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Which factors determine the energy required to raise a flag from the ground to the top of a tall flagpole using a pulley? Does raising the flag slowly or quickly change the amount of work done? If the speed at which the flag is raised is doubled, how does the power requirement change? Explain your answers.
The energy required is determined by the mass of the flag, acceleration due to gravity and vertical height of the flagpole. The work done equals m g h and remains independent of speed or time. However, doubling the speed cuts the time taken in half. Since power is work divided by time, halving the tRead more
The energy required is determined by the mass of the flag, acceleration due to gravity and vertical height of the flagpole. The work done equals m g h and remains independent of speed or time. However, doubling the speed cuts the time taken in half. Since power is work divided by time, halving the time doubles the power requirement.
For more NCERT Solutions of Class 9 Science Exploration Chapter 7 Work, Energy and Simple Machines Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-7/
See lessA basket contains 4 red balls and 5 blue balls. One ball is drawn and laid aside and a second ball is drawn. Draw a tree diagram to represent the possible outcomes and probabilities. Use the tree diagram to answer the following questions: (i) What is the probability of drawing a red ball and then a blue ball? (ii) What is the probability of drawing 2 blue balls?
Total balls = 4 Red + 5 Blue = 9. Since the first ball is not replaced, 8 balls remain for the second pick. Tree diagram paths: First branch splits into Red (4/9) and Blue (5/9). From Red, second branches split into Red (3/8) and Blue (5/8). From Blue, second branches split into Red (4/8) and Blue (Read more
Total balls = 4 Red + 5 Blue = 9. Since the first ball is not replaced, 8 balls remain for the second pick.
Tree diagram paths:
First branch splits into Red (4/9) and Blue (5/9).
From Red, second branches split into Red (3/8) and Blue (5/8).
From Blue, second branches split into Red (4/8) and Blue (4/8).
(i) P(Red then Blue) = (4/9) x (5/8) = 20/72 = 5/18 (approx 0.278).
(ii) P(2 Blue balls) = (5/9) x (4/8) = 20/72 = 5/18 (approx 0.278).
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 7 The Mathematics of Maybe: Introduction to Probability Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-7/
See lessA game of chance consists of spinning an arrow (see Fig. 7.7.) which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 and these are equally likely outcomes. What is the probability that it will point at: (i) 8? (ii) An odd number? (iii) A number greater than 2? (iv) A number less than 9? (v) A multiple of 3?
Total possible outcomes = 8, as the arrow can point to {1, 2, 3, 4, 5, 6, 7, 8}. (i) Favourable outcome = {8}. Probability = 1/8 = 0.125. (ii) Odd numbers = {1, 3, 5, 7}. Probability = 4/8 = 1/2 = 0.5. (iii) Numbers greater than 2 = {3, 4, 5, 6, 7, 8}. Probability = 6/8 = 3/4 = 0.75. (iv) Numbers leRead more
Total possible outcomes = 8, as the arrow can point to {1, 2, 3, 4, 5, 6, 7, 8}.
(i) Favourable outcome = {8}. Probability = 1/8 = 0.125.
(ii) Odd numbers = {1, 3, 5, 7}. Probability = 4/8 = 1/2 = 0.5.
(iii) Numbers greater than 2 = {3, 4, 5, 6, 7, 8}. Probability = 6/8 = 3/4 = 0.75.
(iv) Numbers less than 9 = {1, 2, 3, 4, 5, 6, 7, 8}. Probability = 8/8 = 1 (certain event).
(v) Multiples of 3 = {3, 6}. Probability = 2/8 = 1/4 = 0.25.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 7 The Mathematics of Maybe: Introduction to Probability Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-7/
See lessThe letters of the word ‘PEACE’ are placed on cards. Leela draws a card without looking. (i) What is the probability that it is a P, E or C? (ii) What is the probability that it is not an E?
Total letters in the word PEACE = 5, so total possible outcomes = 5. (i) Favourable letters for 'a P, E or C' are P, E, C, E (1 P, 2 Es and 1 C). Number of favourable outcomes = 1 + 2 + 1 = 4. Probability(P, E or C) = 4 / 5 = 0.8 (or 80%). (ii) Letters that are not E are P, A and C. Number of favourRead more
Total letters in the word PEACE = 5, so total possible outcomes = 5.
(i) Favourable letters for ‘a P, E or C’ are P, E, C, E (1 P, 2 Es and 1 C).
Number of favourable outcomes = 1 + 2 + 1 = 4.
Probability(P, E or C) = 4 / 5 = 0.8 (or 80%).
(ii) Letters that are not E are P, A and C.
Number of favourable outcomes = 3.
Probability(not an E) = 3 / 5 = 0.6 (or 60%).
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 7 The Mathematics of Maybe: Introduction to Probability Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-7/
See lessA tyre company records distances before replacement in 1000 cases. Find the probability that a randomly chosen tyre lasts: (i) Less than 4000 km, (ii) Between 4000 and 14000 km, (iii) More than 14000 km.
Total number of cases = 1000. (i) For tyre lasting less than 4000 km, frequency = 20. Probability = 20 / 1000 = 0.02 (or 2%). (ii) For distance between 4000 and 14000 km, add frequencies (4001 to 9000 km) and (9001 to 14000 km): 210 + 325 = 535. Probability = 535 / 1000 = 0.535 (or 53.5%). (iii) ForRead more
Total number of cases = 1000.
(i) For tyre lasting less than 4000 km, frequency = 20. Probability = 20 / 1000 = 0.02 (or 2%).
(ii) For distance between 4000 and 14000 km, add frequencies (4001 to 9000 km) and (9001 to 14000 km): 210 + 325 = 535. Probability = 535 / 1000 = 0.535 (or 53.5%).
(iii) For tyre lasting more than 14000 km, frequency = 445. Probability = 445 / 1000 = 0.445 (or 44.5%).
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 7 The Mathematics of Maybe: Introduction to Probability Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-7/
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