Question:
A body of volume \(V\) floats on water with \(\frac{1}{3}\) of its volume above the surface. Find the volume of the object above the surface when floating on a liquid of specific gravity 1.5.
A body of volume \(V\) floats on water with \(\frac{1}{3}\) of its volume above the surface. Find the volume of the object above the surface when floating on a liquid of specific gravity 1.5.
Show Hint
For floating bodies, fraction submerged \(V_s/V = \rho_{\text{body}}/\rho_{\text{liquid}}\). Volume above surface = total volume minus submerged volume.
Updated On: Jul 18, 2026
- \(\frac{3V}{8}\)
- \(\frac{4V}{9}\)
- \(\frac{5V}{9}\)
- \(\frac{2V}{3}\)
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Recall Archimedes principle.
A floating body displaces a volume of liquid equal to its weight:
\[ \rho_{\text{liquid}} V_{\text{displaced}} g = \rho_{\text{body}} V g \]
Step 2: Determine fraction submerged in water.
For water (\(\text{SG} = 1\)), \(\frac{2}{3}\) of body is submerged, so fraction submerged \(f_w = 2/3\). Fraction above water = \(1/3\).
Step 3: Apply to liquid of SG = 1.5.
For liquid of density \(\rho_L = 1.5 \rho_{\text{water}}\), let \(V_s\) be submerged volume:
\[ \rho_L V_s = \rho_{\text{body}} V \implies 1.5 \rho_{\text{water}} V_s = \rho_{\text{water}} V \implies V_s = \frac{2}{3} V \cdot \frac{1}{1.5} = \frac{4}{9} V \]
Step 4: Compute volume above surface.
\[ V_{\text{above}} = V - V_s = V - \frac{4V}{9} = \frac{5V}{9} \]
Step 5: Check consistency.
Volume above surface increases as liquid is denser than water. Fraction submerged is less. Calculation consistent.
Step 6: Final conclusion.
Hence, the volume above the surface is:
\[ \boxed{\frac{5V}{9}} \]
A floating body displaces a volume of liquid equal to its weight:
\[ \rho_{\text{liquid}} V_{\text{displaced}} g = \rho_{\text{body}} V g \]
Step 2: Determine fraction submerged in water.
For water (\(\text{SG} = 1\)), \(\frac{2}{3}\) of body is submerged, so fraction submerged \(f_w = 2/3\). Fraction above water = \(1/3\).
Step 3: Apply to liquid of SG = 1.5.
For liquid of density \(\rho_L = 1.5 \rho_{\text{water}}\), let \(V_s\) be submerged volume:
\[ \rho_L V_s = \rho_{\text{body}} V \implies 1.5 \rho_{\text{water}} V_s = \rho_{\text{water}} V \implies V_s = \frac{2}{3} V \cdot \frac{1}{1.5} = \frac{4}{9} V \]
Step 4: Compute volume above surface.
\[ V_{\text{above}} = V - V_s = V - \frac{4V}{9} = \frac{5V}{9} \]
Step 5: Check consistency.
Volume above surface increases as liquid is denser than water. Fraction submerged is less. Calculation consistent.
Step 6: Final conclusion.
Hence, the volume above the surface is:
\[ \boxed{\frac{5V}{9}} \]
Was this answer helpful?
0
0
Top AP EAPCET Physics Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The radii of the two columns in a U - tube are r1 and r2. When a liquid of density p (angle of contact is $ 0{}^\circ $ ) is filled in it, the level difference of the liquid in the two arms is h. The surface tension of the liquid is: (g = acceleration due to gravity)
- A rigid metallic sphere is spinning around its own axis in the absence of external torque. If the temperature is raised, its volume increases by $9 \%$. The change in its angular speed is
View More Questions
Top AP EAPCET Buoyancy and Floatation Questions
- A 20 g copper block is suspended by a vertical spring causing 1 cm elongation over the natural length of the spring. If a beaker of water is placed below the block so that the copper block is completely immersed in the liquid, the elongation of the spring is (Density of copper 9000 kg m\(^{-3}\), Density of water 1000 kg m\(^{-3}\), \( g = 10 \) ms\(^{-2}\)):
- An object of mass \(10\,\text{kg}\) is released from rest in a liquid. If the object moves a distance of \(2\,\text{m}\) while sinking in a time duration of \(1\,\text{s}\), then the mass of the liquid displaced by the submerged object is \((g=10\,\text{m/s}^2)\):
- A block of wood of volume \(V\) floats in water with half of its volume submerged. The same block floats in an oil with \(0.8V\) volume submerged. If the density of water is \(1000\,\text{kg m}^{-3}\), then the density of the oil is
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions