A uniform rod of length $b$ capable of turning about its end which is out of water, rests inclined to the vertical. If the specific gravity of the rod is $\frac{5}{9}$ , find the length of the rod immersed in water.

Option 1 - <math> <mfrac> <mrow> <mrow> <mi>b</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> </math>

Option 2 - <math> <mfrac> <mrow> <mrow> <mi>b</mi> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> </mfrac> </math>

Option 3 - <math> <mfrac> <mrow> <mrow> <mi>b</mi> </mrow> </mrow> <mrow> <mrow> <mn>6</mn> </mrow> </mrow> </mfrac> </math>

Option 4 - <math> <mfrac> <mrow> <mrow> <mi>b</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </mfrac> </math>

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Raj Pandey | Contributor-Level 9

5 months ago

Correct Option - 4

Detailed Solution:

Let $2 x$ length of rod is immersed in water.

$τ_{Hinge} ($ net $) = 0$

$\Rightarrow m g ? \frac{b}{2} s i n ? θ - F_{B} (b - x) s i n ? θ = 0$

$\Rightarrow m g b = 2 (b - x) F_{B}$

$∴ F_{B} = \frac{m g b}{2 (b - x)} = \frac{(A b d) b}{2 (b - x)} g$ , where $d =$ density of material of rod

$_{}$ $F_{B} =$ Buoyant force $\frac{}{} \frac{}{}$

Equate $F_{B}, \frac{A b^{2} d}{2 (b - x)} = 2 A x ρ$ $_{} \frac{^{}}{}$

$\Rightarrow b^{2} = 4 x (b - x) \frac{9}{5}$

$^{} \frac{^{}}{} \frac{}{}$ $\Rightarrow x^{2} - b x + \frac{5 b^{2}}{36} = 0 \Rightarrow x = \frac{b}{6} ∴$ immersed $= \frac{b}{3}$

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A uniform rod of length b capable of turning about its end which is out of water, rests inclined to the vertical. If the specific gravity of the rod is 5 9 , find the length of the rod immersed in water.

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