Physics Class 10 Important Numericals and Solutions FBISE Guess Paper 2024

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April 18, 2024
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FBISE 10th Class - 2024 Guess Paper

Mastering Physics Numericals: Class 10 FBISE Guess Paper 2024

Ace Your Physics Exam with Important Numericals and Solutions

Here are some important numerical problems along with their solutions from each unit:

Unit 10 – Simple Harmonic Motion And Waves:

  • A mass of 0.5 kg is attached to a spring with a spring constant of 50 N/m. Calculate the frequency of oscillation.

Solution: We can calculate the frequency of oscillation using the formula for the natural frequency of a mass-spring system: f = 1 / (2Ï€√(m/k)) where: f is the frequency (in Hz) m is the mass (in kg) k is the spring constant (in N/m) Ï€ (pi) is a mathematical constant approximately equal to 3.14159 Given values: m = 0.5 kg k = 50 N/m Calculation: Plug the values into the formula: f = 1 / (2Ï€√(0.5 kg / 50 N/m)) Simplify the expression: f ≈ 1 / (2Ï€√(0.01)) ≈ 1 / (2Ï€ * 0.1) Solve for f: f ≈ 1 / (0.628) ≈ 1.59 Hz (rounded to two decimal places) Therefore, the frequency of oscillation is approximately 1.59 Hz.

Physics Class 10 Important Numericals and Solutions FBISE Guess Paper 2024

  • A wave has a frequency of 200 Hz and a wavelength of 2 m. Calculate the speed of the wave.

Solution:

We can find the speed of the wave using the following relationship between speed (v), frequency (f), and wavelength (λ):

v = f * λ

where:

  • v is the speed of the wave (in m/s)
  • f is the frequency of the wave (in Hz)
  • λ is the wavelength of the wave (in meters)

Given values:

  • f = 200 Hz
  • λ = 2 m

Calculation:

  1. Plug the values into the formula:

v = 200 Hz * 2 m

  1. Simplify the expression:

v = 400 m/s

Therefore, the speed of the wave is 400 m/s.

Unit 11 – Sound:

  • Calculate the wavelength of a sound wave traveling at a speed of 340 m/s with a frequency of 170 Hz.

Solution: We can calculate the wavelength using the following formula for waves: λ = v / f where: λ (lambda) is the wavelength (in meters) v is the speed of the wave (in meters per second) f is the frequency of the wave (in Hertz) Given values: v = 340 m/s (speed of sound) f = 170 Hz (frequency of the sound wave) Calculation: Plug the values into the formula: λ = 340 m/s / 170 Hz Simplify the expression: λ ≈ 2 meters (rounded to two decimal places) Therefore, the wavelength of the sound wave is approximately 2 meters.

  • A sound wave has a wavelength of 0.5 m and a frequency of 400 Hz. Calculate the speed of sound.

Solution : We can definitely calculate the speed of sound using the same formula for the relationship between speed (v), frequency (f), and wavelength (λ) of waves that we used previously: v = f * λ where: v is the speed of the wave (in m/s) f is the frequency of the wave (in Hz) λ is the wavelength of the wave (in meters) Given values: f = 400 Hz (frequency of the sound wave) λ = 0.5 m (wavelength of the sound wave) Calculation: Plug the values into the formula: v = 400 Hz * 0.5 m Simplify the expression: v = 200 m/s Therefore, the speed of sound in this case is approximately 200 m/s. It's important to remember that the speed of sound can vary depending on the medium it travels through. The value we calculated here is an approximation for the speed of sound in air at room temperature.

Unit 12 – Geometrical Optics:

  • An object is placed 20 cm from a concave mirror whose focal length is 10 cm. Calculate the position and magnification of the image formed.

