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Class 9th Science Chapter 12 Sound – PYQ

Previous year questions (PYQ) with answers for Class 9th Science Chapter 12 Sound, based on the NCERT CBSE syllabus.

Topics covered in Sound

Frequently asked questions

Where can I find previous year questions (PYQ) for Class 9th Science Chapter 12 Sound?

This page lists previous year questions with answers for Sound. Topics covered: Production and propagation of sound, Longitudinal waves, Speed of sound, Reflection of sound, Echo and reverberation, SONAR, Human ear.

Are Sound PYQs free?

Yes, Notebook gives free previous year questions for Class 9 Science.

How to prepare Sound for Class 9 Science exams?

Read the Sound notes first, revise the key topics (Production and propagation of sound, Longitudinal waves, Speed of sound, Reflection of sound, Echo and reverberation, SONAR, Human ear), then practise the important questions and previous year questions (PYQ) for this chapter.

Where can I find Class 9th Science Chapter 12 Sound pyq for quick revision?

You are on the right page: this pyq page for Sound covers Class 9 Science for the CBSE exam 2026-27. Use the tabs on this chapter for notes, important questions and PYQ.

Is Class 9th Science Chapter 12 Sound important for the CBSE exam 2026-27?

Sound is part of the Class 9 Science NCERT CBSE syllabus. Notebook gives notes, important questions and PYQs for it, free.

Also see: Class 9th Science Chapter 12 Sound Notes · Class 9th Science Chapter 12 Sound Important Questions

Sound

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3 Marks Practice PYQ-style (not an authenticated past-paper question)
A constant horizontal force of 12 N acts along a trolley's displacement of 5 m. Find its work using the area under the force–displacement graph. A separate constant resistive force of 4 N acts opposite to the displacement. Find the net work.
Ans: The applied-force graph forms a rectangle of height 12 N and width 5 m. Its area gives work = 12 × 5 = 60 J. Work by the resistive force = −4 × 5 = −20 J. Net work = 60 − 20 = 40 J.
3 Marks Practice PYQ-style (not an authenticated past-paper question)
A pendulum bob is released from one extreme position. Describe its kinetic and gravitational potential energies at the lowest point and at the opposite extreme. Why do its swings gradually become smaller in practice?
Ans: Taking the lowest point as the zero potential-energy level, the bob has zero gravitational potential energy and maximum kinetic energy there. At the opposite extreme, its speed and kinetic energy are momentarily zero, while potential energy is greatest for that swing. Air resistance and friction at the support transform mechanical energy into other forms, mainly thermal energy, so successive swings reach smaller heights.
3 Marks Practice PYQ-style (not an authenticated past-paper question)
A light lever supports a load of 180 N at a distance of 0.20 m from its fulcrum. Effort is applied on the opposite side, 0.90 m from the fulcrum. Assuming both forces act perpendicular to the lever, calculate the balancing effort and mechanical advantage.
Ans: For balance, effort × 0.90 = 180 × 0.20. Hence, effort = 36/0.90 = 40 N. Mechanical advantage = load/effort = 180/40 = 4.5. It is a ratio and has no unit.
3 Marks Practice PYQ-style (not an authenticated past-paper question)
A toy car starts from rest at the top of a smooth track, 1.8 m vertically above its lowest point. Find its speed at the lowest point. Would doubling the car's mass change this speed? Take g = 10 m/s² and neglect friction and wheel rotation.
Ans: Conservation of mechanical energy gives mgh = (1/2)mv². Thus, v² = 2gh = 2 × 10 × 1.8 = 36 m²/s², so v = 6 m/s. Doubling the mass does not change the speed because mass cancels from the equation under the stated assumptions.
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