Section (A)

Q. 2 (a) What is the Tuberculosis and Hepatitis? Explain briefly. (b) Explain the mechanism of Fiber Optic Cable for signal. Explain its construction. (c) Explain the difference between Middle Latitude Cyclones and Tornadoes. (d) What is difference between the lonic and Covalent bonding? Give examples. GSA

Answers

(a) Tuberculosis (TB) and Hepatitis: A CSS Exam Perspective

Introduction:

In the context of the CSS exam in Pakistan, understanding diseases like Tuberculosis (TB) and Hepatitis is crucial as they are significant public health concerns in the country.

Tuberculosis (TB):

Tuberculosis, commonly known as TB, is a bacterial infection caused by Mycobacterium tuberculosis. This disease primarily affects the lungs but can also impact other organs. TB spreads through the air when an infected person coughs, sneezes, or talks, making it highly contagious.

Symptoms of TB:

– Persistent cough

– Chest pain

– Coughing up blood

– Weakness and fatigue

– Weight loss

– Fever

– Night sweats

Diagnosis and Treatment:

Diagnosing TB often involves a combination of clinical evaluation, imaging tests like chest X-rays, and laboratory tests such as sputum analysis. Treatment typically includes a prolonged course of antibiotics to eradicate the bacteria.

Prevention and Control:

Preventing the spread of TB involves early detection, prompt treatment of active cases, and implementing infection control measures in healthcare settings. Vaccination with the Bacille Calmette-Guérin (BCG) vaccine is also a preventive measure against TB.

Hepatitis:

Hepatitis is the inflammation of the liver and can result from various causes, including viral infections, alcohol consumption, and autoimmune diseases.

Types of Viral Hepatitis:

– Hepatitis A: Transmitted through contaminated food and water.

– Hepatitis B and C: Spread through contact with infected blood or bodily fluids, including sexual contact, sharing needles, or from mother to child during childbirth.

Symptoms of Hepatitis:

– Fatigue

– Jaundice (yellowing of the skin and eyes)

– Abdominal pain

– Nausea and vomiting

– Dark urine

Diagnosis and Treatment:

Diagnosing hepatitis involves blood tests to detect viral markers and assess liver function. Treatment varies depending on the type and severity of hepatitis but may include antiviral medications, rest, and lifestyle modifications.

Prevention and Control:

Preventing hepatitis involves vaccination (especially for hepatitis A and B), practicing safe hygiene practices, avoiding high-risk behaviors, and implementing measures to reduce the transmission of the virus in healthcare settings.

Conclusion:

In conclusion, understanding the epidemiology, symptoms, diagnosis, treatment, prevention, and control measures of diseases like Tuberculosis and Hepatitis is essential for aspirants appearing in the CSS exam, as it reflects their awareness of significant public health issues and their ability to address them effectively.

(b) Fiber Optic Cable: Mechanism and Construction

Introduction:

Fiber optic cables are a crucial component of modern telecommunications systems, providing high-speed data transmission over long distances. Understanding their mechanism and construction is essential for comprehending their functionality.

Mechanism of Signal Transmission:

1. Total Internal Reflection: Fiber optic cables transmit signals using light pulses. The core, which is the innermost part of the fiber, carries the light signal. When light enters the core from the surrounding cladding (outer layer), it undergoes total internal reflection, meaning it reflects off the boundary of the core and cladding due to the difference in refractive indices.

2. Light Propagation: The light signal travels through the core of the fiber optic cable, bouncing off the core-cladding interface due to total internal reflection. This process allows the signal to propagate along the length of the cable with minimal loss of signal strength.

3. Single Mode vs. Multimode: Fiber optic cables can be classified as single-mode or multimode based on the number of paths (modes) through which light can travel. Single-mode fibers have a smaller core diameter and allow only one mode of light propagation, enabling longer transmission distances with less signal attenuation. Multimode fibers have a larger core diameter, allowing multiple modes of light propagation, suitable for shorter distances but with higher dispersion.

Construction of Fiber Optic Cable:

1. Core: The core is the central part of the fiber optic cable through which light signals travel. It is typically made of glass or plastic with a high refractive index to facilitate total internal reflection.

2. Cladding: Surrounding the core is the cladding, which has a lower refractive index than the core. This contrast in refractive indices is essential for total internal reflection to occur, confining the light signal within the core.

3. Buffer Coating: To protect the core and cladding, a buffer coating made of a durable material like acrylate or silicone is applied. The buffer coating also provides mechanical strength and flexibility to the fiber optic cable.

4. Strength Members: Strength members such as aramid yarns or fiberglass strands are embedded within the cable to enhance its tensile strength and protect it from stretching or damage during installation or operation.

5. Outer Jacket: Finally, an outer jacket made of materials like PVC (polyvinyl chloride) or LSZH (low smoke zero halogen) encases the entire fiber optic cable, providing additional protection against environmental factors such as moisture, abrasion, and chemical exposure.

Conclusion:

Fiber optic cables leverage the principles of total internal reflection to transmit light signals efficiently over long distances. Their construction involves layers designed to optimize signal transmission, protect the optical fibers, and ensure the cable’s durability and reliability in various operating conditions.

(c) Difference between Middle Latitude Cyclones and Tornadoes:

Middle Latitude Cyclones:

1. Scale and Size: Middle latitude cyclones, also known as extratropical cyclones or mid-latitude cyclones, are large-scale weather systems that typically span hundreds to thousands of kilometers. They are associated with frontal boundaries and are often several times larger in size compared to tornadoes.

