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Standard XII

Physics

Springs

A rectangular...

Question

A rectangular object of length 0.5 m is pulled by two forces, 4N along its length and other 6N making an angle of 60 degrees with the horizontal. The area of cross section of the object is $3 c m^{2}$ . Find the stress acting on the object

2.33 \times 10^{4} N / m^{2}

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2.33 \times 10^{- 4} N / m^{2}

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1.33 \times 10^{4} N / m^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.33 \times 10^{- 4} N / m^{2}

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Solution

The correct option is C $2.33 \times 10^{4} N / m^{2}$
Total force acting along the length/ horizontally = 4 + 6 cos 60 = 7 N
Stress acting on the object along the length or longitudinal stress = p/A= $7 / (3 \times 10^{- 4}) = 2.33 \times 10^{4} N / m^{2}$
The correct option is option(a)

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A rectangular object of length 0.5 m is pulled by two forces, 4N along its length and other 6N making an angle of 60 degrees with the horizontal. The area of cross section of the object is 3cm2. Find the stress acting on the object

The correct option is C 2.33×104N/m2Total force acting along the length/ horizontally = 4 + 6 cos 60 = 7 NStress acting on the object along the length or longitudinal stress = p/A=7/(3×10−4)=2.33×104N/m2The correct option is option(a)

A rectangular object of length 0.5 m is pulled by two forces, 4N along its length and other 6N making an angle of 60 degrees with the horizontal. The area of cross section of the object is $3 c m^{2}$ . Find the stress acting on the object

The correct option is C $2.33 \times 10^{4} N / m^{2}$
Total force acting along the length/ horizontally = 4 + 6 cos 60 = 7 N
Stress acting on the object along the length or longitudinal stress = p/A= $7 / (3 \times 10^{- 4}) = 2.33 \times 10^{4} N / m^{2}$
The correct option is option(a)