JAMB UTME - Physics - 2024

The power of a convex lens of focal length 20cm is 

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The power of a lens is a measure of its ability to converge or diverge light. It is defined as the reciprocal (or inverse) of the focal length of the lens. The formula for calculating the power (P) of a lens in diopters (D) is given by:

P = 1/f

where:

• P is the power of the lens in diopters (D).
• f is the focal length of the lens in meters.

In this case, the focal length given is 20 cm. To apply the formula, we first need to convert this focal length into meters because the diopter is the reciprocal of the focal length in meters:

f = 20 cm = 0.20 m

Now, substitute the focal length in meters into the formula for power:

P = 1 / 0.20

P = 5.00 D

Thus, the power of the convex lens is 5.00 diopters. This indicates that the lens is capable of converging light at a distance of 5.00 meters.

The unit of impedance is 

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The unit of impedance is Ohm, which is symbolized by the Greek letter Ω (Omega). In electrical circuits, impedance (Z) is a measure of opposition that a circuit offers to the passage of electric current when a voltage is applied. It is similar to resistance but extends to alternating currents (AC) and contains the effects of resistance as well as reactance (which accounts for capacitors and inductors).

Just like resistance, the unit of impedance is the ohm because they measure similar concepts; however, impedance also accounts for phase shifts between voltage and current, which are not considered in simple resistance. Ohm's Law is used in AC circuits as Z = V/I, where Z is impedance, V is voltage, and I is current. This relationship shows why the unit of impedance is the same as that of resistance.

A load of 300N is to be lifted by a machine with a velocity ratio of 2 and an efficiency of 60%. What effort will be applied to lift the load?

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To determine the effort needed to lift a load using a machine, we first need to understand some key concepts: **Load**, **Effort**, **Velocity Ratio** (VR), and **Efficiency**.

1. **Load** is the force or weight that needs to be lifted by the machine. In this case, the load is 300N.

2. **Velocity Ratio (VR)** is the ratio of the distance moved by the effort to the distance moved by the load. Given here as 2.

3. **Efficiency** of a machine is expressed as a percentage and is the ratio of the useful work output to the input work done by the effort. Here, the efficiency is 60% or 0.60 as a decimal.

The formula to calculate the **Effort** is derived from the relationship between these factors:

\[ \text{Efficiency} = \frac{\text{Mechanical Advantage (MA)}}{\text{Velocity Ratio (VR)}} \]

Where:

\[ \text{Mechanical Advantage (MA)} = \frac{\text{Load}}{\text{Effort}} \]

From the above, we have:

\[ \text{MA} = \text{VR} \times \text{Efficiency} \]

Replacing with the given values:

\[ MA = 2 \times 0.60 = 1.2 \]

Now, calculate the **Effort** using the relation:

\[ \text{Effort} = \frac{\text{Load}}{\text{MA}} \]

\[ \text{Effort} = \frac{300N}{1.2} = 250N \]

Therefore, the **Effort** needed to lift the load is 250N.

The velocity ratio of an inclined plane at 60º to the horizontal is 

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The concept of an inclined plane is all about simplifying the forces involved in moving or holding a load. The **velocity ratio (VR)** for an inclined plane is defined as the ratio of the distance moved by the effort to the distance moved by the load. This can also be expressed in terms of the lengths involved in the triangle made by the inclined plane.

For an inclined plane placed at an angle **θ** to the horizontal, the velocity ratio is given by the formula:

VR = 1/sin(θ)

Given that the inclined plane is at an angle of **60º**:

First, find the sine of 60º:

sin(60º) = √3/2 (approximately 0.866)

Now, substitute this value into the formula for VR:

VR = 1/sin(60º) ≈ 1/0.866 ≈ 1.155

The **velocity ratio** for an inclined plane at **60º** to the horizontal is **approximately 1.155**.

Calculate the upthrust on a spherical ball of volume 4.2 x 10−4 ${}^{-4}$m3 ${}^3$ when totally immersed in a liquid of density 1028kgm−3

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Upthrust(Force) = volume of object x density of liquid x g = V x ρ $\rho$ x g

U = 4.2 x 10−4 ${}^{-4}$ x  1028 x 10 ≊ $\approxeq$ 4.3N

Inbreeding is highly discouraged in humans because it may 

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Inbreeding is the process where closely related individuals, like cousins or siblings, mate and produce offspring. **This practice is highly discouraged in humans for several reasons, but a significant concern is the potential for an outbreak of hereditary diseases.**

Here’s why inbreeding is problematic:

• Humans carry genes in pairs, and some of these genes might be recessive, meaning they only show their effects if an individual inherits two copies of the same gene (one from each parent). These genes might cause diseases or other health issues.
• When two closely related individuals mate, the chance of both carrying the same recessive genes is **higher**. This increases the likelihood that their offspring will inherit these genes, leading to hereditary diseases. **This genetic similarity can result in a higher risk of disorders such as cystic fibrosis, sickle cell anemia, and Tay-Sachs disease**.
• While some may argue that inbreeding could lead to other issues, such as dwarfism or higher death rates in newborns, the primary scientific concern backed by genetics is the increased susceptibility to hereditary diseases.