2. Formation: These cyclones form in the middle latitudes, typically between 30° and 60° latitude, where temperature gradients between air masses create conditions for their development. They usually form along boundaries between contrasting air masses, such as warm and cold fronts.

3. Duration: Middle latitude cyclones can persist for several days as they move across the Earth’s surface, affecting large regions with their associated weather patterns, including precipitation, wind, and temperature changes.

4. Associated Weather: They are associated with a variety of weather conditions, including rain, snow, thunderstorms, and strong winds. The intensity and type of weather experienced depend on the location relative to the cyclone’s center and the characteristics of the air masses involved.

5. Impact: Middle latitude cyclones can cause significant disruptions to transportation, agriculture, and infrastructure due to their large-scale impacts on weather patterns. However, they are typically less intense than tropical cyclones (hurricanes) but cover larger areas.

Tornadoes:

1. Scale and Size: Tornadoes are small-scale, localized atmospheric phenomena characterized by a rapidly rotating column of air extending from a thunderstorm to the ground. They typically have diameters on the order of tens to hundreds of meters and last for only a few minutes to a couple of hours.

2. Formation: Tornadoes form within severe thunderstorms, known as supercells, which are characterized by strong updrafts and wind shear. The interaction of warm, moist air with cold, dry air can lead to the development of rotating updrafts within these storms, resulting in the formation of tornadoes.

3. Duration: Tornadoes are relatively short-lived compared to middle latitude cyclones, typically lasting for minutes to hours. They often occur within the context of a severe thunderstorm outbreak but can develop and dissipate rapidly.

4. Associated Weather: Tornadoes are associated with intense winds, often exceeding 100 miles per hour (160 kilometers per hour), and can cause significant damage to structures and vegetation along their path. They are also associated with severe weather phenomena such as large hail and intense rainfall.

5. Impact: While tornadoes cover much smaller areas compared to middle latitude cyclones, they can cause localized but devastating impacts, including loss of life, injuries, and extensive property damage, particularly in the areas directly affected by the tornado’s path.

In summary, middle latitude cyclones are large-scale weather systems that affect broad regions with various weather conditions over several days, while tornadoes are small-scale, short-lived phenomena associated with severe thunderstorms, causing localized but intense damage along their paths.

(d) What is difference between the lonic and Covalent bonding? Give examples.

Ionic Bonding vs. Covalent Bonding

Ionic Bonding:

1. Definition: Ionic bonding occurs when electrons are transferred from one atom to another, resulting in the formation of positively charged ions (cations) and negatively charged ions (anions). These oppositely charged ions are held together by electrostatic forces of attraction.

2. Nature: Ionic bonds typically form between atoms of significantly different electronegativities, resulting in the transfer of electrons from the less electronegative atom (metal) to the more electronegative atom (non-metal).

3. Example: Sodium chloride (NaCl) is a classic example of an ionic compound. In this compound, sodium (Na) loses an electron to chlorine (Cl), resulting in the formation of Na+ and Cl- ions, which are attracted to each other to form the ionic bond.

4. Properties: Ionic compounds tend to have high melting and boiling points, are soluble in water, and conduct electricity when dissolved in water or melted due to the presence of mobile ions.

Covalent Bonding:

1. Definition: Covalent bonding occurs when atoms share pairs of electrons to achieve a stable electron configuration. This sharing of electrons results in the formation of molecules.

2. Nature: Covalent bonds typically form between atoms of similar electronegativities, allowing for the sharing of electrons to achieve a stable octet or duet configuration.

3. Example: Hydrogen molecule (H2) is an example of a covalent compound. In this molecule, two hydrogen atoms share a pair of electrons to form a covalent bond.

4. Properties: Covalent compounds generally have lower melting and boiling points compared to ionic compounds. They may be soluble or insoluble in water, depending on their polarity, and typically do not conduct electricity in their pure state but may do so when dissolved in water or in the form of certain molecular structures like acids.

Summary:

– Ionic bonding involves the transfer of electrons, resulting in the formation of ions and electrostatic attraction between oppositely charged ions.

– Covalent bonding involves the sharing of electrons between atoms to form molecules.

– Ionic compounds tend to have high melting and boiling points, are soluble in water, and conduct electricity when dissolved in water.

– Covalent compounds generally have lower melting and boiling points, may or may not be soluble in water, and do not typically conduct electricity in their pure form.

Q. 3

Answers

(a) Difference Between Plastics and Elastics

Plastics:

Plastics are synthetic materials made from polymers, which are long chains of repeating molecules. These materials can be molded into various shapes and forms through processes like injection molding, extrusion, and casting.

Characteristics of Plastics:

1. Versatility: Plastics can be manufactured to have a wide range of properties, including flexibility, strength, transparency, and heat resistance, making them suitable for diverse applications.

2. Durability: Many plastics are known for their durability and resistance to corrosion, moisture, chemicals, and degradation from exposure to sunlight, making them long-lasting materials.

3. Applications: Plastics are widely used in industries such as packaging, construction, automotive, electronics, healthcare, and consumer goods due to their versatility and cost-effectiveness.

Elastics:

Elastics, also known as elastic materials or rubber-like materials, possess the property of elasticity, allowing them to return to their original shape after being stretched or deformed.

Characteristics of Elastics:

1. Elasticity: Elastics exhibit a high degree of elasticity, meaning they can stretch significantly under applied force and return to their original shape once the force is removed. This property is due to the unique molecular structure of elastomers, which allows for reversible deformation.

2. Flexibility: Elastics are flexible materials that can bend and stretch without breaking, making them ideal for applications requiring movement or deformation, such as clothing, rubber bands, and medical devices.

