What's your question?
  1. Innovation alone cannot guarantee business success. A novel product remains useless without commercialization, operational capability, market demand and effective execution. Innovative ventures regularly fail due to poor cash flow management, lack of consumer awareness, weak distribution networks orRead more

    Innovation alone cannot guarantee business success. A novel product remains useless without commercialization, operational capability, market demand and effective execution. Innovative ventures regularly fail due to poor cash flow management, lack of consumer awareness, weak distribution networks or flawed pricing. Lasting success demands marrying great innovation with disciplined financial budgeting, skilled teamwork, reliable logistics and proactive customer relationship management.

     

    For more NCERT Solutions of Class 9 Social Science Part 2 Chapter 6 From Ideas to Startups Question Answer (2026-27)

    https://www.tiwariacademy.com/ncert-solutions/class-9/social-science/part-2-chapter-6/

    See less
    • 77
  2. A business plan is dynamic rather than rigid; it guides initial operations while adapting to ground realities. For example, during the COVID-19 pandemic, dine-in restaurants successfully pivoted to cloud kitchens and app deliveries to survive. An entrepreneur should modify the business plan when conRead more

    A business plan is dynamic rather than rigid; it guides initial operations while adapting to ground realities. For example, during the COVID-19 pandemic, dine-in restaurants successfully pivoted to cloud kitchens and app deliveries to survive. An entrepreneur should modify the business plan when consumer preferences change, disruptive competitors enter the market, government regulations shift or key financial assumptions prove unviable.

     

    For more NCERT Solutions of Class 9 Social Science Part 2 Chapter 6 From Ideas to Startups Question Answer (2026-27)

    https://www.tiwariacademy.com/ncert-solutions/class-9/social-science/part-2-chapter-6/

    See less
    • 80
  3. (i) The large dough ball has volume (4/3)π(6)³ cm³. An individual chapati dough ball typically has a radius of about 1.5 cm, giving volume (4/3)π(1.5)³ cm³. Dividing volumes gives (6 / 1.5)³ = 4³ = 64 chapatis. (ii) Treating the halved fruit as a concentric hemisphere, let outer radius be R and inneRead more

    (i) The large dough ball has volume (4/3)π(6)³ cm³. An individual chapati dough ball typically has a radius of about 1.5 cm, giving volume (4/3)π(1.5)³ cm³. Dividing volumes gives (6 / 1.5)³ = 4³ = 64 chapatis. (ii) Treating the halved fruit as a concentric hemisphere, let outer radius be R and inner cavity radius be r. The edible flesh volume equals (2/3)π(R³ − r³).

     

    For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 14 Math of Space: Surface Area and Volume Question Answer (2026-27)

    https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-14/

    See less
    • 103
  4. The ink reservoir inside a ballpoint pen refill forms a right circular cylinder. Measuring a typical standard refill, the internal radius is approximately r = 0.1 cm (diameter 2 mm) and the column of ink has a height of about h = 10 cm. The volume of ink is given by V = πr²h ≈ 3.14 × (0.1)² × 10 = 0Read more

    The ink reservoir inside a ballpoint pen refill forms a right circular cylinder. Measuring a typical standard refill, the internal radius is approximately r = 0.1 cm (diameter 2 mm) and the column of ink has a height of about h = 10 cm. The volume of ink is given by V = πr²h ≈ 3.14 × (0.1)² × 10 = 0.314 cm³, equivalent to roughly 0.31 millilitres.

     

    For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 14 Math of Space: Surface Area and Volume Question Answer (2026-27)

    https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-14/

    See less
    • 95
  5. (i) New volume is (l + 1)wh = lwh + wh. Change in volume is (lwh + wh) − lwh = wh cubic units, which is option (c). (ii) New volume is π(r + 1)²h = π(r² + 2r + 1)h = πr²h + 2πrh + πh. Subtracting original volume πr²h yields 2πrh + πh cubic units, corresponding to option (e). (iii) Change in volume iRead more

    (i) New volume is (l + 1)wh = lwh + wh. Change in volume is (lwh + wh) − lwh = wh cubic units, which is option (c).

    (ii) New volume is π(r + 1)²h = π(r² + 2r + 1)h = πr²h + 2πrh + πh. Subtracting original volume πr²h yields 2πrh + πh cubic units, corresponding to option (e).

    (iii) Change in volume is (4/3)πr³ − (4/3)π(r − 1)³ = (4/3)π(3r² − 3r + 1) cubic units.

     

    For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 14 Math of Space: Surface Area and Volume Question Answer (2026-27)

    https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-14/

    See less
    • 101