A bob of mass 10 kg is attached to wire 0.3 long. Its breaking stress is 4.8 × 107 Nm 2. The area of cross section of the wire is 10-6 m 2 . The maximum angular velocity with which it can be rotated in a horizontal circle

Question

A bob of mass 10 kg is attached to wire 0.3 long. Its breaking stress is 4.8 × 107 Nm 2. The area of cross section of the wire is 10-6 m 2 . The maximum angular velocity with which it can be rotated in a horizontal circle (A) 8 rad/ sec (B) 4 rad/ sec (C) 2 rad/ sec (D) 1 rad/ sec

Tardigrade Admin · Accepted Answer

Using the relation
Force =breaking stress

\times

area

= (4.8 \times 1 0^{7}) (1 0^{- 6}) = 48 N

...(1)
Using the relation

F = M r ω^{2}

...(2)

M r ω^{2} = 48

10 \times 0.3 \times ω^{2} = 48

ω^{2} = \frac{48}{10 \times 0 \cdot 3} = 16

so

ω^{2} = 16

ω = 4 r a d / sec

Q. A bob of mass 10kg is attached to wire 0.3 long. Its breaking stress is 4.8×107Nm2. The area of cross section of the wire is 10−6m2. The maximum angular velocity with which it can be rotated in a horizontal circle

8 rad/ sec

4 rad/ sec

2 rad/ sec

1 rad/ sec

Using the relation Force =breaking stress × area =(4.8×107)(10−6)=48N ...(1) Using the relation F=Mrω2 ...(2) Mrω2=48 10×0.3×ω2=48 ω2=10×0⋅348​=16 so ω2=16 ω=4rad/sec

Q. A bob of mass $10 k g$ is attached to wire $0.3$ long. Its breaking stress is $4.8 \times 1 0^{7} N m^{2}$ . The area of cross section of the wire is $1 0^{- 6} m^{2} .$ The maximum angular velocity with which it can be rotated in a horizontal circle

Using the relation
Force =breaking stress $\times$ area
$= (4.8 \times 1 0^{7}) (1 0^{- 6}) = 48 N$ ...(1)
Using the relation
$F = M r ω^{2}$ ...(2)
$M r ω^{2} = 48$
$10 \times 0.3 \times ω^{2} = 48$
$ω^{2} = \frac{48}{10 \times 0 \cdot 3} = 16$
so $ω^{2} = 16$
$ω = 4 r a d / sec$