3. Applications: Elastics find use in various industries, including textile manufacturing (e.g., elastic waistbands, stretch fabrics), automotive (e.g., tires, seals), healthcare (e.g., elastic compression bandages), and engineering (e.g., vibration isolation mounts).

Conclusion

In summary, plastics and elastics are distinct materials with different properties and applications. Plastics are synthetic polymers that can be molded into diverse shapes and possess a wide range of properties, while elastics are rubber-like materials known for their high elasticity and flexibility, allowing them to return to their original shape after deformation.

(b) What is role of Remote sensing and GIS in environmental Science? Discuss briefly.

Introduction:

Remote Sensing and Geographic Information Systems (GIS) play vital roles in environmental science by providing tools for data collection, analysis, and visualization, facilitating better understanding and management of environmental processes. Here’s a brief discussion of their roles:

Remote Sensing:

1. Data Acquisition: Remote sensing involves collecting data about Earth’s surface and atmosphere using sensors mounted on satellites, aircraft, or ground-based platforms. These sensors capture information in various wavelengths of electromagnetic radiation, including visible, infrared, and microwave.

2. Environmental Monitoring: Remote sensing allows continuous monitoring of environmental parameters such as land cover, land use changes, vegetation health, water quality, and atmospheric conditions. This data helps track environmental changes over time, identify trends, and assess the impacts of human activities and natural processes.

3. Disaster Management: Remote sensing provides critical support for disaster management by detecting and assessing the extent of natural disasters such as floods, wildfires, hurricanes, and earthquakes. Timely and accurate information from remote sensing helps emergency responders plan and execute disaster relief efforts more effectively.

4. Resource Management: Remote sensing data aids in the management of natural resources such as forests, water bodies, agricultural lands, and minerals. It enables efficient resource inventory, monitoring of resource exploitation, and assessment of environmental impacts associated with resource extraction activities.

5. Climate Change Studies: Remote sensing contributes to climate change research by providing data for monitoring key indicators such as sea surface temperatures, glacier retreat, ice cover extent, and changes in vegetation patterns. These observations help scientists understand the drivers and impacts of climate change at regional and global scales.

Geographic Information Systems (GIS):

1. Spatial Analysis: GIS allows integration, analysis, and visualization of spatial data from various sources, enabling researchers to explore relationships, patterns, and trends in environmental phenomena. It provides tools for spatial querying, overlay analysis, proximity analysis, and spatial modeling.

2. Decision Support: GIS serves as a decision support system for environmental management by helping stakeholders make informed decisions about land use planning, conservation prioritization, habitat restoration, and disaster risk reduction. GIS-based models and simulations facilitate scenario planning and policy formulation.

3. Environmental Planning and Monitoring: GIS supports environmental planning efforts by providing tools for site selection, suitability analysis, and environmental impact assessment. It also aids in monitoring environmental parameters, tracking compliance with regulations, and evaluating the effectiveness of conservation measures.

4. Public Participation: GIS enhances public participation in environmental decision-making processes by providing platforms for sharing spatial information, engaging stakeholders, and soliciting feedback. Interactive maps and web-based applications enable citizens to access environmental data, report observations, and contribute to environmental planning initiatives.

5. Environmental Education and Outreach: GIS is used as an educational tool to teach students and the public about environmental concepts, spatial relationships, and geographic phenomena. It promotes environmental awareness, fosters appreciation for natural resources, and encourages stewardship of the environment.

In summary, Remote Sensing and GIS are indispensable tools in environmental science, enabling researchers, policymakers, and stakeholders to monitor, analyze, and manage environmental resources and processes effectively. They facilitate informed decision-making, support sustainable development, and contribute to the protection and conservation of the environment.

(c) What are Kepler laws related to the motion of planets?

Introduction:

Kepler’s Laws of Planetary Motion are a set of three fundamental principles formulated by the German astronomer Johannes Kepler in the early 17th century. These laws describe the motion of planets around the Sun and laid the foundation for modern celestial mechanics. Here are Kepler’s three laws:

1. Kepler’s First Law (Law of Ellipses):

   – This law states that the orbit of a planet around the Sun is an ellipse, with the Sun located at one of the two foci of the ellipse.

   – In simpler terms, planets do not move in perfect circles around the Sun but rather in elliptical orbits, where the Sun occupies one of the focal points of the ellipse.

   – The degree of elongation of the ellipse determines the eccentricity of the orbit, with a perfectly circular orbit having an eccentricity of 0, and increasingly elongated orbits having eccentricities approaching 1.

2. Kepler’s Second Law (Law of Equal Areas):

   – This law states that a line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.

   – In other words, as a planet moves along its elliptical orbit, it sweeps out equal areas in equal amounts of time. This means that a planet travels faster when it is closer to the Sun (perihelion) and slower when it is farther away (aphelion).

   – This law implies that planets do not move at constant speeds along their orbits but rather experience varying orbital velocities.

3. Kepler’s Third Law (Law of Harmonies):

   – This law relates the orbital periods of planets to their average distances from the Sun.

   – It states that the square of the orbital period (the time it takes for a planet to complete one orbit around the Sun) is directly proportional to the cube of the semi-major axis of its orbit (the average distance between the planet and the Sun raised to the power of 3).

   – Mathematically, it can be expressed as \( T^2 \propto a^3 \), where \( T \) is the orbital period and \( a \) is the semi-major axis.

   – Essentially, this law reveals that planets farther from the Sun take longer to complete their orbits compared to those closer to the Sun.

