A metal wire of length 0.5 m and cross-sectional area $10^{-4} \mathrm{~m}^{2}$ has breaking stress $5 \times 10^{8} \mathrm{Nm}^{-2}$. A block of 10 kg is attached at one end of the string and is rotating in a horizontal circle. The maximum linear velocity of block will be \_\_\_\_ $\mathrm{ms}^{-1}$.

JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Laws of Motion

A metal wire of length 0.5 m and cross-sectional area

1 0^{- 4} m^{2}

has breaking stress

5 \times 1 0^{8} Nm^{- 2}

. A block of 10 kg is attached at one end of the string and is rotating in a horizontal circle. The maximum linear velocity of block will be ____

ms^{- 1}

Answer: 50

Solution

For circular motion, the tension provides the centripetal force:

T = \frac{m v ^{2}}{ℓ}

The maximum tension is limited by the breaking stress:

T_{ma x} = Breaking stress \times Area = 5 \times 1 0^{8} \times 1 0^{- 4} = 5 \times 1 0^{4} N

Setting

T_{ma x} = \frac{m v ^{2}}{ℓ}

5 \times 1 0^{4} = \frac{10 \times v ^{2}}{0.5}

5 \times 1 0^{4} = 20 v^{2}

v^{2} = 2500 ⟹ v = 50 m/s

The answer is 50.

JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Integer answer

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