Properties Of Matter Ques 35
- A uniform rod of length $L$ and density $\rho$ is being pulled along a smooth floor with a horizontal acceleration $\alpha$ (see figure). The magnitude of the stress at the transverse cross-section through the mid-point of the rod is
(1993, 1M)
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Answer:
Correct Answer: 35.$(\frac{\rho L \alpha }{2})$
Solution:
- Let $A$ be the area of cross-section of the rod. FBD of rod at mid-point
Mass $m=$ volume $\times$ density
$ =\left(\frac{L}{2} \cdot A\right) \rho $
$\therefore \quad T=m \alpha=\left(\frac{L}{2} A \rho \alpha\right)$
$\therefore$ Stress $=\frac{T}{A}=\frac{1}{2} \rho \alpha L$
BETA