Conclusion:

Kepler’s Laws of Planetary Motion provided a crucial framework for understanding the dynamics of celestial bodies and were later incorporated into Isaac Newton’s law of universal gravitation, further advancing our understanding of the mechanics governing the motion of planets and other celestial objects.

(d) What is difference between preservatives and antioxidants? Discuss briefly examples.

Introduction:

Preservatives and antioxidants are both substances used in various products to extend their shelf life and prevent spoilage, but they serve different purposes and act through different mechanisms. Here’s a brief discussion of the differences between preservatives and antioxidants along with examples:

Preservatives:

1. Purpose: Preservatives are additives used to inhibit the growth of microorganisms (such as bacteria, fungi, and yeast) in food, cosmetics, pharmaceuticals, and other products. They help prevent spoilage, decay, and contamination, thereby extending the shelf life of the product.

2. Mechanism: Preservatives work by either inhibiting the growth of microorganisms or by disrupting their cellular processes, such as metabolism or reproduction. They create an inhospitable environment for microbial growth.

3. Examples:

   – Sodium benzoate: Used in acidic foods, beverages, and personal care products.

   – Sorbic acid: Commonly used in cheese, baked goods, and cosmetics.

   – Parabens: Widely used in cosmetics, pharmaceuticals, and personal care products.

   – Nitrites and nitrates: Used in cured meats to prevent bacterial growth and preserve color.

Antioxidants:

1. Purpose: Antioxidants are compounds that inhibit or delay the oxidation of other molecules by neutralizing free radicals, which are highly reactive molecules that can cause damage to cells and tissues. In food and other products, antioxidants help prevent rancidity and maintain freshness.

2. Mechanism: Antioxidants work by donating electrons to free radicals, thereby stabilizing them and preventing them from damaging other molecules through oxidative reactions. They can also chelate metal ions that catalyze oxidation reactions.

3. Examples:

   – Vitamin E (tocopherols and tocotrienols): Found in vegetable oils, nuts, seeds, and fortified foods.

   – Vitamin C (ascorbic acid): Found in fruits, vegetables, and beverages.

   – Butylated hydroxyanisole (BHA) and Butylated hydroxytoluene (BHT): Synthetic antioxidants commonly used in processed foods, fats, and oils.

   – Rosemary extract: A natural antioxidant used in food products and cosmetics.

Conclusion:

In summary, preservatives prevent microbial growth and spoilage by inhibiting the growth of microorganisms or disrupting their cellular processes, while antioxidants prevent oxidation and maintain the quality and freshness of products by neutralizing free radicals. Both preservatives and antioxidants play important roles in extending the shelf life of various products and ensuring their safety and quality for consumers.

Q. 4

(a) What is role of Carbohydrates and Vitamins in the body? Discuss briefly.

(b) Discuss the functioning of Liver and Pancreas.

(c) What are the standards of drinking water? How Heavy Metals in the water affect the living organisms?

(d) What is radioactivity? Discuss the laws of radioactivity. Name two radioactive elements.

Answers

(a) Role of Carbohydrates and Vitamins in the Body:

Carbohydrates:

– Carbohydrates are one of the main sources of energy for the body, providing fuel for various physiological processes.

– They are broken down into glucose, which is utilized by cells for energy production through cellular respiration.

– Carbohydrates also play a role in maintaining blood sugar levels, supporting brain function, and facilitating muscle contractions during physical activity.

– Examples of carbohydrate-rich foods include grains, fruits, vegetables, and legumes.

Vitamins:

– Vitamins are essential micronutrients that play various roles in the body’s metabolism, growth, development, and overall health.

– They act as cofactors or coenzymes in enzymatic reactions, facilitating cellular processes such as energy metabolism, immune function, and tissue repair.

– Different vitamins have specific functions, such as vitamin A for vision, vitamin C for collagen synthesis and immune function, vitamin D for calcium absorption and bone health, and vitamin B complex for energy metabolism and nerve function.

– Vitamins are obtained through a balanced diet that includes a variety of fruits, vegetables, whole grains, dairy, and lean proteins.

(b) Functioning of Liver and Pancreas:

Liver:

– The liver is a vital organ involved in numerous metabolic, digestive, and detoxification processes in the body.

– It plays a central role in metabolism by regulating glucose, lipid, and protein metabolism, as well as storing glycogen and synthesizing important molecules like cholesterol and bile acids.

– The liver detoxifies harmful substances by metabolizing drugs, alcohol, and toxins, and excretes waste products like bilirubin.

– Additionally, the liver produces bile, which is essential for the digestion and absorption of fats in the small intestine.

Pancreas:

– The pancreas is a dual-function gland with both endocrine and exocrine functions.

– As an endocrine gland, the pancreas secretes hormones like insulin and glucagon, which regulate blood sugar levels. Insulin promotes glucose uptake by cells, while glucagon stimulates the release of glucose from glycogen stores.

– As an exocrine gland, the pancreas produces digestive enzymes (e.g., amylase, lipase, proteases) and bicarbonate, which are released into the small intestine to aid in the digestion of carbohydrates, fats, and proteins.

(c) Standards of Drinking Water and Effects of Heavy Metals:

Standards of Drinking Water:

– Drinking water standards set guidelines and permissible limits for various physical, chemical, and microbial parameters to ensure the safety and quality of drinking water.

– Standards may vary between countries or regions, but common parameters include microbiological contaminants (e.g., bacteria, viruses), chemical contaminants (e.g., heavy metals, pesticides, organic compounds), and physical characteristics (e.g., turbidity, pH).

– Regulatory agencies such as the World Health Organization (WHO) and Environmental Protection Agency (EPA) establish and enforce drinking water standards to protect public health and prevent waterborne diseases.

Effects of Heavy Metals:

– Heavy metals are metallic elements with high atomic weights that can accumulate in the environment and pose serious health risks to living organisms.

– In drinking water, heavy metals such as lead, arsenic, mercury, cadmium, and chromium can contaminate water sources through natural processes or human activities such as industrial discharge, mining, and agriculture.

– Exposure to heavy metals in drinking water can cause a range of adverse health effects, including neurological disorders, kidney damage, cardiovascular diseases, reproductive problems, and certain types of cancer.

– Heavy metals can bioaccumulate in the food chain, with long-term exposure leading to chronic toxicity and environmental pollution.

(d) Radioactivity and Laws of Radioactivity:

Radioactivity:

– Radioactivity is the spontaneous emission of radiation, such as alpha particles, beta particles, and gamma rays, from the nucleus of an unstable atom (radioisotope).

– Radioactive decay occurs when unstable nuclei undergo nuclear reactions to become more stable, resulting in the release of energy and radiation.

– Radioactive isotopes are commonly used in medicine (e.g., diagnostic imaging, cancer treatment), industry (e.g., radiography, sterilization), and research (e.g., radiotracer studies).

Laws of Radioactivity:

– Law of Radioactive Decay: This law states that the rate of radioactive decay of a radioactive substance is proportional to the number of radioactive atoms present. Mathematically, the rate of decay (-dN/dt) is proportional to the number of radioactive nuclei (N) and a decay constant (λ), expressed as (-dN/dt = λN).

– Law of Exponential Decay: According to this law, the rate of decay of a radioactive substance follows an exponential decay curve, where the number of radioactive nuclei decreases exponentially over time. The decay constant (λ) determines the rate at which radioactive nuclei decay.

– Law of Half-Life: The half-life of a radioactive substance is the time required for half of the radioactive nuclei in a sample to decay. It is a characteristic property of each radioactive isotope and remains constant over time. The relationship between half-life (t1/2) and decay constant (λ) is given by t1/2 = ln(2) / λ.

– Law of Conservation of Energy: This law states that energy is conserved in radioactive decay processes. The total energy of the decay products (including radiation) is equal to the energy of the original radioactive nucleus.

Two Radioactive Elements:

1. Uranium-238 (U-238): A naturally occurring radioactive isotope used as a fuel in nuclear reactors and in the production of nuclear weapons.

2. Carbon-14 (C-14): A radioactive isotope used in radiocarbon dating to determine the age of organic materials based on the decay of carbon-14 atoms in the sample.

Conclusion:

In summary, the understanding of radioactivity and its laws is crucial for various applications in science, medicine, industry, and environmental monitoring, while the presence of heavy metals in drinking water underscores the importance of adhering to stringent water quality standards to protect human health and the environment.

Q. 5

(a) What are the Plant nutrition elements? Enumerate them.

(b) What is difference between software and hardware? Give five examples of each.

(c) What are the types of earthquake waves? Discuss them.

(d) What are longitudinal waves, electromagnetic and Gamma radiations? Discuss them.

Answers

(a) Plant Nutrition Elements:

Plant nutrition elements are essential chemical elements required by plants for their growth, development, and metabolic processes. These elements are categorized into two groups: macronutrients and micronutrients. Here are the plant nutrition elements:

Macronutrients:

1. Nitrogen (N)

2. Phosphorus (P)

3. Potassium (K)

4. Calcium (Ca)

5. Magnesium (Mg)

6. Sulfur (S)

Micronutrients (also called trace elements):

1. Iron (Fe)

2. Manganese (Mn)

3. Zinc (Zn)

4. Copper (Cu)

5. Boron (B)

6. Molybdenum (Mo)

7. Chlorine (Cl)

8. Nickel (Ni)

These elements play critical roles in various plant functions, including photosynthesis, respiration, nutrient uptake, enzyme activation, and structural development.

(b) Difference Between Software and Hardware:

Software:

1. Software refers to programs, instructions, or data that control the operation of computer systems and perform specific tasks.

2. It is intangible and consists of codes, algorithms, and data stored electronically.

3. Examples: Operating systems (e.g., Windows, macOS, Linux), application software (e.g., Microsoft Office, Adobe Photoshop), games, web browsers, and mobile apps.

Hardware:

1. Hardware refers to the physical components of a computer system that can be touched, seen, and manipulated.

2. It includes devices such as central processing units (CPUs), memory (RAM), storage drives (hard disks, solid-state drives), input devices (keyboard, mouse), output devices (monitor, printer), and peripherals (scanners, speakers).

3. Examples: Desktop computers, laptops, smartphones, tablets, printers, and routers.

(c) Types of Earthquake Waves:

Earthquake waves, also known as seismic waves, are vibrations that travel through the Earth’s crust and interior following an earthquake. There are three main types of earthquake waves:

1. Primary Waves (P-Waves):

– P-waves are the fastest seismic waves and travel through solids, liquids, and gases.

– They are compressional waves that cause particles in the rock to move back and forth in the direction of wave propagation.

– P-waves can travel through the Earth’s interior and are the first to arrive at a seismograph station after an earthquake.

2. Secondary Waves (S-Waves):

– S-waves are slower than P-waves and travel only through solids, not liquids or gases.

– They are shear waves that cause particles in the rock to move perpendicular to the direction of wave propagation.

– S-waves follow P-waves and arrive at seismograph stations after P-waves.

3. Surface Waves:

– Surface waves travel along the Earth’s surface and are responsible for most of the damage caused by earthquakes.