Solution: Solving for the Image Position and Magnification We can use the mirror formula to calculate the image position (v) for a concave mirror: 1/v + 1/u = 1/f where: v is the image distance (measured from the mirror) u is the object distance (measured from the mirror, positive for in front of the mirror) f is the focal length of the mirror (positive for concave mirrors) Given values: u = -20 cm (object is in front of the mirror, so we use a negative sign) f = 10 cm Calculation: Plug the values into the formula: 1/v + 1/-20 cm = 1/10 cm Simplify the equation: 1/v = 1/10 cm - 1/(-20) cm = (1 - 5) / 20 cm = -2 / 20 cm Take the reciprocal of both sides to find v: v = -20 cm / -2 = 10 cm Interpretation: The negative sign for the object distance indicates the object is placed in front of the mirror. The positive value of the image distance (v = 10 cm) tells us the image is formed behind the mirror and is 10 cm away from the mirror's surface. Calculating Magnification Magnification (M) is the ratio of the image height (h') to the object height (h): M = h' / h For a concave mirror, with an object placed between the focal point and the mirror (as in this case), the image is virtual, upright, and smaller than the object. Therefore, the magnification will have a positive value between 0 and 1. In FBISE exams, you might be asked to find the image height based on the object height and the magnification. However, without information on the object height, we can only determine the relative size using the sign of the magnification. Summary: Image position (v): 10 cm (formed behind the mirror) Magnification (M): Positive value between 0 and 1 (indicating a virtual, upright, and smaller image compared to the object)

  • A convex lens has a focal length of 15 cm. An object is placed 30 cm from the lens. Calculate the position and size of the image formed.

Solution: When a convex lens with a focal length of 15 cm is used with an object placed 30 cm from the lens, the image formed is a real image and has the same size as the object. Here's how we can find the image position and size: Finding the Image Distance: We can use the lens formula to calculate the image distance (v) where f is the focal length (15 cm) and u is the object distance (30 cm). 1/v - 1/u = 1/f Substituting the values: 1/v - 1/30 = 1/15 1/v = 1/15 + 1/30 = 1/10 v = 10 cm Therefore, the image is formed 10 cm behind the lens. Finding the Image Size: The magnification (m) of the image can be calculated using the formula: m = v / u Here, v is 10 cm and u is 30 cm. m = 10 cm / 30 cm = 1/3 Since the magnification is 1/3, the image will be 1/3 times the size of the object. In this case, the image will have the same size as the object because 1 is equal to 3/3. Summary: The image is formed 10 cm behind the lens. The image is a real image. The image has the same size as the object. https://fbisenotes2024.blogspot.com/2024/03/physics-ssc-ii-class-10-exam-2024.html










Unit 13 – Electrostatics:

  • Two charges of +5 μC and -8 μC are placed 10 cm apart. Calculate the force between them.

Solution:

We can calculate the force between the two charges using Coulomb's law.

Here's what we know:

  • Charge 1 (q1) = +5 μC (positive 5 microCoulombs)
  • Charge 2 (q2) = -8 μC (negative 8 microCoulombs)
  • Distance between the charges (d) = 10 cm (convert to meters for the calculation: 0.1 m)

Coulomb's law states that the force (F) between two point charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. It can be expressed by the following equation:

F = k * |q1 * q2| / d^2

where:

  • F is the force between the charges (Newtons)
  • k is the Coulomb's constant (approximately 8.988 x 10^9 N⋅m²⋅C⁻²)
  • q1 and q2 are the magnitudes of the charges (Coulombs)
  • d is the distance between the charges (meters)

Steps to solve:

  1. Convert microCoulombs to Coulombs:

    • q1 = 5 μC * (1 x 10^-6 C/μC) = 5 x 10^-6 C
    • q2 = -8 μC * (1 x 10^-6 C/μC) = -8 x 10^-6 C

  2. Plug the values into Coulomb's law:

    • F = 8.988 x 10^9 N⋅m²⋅C⁻² * |(5 x 10^-6 C) * (-8 x 10^-6 C)| / (0.1 m)^2

  3. Calculate the force:

    • F ≈ 35.952 N (Newtons)

Since the charges have opposite signs (one positive and one negative), the force between them will be attractive.

Therefore, the force between the two charges is approximately 35.952 Newtons.

  • Three charges of +2 μC, -3 μC, and +4 μC are placed at the vertices of an equilateral triangle of side 10 cm. Calculate the net force on the +2 μC charge.

Solution: Symmetry cancels horizontal forces. The -3 μC charge above exerts a downward force due to the triangle's height. Therefore, the net force on the +2 μC charge is roughly downwards (negative y-direction).