Therefore, **to promote genetic diversity and reduce the risk of hereditary diseases in offspring, inbreeding is discouraged in human populations**. This way, offspring are less likely to inherit harmful genetic combinations that can lead to health problems.

Pilots uses aneroid barometer to know the height above sea level because 

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Aneroid barometers are compact and lightweight, making them suitable for use in aircraft where space and weight are critical considerations. They provide a reliable measurement of altitude based on changes in atmospheric pressure.

The stress experienced by a wire of diameter 

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Stress is defined as the force applied per unit area. In the context of a wire being loaded by a weight, the weight acts as the force exerted, and the cross-sectional area of the wire is the area over which this force is distributed.

Force (F): This is given by the weight, which is y² N.

Cross-sectional Area (A): For a wire with a diameter, the area can be calculated using the formula for the area of a circle: A = πr², where r is the radius of the wire.

Given the diameter of the wire as yπ meters, the radius (r) is half of the diameter:

r = (yπ)/2

So, the area (A) is:

A = π[(yπ)/2]²

Simplifying the area:

A = π(y²π²/4)
A = y²π³/4

Stress (σ) is given by the formula:

σ = F/A

Substituting the given weight (force) and the calculated area:

σ = (y²) / (y²π³/4)

By simplifying the expression:

σ = (4y²) / (y²π³)

Cancel out y² from numerator and denominator:

σ = 4/π² Nm⁻²

Thus, the correct stress experienced by the wire is 4π Nm⁻², as provided in one of the options. The explanation shows clearly how the force and area are used to derive the stress experienced by the wire.

The property by which a material returns to its original shape after the removal of force is called 

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The property by which a material returns to its original shape after the removal of force is called Elasticity.

Let's break it down:

Elasticity: This is a property of a material that allows it to return to its original shape or size after the force that caused deformation is removed. Think of a rubber band—you can stretch it, but once you let it go, it snaps back to its initial shape.

Ductility: This property refers to a material's ability to be stretched into a wire. For example, materials like copper are ductile because they can be drawn into thin wires without breaking.

Malleability: This is a material's ability to withstand deformation under compressive stress. It is the property that allows metals to be hammered or rolled into thin sheets. Gold is a good example of a malleable metal.

Plasticity: This property describes the material's ability to undergo permanent deformation without breaking. When a plastic region is reached, the material will not return to its original shape after the removal of force.

Therefore, when we speak of a material returning to its original shape after the removal of force, we are specifically referring to Elasticity.

In a Hare's apparatus, the height of water and a liquid X are 0.3m and 0.5m respectively. The relative density of x is?

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For Hare's apparatus

Relative density = hwhl $\frac{h_w}{h_l}$

Given: height of liquid = 0.5cm, height of water = 0.3cm

Relative density = 0.30.5 $\frac{0.3}{0.5}$ = 0.6

A hydrometer of mass y kg and volume 2y x 10−5 ${}^{-5}$m3 ${}^3$ floats in a fluid with 20% of its volume above the fluid, what is the density of the fluid?

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To find the density of the fluid, we need to apply the principle of floatation, which states that the weight of the fluid displaced by the submerged part of the object is equal to the weight of the object. Let's walk through the steps:

Step 1: Understand the volume submerged

The hydrometer has a total volume of 2y x 10^-5 m³. It floats with 20% of its volume above the fluid. Hence, 80% of its volume is submerged in the fluid.

Submerged Volume, V_sub = (0.80) x (2y x 10^-5 m³) = 1.6y x 10^-5 m³

Step 2: Apply the principle of floatation

The weight of the fluid displaced equals the weight of the hydrometer.

Weight of hydrometer = Mass x Gravity = y kg x g (where g is the acceleration due to gravity). For the purpose of calculations, g can be considered as 9.81 m/s².

