Question:
A thin metal disc of radius \(r\) floats on water surface and bends the surface downwards along the perimeter making an angle \(\theta\) with vertical edge of the disc. If the disc displaces a weight of water \(W\) and surface tension of water is \(T\), then the weight of metal disc is
A thin metal disc of radius \(r\) floats on water surface and bends the surface downwards along the perimeter making an angle \(\theta\) with vertical edge of the disc. If the disc displaces a weight of water \(W\) and surface tension of water is \(T\), then the weight of metal disc is
Show Hint
Always analyze the geometry of the "bend" to determine the direction of the surface tension force. Since the disc bends the surface downwards, the elastic nature of the water surface tries to pull it upwards to restore flatness.
Updated On: May 13, 2026
- \(2\pi r T + W\)
- \(2\pi r \cos\theta - W\)
- \(2\pi r T \cos\theta + W\)
- \(W - 2\pi r T \cos\theta\)
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Concept:
For a body floating in equilibrium on a liquid surface, the net force acting on it must be zero. The downward force is the weight of the object. The upward forces opposing it consist of the buoyant force (upthrust) and the vertical component of the surface tension force.
Step 1: Identifying the forces in equilibrium.
Downward force:
• Weight of the metal disc \(= W_{\text{disc}}\) Upward forces:
• Buoyant force (weight of water displaced) \(= W\)
• Total upward force due to surface tension \(= F_T\) Equating downward and upward forces: \[ W_{\text{disc}} = W + F_T \]
Step 2: Calculating the upward force due to surface tension (\(F_T\)).
Surface tension \(T\) acts tangentially to the liquid surface. Since the surface bends downwards, the liquid pulls back upwards on the disc. The force acts along the entire perimeter of the disc (\(L = 2\pi r\)). The angle made with the vertical edge is \(\theta\). Thus, the vertical component of the surface tension force per unit length is \(T \cos\theta\). Total upward force from surface tension: \[ F_T = \text{Perimeter} \times \text{Vertical component of } T \] \[ F_T = (2\pi r)(T \cos\theta) \]
Step 3: Finding the total weight of the disc.
Substitute \(F_T\) into the equilibrium equation: \[ W_{\text{disc}} = W + 2\pi r T \cos\theta \]
Step 4: Selecting the correct option.
Rearranging the terms to match the options: \[ \boxed{2\pi r T \cos\theta + W} \] Therefore, the correct option is: \[ \boxed{\text{Option (C)}} \]
Step 1: Identifying the forces in equilibrium.
Downward force:
• Weight of the metal disc \(= W_{\text{disc}}\) Upward forces:
• Buoyant force (weight of water displaced) \(= W\)
• Total upward force due to surface tension \(= F_T\) Equating downward and upward forces: \[ W_{\text{disc}} = W + F_T \]
Step 2: Calculating the upward force due to surface tension (\(F_T\)).
Surface tension \(T\) acts tangentially to the liquid surface. Since the surface bends downwards, the liquid pulls back upwards on the disc. The force acts along the entire perimeter of the disc (\(L = 2\pi r\)). The angle made with the vertical edge is \(\theta\). Thus, the vertical component of the surface tension force per unit length is \(T \cos\theta\). Total upward force from surface tension: \[ F_T = \text{Perimeter} \times \text{Vertical component of } T \] \[ F_T = (2\pi r)(T \cos\theta) \]
Step 3: Finding the total weight of the disc.
Substitute \(F_T\) into the equilibrium equation: \[ W_{\text{disc}} = W + 2\pi r T \cos\theta \]
Step 4: Selecting the correct option.
Rearranging the terms to match the options: \[ \boxed{2\pi r T \cos\theta + W} \] Therefore, the correct option is: \[ \boxed{\text{Option (C)}} \]
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 Fluid Mechanics Questions
- In a water tank, an air bubble rises from the bottom to the top surface of the water. If the depth of the water in the tank is 7.28 m and atmospheric pressure is 10 m of water, then the ratio of the radii of the bubble at the bottom of the tank and at the top surface of the water is (Temperature of the water in the tank is constant)
- A small solid sphere of mass 10 g and density \( 2600 \, \text{kg/m}^3 \) is dropped into a long vertical column of glycerine. When the sphere attains terminal velocity, the magnitude of the viscous force acting on the sphere is
(Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \), Density of glycerine \( = 1300 \, \text{kg/m}^3 \))
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