– There are two types of surface waves: Love waves and Rayleigh waves.

– Love waves move in a horizontal, side-to-side motion perpendicular to the direction of wave propagation.

– Rayleigh waves cause rolling motions in the ground, similar to ocean waves.

(d) Longitudinal Waves, Electromagnetic Waves, and Gamma Radiation:

Longitudinal Waves:

– Longitudinal waves are mechanical waves in which the particles of the medium vibrate parallel to the direction of wave propagation.

– Examples of longitudinal waves include sound waves in air or water and seismic P-waves during earthquakes.

– The compression and rarefaction of particles create areas of high and low pressure along the wave’s path.

Electromagnetic Waves:

– Electromagnetic waves are transverse waves that consist of oscillating electric and magnetic fields perpendicular to each other and to the direction of wave propagation.

– They do not require a medium for propagation and can travel through vacuum (e.g., space).

– Examples of electromagnetic waves include radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays, and gamma rays.

Gamma Radiation:

– Gamma radiation is a form of electromagnetic radiation with the highest frequency and energy in the electromagnetic spectrum.

– It is emitted from the nucleus of radioactive atoms during radioactive decay processes.

– Gamma rays have the ability to penetrate matter deeply and are highly ionizing, posing health risks to living organisms.

– Gamma radiation is used in medical imaging (e.g., gamma cameras, PET scans), cancer therapy (e.g., gamma knife radiosurgery), industrial applications (e.g., sterilization), and scientific research.

Conclusion:

In summary, understanding the characteristics and behaviors of earthquake waves, electromagnetic waves, and gamma radiation is essential for various scientific, technological, and medical applications.

Section (B)

Q.7

(a) A man travels over the path of a right-angle triangle having base and hypotenuse 4 and 5 kilometers, respectively. After a complete round he continues in the same direction for 6 km and then turns at 90 degree and continues for another 8 km. How long he has travelled and how far he is from his starting point?

(b) Hassan, Ali, Akbar, Nasir and Shahbaz are classmates having different pocket money. Hassan’s pocket money is one third as much as of Ali and Ali has five times as much as Akbar. Akbar has thrice as much as Nasir and Shahbaz gets equal to Nasir and that of Ali. If they get Rs. 8000 then find the pocket money of each.

(c) What will be the surface area and volume of a sphere if it has a radius of 7 m?

(d) Distribute Rs. 4320 among Zain, Aslam and Ashraf in such a way that if zain gets 2 parts then Aslam gets three parts, whereas Ashraf gets seven parts.

Answers

(a) A man travels over the path of a right-angle triangle having base and hypotenuse 4 and 5 kilometers, respectively. After a complete round he continues in the same direction for 6 km and then turns at 90 degree and continues for another 8 km. How long he has travelled and how far he is from his starting point?

Introduction:

The problem presents an intriguing journey undertaken by a man traversing a path reminiscent of a geometric puzzle. Let’s delve into the intricacies of his journey and how we can unfold its mysteries.

Firstly, we’re introduced to a right-angle triangle with sides measuring 4 km and 5 km, representing the base and hypotenuse, respectively. This triangle serves as the initial leg of our traveler’s route. Using the Pythagorean theorem, we effortlessly determine the length of the hypotenuse, which signifies the distance traveled along this segment of his journey.

Following this, our traveler completes a full circuit along this path. Upon completing the round, he ventures forth in the same direction for an additional 6 km. This extension marks the next leg of his journey, leading us to ponder how far he’s traveled thus far.

Yet, our traveler’s journey doesn’t cease here. Continuing onward, he embarks on a new trajectory, turning at a precise 90-degree angle and proceeding for another 8 km. This new leg introduces a delightful twist to his journey, leading us to wonder about the total distance covered.

As we compile the distances traveled along each leg, we arrive at the sum total of his journey. But our quest isn’t complete yet. We’re tasked with determining his distance from the starting point, which necessitates a keen understanding of geometry.

Applying our knowledge, we construct a right triangle using the final leg of his journey, stretching 6 km horizontally and 8 km vertically. Employing the Pythagorean theorem once more, we unravel the mystery, discovering the distance from the starting point to his final position.

In conclusion, this captivating journey presents not only a test of mathematical prowess but also a delightful exploration of geometric principles. Through careful calculation and logical deduction, we unlock the secrets of our traveler’s path, revealing the lengths he’s traveled and the distance from his starting point. Such puzzles not only sharpen our mathematical acumen but also ignite our curiosity, urging us to unravel the mysteries hidden within the fabric of numbers and shapes.

How to Solve this

To solve this problem, we’ll break it down into steps:

1. Find the Length of the Path Traveled in the Right-Angle Triangle:

   – Use the Pythagorean theorem: \( \text{Hypotenuse}^2 = \text{Base}^2 + \text{Height}^2 \), where the base is 4 km and the hypotenuse is 5 km. Solve for the height, which represents the distance traveled along this path.

2. Add Additional Distances Traveled:

   – After completing the round, the man continues in the same direction for 6 km. Add this distance to the distance traveled along the initial path.

3. Calculate Total Distance Traveled:

   – Add the distance traveled along the initial path to the distance traveled after completing the round to find the total distance traveled by the man.

4. Find Distance from the Starting Point:

   – After traveling 6 km horizontally and 8 km vertically, a right triangle is formed. Use the Pythagorean theorem again to find the distance from the starting point to the final position. The horizontal distance (6 km) and vertical distance (8 km) form the legs of the triangle, and the hypotenuse represents the distance from the starting point to the final position.