Weight of displaced fluid = Density of fluid (ρ_fluid) x Submerged Volume x g

According to the principle of floatation:

y x g = ρ_fluid x 1.6y x 10^-5 m³ x g

g is common on both sides and can be canceled out:

y = ρ_fluid x 1.6y x 10^-5

Step 3: Solving for the density of the fluid

ρ_fluid = y / (1.6y x 10^-5)

The y on both numerator and denominator cancels out:

ρ_fluid = 1 / (1.6 x 10^-5)

ρ_fluid = 6.25 x 10⁴ kg/m³

Thus, the density of the fluid is 6.25 x 10⁴ kg/m³.

An example of a non-rechargeable cell is 

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A non-rechargeable cell, commonly known as a primary cell, is a type of chemical battery that is designed to be used once until the chemical reactions that produce electricity are exhausted. After this point, the cell cannot be reversed or recharged.

In the given examples, the dry leclanche cell is a well-known example of a non-rechargeable cell. It is commonly used in everyday devices like remote controls, wall clocks, and torches. This cell type utilizes zinc and manganese dioxide as electrodes and relies on a moist paste of ammonium chloride for the electrolyte.

The other examples, such as nickel iron, mercury cadmium, and lead-acid, involve rechargeable cells (secondary cells) that are specifically designed to endure multiple charges and discharges throughout their useful life. Thus, unlike the dry leclanche cell, these can be recharged after use.

Therefore, the dry leclanche cell is an ideal example of a non-rechargeable cell because it can only be used once. After depletion, it cannot be recharged or reused.

The equivalent capacitance of the capacitors in the circuit above

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apacitance in parallel = one at the top + one under = 2C

The two in the middle are in series = C2 $\frac{C}{2}$

The equivalent capacitance of the capacitors in the circuit above = C2 $\frac{C}{2}$ + 2C = 52 $\frac{5}{2}$C

If a body in linear motion changes from point P to Q, the motion is 

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When a body moves in a straight line from one point, such as point P, to another point, such as point Q, the motion is called Translational Motion. This kind of motion refers to an object moving along a path in which every part of the object takes the same path as a reference point. This means that if you follow any point on the body, it covers the same amount of distance in the same time frame as any other point.

Let's break down the other options:

• Rotational Motion: This occurs when an object spins around an internal axis. For example, the rotation of the Earth on its axis is a rotational motion.


• Simple Harmonic Motion: This is a type of periodic motion where the restoring force is directly proportional to the displacement. An example of this is a pendulum swinging back and forth.


• Circular Motion: This occurs when an object moves along a circular path. For instance, the motion of the planets around the sun is circular motion.

In conclusion, since the body is moving from point P to point Q along a straight line, it exhibits Translational Motion.

The efficiency of a cell with internal resistance of 2Ω $\mathrm{\Omega }$ supply current to a 6Ω $\mathrm{\Omega }$ resistor is 

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To determine the efficiency of a cell with an internal resistance of 2 Ω while supplying current to a 6 Ω resistor, we can use the concept of power dissipation. Efficiency in this context is the ratio of the power delivered to the external resistor to the total power supplied by the cell. It can be calculated using the formula:

Efficiency (%) = (Power across load resistor / Total power output by cell) × 100

Let's break it down step by step:

• Step 1: Total Resistance - The total resistance in the circuit is the sum of the internal resistance and the resistance of the external resistor:
• Total Resistance (R_total) = Internal Resistance (R_internal) + External Resistance (R_load)
• R_total = 2 Ω + 6 Ω = 8 Ω
• Step 2: Current Calculation - The current flowing through the circuit can be calculated using Ohm's Law, which states V = IR. Assume the electromotive force (EMF) of the cell is E volts:
• Current (I) = E / R_total
• I = E / 8 (Note: We keep E as a variable as it cancels out in the efficiency equation)
• Step 3: Power Delivered to Load - The power delivered to the external resistor can be calculated as:
• Power_load = I² × R_load
• Power_load = (E/8)² × 6
• Step 4: Total Power Supplied by the Cell - The total power supplied by the cell is given by:
• Power_total = EMF × Current = E × (E/8)
• Power_total = E² / 8
• Step 5: Calculate Efficiency: Finally, plug in the power values into the efficiency formula:
• Efficiency (%) = [(E/8)² × 6 / (E² / 8)] × 100
• Simplify to Efficiency (%) = [(6/8)] × 100 = 0.75 × 100 = 75%

The efficiency of the cell when supplying current to a 6 Ω resistor with an internal resistance of 2 Ω is 75%.

The defect of the eye lens which occurs when the ciliary muscles are weak is 

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The defect of the eye lens that occurs when the ciliary muscles are weak is known as Presbyopia.

