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Choose the correct options and explain the reason for the correct and incorrect options. Which among the following mixtures show the Tyndall Effect? A mixture of: (a) air and dust particles (b) copper sulfate and water (c) starch and water (d) acetone and water (i) a and b (ii) b and d (iii) a and c (iv) c and d
Option iii is correct because air containing dust particles and starch mixed with water can scatter light and show the Tyndall effect. The chapter explains that particles in a colloid or suspension scatter light, making its path visible. Copper sulfate solution and acetone with water are solutions,Read more
Option iii is correct because air containing dust particles and starch mixed with water can scatter light and show the Tyndall effect. The chapter explains that particles in a colloid or suspension scatter light, making its path visible. Copper sulfate solution and acetone with water are solutions, so their particles do not scatter light sufficiently to make the light beam visible clearly to the observer.
For more NCERT Solutions of Class 9 Science Exploration Chapter 5 Exploring Mixtures and their Separation Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-5/
See lessTwo cars A and B start moving with a constant acceleration from rest, in a straight line. Car A attains a velocity of 5 m s⁻¹ in 5 s. Car B attains a velocity of 3 m s⁻¹ in 10 s. Plot the velocity-time graphs for both the cars in the same graph. Using the graph, calculate the displacement in the two time intervals mentioned (Hint: Calculate the acceleration in both cases). Then calculate their velocities at five instants of time to plot the graph.
For car A, acceleration = (5−0)/5 = 1 m s⁻². Its displacement in 5 s = ½ × 1 × 5² = 12.5 m. For car B, acceleration = (3−0)/10 = 0.3 m s⁻². Its displacement in 10 s = ½ × 0.3 × 10² = 15 m. Thus, both velocity-time graphs are straight lines starting from the origin, with slopes 1 and 0.3 m s⁻², respeRead more
For car A, acceleration = (5−0)/5 = 1 m s⁻². Its displacement in 5 s = ½ × 1 × 5² = 12.5 m. For car B, acceleration = (3−0)/10 = 0.3 m s⁻². Its displacement in 10 s = ½ × 0.3 × 10² = 15 m. Thus, both velocity-time graphs are straight lines starting from the origin, with slopes 1 and 0.3 m s⁻², respectively.
For more NCERT Solutions of Class 9 Science Exploration Chapter 4 Describing Motion Around Us Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-4/
See lessOn entering a state highway, a car continues to move with a constant velocity of 6 m s⁻¹ for 2 minutes and then accelerates with a constant acceleration 1 m s⁻² for 6 seconds. Find the displacement of the car on the state highway in the 2 min 6 s time interval by drawing a velocity-time graph for its motion.
For the first 2 minutes = 120 s, the car moves at 6 m s⁻¹. Therefore, displacement = 6 × 120 = 720 m. During the next 6 s, initial velocity is 6 m s⁻¹ and acceleration is 1 m s⁻². Displacement = ut + ½at² = 6 × 6 + ½ × 1 × 36 = 54 m. Hence, total displacement = 720 + 54 = 774 m. For more NCERRead more
For the first 2 minutes = 120 s, the car moves at 6 m s⁻¹. Therefore, displacement = 6 × 120 = 720 m. During the next 6 s, initial velocity is 6 m s⁻¹ and acceleration is 1 m s⁻². Displacement = ut + ½at² = 6 × 6 + ½ × 1 × 36 = 54 m. Hence, total displacement = 720 + 54 = 774 m.
For more NCERT Solutions of Class 9 Science Exploration Chapter 4 Describing Motion Around Us Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-4/
See lessA girl is preparing for her first marathon by running on a straight road. She uses a smartwatch to calculate her running speed at different intervals. The graph (Fig. 4.31) depicts her velocity versus time. Estimate the distance she ran based on the graph.
The distance travelled by the girl is given by the area under the velocity-time graph. We divide the graph into convenient trapeziums and estimate their areas using the velocities at different times. Adding these areas gives the approximate distance covered during the run. From the values shown in FRead more
The distance travelled by the girl is given by the area under the velocity-time graph. We divide the graph into convenient trapeziums and estimate their areas using the velocities at different times. Adding these areas gives the approximate distance covered during the run. From the values shown in Fig. 4.31, the girl ran approximately 45 km. Therefore, her estimated distance travelled is about 45 km.
For more NCERT Solutions of Class 9 Science Exploration Chapter 4 Describing Motion Around Us Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-4/
See lessThe velocity-time graph from 0 s to 120 s for a cyclist is shown in Fig. 4.30. Shade the areas (in different colours) representing the displacement of the cyclist (i) while cyclist is moving with constant velocity. (ii) when the velocity of cyclist is decreasing. Also, calculate the displacement and average acceleration in the 120 s time interval.
From the velocity-time graph: Constant velocity (20–100 s): displacement = area of rectangle = 3 × 80 = 240 m. Decreasing velocity (100–120 s): displacement = area of trapezium = ½(3+2) × 20 = 50 m. Initial part (0–20 s): displacement = area of triangle = ½ × 20 × 3 = 30 m. Therefore, total displaceRead more
From the velocity-time graph:
Therefore, total displacement = 30 + 240 + 50 = 320 m. Average acceleration = (2−0)/120 = 0.0167 m s⁻².
For more NCERT Solutions of Class 9 Science Exploration Chapter 4 Describing Motion Around Us Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-4/
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