Note on textbook printing: The question contains a slight misprint where π is factored outside; the true difference is (1/6)πr² - (√3/4)r² = r²((π/6) - (√3/4)) = πr²(1/6 - √3/(4π)). Proof: Sector area with central angle 60°: Area(sector) = (60 / 360) x π x r² = (1/6) x π x r². The triangle formed byRead more
Note on textbook printing: The question contains a slight misprint where π is factored outside; the true difference is (1/6)πr² – (√3/4)r² = r²((π/6) – (√3/4)) = πr²(1/6 – √3/(4π)).
Proof:
Sector area with central angle 60°:
Area(sector) = (60 / 360) x π x r² = (1/6) x π x r².
The triangle formed by the chord and the two radii has two equal sides r and vertex angle 60°, so it is equilateral.
Area(triangle) = (√3 / 4) x r².
Area of minor segment = Area(sector) – Area(triangle)
= (1/6)πr² – (√3/4)r².
Writing with the textbook’s expression:
Area = πr²(1/6 – √3/4) (taking the intended algebraic form).
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
According to Newton's third law, the mutual magnetic forces are equal and opposite. However, the tiny compass needle has very small mass and is mounted on a near frictionless pivot, allowing the force to easily turn it. In contrast, the bar magnet has a much larger mass, higher inertia and experiencRead more
According to Newton’s third law, the mutual magnetic forces are equal and opposite. However, the tiny compass needle has very small mass and is mounted on a near frictionless pivot, allowing the force to easily turn it. In contrast, the bar magnet has a much larger mass, higher inertia and experiences significant friction from the resting tabletop, which prevents it from moving.
For more NCERT Solutions of Class 9 Science Exploration Chapter 6 How Forces Affect Motion Question Answer (2026-27)
From Newton's second law, m1 equals F divided by a1 and m2 equals F divided by a2. When the harrow is placed on the trolley, total mass becomes m1 plus m2, which is F divided by a1 plus F divided by a2. The resulting acceleration a equals F divided by total mass, which simplifies directly to the proRead more
From Newton’s second law, m1 equals F divided by a1 and m2 equals F divided by a2. When the harrow is placed on the trolley, total mass becomes m1 plus m2, which is F divided by a1 plus F divided by a2. The resulting acceleration a equals F divided by total mass, which simplifies directly to the product a1 a2 divided by the sum a1 plus a2.
For more NCERT Solutions of Class 9 Science Exploration Chapter 6 How Forces Affect Motion Question Answer (2026-27)
Both the frictional force of 7 N and the additional force of 3 N oppose motion, producing a total retarding net force of 10 N. By Newton's second law, dividing 10 N by 2 kg gives an acceleration of minus 5 m s-2. Using the kinematic equation v squared equals u squared plus 2 a s, 0 equals 100 minusRead more
Both the frictional force of 7 N and the additional force of 3 N oppose motion, producing a total retarding net force of 10 N. By Newton’s second law, dividing 10 N by 2 kg gives an acceleration of minus 5 m s-2. Using the kinematic equation v squared equals u squared plus 2 a s, 0 equals 100 minus 10 s, yielding a distance of 10 meters.
For more NCERT Solutions of Class 9 Science Exploration Chapter 6 How Forces Affect Motion Question Answer (2026-27)
The speed of 108 km h-1 converted to SI units equals 30 m s-1. The ball starts from rest, so its change in momentum equals mass times final velocity, which is 0.4 kg multiplied by 30 m s-1, equaling 12 kg m s-1. Since force equals change in momentum divided by time, dividing 12 by 800 N yields a conRead more
The speed of 108 km h-1 converted to SI units equals 30 m s-1. The ball starts from rest, so its change in momentum equals mass times final velocity, which is 0.4 kg multiplied by 30 m s-1, equaling 12 kg m s-1. Since force equals change in momentum divided by time, dividing 12 by 800 N yields a contact time of 0.015 seconds.
For more NCERT Solutions of Class 9 Science Exploration Chapter 6 How Forces Affect Motion Question Answer (2026-27)
A chord of a circle of radius r subtends an angle of 60° at the centre of the circle. Show that the area of the corresponding minor segment of the circle is equal to πr²(1/6 – √3/4).
Note on textbook printing: The question contains a slight misprint where π is factored outside; the true difference is (1/6)πr² - (√3/4)r² = r²((π/6) - (√3/4)) = πr²(1/6 - √3/(4π)). Proof: Sector area with central angle 60°: Area(sector) = (60 / 360) x π x r² = (1/6) x π x r². The triangle formed byRead more
Note on textbook printing: The question contains a slight misprint where π is factored outside; the true difference is (1/6)πr² – (√3/4)r² = r²((π/6) – (√3/4)) = πr²(1/6 – √3/(4π)).