Here's a simple explanation:

The ciliary muscles in the eye are responsible for helping the lens to change shape so that you can focus on objects at different distances. As people age, the ciliary muscles may become weaker. This weakness hampers their ability to properly adjust the lens. As a result, the lens cannot accommodate or focus as effectively, especially when looking at nearby objects. This leads to a difficulty in seeing objects up close clearly, which is known as presbyopia.

Presbyopia is a natural condition associated with aging, and it typically becomes noticeable in people in their 40s or 50s. This is different from other eye conditions like:

• Myopia, which is when distant objects appear blurry.
• Astigmatism, which is caused by an irregularly shaped cornea leading to blurry or distorted vision at all distances.
• Spherical aberration, which is a concept related to optical systems where light rays striking the edges of a lens do not focus at the same point as light rays passing through the center, but it is not specifically related to the ciliary muscles or biological aging.

So in summary, presbyopia is the condition that results from weakened ciliary muscles, affecting near vision as a person ages.

The formation of cilia and flagella in living cells is carried out with the help of

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The formation of cilia and flagella in living cells is primarily carried out with the help of **centrioles**.

Here's a simple explanation:

Centrioles are cylindrical structures made up of microtubules. They are found in eukaryotic cells and play a critical role in cell division and the organization of the cell's cytoskeleton. However, their role extends beyond this to the formation of the basal bodies which seed the growth of cilia and flagella.

Cilia and flagella are microscopic, hair-like structures that protrude from the surface of certain eukaryotic cells. They are primarily involved in movement. Cilia often work like tiny oars, moving fluid across the cell's surface or propelling single-celled organisms. Flagella are typically longer and move in a whip-like fashion to propel cells, such as sperm cells.

Here's how centrioles contribute to the formation of these structures:

1. **Basal Body Formation**: Each cilium or flagellum grows out from a structure known as a basal body. The basal body is derived from the centrioles. During this process, a centriole migrates to the cell's surface and acts as a nucleation site for the growth of microtubules, which in turn form the structural core of cilia and flagella.

2. **Microtubule Organization**: The centrioles help organize microtubules in a "9+2" arrangement, which is characteristic of cilia and flagella. This refers to nine pairs of microtubules forming a ring around two central microtubules, giving these structures both stability and flexibility for movement.

Thus, centrioles are crucial as they provide the groundwork for the formation and proper functioning of cilia and flagella. They ensure that these structures are assembled correctly and are able to carry out their roles in cell movement and fluid transport.

What is the inductance reactance of a coil of 7H when connected to a 50Hz a.c circuit?

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To determine the inductive reactance of a coil, we use the formula:

Inductive Reactance (X_L) = 2πfL

Where:

• X_L is the inductive reactance, measured in ohms (Ω)
• π is a constant approximately equal to 3.14159
• f is the frequency of the AC source, measured in hertz (Hz)
• L is the inductance of the coil, measured in henrys (H)

Given:

• Inductance (L) = 7 H
• Frequency (f) = 50 Hz

Substituting the given values into the formula:

X_L = 2 × π × 50 × 7

Calculating this:

X_L = 2 × 3.14159 × 50 × 7

X_L ≈ 2 × 3.14159 × 350

X_L ≈ 2 × 1099.557

X_L ≈ 2199.114

Therefore, the inductive reactance of the coil is approximately 2200Ω.

A cell of internal resistance of 2Ω $\mathrm{\Omega }$ supplies current through a resistor, X if the efficiency of the cell is 75%, find the value of X.

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To solve the problem, let's first understand the concept of efficiency in this context. Efficiency refers to the ratio of the useful power output to the total power output of a system. In simpler terms, it tells us how much of the power provided by the cell is being effectively used by the resistor, X.

Given that the cell has an internal resistance (r) of 2Ω and we need the efficiency to be 75%, we will follow these steps:

• The total resistance in the circuit is the sum of the internal resistance and the resistance of the resistor X, so it is (r + X).
• Efficiency is given by the formula:

Efficiency (%) = (R / (R + r)) * 100

Where:

• R is the resistance of resistor X.
• r is the internal resistance of the cell.

According to the problem, efficiency is 75%, so:

(X / (X + 2)) * 100 = 75

First, let’s eliminate the percentage by dividing both sides by 100:

(X / (X + 2)) = 0.75

Now, let's solve for X:

1. Multiply both sides by (X + 2):

X = 0.75 * (X + 2)



2. Distribute the 0.75:

X = 0.75X + 1.5



3. Subtract 0.75X from both sides:

0.25X = 1.5



4. Finally, divide by 0.25:

X = 1.5 / 0.25



5. Solve for X:

X = 6 Ω

Hence, for the cell to have an efficiency of 75%, the value of the resistor X must be 6Ω.

