A steel wire having a radius of 2-0 mm- carrying a load of 4 kg- is hanging from a ceiling- Given that g -3-1 - ms -2- what will be the tensile stress that would be developed in the wire-A- 4-8 - 106 Nm -2B- 5-2 - 106 Nm -2C- 6-2 - 106 Nm -2D- 3-1 - 106 Nm -2

Tensile stress in wire will be = Tensile forceCross sectional Area = mgπ R^2 = 4× 3.1ππ× (2× 10^ 3)^2 Nm^ 2 = 3.1× 10^6 Nm^ 2. Hence, option (D) is correct. ...

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Standard XII

Physics

Excess Pressure in Bubbles

A steel wire ...

Question

A steel wire having a radius of $2.0 m m$ , carrying a load of $4 k g$ , is hanging from a ceiling. Given that $g = 3.1 π m s^{- 2}$ , what will be the tensile stress that would be developed in the wire?

4.8 \times 10^{6} N m^{- 2}

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5.2 \times 10^{6} N m^{- 2}

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6.2 \times 10^{6} N m^{- 2}

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3.1 \times 10^{6} N m^{- 2}

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Solution

The correct option is D $3.1 \times 10^{6} N m^{- 2}$
Tensile stress in wire will be
$= \frac{Tensile force}{Cross-sectional Area}$

$= \frac{m g}{π R^{2}} = \frac{4 \times 3.1 π}{π \times 4 \times 10^{- 6}} N m^{- 2} = 3.1 \times 10^{6} N m^{- 2}$ .

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A steel wire having a radius of 2.0 mm, carrying a load of 4 kg, is hanging from a ceiling. Given that g=3.1π ms−2, what will be the tensile stress that would be developed in the wire?

The correct option is D 3.1×106 Nm−2Tensile stress in wire will be =Tensile forceCross-sectional Area =mgπR2=4×3.1ππ×4×10−6Nm−2=3.1×106 Nm−2.

A steel wire having a radius of $2.0 m m$ , carrying a load of $4 k g$ , is hanging from a ceiling. Given that $g = 3.1 π m s^{- 2}$ , what will be the tensile stress that would be developed in the wire?

The correct option is D $3.1 \times 10^{6} N m^{- 2}$
Tensile stress in wire will be
$= \frac{Tensile force}{Cross-sectional Area}$

$= \frac{m g}{π R^{2}} = \frac{4 \times 3.1 π}{π \times 4 \times 10^{- 6}} N m^{- 2} = 3.1 \times 10^{6} N m^{- 2}$ .