Unit 14 – Current Electricity:

  • A resistor of 20 ohms is connected to a 12V battery. Calculate the current passing through the resistor.

Solution:

Use Ohm's Law to calculate the current passing through the resistor.

Ohm's Law states that the current (I) flowing through a conductor is directly proportional to the voltage (V) across the conductor and inversely proportional to its resistance (R). It can be expressed by the following equation:

I = V / R

where:

  • I is the current (in amperes)
  • V is the voltage (in volts)
  • R is the resistance (in ohms)

Given values:

  • Voltage (V) = 12 V
  • Resistance (R) = 20 ohms

Calculate the current (I):

I = 12 V / 20 ohms = 0.6 amperes

Therefore, the current passing through the resistor is 0.6 amperes (or 0.6 A).

  • Calculate the resistance of a wire if a current of 2A flows through it when a voltage of 10V is applied across it.

Solution:

Use Ohm's Law again to calculate the resistance of the wire.

As you mentioned earlier, Ohm's Law states the relationship between voltage (V), current (I), and resistance (R) of a conductor:

I = V / R

We know the current (I) is 2A and the voltage (V) is 10V. We need to find the resistance (R).

Rearrange the equation to solve for R:

R = V / I

Plug in the known values:

R = 10V / 2A = 5 ohms

Therefore, the resistance of the wire is 5 ohms.

Unit 15 – Electromagnetism:

  • Calculate the magnetic field strength at a point located 10 cm away from a straight current-carrying wire if the current flowing through the wire is 5A.

Solution:

The magnetic field strength (B) around a long, straight current-carrying wire can be calculated using the following equation:

B = μ₀ * I / (2Ï€ * d)

where:

  • B is the magnetic field strength (Tesla, T)
  • μ₀ (mu-nought) is the permeability of free space (approximately 4Ï€ x 10⁻⁷ Tm/A)
  • I is the current flowing through the wire (Amperes, A)
  • d is the distance between the point and the wire (meters, m)

Given values:

  • I = 5 A (current)
  • d = 10 cm (distance) = 0.1 m (convert centimeters to meters)

Calculating the magnetic field strength:

  1. Plug the values into the equation: B ≈ 4Ï€ x 10⁻⁷ Tm/A * 5 A / (2Ï€ * 0.1 m)

  2. Simplify: B ≈ 1 x 10⁻⁵ T

Therefore, the magnetic field strength at a point 10 cm away from the wire is approximately 1 x 10⁻⁵ Tesla (10 microTesla).

  • A coil with 100 turns carries a current of 2A. Calculate the magnetic field strength inside the coil if its radius is 5 cm.

Solution:

Absolutely! We can calculate the magnetic field strength inside a solenoid (coil) using the following equation:

B = μ₀ * n * I / (2 * Ï€ * r)

where:

  • B is the magnetic field strength (Tesla, T)
  • μ₀ (mu-nought) is the permeability of free space (approximately 4Ï€ x 10⁻⁷ Tm/A)
  • n is the number of turns in the coil
  • I is the current flowing through the wire (Amperes, A)
  • r is the radius of the coil (meters, m)

Given values:

  • n = 100 (number of turns)
  • I = 2 A (current)
  • r = 5 cm (radius) = 0.05 m (convert centimeters to meters)

Calculating the magnetic field strength:

  1. Plug the values into the equation: B ≈ 4Ï€ x 10⁻⁷ Tm/A * 100 * 2 A / (2 * Ï€ * 0.05 m)

  2. Simplify: B ≈ 0.00025 T

Therefore, the magnetic field strength inside the coil is approximately 0.00025 Tesla.

Unit 16 – Basic Electronics:

  • Calculate the total resistance in a circuit consisting of resistors of 10 ohms, 20 ohms, and 30 ohms connected in series.

Solution:

When resistors are connected in series, their resistances simply add up. Therefore, the total resistance (R_total) in this circuit is:

R_total = R1 + R2 + R3

where:

  • R1 = 10 ohms
  • R2 = 20 ohms
  • R3 = 30 ohms

Calculation:

R_total = 10 ohms + 20 ohms + 30 ohms = 60 ohms

Therefore, the total resistance in the circuit is 60 ohms.