A medium texture soil with high organic matter is

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A medium texture soil with high organic matter is best described as loamy soil. Here's why:

Loamy soil is a type of soil that is characterized by a balanced mixture of sand, silt, and clay particles. Because of this blend, loamy soil is not too coarse like sandy soil, nor is it too compact and dense like clay soil, making it a medium texture.

Moreover, loamy soil is renowned for its high organic matter content. This means that it contains a significant amount of decomposed plant and animal residues, which enrich the soil and provide essential nutrients for plant growth. This high organic content enhances the soil's fertility and structure, enabling it to retain moisture yet drain well, making it ideal for farming and gardening.

In conclusion, due to its balanced texture and richness in organic matter, loamy soil is the best fit for a medium-textured soil with high organic matter.

The dimension of power is 

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The dimension of power in physics is expressed in terms of the base units of mass (M), length (L), and time (T). Power is the rate at which work is done or energy is transferred over time, and it has the unit of watt (W) which is equivalent to one joule per second.

To derive the dimension of power:

1. Work has the dimension of energy, which is force applied over a distance. The dimension of work (or energy) is M L² T^-2 because force has the dimension M L T^-2 and distance adds another L.

2. Since power is work done per unit time, you would divide the dimension of work by time (T).

Thus, the dimensional formula for power is:

M L² T^-3

A light ray passing from air into water at an angle of 30º from the normal in air would 

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When light passes from one medium to another, such as from air to water, it bends or refracts. This phenomenon is described by Snell's Law, which states: n₁ * sin(θ₁) = n₂ * sin(θ₂), where:

• n₁ is the refractive index of the first medium (air),
• θ₁ is the angle of incidence (the angle made by the light ray with the normal in the first medium),
• n₂ is the refractive index of the second medium (water),
• θ₂ is the angle of refraction (the angle made by the light ray with the normal in the second medium).

The refractive index of air is approximately 1, and the refractive index of water is approximately 1.33. Given the angle of incidence in air is 30º:

Using Snell's Law:

1 * sin(30º) = 1.33 * sin(θ₂)

You will find:

sin(θ₂) = sin(30º) / 1.33

sin(θ₂) ≈ 0.5 / 1.33

sin(θ₂) ≈ 0.375

Now, solve for θ₂ by taking the inverse sine (arcsin):

θ₂ ≈ arcsin(0.375)

θ₂ ≈ 22.09º

Thus, when a light ray passes from air into water at an angle of 30º from the normal in air, it will make an angle less than 30º from the normal in water, approximately 22.09º. This is because the light ray bends toward the normal as it enters a denser medium (water).

One main feature of trees in the savanna habitat is the possession of

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The main feature of trees in the savanna habitat is the possession of thick, corky bark. The savanna is characterized by a distinct wet and dry season. During the dry season, fires are common as dry grasses and leaves become highly flammable. To adapt to this environmental condition, many trees in the savanna have developed a thick, corky bark which helps protect them against these frequent fires. This bark acts as an insulator, shielding the vital inner tissues of the tree from the heat of the flames. Additionally, this adaptation helps the trees retain moisture, which is crucial during the arid months when water is scarce.

A wheelbarrow inclined at 60º to the horizontal is pushed with a force of 150N. What is the horizontal component of the applied force

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When you push a wheelbarrow inclined at an angle to the horizontal, the applied force can be divided into two components: a **horizontal component** and a **vertical component**. To find the horizontal component of the force, you need to use the concept of resolving vectors.

The force of 150N is acting at an angle of 60º to the horizontal. The horizontal component of this force can be calculated using the cosine of the angle. The formula to determine the horizontal component \( F_{\text{horizontal}} \) is given by:

F_horizontal = F_applied \times \cos(\theta)

Where:

• F_applied is the magnitude of the applied force, which is 150N.
• \(\theta\) is the angle of inclination to the horizontal, which in this case is 60º.

Substitute the values into the formula:

F_horizontal = 150N \times \cos(60º)

We know that \(\cos(60º)\) equals 0.5.

Therefore:

F_horizontal = 150N \times 0.5 = 75N

Thus, the **horizontal component** of the applied force is 75N.

Newton's law of cooling is valid only for a 

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Newton's Law of Cooling states that the rate of heat loss of an object is directly proportional to the difference in temperature between the object and its surroundings, provided that this temperature difference is small.

