A body of mass $\mathrm{m}=10 \mathrm{~kg}$ is attached to one end of a wire of length 0.3 m. The maximum angular speed (in $\mathrm{rad} \mathrm{s}^{-1}$) with which it can be rotated about its other end in space station is (Breaking stress of wire $=4.8 \times 10^{7} \mathrm{Nm}^{-2}$ and area of cross-section of the wire $=10^{-2} \mathrm{~cm}^{2}$) is

JEE Main · 2020 · 09 Jan 2020 · Shift 1 · Laws of Motion

A body of mass

m = 10 kg

is attached to one end of a wire of length 0.3 m. The maximum angular speed (in

rad s^{- 1}

) with which it can be rotated about its other end in space station is (Breaking stress of wire

= 4.8 \times 1 0^{7} Nm^{- 2}

and area of cross-section of the wire

= 1 0^{- 2} cm^{2}

) is

Answer: 4

Solution

In the weightless station the wire can give all of its breaking tension to the circular motion. That maximum tension equals breaking stress times area,

T_{m a x} = 4.8 \times 1 0^{7} \times 1 0^{- 6} = 48

N, where the cross-section

1 0^{- 2} cm^{2}

1 0^{- 6} m^{2}

. Setting

T_{m a x} = m ω^{2} r

gives

48 = 10 \times ω^{2} \times 0.3

, so

ω^{2} = 16

and the greatest angular speed is

ω = 4 rad s^{- 1}

JEE Main · 2020 · 09 Jan 2020 · Shift 1 · Integer answer

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