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Q.1 A solid block of volume V=10-3m3 and density d=800kg/m3 is tied to one end of a string, the other end of which is tied to the bottom of the vessel. The vessel contains 2 immiscible liquids of densities 1 =1000kg/m3 and 2 =1500kg/m3 . The solid block is immersed with 2/5th of its volume in the liquid of higher density & 3/5th in the liquid of lower density. The vessel is placed in an elevator which is moving up with an acceleration of a=g/2. Find the tension in the string. [g=10m/s2 ]

MANAV AGARWAL , 7 Years ago
Grade 11
anser 1 Answers
Askiitians Tutor Team

Last Activity: 13 Days ago

To determine the tension in the string when the solid block is submerged in two immiscible liquids while the elevator is accelerating upwards, we need to analyze the forces acting on the block. Let's break this down step by step.

Understanding the Forces at Play

When the elevator accelerates upwards, the effective gravitational acceleration acting on the block increases. The effective acceleration can be calculated as:

  • Effective acceleration (a') = g + a

Given that the acceleration of the elevator is a = g/2, we can substitute the values:

  • a' = g + g/2 = 10 m/s² + 5 m/s² = 15 m/s²

Calculating the Buoyant Forces

The block is partially submerged in two different liquids. We need to calculate the buoyant force exerted by each liquid separately. The buoyant force can be calculated using Archimedes' principle, which states that the buoyant force is equal to the weight of the fluid displaced by the submerged part of the object.

Let's denote the volume of the block as V = 10-3 m3 and its density as d = 800 kg/m3. The volume submerged in each liquid is:

  • Volume in liquid 1 (ρ1 = 1500 kg/m³): V1 = (2/5) * V = (2/5) * 10-3 = 4 * 10-4 m3
  • Volume in liquid 2 (ρ2 = 1000 kg/m³): V2 = (3/5) * V = (3/5) * 10-3 = 6 * 10-4 m3

Now, we can calculate the buoyant forces:

  • Buoyant force from liquid 1 (Fb1): Fb1 = ρ1 * V1 * a' = 1500 kg/m3 * 4 * 10-4 m3 * 15 m/s² = 9 kg·m/s² = 9 N
  • Buoyant force from liquid 2 (Fb2): Fb2 = ρ2 * V2 * a' = 1000 kg/m3 * 6 * 10-4 m3 * 15 m/s² = 9 kg·m/s² = 9 N

Calculating the Weight of the Block

The weight of the block (W) can be calculated using the formula:

  • Weight (W) = density * volume * effective acceleration = d * V * a' = 800 kg/m3 * 10-3 m3 * 15 m/s² = 12 N

Finding the Tension in the String

Now that we have the weight of the block and the buoyant forces, we can find the tension (T) in the string using the equilibrium of forces. The net force acting on the block can be expressed as:

  • T = W - (Fb1 + Fb2)

Substituting the values we calculated:

  • T = 12 N - (9 N + 9 N) = 12 N - 18 N = -6 N

Since tension cannot be negative, this indicates that the buoyant forces exceed the weight of the block, meaning the block would float if not restrained by the string. Therefore, the tension in the string must be zero when the block is fully submerged and floating.

Final Thoughts

In this scenario, the block is effectively buoyed up by the two liquids, and the tension in the string is zero because the buoyant forces acting on the block surpass its weight. This example illustrates the importance of understanding buoyancy and how it interacts with forces in a dynamic environment, such as an accelerating elevator.

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