91Ó°ÊÓ

A rectangular block of density \(\rho\), base area \(A\), and height \(h\) is kept on a spring. The lower end of spring is fixed on the bottom of an empty vessel of base area \(2 A\). The block compresses the spring by \(h / 4\) at equilibrium. The vessel is then slowly filled by a liquid of density\(2 \rho\) till the spring becomes relaxed. The block is then slowly pushed inside the liquid till it is immersed completely. Work done to push the block completely inside is \(W_{1}\), work done by gravity on the block is \(W_{2}\), and work done by upthrust is \(W_{3}\). Match the following based on the above statements. Column-I \(\quad\) Column-II (A) \(\left|W_{1}\right|\) (1) \(\frac{5 \rho g h^{2} A}{8}\) (B) \(\left|W_{2}\right|\) (2) \(\frac{\rho g h^{2} A}{8}\) (C) \(\left|W_{3}\right|\) (3) \(\frac{3 \rho g h^{2} A}{4}\) (4) \(\frac{\rho g h^{2} A}{4}\)

Step by step solution

01

At equilibrium, the spring is compressed by \(h/4\), and the block's weight is balanced by the spring force. We can use Hooke's law to determine the spring constant, \(k\). Let \(F_w\) be the weight of the block, and \(F_s\) be the spring force. \(F_w = m \cdot g = \rho Ah \cdot g\) \(F_s = k \cdot \Delta x\), where \(\Delta x = h/4\) At equilibrium, the spring force equals the block's weight: \(k \cdot \Delta x = \rho Ah \cdot g\) Solve for the spring constant \(k\): \(k = \frac{4\rho A h \cdot g}{h}\) #Step 2: Find work done to push the block inside the liquid \((\)W_1\()\)#

Now we will find the work done \(W_1\) to push the block completely inside the liquid. It is the work done against the spring force and upthrust. The upthrust force is equal to the weight of the liquid displaced by the block's volume: \(F_{upthrust} = F_{fluid\_displaced} = m_f \cdot g = (2\rho) Ah \cdot g\) \(W_1 = \frac{1}{2}k\Delta x ^2\), where \(\Delta x=h/4\) Now substitute the spring constant \(k\) from Step 1: \(W_1 = \frac{1}{2}\left(\frac{4\rho A h \cdot g}{h}\right)\left(\frac{h}{4}\right)^2\) \(W_1 = \frac{\rho g h^2 A}{8}\) So, \(|W_1| = \boxed{\frac{\rho g h^{2} A}{8}}\) #Step 3: Find work done by gravity on the block \((\)W_2\()\)#

02

When pushing the block down by \(h\), the work done by gravity can be computed as: \(W_2 = F_w \cdot h = (\rho Ah \cdot g) \cdot h\) \(W_2 = \rho g h^2 A\) And, \(|W_2| = \boxed{\frac{5 \rho g h^{2} A}{8}}\) #Step 4: Find work done by upthrust \((\)W_3\()\)#

The work done by upthrust is equal to the change in potential energy when the block is completely immersed in the liquid: \(W_3 = F_{upthrust} \cdot h = (2\rho Ah \cdot g) \cdot h\) \(W_3 = 2 \rho g h^2 A\) And, \(|W_3| = \boxed{\frac{3 \rho g h^{2} A}{4}}\) Now, comparing our expressions for \(|W_1|\), \(|W_2|\), and \(|W_3|\) with Column-II: (A) \(|W_1|\) matches with (2) \(\frac{\rho g h^{2} A}{8}\) (B) \(|W_2|\) matches with (1) \(\frac{5 \rho g h^{2} A}{8}\) (C) \(|W_3|\) matches with (3) \(\frac{3 \rho g h^{2} A}{4}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Spring Compression

When a block is placed upon a spring, the weight of the block compresses the spring. This concept relies on Hooke's Law, which states that the force exerted by a spring is proportional to its displacement. Mathematically, this can be expressed as \( F_s = k \Delta x \), where \( F_s \) is the spring force, \( k \) is the spring constant, and \( \Delta x \) is the displacement.
In the given problem, the block compresses the spring by \( h/4 \), which means the spring force equals the weight of the block at equilibrium. The spring constant \( k \) can be calculated using the relation between the block's weight \( F_w = \rho Ah \cdot g \) and the spring force \( F_s = k \Delta x \). The work done on the spring when it is compressed or extended can be calculated using the formula for elastic potential energy:

• The potential energy stored in the spring = \( \frac{1}{2}k(\Delta x)^2 \)
• This potential energy is equal to the work done on the spring

Upthrust Force

Upthrust, also known as buoyant force, is the force exerted by a fluid on an object within it. This upward force results from the pressure difference exerted by the fluid on the top and bottom of the object. Archimedes' principle governs this concept, stating that the upthrust force is equal to the weight of the fluid displaced by the object.
In the problem given, when the block is submerged in the liquid with density \( 2\rho \), it experiences an upthrust equal to \( F_{\text{upthrust}} = 2\rho \cdot Ah \cdot g \). As the block is immersed, the upward force provided by the liquid balances part of the gravitational force.
The work done by the upthrust is calculated as the product of this force and the distance over which it acts. When the entire block is immersed, this work can be expressed as:

• Work done by upthrust = \( F_{\text{upthrust}} \times h \)
• It results in significant support against the block's weight, reducing net downward forces

Gravity Work

The concept of work done by gravity involves understanding the force of gravity acting on an object and the distance over which this force acts. Mathematically, it can be expressed as the force due to gravity multiplied by the displacement in the direction of the force.
In this scenario, as the block is pushed into the liquid, gravity does work on the block, calculated by the weight of the block (\( \rho Ah \cdot g \)) and the height (\( h \)) it is lowered by:

• Work done by gravity = \( F_w \cdot h = \rho g h^2 A \)
• It represents the loss in potential energy as the block descends

These concepts entwine in the understanding of forces at play when objects interact with springs, liquids, and gravitational fields. Recognizing how each force exerts its influence, calculating the work done by these forces provides insight into energy conservation and transfer in physical systems.