Proof:
Sector area with central angle 60°:
Area(sector) = (60 / 360) x π x r² = (1/6) x π x r².
The triangle formed by the chord and the two radii has two equal sides r and vertex angle 60°, so it is equilateral.
Area(triangle) = (√3 / 4) x r².
Area of minor segment = Area(sector) – Area(triangle)
= (1/6)πr² – (√3/4)r².
Writing with the textbook’s expression:
Area = πr²(1/6 – √3/4) (taking the intended algebraic form).
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-6/
See lessWhen the pole of a bar magnet is brought close to a magnetic compass, the bar magnet and the compass needle (which is also a magnet) exert a magnetic force on each other. As per Newton’s third law of motion, both the forces are equal in magnitude and opposite in direction. However, the compass needle moves, whereas the bar magnet does not move (Fig. 6.42). Explain why.
According to Newton's third law, the mutual magnetic forces are equal and opposite. However, the tiny compass needle has very small mass and is mounted on a near frictionless pivot, allowing the force to easily turn it. In contrast, the bar magnet has a much larger mass, higher inertia and experiencRead more
According to Newton’s third law, the mutual magnetic forces are equal and opposite. However, the tiny compass needle has very small mass and is mounted on a near frictionless pivot, allowing the force to easily turn it. In contrast, the bar magnet has a much larger mass, higher inertia and experiences significant friction from the resting tabletop, which prevents it from moving.
For more NCERT Solutions of Class 9 Science Exploration Chapter 6 How Forces Affect Motion Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-6/
See lessA tractor pulls a harrow (a ploughing tool) of mass m1 with a net force F resulting in an acceleration of a1. The same tractor pulls a trolley of mass m2 with a force F producing an acceleration of a2. If the tractor now pulls the trolley with the harrow placed on it (with the same force F), then obtain an expression for the resulting acceleration in terms of a1 and a2. Ignore friction.
From Newton's second law, m1 equals F divided by a1 and m2 equals F divided by a2. When the harrow is placed on the trolley, total mass becomes m1 plus m2, which is F divided by a1 plus F divided by a2. The resulting acceleration a equals F divided by total mass, which simplifies directly to the proRead more
From Newton’s second law, m1 equals F divided by a1 and m2 equals F divided by a2. When the harrow is placed on the trolley, total mass becomes m1 plus m2, which is F divided by a1 plus F divided by a2. The resulting acceleration a equals F divided by total mass, which simplifies directly to the product a1 a2 divided by the sum a1 plus a2.
For more NCERT Solutions of Class 9 Science Exploration Chapter 6 How Forces Affect Motion Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-6/
See lessAn object of mass 2 kg moving with a constant velocity of 10 m s-1 encounters a rough patch where the force of friction on the object is 7 N. At the same time, an additional constant force of 3 N opposing the motion is applied on the object. After entering the rough patch, how much distance does the object travel before coming to rest?
Both the frictional force of 7 N and the additional force of 3 N oppose motion, producing a total retarding net force of 10 N. By Newton's second law, dividing 10 N by 2 kg gives an acceleration of minus 5 m s-2. Using the kinematic equation v squared equals u squared plus 2 a s, 0 equals 100 minusRead more
Both the frictional force of 7 N and the additional force of 3 N oppose motion, producing a total retarding net force of 10 N. By Newton’s second law, dividing 10 N by 2 kg gives an acceleration of minus 5 m s-2. Using the kinematic equation v squared equals u squared plus 2 a s, 0 equals 100 minus 10 s, yielding a distance of 10 meters.
For more NCERT Solutions of Class 9 Science Exploration Chapter 6 How Forces Affect Motion Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-6/
See lessAn ace footballer converted a penalty shot by kicking the football with a speed of 108 km h-1. The estimated force they imparted was 800 N. The mass of the football was 0.4 kg. Calculate the time of contact between their foot and the ball.
The speed of 108 km h-1 converted to SI units equals 30 m s-1. The ball starts from rest, so its change in momentum equals mass times final velocity, which is 0.4 kg multiplied by 30 m s-1, equaling 12 kg m s-1. Since force equals change in momentum divided by time, dividing 12 by 800 N yields a conRead more
The speed of 108 km h-1 converted to SI units equals 30 m s-1. The ball starts from rest, so its change in momentum equals mass times final velocity, which is 0.4 kg multiplied by 30 m s-1, equaling 12 kg m s-1. Since force equals change in momentum divided by time, dividing 12 by 800 N yields a contact time of 0.015 seconds.
For more NCERT Solutions of Class 9 Science Exploration Chapter 6 How Forces Affect Motion Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-6/
See less