  • Three resistors of 5 ohms, 10 ohms, and 15 ohms are connected in parallel. Calculate the total resistance.

Solution:

Unlike resistors in series where resistances simply add up, resistors in parallel require a different formula to calculate the total resistance. Here's how to find the total resistance for resistors connected in parallel:

1/R_total = 1/R1 + 1/R2 + 1/R3

where:

  • R_total is the total resistance
  • R1, R2, and R3 are the individual resistances

Given values:

  • R1 = 5 ohms
  • R2 = 10 ohms
  • R3 = 15 ohms

Calculate the total resistance:

  1. Substitute the values: 1/R_total = 1/5 ohms + 1/10 ohms + 1/15 ohms

  2. Take the reciprocal of both sides to find R_total: R_total = 1 / (1/5 + 1/10 + 1/15)

  3. Find a common denominator for the fractions (here, least common multiple is 30): R_total = 1 / ((6 + 3 + 2) / 30) R_total = 1 / (11 / 30)

  4. Simplify: R_total = 30 / 11 ohms

Therefore, the total resistance of the three resistors connected in parallel is approximately 2.73 ohms.

Unit 17 – Information And Communication Technology:

  • If a computer system has a clock speed of 2.5 GHz, how many cycles does it complete in one second?

Solution:

A clock speed of 2.5 GHz translates to 2.5 billion cycles per second. Here's the breakdown:

  • GHz: Gigahertz, a unit denoting billions of cycles per second (Hz).
  • 2.5 GHz: This signifies 2.5 Giga cycles per second.
  • Conversion to cycles per second: Since Giga signifies billion, we can multiply 2.5 GHz by 1 billion (10^9) to get the actual cycles per second.

Therefore, the computer system completes 2.5 GHz * 10^9 cycles/GHz = 2.5 x 10^9 cycles per second.

  • A file has a size of 5 MB. If the download speed is 1 Mbps, how long will it take to download the file?

Solution:

Here's how to calculate the download time:

  1. Convert file size to bits: We need to match the units of file size and download speed. Since download speed is in Mbps (megabits per second), we should convert the file size from megabytes (MB) to megabits (Mb).

One megabyte (MB) is equal to eight megabits (Mb) because there are eight bits in a byte. Therefore, a 5 MB file size is equivalent to:

File size (bits) = 5 MB * 8 Mb/MB = 40 Mb

  1. Download time calculation: We can use the formula:

Download time (seconds) = File size (bits) / Download speed (bits/second)

Here, the file size is 40 Mb and the download speed is 1 Mbps (which is equal to 1 million bits per second or 1 Mb/s).

Download time (seconds) = 40 Mb / 1 Mb/s = 40 seconds

Therefore, it will take approximately 40 seconds to download the 5 MB file at a speed of 1 Mbps.

Unit 18 – Atomic And Nuclear Physics:

  • Calculate the energy released in the fission of 1 kg of uranium-235 if each fission event releases 200 MeV of energy.

Solution:

Here's how to calculate the energy released in the fission of 1 kg of uranium-235:

  1. Number of Fission Events: We need to find the number of fission events that occur when 1 kg of uranium-235 undergoes fission. To do this, we'll need some additional information:

    • Avogadro's number: The number of atoms in one mole of a substance (approximately 6.022 x 10^23 atoms/mole).
    • Molar mass of uranium-235: The mass of one mole of uranium-235 atoms (approximately 235 grams/mole).

  2. Mass Conversion: We are given the mass of uranium-235 in kilograms (kg). However, Avogadro's number is in terms of moles. Therefore, we need to convert 1 kg to grams:

    1 kg * 1000 grams/kg = 1000 grams

  3. Number of Uranium-235 Atoms: Now, we can find the number of uranium-235 atoms in 1000 grams using Avogadro's number and the molar mass:

    Number of atoms = (1000 grams) / (235 grams/mole) * (6.022 x 10^23 atoms/mole)

  4. Assuming Complete Fission: For this calculation, let's assume that every uranium-235 atom undergoes fission. This might not be entirely true in a real reactor, but it's a good starting point.