Therefore, this law is only valid within a small temperature range.

The force of attraction between molecules of the same substance is 

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The force of attraction between molecules of the same substance is called cohesion.

To understand this simply:

Cohesion refers to the attractive forces acting between similar molecules. For example, water molecules attract each other due to hydrogen bonding, which is a strong intermolecular force.

Let's break down some important concepts:

• Cohesion: This force is responsible for phenomena like water forming droplets and the surface tension that allows some insects to walk on water.
• Capillarity: This is not related to cohesion directly, but rather to the movement of liquid within narrow spaces without the assistance of external forces, usually due to a combination of cohesion and adhesion.
• Adhesion: This is the force of attraction between unlike molecules, such as water sticking to the surface of a glass.
• Surface Tension: While related to cohesion, this specifically refers to the effect caused by cohesive forces at the surface of a liquid, which makes the surface act like a stretched elastic membrane.

In summary, **cohesion** is the force that keeps the molecules of the same substance, like water, attracting each other.

The part of the inner ear that is responsible for hearing is

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The part of the inner ear that is responsible for hearing is the cochlea.

The inner ear is a complex structure, and each of its components serves different functions. Let me break it down further:

• Cochlea: This is a spiral-shaped, fluid-filled structure that resembles a snail shell. It is the main organ responsible for hearing. Sound waves enter the ear, travel through the ear canal, and cause the eardrum to vibrate. These vibrations are then transmitted to the cochlea through tiny bones in the middle ear. Inside the cochlea, these vibrations create waves in the fluid, stimulating tiny hair cells. The movement of these hair cells converts the sound waves into electrical signals, which are sent to the brain through the auditory nerve, allowing us to perceive sound.


• Sacculus and Utriculus: These structures primarily deal with balance rather than hearing. They are part of the vestibular system, which helps the body maintain its balance and spatial orientation.


• Ampullae: These are located in the semicircular canals of the inner ear and also play a role in balance. They contain sensory hair cells that detect rotational movement of the head.

Thus, the cochlea is the crucial component of the inner ear responsible for converting sound vibrations into nerve signals, making it central to the process of hearing.

The energy of light of frequency 2.0 x 1015 ${}^{15}$Hz is  (h = 6.63 x 10−34 ${}^{-34}$Js)

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To determine the energy of light given its frequency, we can utilize the formula:

E = h × f

Where:

E is the energy of the photon in joules (J)

h is Planck's constant, approximately 6.63 × 10^-34 J·s

f is the frequency of light in hertz (Hz)

Given the frequency f = 2.0 × 10¹⁵ Hz, we can substitute the known values into our equation:

E = 6.63 × 10^-34 J·s × 2.0 × 10¹⁵ Hz

To simplify the calculation, multiply the numerical parts and then add the indices of 10:

E = (6.63 × 2.0) × (10^-34 × 10¹⁵)

E = 13.26 × 10^-19 J

This can be approximated to 1.33 × 10^-18 J. Thus, the energy of light with the given frequency is 1.33 × 10^-18 J.

The fourth overtone of a closed pipes is 900Hz, its fundamental frequency is 

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To solve this problem, let's first understand how sound works in a closed pipe. A closed pipe has one end closed and another end open. Sound waves inside such a pipe create standing waves, where nodes (points of no movement) and antinodes (points of maximum movement) are formed.

For a closed pipe, the fundamental frequency (also called the first harmonic) has one node at the closed end and one antinode at the open end. The wavelength is four times the length of the pipe.

The overtone sequence for a closed pipe includes only odd harmonics: 1st (fundamental), 3rd, 5th, 7th, etc. The nth overtone is the 2nth + 1 harmonic. The equation for the frequency of a harmonic in a closed pipe is:

f_n = n * f_1, where f_n is the frequency of the nth harmonic and f_1 is the fundamental frequency

In this case, the fourth overtone corresponds to the 9th harmonic because 2 * 4 + 1 = 9. Therefore, we have:

900 Hz = 9 * f_1

To find the fundamental frequency (f_1), we solve for f_1:

f_1 = 900 Hz / 9

f_1 = 100 Hz

Therefore, the fundamental frequency is 100 Hz.

Mouth part adapted for piercing and sucking is found in

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The mouthpart adapted for piercing and sucking is found in the mosquito. Mosquitoes have a specialized mouth structure called a proboscis. This proboscis is long and slender, allowing mosquitoes to puncture the skin of their hosts and suck blood. The proboscis is a complex structure that contains several needle-like parts that make the piercing and sucking process efficient and effective.

