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A uniform rod of length L and density $\rho$  is being pulled along A smooth floor with a horizontal acceleration $\alpha$ what is the magnitude of the stress at the transverse cross-section through the midpoint of the rod?

Option: 1

\frac{1}{3} \mathrm{L} \rho \alpha


Option: 2

\frac{2}{3} \mathrm{L} \rho \alpha


Option: 3

\mathrm{L} \rho \alpha


Option: 4

\frac{1}{2} \mathrm{L} \rho \alpha


Answers (1)

best_answer

 

 

 

Let cross-sectional area of the rod be A.

So, mass of rod =\rho(AL)

Now, we have to find the tension at the mid point of the rod.

\begin{aligned} &T=\frac{\rho A L}{2}(\alpha)\\ &\text { stress }=\frac{T}{a r e a}=\frac{\rho \alpha L}{2} \end{aligned}

Correct option is (4).

Posted by

Nehul

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