Question
Download Solution PDFThe minute hand of a watch is 2 cm long. How far does its tip move in 45 minutes? (use π = 3.14)
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFGiven:
The minute hand of a watch is 2 cm long.
Formula Used:
Arc length, l = r × θ
l = arc length (in radian)
r = radius
θ = central angle in radian
1° = πc/180
Calculation:
We know that,
Angle traced by the minute hand in 60 min = (2π)c
Angle traced by the minute hand in 45 min = \((\frac{2π}{60} × 45)^c = (\frac{3 π}{2})^c\)
⇒ r = 2 cm and θ = \((\frac{3 π}{2})^c\)
⇒ l = r × θ
⇒ l = \((2 × \frac{3 π}{2})\)
⇒ l = 3 π
⇒ l = 3 × 3.14
⇒ l = 9.42
∴ It tip moves 9.42 in 45 minutes.
Last updated on Jun 19, 2025
-> The AAI ATC Exam 2025 will be conducted on July 14, 2025 for Junior Executive..
-> AAI JE ATC recruitment 2025 application form has been released at the official website. The last date to apply for AAI ATC recruitment 2025 is May 24, 2025.
-> AAI JE ATC 2025 notification is released on April 4, 2025, along with the details of application dates, eligibility, and selection process.
-> A total number of 309 vacancies are announced for the AAI JE ATC 2025 recruitment.
-> This exam is going to be conducted for the post of Junior Executive (Air Traffic Control) in the Airports Authority of India (AAI).
-> The Selection of the candidates is based on the Computer-Based Test, Voice Test and Test for consumption of Psychoactive Substances.
-> The AAI JE ATC Salary 2025 will be in the pay scale of Rs 40,000-3%-1,40,000 (E-1).
-> Candidates can check the AAI JE ATC Previous Year Papers to check the difficulty level of the exam.
-> Applicants can also attend the AAI JE ATC Test Series which helps in the preparation.