A load of mass M kg is suspended from a steel wire of length 2 m and radius 1.0 mm in Searle's appar

Q: Why does the extension decrease when immersed in liquid?

Because the liquid exerts an upward buoyant force on the load, reducing the net downward force (apparent weight) acting on the wire. Less force means less extension.

Q: What is relative density and how is it used here?

Relative density is the ratio of the density of a substance to the density of water. The ratio of the relative densities of the liquid and the load gives the fraction of the weight that is supported by buoyancy.

Q: Does the Young's modulus of the wire change when immersed?

No, Young's modulus is a material property of the steel wire and remains constant regardless of the surrounding medium.

Q: Why didn't we need the length and radius of the wire for the final calculation?

Because we are comparing two states of the same wire. The length, radius, and Young's modulus remain constant and cancel out when we take the ratio of the new extension to the old extension.

Q: What if the liquid was denser than the material of the load?

If the liquid were denser, the buoyant force would be greater than the weight of the load. The load would float, the tension in the wire would become zero, and the extension would be zero.

Question

A load of mass $M 	ext{ kg}$ is suspended from a steel wire of length $2 	ext{ m}$ and radius $1.0 	ext{ mm}$ in Searle's apparatus experiment. The increase in length produced in the wire is $4.0 	ext{ mm}$. Now, the load is fully immersed in a liquid of relative density $2$. The relative density of the material of load is $8$. The new value of increase in length of the steel wire is

Options:

(A) zero

(B) $5.0 	ext{ mm}$

(C) $4.0 	ext{ mm}$

(D) $3.0 	ext{ mm}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are in a physics lab, looking at Searle's apparatus. A heavy mass $M$ is suspended from a sturdy steel wire. Gravity is pulling it down with a force of $Mg$.

This downward pull creates tension in the wire, causing it to stretch. According to **Hooke's Law** and the definition of Young's modulus, the extension $\Delta l$ is directly proportional to the applied force.

Initially, this force $Mg$ causes an extension of $4.0$ mm.

### The Plot Twist: Enter the Liquid

Now, we introduce a twist. We take a beaker of liquid and completely submerge the hanging mass into it.

The moment the mass enters the liquid, **Archimedes' principle** comes into play. The liquid exerts an upward buoyant force, $F_B$, on the mass.

This buoyant force acts like an invisible hand, pushing the mass up and effectively reducing the downward pull on the wire.

### The Master Equation

The new effective force stretching the wire, often called the apparent weight $W'$, is the true weight minus the buoyant force:

$$W' = Mg - F_B$$

But how big is this buoyant force? We know that $F_B$ equals the weight of the displaced liquid, which is the volume of the mass $V$ times the density of the liquid $ho_l$ times $g$.

$$F_B = V ho_l g$$

Since the volume of the mass is its mass divided by its density ($V = \frac{M}{ho_b}$), we can rewrite the buoyant force as:

$$F_B = \left(\frac{M}{ho_b}ight) ho_l g = Mg \left(\frac{ho_l}{ho_b}ight)$$

### Final Calculation

This is where the magic happens. We are given the relative densities! The liquid has a relative density of $2$, and the block has a relative density of $8$.

Their ratio is simply $\frac{2}{8} = \frac{1}{4}$. Substituting this back into our apparent weight equation:

$$W' = Mg - Mg\left(\frac{1}{4}ight) = \frac{3}{4}Mg$$

The effective force is now exactly three-fourths of the original force. Because the extension is directly proportional to the force, the new extension $\Delta l_2$ must also be three-fourths of the original extension $\Delta l_1$.

$$\Delta l_2 = \frac{3}{4} \Delta l_1 = \frac{3}{4} 	imes 4.0 	ext{ mm} = 3.0 	ext{ mm}$$

And there we have it! The liquid's buoyant support reduces the stretch in the wire to **$3.0$ mm**. A beautiful interplay of elasticity and fluid mechanics.

Select Answer:

Visualized Solution

Initial Extension $Δ l_{1}$

Immersing in Liquid

Buoyant Force $F_{B}$

Apparent Weight $W^{'}$

Density Ratio

New Effective Force

New Extension $Δ l_{2}$

Final Calculation

Conclusion

The Sigma Insight: Young's Modulus, Bulk Modulus and Modulus of Rigidity

Analyzing the Setup

The Plot Twist

The Master Equation

Final Calculation

Similar Questions

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The value of tension in a long thin metal wire has been changed from $T_{1}$ to $T_{2}$ . The lengths of the metal wire at two different values of tension $T_{1}$ and $T_{2}$ are $l_{1}$ and $l_{2}$ , respectively. The actual length of the metal wire is

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Select Answer:

Visualized Solution

Initial Extension .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δl1​

Immersing in Liquid

Buoyant Force .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } FB​

Apparent Weight .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } W′

Density Ratio

New Effective Force

New Extension .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δl2​

Final Calculation

Conclusion

The Sigma Insight: Young's Modulus, Bulk Modulus and Modulus of Rigidity

Analyzing the Setup

The Plot Twist

The Master Equation

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The length of metallic wire is l1​ when tension in it is T1​. It is l2​ when the tension is T2​. The original length of the wire will be

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Buoyant Force $F_{B}$

Apparent Weight $W^{'}$

New Extension $Δ l_{2}$

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