Question 50Asked in CTET December 2024 Paper 2 Maths & Science, 14 Dec 2024

The distance covered (in metres) by a wheel of radius 25 cm in 350 rounds is (use π = 22/7) :त्रिज्या 25 cm वाले एक पहिए द्वारा 350 चक्कर लगाने में तय की गई दूरी (मीटरों में) है (π = 22/7 का उपयोग कीजिए) :

त्रिज्या 25 cm वाले एक पहिए द्वारा 350 चक्कर लगाने में तय की गई दूरी (मीटरों में) है (π = 22/7 का उपयोग कीजिए) :
  1. A11001100
  2. B275275
  3. C550550
  4. D825825
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Correct answer: (C) 550550

Explanation

Circumference = 2 × 22/7 × 25 = 1100/7 cm. Distance = 350 × 1100/7 = 55000 cm = 550 m.परिधि = 2 × 22/7 × 25 = 1100/7 cm। दूरी = 350 × 1100/7 = 55000 cm = 550 m।

From CTET December 2024 Paper 2 Maths & Science, 14 Dec 2024 — Previous Year Paper

This is question 50 of CTET December 2024 Paper 2 Maths & Science, 14 Dec 2024 — Previous Year Paper. See all questions of this paper with answers, or attempt the paper as a timed online test to get your score and all-India rank.