box with a force of 2000N. If the box is moving with a uniform velocity, then the force of friction acting on the box is equal and opposite to the pushing force applied by the boy. We can calculate the force of friction using the formula: frictional force = coefficient of friction x normal force where the normal force is the force exerted by the floor on the box in a direction perpendicular to the floor. Since the box is not moving up or down, the normal force is equal to the weight of the box. The weight of the box can be calculated using the formula: weight = mass x gravity where mass is the mass of the box and gravity is the acceleration due to gravity (9.8 m/s^2). So, the weight of the box is: weight = 500 kg x 9.8 m/s^2 = 4900 N The force of friction is equal to the pushing force of 2000N, so we can set these two equal to each other and solve for the coefficient of friction: frictional force = 2000N coefficient of friction x normal force = 2000N coefficient of friction x 4900N = 2000N coefficient of friction = 2000N / 4900N = 0.408 So, the coefficient of friction between the box and the floor is approximately 0.4. Therefore, the correct answer is 0.4.
Tambaya 32 Rahoto
The volume of a stone having an irregular shape can be determined using?
Bayanin Amsa
The volume of a stone with an irregular shape can be determined using a measuring cylinder. A measuring cylinder is a glass or plastic container with a narrow cylindrical shape and markings on the side to indicate the volume it contains. To determine the volume of an irregularly shaped stone, you would fill the measuring cylinder with water, carefully lower the stone into the water, and note the increase in the volume of the water. The difference in the volume of the water before and after the stone was added is equal to the volume of the stone. The meter rule, vernier calliper, and micrometer screw gauge are all measuring instruments, but they are not designed to measure the volume of irregularly shaped objects. The meter rule is a measuring tool used for measuring length. The vernier calliper is used for measuring the diameter of objects, and the micrometer screw gauge is used for precise measurements of small distances.
Tambaya 33 Rahoto
Which of the following is/are the limitations to the Rutherford's atomic models?
I. It is applicable when energy is radiated as electrons are revolving
II. It is applicable when energy is radiated in a continuous mode
III. It is applicable to an atom with only one electron in the other shell
Bayanin Amsa
Rutherford assumed that (I) energy is radiated when electrons are revolving (II) energy is radiated in a continuous mode. These are limitations of Rutherford's model
Tambaya 34 Rahoto
Lamps in domestic lightings are usually in
Bayanin Amsa
Lamps in domestic lighting are usually connected in parallel. This means that each lamp is connected directly to the power supply, rather than being connected in a series or divergent or convergent configuration. In a parallel configuration, each lamp operates independently of the others, and if one lamp fails, the other lamps will continue to function. This is an important feature for domestic lighting, as it ensures that a single lamp failure will not leave the entire room in darkness. Additionally, in a parallel configuration, each lamp can be controlled independently, for example by a switch or dimmer, without affecting the operation of the other lamps. This allows for greater flexibility in lighting design and control. In summary, lamps in domestic lighting are usually connected in parallel because it allows for independent operation of each lamp and ensures that a single lamp failure does not affect the operation of the others.
Tambaya 35 Rahoto
The lead-acid accumulator consists of
Bayanin Amsa
- the positive pole is lead peroxide (PbO2
)
- the negative pole is head
- the electrolyte is H2
SO4
Tambaya 36 Rahoto
Which of the following bodies, each with centre of gravity G, lying on a horizontal table, is/are in unstable equilibrium?
Bayanin Amsa
- I and II are in neutral equilibrium. They will roll continuously on the table
- III is a body with high centre of gravity (unstable)
- IV is a body with high centre of gravity (stable)
Tambaya 37 Rahoto
One newton × One meter equals?
Bayanin Amsa
One newton times one meter is equal to one Joule. A newton is the unit of measurement for force, and a meter is the unit of measurement for distance. When force is applied over a distance, work is done, which is measured in Joules. Therefore, one newton multiplied by one meter results in one Joule of work done. The other options listed (one water, one ampere, one kilogram) are not correct units of measurement for this calculation.
Tambaya 39 Rahoto
The diagram shows a uniform meter rule AB which balances horizontally at the 90cm mark when a mass of 0.2kg is suspended from B. Calculate the mass of the meter rule.
Bayanin Amsa
Mr
(90 - 50) = 0.2(100 - 90)
40Mr
= 0.2 × 10
Mr
= 240
= 0.05kg
Tambaya 40 Rahoto
A cone is in unstable equilibrium has its potential energy
Bayanin Amsa
In unstable equilibrium, potential energy decreases as the height decreases.
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