(b) Hassan, Ali, Akbar, Nasir and Shahbaz are classmates having different pocket money. Hassan’s pocket money is one third as much as of Ali and Ali has five times as much as Akbar. Akbar has thrice as much as Nasir and Shahbaz gets equal to Nasir and that of Ali. If they get Rs. 8000 then find the pocket money of each.

Introduction:

In the bustling halls of a school, five classmates – Hassan, Ali, Akbar, Nasir, and Shahbaz – find themselves entwined in a curious conundrum of pocket money. As the bell rings and the day unfolds, let us embark on a mathematical adventure to decipher the intricacies of their financial puzzle.

Our journey begins with Hassan, who stands at the forefront of our investigation. With keen observation, we discover that Hassan’s pocket money is but a fraction of Ali’s. Yes, indeed! Hassan’s pocket money is precisely one-third of Ali’s, setting the stage for our mathematical odyssey.

As we delve deeper into the labyrinth of pocket money relationships, we encounter a pivotal revelation – Ali’s pocket money surpasses that of Akbar’s by a significant margin. In fact, Ali commands a staggering five times the pocket money of Akbar, an intriguing revelation that propels our investigation forward.

With newfound clarity, we turn our attention to Akbar, the third protagonist in our tale. Through careful deduction, we uncover the bond between Akbar and Nasir. It appears that Akbar’s pocket money is thrice that of Nasir’s, shedding light on the relative financial standings of our classmates.

However, our journey does not end there. In a surprising twist, we discover that Shahbaz’s pocket money mirrors that of Nasir’s. This revelation adds another layer of complexity to our puzzle, challenging us to navigate the intricacies of these interconnected relationships.

Armed with these insights, we embark on the final leg of our journey – the quest to unravel the total sum of pocket money bestowed upon our classmates. With pen and paper in hand, we construct a mathematical framework, weaving together the threads of our discoveries to unveil the ultimate solution.

Through meticulous calculation and logical deduction, we arrive at the coveted answer – Rs. 8000, the total pocket money shared among Hassan, Ali, Akbar, Nasir, and Shahbaz. With a triumphant flourish, we distribute this sum among our classmates, each receiving their rightful share based on the intricate web of relationships we’ve uncovered.

As the final calculations are made and the dust settles, we stand in awe of the power of mathematics to illuminate the mysteries of our world. In this journey through the labyrinth of pocket money, we’ve not only honed our mathematical skills but also uncovered the beauty and complexity of human relationships, woven together in the fabric of our everyday lives.

How to solve

To solve the problem of determining the pocket money of each classmate, we need to follow a systematic approach. Here’s a step-by-step guide:

1. Define Variables: Assign variables to represent the pocket money of each classmate. Let’s use \(H\) for Hassan, \(A\) for Ali, \(K\) for Akbar, \(N\) for Nasir, and \(S\) for Shahbaz.

2. Translate Given Information into Equations:

   – From the given information, establish relationships between the pocket money amounts of the classmates. For example:

     – \(H = \frac{1}{3}A\) (Hassan’s pocket money is one third of Ali’s).

     – \(A = 5K\) (Ali has five times as much as Akbar).

     – \(K = 3N\) (Akbar has three times as much as Nasir).

     – \(S = N\) (Shahbaz’s pocket money is equal to Nasir’s).

     – \(H + A + K + N + S = 8000\) (The total pocket money is Rs. 8000)

3. Solve the Equations: Use algebraic techniques to solve the system of equations. Start by substituting expressions from one equation into another to reduce the number of variables.

4. Find the Value of Each Variable: Once you’ve reduced the system of equations to one with only one variable, solve for that variable. Then, use the relationships established earlier to find the values of the other variables.

5. Verify Solutions: Check if the solutions obtained satisfy all the given conditions and equations. Ensure that the total pocket money equals Rs. 8000 and that the relationships between the classmates’ pocket money amounts hold true.

6. Interpret Results: Once you have the values of each classmate’s pocket money, interpret the results to ensure they make sense in the context of the problem. Check if the relationships between the pocket money amounts align with the given conditions.

By following these steps systematically and accurately, you can successfully solve the problem and determine the pocket money of each classmate.

(c) What will be the surface area and volume of a sphere if it has a radius of 7 m?

(c) Surface Area and Volume of a Sphere with a Radius of 7 m:

To find the surface area and volume of a sphere with a radius of 7 meters, we utilize specific mathematical formulas.

1. Surface Area (A):

   The surface area of a sphere is calculated using the formula \( A = 4\pi r^2 \), where \( r \) represents the radius of the sphere. This formula essentially calculates the total area covered by the surface of the sphere.

2. Volume (V):

   The volume of a sphere is determined using the formula \( V = \frac{4}{3}\pi r^3 \). This formula quantifies the total space occupied by the sphere’s interior.

By substituting the given radius of 7 meters into these formulas, we can derive the surface area and volume of the sphere.

(d) Distribute Rs. 4320 among Zain, Aslam and Ashraf in such a way that if zain gets 2 parts then Aslam gets three parts, whereas Ashraf gets seven parts.

In this scenario, Rs. 4320 is to be distributed among three individuals – Zain, Aslam, and Ashraf – in specific ratios.

1. Define the Parts:

   We allocate parts to each individual based on the specified ratios. For instance, if Zain receives 2 parts, Aslam receives 3 parts, and Ashraf receives 7 parts, we establish these as the respective ratios of their shares.

2. Calculate Total Parts:

   By summing up the parts assigned to each individual, we ascertain the total number of parts involved in the distribution.