  5. Total Energy Released: If each fission event releases 200 MeV of energy, and we have the number of fission events from step 3, we can find the total energy released:

    Total Energy (MeV) = Number of fission events * Energy per fission event

Important Note: MeV (Megaelectronvolt) is a unit of energy on an atomic scale. While the answer will be in MeV, it's often useful to convert it to a more common energy unit like Joules (J) for practical applications.

Conversion to Joules (Optional):

1 Joule (J) is equal to 1.602 x 10^-13 MeV. Therefore, to convert the total energy from MeV to Joules, multiply by this conversion factor.

  • A sample of a radioactive material has an initial activity of 4000 disintegrations per second. After 10 days, the activity decreases to 1000 disintegrations per second. Calculate the half-life of the material.

Solution:

We can use the concept of half-life and activity of radioactive materials to solve this problem. Here's how:

Given information:

  • Initial activity (Ao) = 4000 disintegrations per second
  • Final activity (A) = 1000 disintegrations per second
  • Time elapsed (t) = 10 days

Formula for Activity Decay:

The activity (A) of a radioactive material at any given time (t) can be related to its initial activity (Ao) and half-life (T₂) using the following formula:

A = Ao * (1/2)^(t / T₂)

Finding the Half-Life (T₂):

  1. We can rearrange the formula to solve for the half-life (T₂):

    T₂ = t * (log(Ao / A)) / log(2)

  2. Plug in the known values:

    T₂ = 10 days * (log(4000 / 1000)) / log(2)

  3. Calculate the half-life:

    T₂ ≈ 3.32 days (approximately)

Therefore, the half-life of the radioactive material is approximately 3.32 days.

These numericals should provide a good practice for understanding and applying the concepts covered in each unit.

Unit 10: Simple Harmonic Motion and Waves

Problem: A spring with a spring constant of 200 N/m stretches 0.2 meters when a mass is attached. What is the frequency of oscillation of the mass?

Solution: Identify the formula: f = 1 / (2Ï€√(m/k)) where f is frequency, m is mass, and k is spring constant. We don't have the mass in this problem. FBISE problems might provide the mass or ask you to calculate it using the spring constant and stretch distance (Hooke's Law). Assuming a mass of 0.1 kg (adjust based on the problem): f ≈ 1 / (2Ï€√(0.1 kg / 200 N/m)) ≈ 5 Hz.

Resources: Look for FBISE past papers or past paper solutions focusing on Simple Harmonic Motion.

Unit 11: Sound

Problem: A sound wave travels at a speed of 343 m/s and has a frequency of 440 Hz. What is the wavelength of the sound wave?

Solution: Identify the formula: λ = v/f where λ is wavelength, v is sound speed, and f is frequency. Plug the values: λ = 343 m/s / 440 Hz ≈ 0.78 meters.

Resources: Utilize FBISE-approved Class 10 Physics textbooks that often have solved problems specific to the FBISE curriculum.

Unit 12: Geometrical Optics

Problem: An object is placed 15 cm in front of a convex mirror with a focal length of 10 cm. Calculate the position and nature of the image formed.

Solution: Identify the formula (mirror formula) and interpret the signs for object distance and image distance. Solve for the image distance. FBISE problems might ask you to determine the nature of the image (virtual/real, upright/inverted) based on the sign of the image distance.

Resources: Search for online resources or past paper solutions that explain solving mirror and lens problems using FBISE-specific conventions.

Unit 13: Electrostatics

Problem: Two point charges, one with +6 μC and the other with -8 μC, are placed 20 cm apart. Calculate the force between them.

Solution: Identify the formula: F = k * |q1 * q2| / r², where F is the force, k is Coulomb's constant (approximately 9x10^9 N⋅m²/C²), q1 and q2 are the charges, and r is the distance between them (convert cm to m: 0.2 m). Plug the values and calculate the force. FBISE problems might ask you to determine the nature of the force (attractive/repulsive) based on the signs of the charges. Expected Answer: F ≈ 1.44 N (repulsive)

Resources: Look for FBISE past papers or solutions that cover Electrostatics and Coulomb's Law.

Unit 14: Current Electricity

Problem: A circuit has three resistors connected in series: 12 ohms, 18 ohms, and 25 ohms. If a current of 0.5 A flows through the circuit, calculate the total voltage across the circuit.

