A ball of mass 0.12 kg is being whirled in a horizontal circle at the end of string 0.5 m long. It incapable of making 231 revolutions in one minute. The breaking tension of the string is

Question

A ball of mass 0.12 kg is being whirled in a horizontal circle at the end of string 0.5 m long. It incapable of making 231 revolutions in one minute. The breaking tension of the string is (A) 3N (B) 15.1 N (C) 31.5 N (D) 35.1 N

Tardigrade Admin · Accepted Answer

When body performs circular motion, it is acted upon by a centripetal force the magnitude of which is given by

F = \frac{m v ^{2}}{r}

where,

m

is mass,

v

the velocity and

r

the radius.

Since,

v = r ω

∴ F = m r ω^{2}

Given,

m = 0.12 k g, r = 0.5 m,

ω = 231 r p m = \frac{2 π \times 231}{60} r a d / s = 24.2 r a d / s

∴ F = 0.12 \times 0.5 \times (24.2)^{2} = 35.1 N

Q. A ball of mass 0.12kg is being whirled in a horizontal circle at the end of string 0.5m long. It incapable of making 231 revolutions in one minute. The breaking tension of the string is

3N

15.1 N

31.5 N

35.1 N

When body performs circular motion, it is acted upon by a centripetal force the magnitude of which is given by F=rmv2​ where, m is mass, v the velocity and r the radius. Since, v=rω ∴F=mrω2 Given, m=0.12kg,r=0.5m, ω=231rpm=602π×231​rad/s=24.2rad/s ∴F=0.12×0.5×(24.2)2=35.1N

Q. A ball of mass $0.12 k g$ is being whirled in a horizontal circle at the end of string $0.5 m$ long. It incapable of making $231$ revolutions in one minute. The breaking tension of the string is