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A uniform steel bar of cross-sectional a...

A uniform steel bar of cross-sectional area A and length L is suspended, so that it hangs vertically. The stress at the middle point of the bar is (`rho` is the density of stell)

A

`(L)/(2A) rho g`

B

`(L rho g)/(2)`

C

`(LA)/(rho g)`

D

`L rho g`

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The correct Answer is:
To find the stress at the middle point of a uniform steel bar of cross-sectional area \( A \) and length \( L \) that is suspended vertically, we can follow these steps: ### Step 1: Understand the Concept of Stress Stress is defined as the force applied per unit area. Mathematically, it is expressed as: \[ \text{Stress} = \frac{\text{Force}}{\text{Area}} \] ### Step 2: Identify the Force at the Midpoint When the bar is suspended, the force acting on the midpoint is due to the weight of the lower half of the bar. The weight of an object is given by: \[ \text{Weight} = \text{mass} \times g \] where \( g \) is the acceleration due to gravity. ### Step 3: Calculate the Mass of the Lower Half of the Bar The mass of the entire bar can be expressed in terms of its density \( \rho \) and volume. The volume \( V \) of the bar is given by: \[ V = A \times L \] Thus, the mass \( m \) of the entire bar is: \[ m = \rho \times V = \rho \times A \times L \] The mass of the lower half of the bar is: \[ m_{\text{lower}} = \frac{m}{2} = \frac{\rho \times A \times L}{2} \] ### Step 4: Calculate the Force at the Midpoint The force acting at the midpoint due to the weight of the lower half of the bar is: \[ F = m_{\text{lower}} \times g = \left(\frac{\rho \times A \times L}{2}\right) \times g = \frac{\rho \times A \times L \times g}{2} \] ### Step 5: Substitute into the Stress Formula Now, we can substitute the force \( F \) into the stress formula: \[ \text{Stress} = \frac{F}{A} = \frac{\frac{\rho \times A \times L \times g}{2}}{A} \] ### Step 6: Simplify the Expression The area \( A \) cancels out: \[ \text{Stress} = \frac{\rho \times L \times g}{2} \] ### Final Answer Thus, the stress at the midpoint of the bar is: \[ \text{Stress} = \frac{\rho \times L \times g}{2} \]

To find the stress at the middle point of a uniform steel bar of cross-sectional area \( A \) and length \( L \) that is suspended vertically, we can follow these steps: ### Step 1: Understand the Concept of Stress Stress is defined as the force applied per unit area. Mathematically, it is expressed as: \[ \text{Stress} = \frac{\text{Force}}{\text{Area}} \] ...
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MHTCET PREVIOUS YEAR PAPERS AND PRACTICE PAPERS-ELASTICITY -EXERCISE 2
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  8. A steel wire of length 20 cm and uniform cross-sectional 1 mm^(2) is t...

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  10. A stress of 10^(6) Nm^(-2) is required for breaking a material. If the...

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  11. The temperature of a wire length 1 m and area of cross-section 1 cm^(2...

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  12. Two wires of copper having the length in the ratio 4 : 1 and their rad...

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  13. A steel wire of length 20 cm and uniform cross-sectional area of 1 mm^...

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  14. A force of 20 N is applied at one end of a wire of length 2 m having a...

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  15. Two wires of same diameter of the same material having the length l an...

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  16. A 5 metre long wire is fixed to the ceiling. A weight of 10 kg is hung...

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  17. The stress versus strain graphs for wires of two materials A and B are...

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  18. The load versus elongation graph for four wire of the same material is...

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  19. The strain stress curves of three wires of different materials are sho...

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  20. The density of a metal at normal pressure is rho. Its density when it ...

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