Question

A body of mass 300gm is tied to one end of string of length 50 cm and revolved in horizontal circle with the other end fixed in hand if the breaking tension in string is 486 dyne what is the maximum speed with which the body can be revolved

Answer 1

Answer:

Answer:

Explanation:

=> For equator acceleration is centripetal acceleration  :

a_c = Rω² = 20 * 10³ (1)²

a_c = 20 * 10³ m/s²

=> maximum tension will be centripetal force on the mass -

T = ${mrω}^2$    [But, ω = 2πf]

and f = 200 rotation per minute = 200 / 60 = 10/3 rotation per second.

T = ( 100 * 10³ ) (2) (20π/3)²

T = 400π² * 2 * 10⁻¹ / 9

= 89.42 N

Thus, The maximum tensile strength of the string is approx 89.42 N.......