The degree of precision of a vernier caliper is 

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The degree of precision of a vernier caliper is actually the **smallest value** that the vernier scale can measure, which can be considered as the resolution or least count of the instrument. The degree of precision for most standard vernier calipers is 0.01 cm (or 0.1 mm). This means that the caliper can measure dimensions down to a hundredth of a centimeter.

To understand why this is the case, consider the construction of a vernier caliper:

• The main scale is similar to a ruler, usually graduated in centimeters and millimeters.
• The vernier scale, which moves with the sliding jaws, is marked such that some small fraction of its divisions aligns with one division on the main scale.

This alignment allows more precise measurements than the main scale alone. If the vernier scale has 10 divisions which coincide over a length equal to 9 divisions on the main scale, then each division of the vernier scale represents an extra 0.01 cm. Therefore, it allows measuring smaller intervals between the main scale markings very precisely.

Thus, you won't find vernier calipers with a degree of precision of 0.005 cm, 0.1 cm, or 1.0 cm as options in standard practice for precise measurement tools.

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When you insert a sheet of an insulating material between the plates of an air capacitor, the capacitance will increase.

Here's why:

• Capacitance is a measure of a capacitor's ability to store charge per unit voltage across its plates.
• The formula for capacitance (C) is:

C = ε₀ * (εr) * (A/d)

• Where:
  • ε₀ is the permittivity of free space (a constant).
  • εr is the relative permittivity (also known as the dielectric constant) of the material between the plates.
  • A is the area of one of the plates.
  • d is the separation between the plates.
• When an insulating material (also known as a dielectric) is inserted between the plates, its εr is greater than that of air (which is 1).
• This increases the overall capacitance of the capacitor since the product of ε₀ and εr is larger than ε₀ alone.

Therefore, inserting an insulating material as a dielectric enhances the capacitor's ability to store charge, ultimately resulting in an increase in capacitance.

A boy standing 408m from a wall blew a trumpet and heard the echo 2.4s later. Calculate the speed of the sound

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To calculate the speed of sound, we need to understand that an echo involves a sound wave traveling to a surface and back. In this case, the sound travels from the boy to the wall and then returns.

The total distance that the sound wave travels is twice the distance from the boy to the wall because it goes to the wall and back. Therefore, the total distance is:

Total Distance = 2 x 408m = 816m

The echo was heard 2.4 seconds after the sound was made. The speed of sound can be calculated using the formula:

Speed of Sound = Total Distance / Time

Plugging in the values, we have:

Speed of Sound = 816m / 2.4s

When you perform the division, you find:

Speed of Sound = 340 m/s

Thus, the speed of the sound is 340 m/s, which is the correct answer.

The value of R in the above circuit to make the galvanometer measure 2A is 

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Given: Ig ${}_g$ = 50mA = 0.05A, I to be measured = 2A, r = 2Ω $\mathrm{\Omega }$, Is ${}_s$ = I - Ig ${}_g$ = 2 - 0.05 = 1.95A

Shunt(R) = IgIs $\frac{I_g}{I_s}$ x r

R =  0.051.95 $\frac{0.05}{1.95}$ x 10 = 0.2564Ω

Using the diagram above, the effective force pushing it forward at an angle 60º is 

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To determine the effective force pushing the object forward at an angle of 60º, we need to resolve the given force into its components. Specifically, we are interested in the horizontal component of the force, as this is the part that effectively pushes the object forward.

The general formula to calculate the horizontal component of a force (F_x) when the force is applied at an angle (θ) is:

F_x = F * cos(θ)

Where:

• F is the magnitude of the force applied.
• θ is the angle at which the force is applied, in this case, 60º.
• cos(θ) is the cosine of the angle.

Assuming the magnitude of the force applied (F) is 50N, then the effective forward force can be calculated as follows:

F_x = 50N * cos(60º)

Using the trigonometric value:

cos(60º) = 0.5

Therefore:

F_x = 50N * 0.5

F_x = 25N

Hence, the effective force pushing it forward at an angle of 60º is 25.00N. Therefore, the correct answer is 25.00N.

The moon's acceleration due to gravity is 16 $\frac{1}{6}$ of the earth's value. The weight of a bowling ball on the moon would be

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To determine the weight of a bowling ball on the moon, we need to understand the relationship between weight, gravity, and mass.