Solution: Identify the formula for total resistance in series: R_total = R1 + R2 + R3. Calculate the total resistance. Apply Ohm's Law: V = I * R_total (V is voltage). Plug the values and solve for the voltage.

Resources: Utilize FBISE-approved Class 10 Physics textbooks or online resources that cover series and parallel circuits and Ohm's Law in the context of the FBISE curriculum.

Unit 15: Electromagnetism

Problem: A straight wire carries a current of 2 A. Calculate the magnetic field strength at a point 5 cm away from the wire.

Solution: Identify the formula for magnetic field around a straight current-carrying wire: B = 2πμ₀ * I / r, where B is the magnetic field strength, μ₀ is permeability constant (approximately 4Ï€x10^-7 T⋅m/A), I is the current, and r is the distance from the wire (convert cm to m: 0.05 m). Plug the values and calculate the magnetic field strength.

Resources: Search for online resources or past paper solutions that explain magnetic field due to a current-carrying wire relevant to the FBISE syllabus.

Unit 16: Basic Electronics

Problem: A transistor circuit has a voltage of 6 V applied across a 1 kΩ resistor. Calculate the current flowing through the resistor.

Solution: Apply Ohm's Law: I = V / R (I is current). Plug the values and solve for the current.

Resources: While FBISE might not delve into complex transistor circuits at this level, look for resources that cover basic applications of Ohm's Law in electronic components like resistors.

Unit 17: Information and Communication Technology

Problem: A modem has a download speed of 10 Mbps (megabits per second). How long does it take to download a file with a size of 20 MB (megabytes)?

Solution:

Convert file size to bits: 20 MB * 8 bits/byte = 160 million bits. Calculate download time: Time = File size (bits) / Download speed (bits/second). Convert time to seconds (if necessary). Resources: Online resources or past papers might cover basic data transfer calculations related to bits, bytes, and data transfer rates.

Unit 18: Atomic and Nuclear Physics

Problem: A radioactive isotope has a half-life of 5 days. If the initial activity is 1000 disintegrations per second, what is the activity after 15 days?

Solution: Identify the concept of half-life and its relation to activity decay. FBISE problems might ask you to calculate the number of half-life periods elapsed within a time interval. Calculate the number of half-life periods based on the given half-life and time interval. Apply the concept of exponential decay to find the final activity after the specified time.

Resources: Look for FBISE-approved textbooks or online resources that cover radioactive decay and half-life calculations relevant to the FBISE curriculum.

Remember, these are general examples. The specific problems you encounter in FBISE exams might differ slightly.

Topic: Education, Physics, Numericals, Class 10, FBISE Guess Paper 2024

FAQs:

  1. What's included in this blog post? This blog post provides a comprehensive collection of important numerical problems along with detailed solutions from all units of the Class 10 Physics syllabus, following the FBISE Guess Paper 2024.

  2. Why are numerical problems important in physics exams? Numerical problems in physics exams assess students' understanding of fundamental concepts and their ability to apply mathematical principles to real-world scenarios. They help reinforce learning and develop problem-solving skills.

  3. How can I use this blog post to prepare for my physics exam? You can use this blog post as a supplementary study resource to practice solving numerical problems across various topics covered in the Class 10 Physics curriculum. Work through each numerical problem, compare your solutions with the provided answers, and identify areas where you may need further clarification or practice.

  4. Are these numericals relevant to the FBISE exam pattern? Yes, the numerical problems included in this blog post are aligned with the FBISE Guess Paper 2024 for Class 10 Physics. They cover all units of the syllabus and are designed to help you prepare effectively for your upcoming exams.

  5. How should I approach practicing these numericals? Start by reading each numerical problem carefully, understand the given data, and identify the concepts involved. Then, apply the relevant formulas and principles to solve the problem step by step. Finally, compare your solution with the provided solution to verify your understanding and accuracy.

Tags: Physics, Numericals, Class 10, FBISE, Guess Paper, 2024, Simple Harmonic Motion, Waves, Sound, Geometrical Optics, Electrostatics, Current Electricity, Electromagnetism, Basic Electronics, Information and Communication Technology, Atomic Physics, Nuclear Physics.

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