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Young's Modulus of Elasticity

One a uniform...

Question

One a uniform rod of mass $m_{1}$ and cross-sectional area A is hung from a ceiling. The other end of bar is supporting mass $m_{2}$ The stress at the midpoint is

\frac{g (m_{2} + 2 m_{1})}{2 A}

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\frac{g (m_{2} + m_{1})}{2 A}

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\frac{g (2 m_{2} + m_{1})}{2 A}

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\frac{g (m_{2} + m_{1})}{A}

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Solution

The correct option is A $\frac{g (m_{2} + 2 m_{1})}{2 A}$
$\begin{matrix} A c c o r d i n g t o q e s t i o n . . . . . . . . . . . . . . . . . . W e i g h t o f s u s p e n d e d m a s s : W_{1} = m_{1} g o n l y h a l f o f t h e w e i g h t o f t h e r o d a c t i n g a t t h e m i d p o i n t, W_{2} = \frac{m_{2} g}{2} s t r e e s a t t h e m i d p o i n t : S t r e s s = \frac{W_{1} + W_{2}}{A} \Rightarrow \frac{m_{1} g + \frac{m_{2} g}{2}}{A} \Rightarrow \frac{\frac{2 m_{1} g + m_{2} g}{2}}{A} ∴ \frac{g (2 m_{1} + m_{2})}{2 A} s o t h a t t h e c o r r e c t o p t i o n i s A . \end{matrix}$

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One a uniform rod of mass m1 and cross-sectional area A is hung from a ceiling. The other end of bar is supporting mass m2 The stress at the midpoint is

The correct option is A g(m2+2m1)2AAccordingtoqestion..................Weightofsuspendedmass:W1=m1gonlyhalfoftheweightoftherodactingatthemidpoint,W2=m2g2streesatthemidpoint:Stress=W1+W2A⇒m1g+m2g2A⇒2m1g+m2g2A∴g(2m1+m2)2AsothatthecorrectoptionisA.

One a uniform rod of mass $m_{1}$ and cross-sectional area A is hung from a ceiling. The other end of bar is supporting mass $m_{2}$ The stress at the midpoint is