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Standard XII

Physics

Angular Velocity

A small but h...

Question

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

4 r a d / s

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8 r a d / s

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10 r a d / s

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32 r a d / s

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Solution

The correct option is A $4 r a d / s$
$m r ω^{2} = B r e a k i n g s t r e s s \times C r o s s - s e c t i o n a l a r e a$
$10 \times 0.3 ω^{2} = 4.8 \times 10^{7} \times 10^{- 6} = 48$
or $ω^{2} = \frac{48}{3} = 16$ or $ω = 4 r a d / s$ .

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A bob of mass 10 kg is attached to wire 0.3 m long. Its breaking stress is $4.8 \times 10^{7} N / m 2$ . 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

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A small but heavy block of mass 10 kg is attached to a wire 0.3 m long. Its breaking stress is 4.8×107N/m2. The area of the cross section of the wire is 10−6m2. The maximum angular velocity with which the block can be rotated in the horizontal circle is

The correct option is A 4rad/smrω2=Breaking stress×Cross−sectional area10×0.3ω2=4.8×107×10−6=48or ω2=483=16 or ω=4 rad/s.

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

The correct option is A $4 r a d / s$
$m r ω^{2} = B r e a k i n g s t r e s s \times C r o s s - s e c t i o n a l a r e a$
$10 \times 0.3 ω^{2} = 4.8 \times 10^{7} \times 10^{- 6} = 48$
or $ω^{2} = \frac{48}{3} = 16$ or $ω = 4 r a d / s$ .