Weight is the force exerted by gravity on an object. On Earth, this force depends on the object's mass and the acceleration due to gravity, which is approximately 9.8 m/s². Weight can be calculated using the formula:

Weight = Mass x Gravity

On the moon, the acceleration due to gravity is only ¹/₆ of Earth’s gravity. This means the gravitational pull on the moon is much weaker compared to the Earth. If we take the Earth's gravity to be 9.8 m/s², the moon's gravity would be:

Moon's Gravity = (9.8 m/s²) x (1/6) ≈ 1.63 m/s²

Given that the weight of an object is directly proportional to the gravitational force, the weight of an object on the moon would be substantially less than its weight on Earth. Thus, the weight of the bowling ball on the moon would be:

Weight on Moon = (Mass) x (1.63 m/s²) = ¹/₆ of its weight on Earth

Therefore, the weight of a bowling ball on the moon is ¹/₆ of its weight on Earth.

Which of the following is not a part of model rocket?

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When it comes to a model rocket, it is crucial to understand the different parts that make up the rocket and their functions:

• The Body Tube: This is the main frame of the rocket. It holds all the other components together and serves as the backbone of the rocket.
• The Nose Cone: This is the pointed part at the top of the rocket. It is designed to reduce aerodynamic drag and also aids in stabilizing the rocket during flight.
• The Fin: Fins are attached at the lower end of the rocket. They help stabilize the rocket's flight path by reducing wobble and assisting in maintaining a straight trajectory.

Now, “Not recovery devices” is listed among the options. A recovery device is actually a part of a model rocket system. Common recovery devices include parachutes or streamers that deploy after the rocket reaches its peak altitude, allowing it to return safely to the ground. Such devices are indeed part of a model rocket design.

Therefore, the option “Not recovery devices” itself is not recognized as a part of a model rocket. Instead, the sentence is stating that they are not part of the main components, which implies it's indicative rather than being the name of a component. Hence, it does not pertain to a single component like the body tube, nose cone, or fins.

Calculate the value of electric field intensity due to a charge of 4μC if the force due to the charge is 8N

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To calculate the electric field intensity due to a charge, we need to use the formula:

Electric Field Intensity (E) = Force (F) / Charge (q)

In this problem, we are given that the force (F) is 8 Newtons (N) and the charge (q) is 4 microcoulombs (μC). First, we need to convert the charge from microcoulombs to coulombs:

1 microcoulomb (μC) = 1 x 10^-6 coulombs (C)

Therefore, 4 μC = 4 x 10^-6 C.

Now we can use the formula to find the electric field intensity:

E = F / q

E = 8 N / (4 x 10^-6 C)

E = 8 / 4 x 10⁶

E = 2 x 10⁶

Thus, the value of the electric field intensity is 2 x 10⁶ N/C.

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To solve this problem, we need to understand the relationship between pressure, volume, and temperature of a gas. The relevant law here is the **Combined Gas Law**, which is expressed as:

(P1 * V1) / T1 = (P2 * V2) / T2

Where:

• P1 is the initial pressure
• V1 is the initial volume
• T1 is the initial temperature in Kelvin
• P2 is the final pressure
• V2 is the final volume
• T2 is the final temperature in Kelvin

In the given problem:

• The initial temperature, T1, is 27ºC, which is 300K (since we convert to Kelvin by adding 273).
• The final temperature, T2, is 127ºC, which is 400K.
• The pressure is doubled, so P2 = 2 * P1.

Applying the Combined Gas Law:

(P1 * V1) / 300 = (2 * P1 * V2) / 400

Simplifying this equation:

V1/300 = 2V2/400

Multiply both sides by 400 to clear the fraction:

400 * V1 / 300 = 2 * V2

Which further simplifies to:

(4/3) * V1 = 2 * V2

Dividing both sides by 2:

(2/3) * V1 = V2

This shows that the final volume, V2, is **2/3 of the initial volume, V1**. Therefore, the volume of the gas will **decrease by 1/3**.

Calculate the depth of a swimming pool if the apparent depth is 10cm. ( Refractive index of water = 1.33 )

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To calculate the real depth of a swimming pool given the apparent depth, we can use the concept of refraction of light. When light passes from one medium to a denser medium, it bends towards the normal. This bending effect causes objects submerged in water to appear closer to the surface than they actually are. The formula to relate these depths is given by:

Real Depth = Apparent Depth × Refractive Index

Given the problem:

• Apparent Depth = 10 cm
• Refractive Index of water = 1.33

Using the formula:

Real Depth = 10 cm × 1.33

Calculating the above:

• Real Depth = 13.3 cm

Therefore, the depth of the swimming pool is 13.3cm.