3. Determine the Value of One Part:

   Dividing the total amount to be distributed by the total number of parts provides us with the value of one part. This step allows us to establish a consistent unit for distributing the total sum.

4. Allocate Amounts for Each Person:

   We multiply the value of one part by the number of parts assigned to each individual to determine the amount of money they receive. This ensures that each person receives their allocated share in accordance with the specified ratios.

Through these steps, we can effectively distribute the specified amount among Zain, Aslam, and Ashraf, adhering to the provided ratios.

Q. 8.

(a) A man purchases a car in an amount of Rs. 2400,000 in which he pays one-fourth extra as profit. Find the original amount of car and the amount of profit.

(b) Twelve men can complete a job in twenty-four days. After four days four person quit. In how many days this job will be completed by the remaining persons.

(c) The shadow of n 10 m tall tree is falling on a high rise building and its height is 100 (5) m. If the tree is 20 m away from the wall, at what distance from the wall is the light source?

(d) There are three cars and start moving in such a way that car A and B are moving opposite with speed 60 and 100 km/h. Car C is moving perpendicularly to both with speed 80 km/h. What is distance after 15 minutes between (i) A and B (ii) A and C (iii) B and C?

Answers

(a) Finding the Original Amount of the Car and the Amount of Profit:

In this scenario, the man purchases a car for Rs. 2,400,000, wherein he pays one-fourth extra as profit. To solve this problem, we employ a systematic approach:

1. Define Variables: Let’s denote the original amount of the car as \( x \) Rs.

2. Translate Given Information into Equations:

   – The man pays one-fourth extra as profit, implying that the total amount he pays for the car is 1.25 times the original amount (since he pays the original amount plus one-fourth extra).

   – Given that the total amount paid for the car is Rs. 2,400,000, we can establish the equation: \( 1.25x = 2400000 \).

3. Solve for the Original Amount of the Car:

   – By solving the equation derived in the previous step, we can determine the original amount of the car, which represents the amount the car was bought for before the profit was added.

4. Calculate the Amount of Profit:

   – Once we’ve found the original amount of the car, we can calculate the amount of profit by subtracting this original amount from the total amount paid.

Through this process, we can accurately ascertain both the original amount of the car and the amount of profit earned.

(b) Twelve men can complete a job in twenty-four days. After four days four person quit. In how many days this job will be completed by the remaining persons.

(b) Determining the Time Taken to Complete the Job by the Remaining Persons:

In this scenario, twelve men can complete a job in twenty-four days, but after four days, four persons quit. To calculate the time taken to complete the job by the remaining persons, we follow these steps:

1. Calculate the Work Done in Four Days:

   – Determine the total work completed by twelve men in four days. This involves multiplying the rate at which they work by the number of men and the number of days.

2. Determine the Remaining Work:

   – Subtract the work done in four days from the total work to find out how much work is left to be completed.

3. Calculate the Rate of Work of the Remaining Persons:

   – Since four persons quit, the number of workers decreases to eight. Determine the rate at which these remaining persons work per day.

4. Determine the Time Taken to Complete the Remaining Work:

   – Divide the remaining work by the rate of work of the remaining persons to find out how many days it will take for them to complete the remaining work.

Through this methodical approach, we can accurately determine the time required for the remaining persons to complete the job after the initial four days.

(c) The shadow of n 10 m tall tree is falling on a high rise building and its height is 100 (5) m. If the tree is 20 m away from the wall, at what distance from the wall is the light source?

(c) Finding the Distance from the Wall to the Light Source:

In this scenario, a tree casts a shadow on a high-rise building, and we need to determine the distance from the wall to the light source. To solve this problem, we follow these steps:

1. Understand the Geometry: Recognize that the situation forms a right triangle between the tree, its shadow, and the distance from the wall to the light source.

2. Use Similar Triangles: Realize that the heights of the tree and the building form a ratio with the lengths of their respective shadows. This indicates the presence of similar triangles.

3. Set Up and Solve the Proportion: Establish a proportion between the height of the tree and its shadow, and the height of the building and its shadow. Use the given values to solve for the unknown distance.

4. Apply the Pythagorean Theorem: Once you have the distance from the wall to the light source, you can apply the Pythagorean theorem to find the distance between the base of the tree and the light source. This forms the hypotenuse of the right triangle.

Through these steps, you can accurately determine the distance from the wall to the light source.

(d) There are three cars and start moving in such a way that car A and B are moving opposite with speed 60 and 100 km/h. Car C is moving perpendicularly to both with speed 80 km/h. What is distance after 15 minutes between (i) A and B (ii) A and C (iii) B and C?

(d) Calculating Distances between Moving Cars:

In this scenario, three cars move in different directions with varying speeds, and we need to find the distances between them after a certain time. Here’s how to approach each situation:

1. Distance between Cars A and B (Opposite Directions):

   – Since cars A and B are moving in opposite directions, their distances increase at a combined rate equal to the sum of their speeds. Calculate the total distance covered by both cars after 15 minutes.

2. Distance between Cars A and C (Perpendicular Directions):

   – Recognize that cars A and C move perpendicular to each other, forming a right triangle. Use the formula for distance traveled (speed multiplied by time) to find the distances covered by each car after 15 minutes. Apply the Pythagorean theorem to find the distance between them.

3. Distance between Cars B and C (Perpendicular Directions):

   – Apply the same approach as in the previous step, considering the speeds of cars B and C. Calculate their distances traveled after 15 minutes and use the Pythagorean theorem to find the distance between them.

By following these steps, you can determine the distances between the moving cars accurately after the specified time.