WBJEE · Physics · Mechanical Properties of Fluids
A uniform rod is suspended horizontally from its mid-point. A piece of metal whose weight is \(w\) is suspended at a distance \(l\) from the mid-point. Another weight \(W_{1}\) is suspended on the other side at a distance \(l_{1}\) from the mid-point to bring the rod to a horizontal position. Wben \(w\) is completely immersed in water, \(w_{1}\) needs to be kept at a distance \(l_{2}\) from the mid-point to get the rod back into horizontal position. The specific gravity of the metal piece is
- A \(\frac{w}{w_{1}}\)
- B \(\frac{w l_{1}}{w l-w_{1} l_{2}}\)
- C \(\frac{L_{1}}{l_{1}-l_{2}}\)
- D \(\frac{l_{1}}{l_{2}}\)
Answer & Solution
Correct Answer
(C) \(\frac{L_{1}}{l_{1}-l_{2}}\)
Step-by-step Solution
Detailed explanation
\[ W-F_{B}=W\left(1-\frac{1}{\rho}\right) \] where \(\rho=\) specific gravity \(W\left(1-\frac{1}{\rho}\right)=W_{1} b\) \(\left(1-\frac{1}{\rho}\right)=\frac{W_{1} L_{2}}{W}\) \(\left(1-\frac{1}{\rho}\right)=\frac{W_{1} b_{2}}{W_{1} L_{1}}\)…
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