🎵 Chapter 13: Sound – Constructed Response Questions
Prepared by Muhammad Tayyab, Subject Specialist Physics, Govt Christian High School Daska. Based on PECTAA 2026 syllabus (National Curriculum 2023).
📖 What's Inside: This section covers constructed response questions from Chapter 13 Sound including tuning fork & surface area, bell jar experiment, echo distance calculation, pitch & frequency relationship, acoustic protection with soft materials, and canyon echo calculation. Each question is presented with the exact exam-ready answer as per the official PECTAA 2026 Physics curriculum. Perfect for Punjab Boards (Lahore, Gujranwala, Multan, etc.) and all BISE boards across Pakistan.
📚 Related Resources – Chapter 13: Sound
Sound covers waves, echo, ultrasound, SONAR, pitch, loudness, and acoustics. Includes solved examples and numerical problems.
📖 Constructed Response Questions & Answers (PECTAA 2026)
13.1 Why does a tuning fork placed on a wooden table produce a louder sound than when held in the air? What does this tell us about the role of surface area in sound propagation?
Reason: When a vibrating tuning fork is placed on a wooden table, it forces the large tabletop to vibrate.
Role of Surface Area: The loudness of sound is also influenced by the size of the vibrating surface. A larger vibrating surface area forces more air to vibrate, which increases the loudness of the sound.
Reason: When a vibrating tuning fork is placed on a wooden table, it forces the large tabletop to vibrate.
Role of Surface Area: The loudness of sound is also influenced by the size of the vibrating surface. A larger vibrating surface area forces more air to vibrate, which increases the loudness of the sound.
13.2 If sound needs a medium to travel, how does the bell jar experiment prove this requirement? What happens when air is removed, and why?
Effect of Removing Air: When air is completely pumped out of the bell jar, the sound of the ringing bell inside dies out and cannot be heard.
Reason: This happens because "sound is a MECHANICAL WAVE" and "requires a medium (like air, water, or solids) to travel through." In a vacuum, there are no particles to vibrate, which proves that sound cannot travel without a material medium.
Effect of Removing Air: When air is completely pumped out of the bell jar, the sound of the ringing bell inside dies out and cannot be heard.
Reason: This happens because "sound is a MECHANICAL WAVE" and "requires a medium (like air, water, or solids) to travel through." In a vacuum, there are no particles to vibrate, which proves that sound cannot travel without a material medium.
13.3 A person hears his echo when shouting near a tall building. Why must the building be at least 17 metres away for the echo to be heard clearly?
Brain Sensation Time: "The human brain retains sound for about \(0.1\) seconds, so to hear a distinct echo, the reflected sound must reach us after at least 0.1 seconds."
Distance Calculation: Given the speed of sound in air (\(340\ ms^{-1}\)), the total two-way distance covered by sound in \(0.1\ s\) must be at least \(34\ metres\). Therefore, the single-way distance to the reflecting surface must be at least:
\[ Distance = \frac{34\ m}{2} = 17\ m \]
Brain Sensation Time: "The human brain retains sound for about \(0.1\) seconds, so to hear a distinct echo, the reflected sound must reach us after at least 0.1 seconds."
Distance Calculation: Given the speed of sound in air (\(340\ ms^{-1}\)), the total two-way distance covered by sound in \(0.1\ s\) must be at least \(34\ metres\). Therefore, the single-way distance to the reflecting surface must be at least:
\[ Distance = \frac{34\ m}{2} = 17\ m \]
13.4 Why do voices of men sound deeper than those of women or children, even when they are speaking at the same intensity? Explain using the relationship between frequency and pitch.
Pitch and Frequency Relationship: Pitch is directly related to frequency; higher frequency results in higher pitch, while lower frequency produces lower pitch.
Exact Reason: The voices of women and children have a higher frequency, making them shrill and high-pitched, whereas the voices of men have a lower frequency, giving them a deeper and lower-pitched sound.
Pitch and Frequency Relationship: Pitch is directly related to frequency; higher frequency results in higher pitch, while lower frequency produces lower pitch.
Exact Reason: The voices of women and children have a higher frequency, making them shrill and high-pitched, whereas the voices of men have a lower frequency, giving them a deeper and lower-pitched sound.
13.5 In what ways do soft materials like carpets or curtains improve the acoustics of a room? How do they affect echoes and sound quality?
Echo and Absorption: Acoustic protection reduces unwanted sound using soft, porous materials like carpets and curtains, which absorb sound and minimize echoes.
Sound Quality: By absorbing excessive sound energy, they control multiple reflections, known as reverberation, which can distort sound. This balance prevents the sound from becoming garbled, thereby improving overall clarity and quality.
Echo and Absorption: Acoustic protection reduces unwanted sound using soft, porous materials like carpets and curtains, which absorb sound and minimize echoes.
Sound Quality: By absorbing excessive sound energy, they control multiple reflections, known as reverberation, which can distort sound. This balance prevents the sound from becoming garbled, thereby improving overall clarity and quality.
13.6 When a person claps near a canyon and hears the echo after a few seconds, how can they calculate the distance to the canyon? What principle is used in this method?
Principle Used: This method is based on the principle of the Reflection of sound (Echo).
Method of Calculation: By measuring the total time \(t\) taken for the echo to return and knowing the speed of sound in air \(v\), the one-way distance \(d\) to the canyon is calculated using the formula:
\[ Distance = \frac{Speed\ of\ Sound \times Time}{2} \]
\[ d = \frac{v \times t}{2} \]
Principle Used: This method is based on the principle of the Reflection of sound (Echo).
Method of Calculation: By measuring the total time \(t\) taken for the echo to return and knowing the speed of sound in air \(v\), the one-way distance \(d\) to the canyon is calculated using the formula:
\[ Distance = \frac{Speed\ of\ Sound \times Time}{2} \]
\[ d = \frac{v \times t}{2} \]
📐 Key Formulas – Sound
Wave Equation: \( v = f\lambda \)
Time Period: \( T = \frac{1}{f} \)
SONAR / Echo Distance: \( d = \frac{v \times t}{2} \)
Distance (lightning): \( S = vt \)
Unit 10Thermal Physics
Unit 11Transfer of Thermal Energy
Unit 12Waves
Unit 13✨ Sound
Unit 14Light
Unit 15Electrostatics
Unit 16Electricity
Unit 17Electromagnetism
Unit 18Electromagnetic Induction & EM Waves
Unit 19Electronics
Unit 20Atomic and Nuclear Physics
Unit 21Space and Environment
📖 Complete syllabus coverage for Class 10 Physics (PECTAA 2026) – Units 10 to 21
💡 Exam Tip:
For board exams, construct detailed responses with clear reasoning, include relevant formulas with units, and relate to real-life applications. These constructed response questions follow the PECTAA 2026 pattern and are prepared by Subject Specialist Muhammad Tayyab.
Created by Hira Science Academy | Aligned with PECTAA 2026 Syllabus