The spring balance A reads 2 kg with a block m suspended from it. A balance B reads 5 kg when a beak

Q: Why doesn't the reading of balance B remain 5 kg since the block is suspended from spring balance A?

Even though the block is suspended from balance A, it interacts with the liquid via the buoyant force. By Newton's Third Law, the upward buoyant force on the block has an equal and downward reaction force on the liquid, which presses down on balance B, increasing its reading.

Q: Does the density of the block affect the change in the readings?

The density of the block determines its volume for a given mass, which in turn affects the magnitude of the buoyant force $F_B$. However, as long as the block is denser than the liquid (so it sinks and can be submerged), $F_B$ will be positive, meaning balance A's reading will always decrease and balance B's reading will always increase.

Q: What would happen if the block touched the bottom of the beaker?

If the block touched the bottom, the normal force from the bottom of the beaker would support some of its weight. In this case, the tension in spring balance A would decrease even further (or go to zero), and balance B would read the combined weight of the beaker, liquid, and block ($7\text{ kg}$).

Q: Is the increase in balance B's reading exactly equal to the decrease in balance A's reading?

Yes! The decrease in balance A's reading is equal to the buoyant force $F_B$ ($T = mg - F_B$). The increase in balance B's reading is also equal to the reaction force $F_B'$ ($N = W_B + F_B$). Since $F_B = F_B'$, the loss in weight on balance A is exactly gained by balance B, conserving the total weight of the system.

Q: What if the liquid was highly viscous? Would the readings change over time?

No, because the system is in static equilibrium. Viscosity only affects fluid flow and dynamic forces (like drag) when there is relative motion. In static equilibrium, only hydrostatic pressure and buoyancy play a role, so viscosity has no effect on the final steady readings.

Question

The spring balance $A$ reads $2	ext{ kg}$ with a block $m$ suspended from it. A balance $B$ reads $5	ext{ kg}$ when a beaker with liquid is put on the pan of the balance. The two balances are now so arranged that the hanging mass is inside the liquid in the beaker as shown in the figure. In this situation

Options:

(A) the balance $A$ will read more than $2	ext{ kg}$

(B) the balance $B$ will read more than $5	ext{ kg}$

(C) the balance $A$ will read less than $2	ext{ kg}$ and $B$ will read more than $5	ext{ kg}$

(D) the balances $A$ and $B$ will read $2	ext{ kg}$ and $5	ext{ kg}$ respectively

Sigma Percentile · Accepted Answer

### Analyzing the Setup Imagine you are standing in a physics lab, looking at two separate, beautifully simple instruments: a **spring balance $A$** hanging from the ceiling, holding a block of mass $m$, and a **pan balance $B$** on the table, supporting a beaker filled with liquid. Initially, before any interaction takes place, the spring balance $A$ reads exactly the weight of the block: $$W_A = mg = 2 ext{ kg-wt}$$ Meanwhile, the pan balance $B$ reads the weight of the beaker and the liquid: $$W_B = 5 ext{ kg-wt}$$ Now, we perform a classic experiment: we lower the spring balance so that the block $m$ is completely submerged in the liquid, without touching the bottom or sides of the beaker. What happens to the readings of both balances? ### The Role of Archimedes' Principle As the block enters the liquid, it displaces a volume of fluid equal to its own submerged volume. According to **Archimedes' Principle**, the fluid fights back by exerting an upward buoyant force (upthrust) $F_B$ on the block: $$F_B = ho_{ ext{liquid}} V_{ ext{submerged}} g$$ This upward force acts as a helping hand, partially supporting the weight of the block. Let's look at the equilibrium of the block $m$. The forces acting on it are: 1. The downward force of gravity: $mg = 2 ext{ kg-wt}$ 2. The upward tension $T$ from the spring balance 3. The upward buoyant force $F_B$ from the liquid Since the block is in static equilibrium, these forces must balance perfectly: $$T + F_B = mg$$ Solving for the tension $T$, which is precisely what the spring balance $A$ measures, we get: $$T = mg - F_B$$ Since $F_B > 0$, it is mathematically clear that: $$T < mg \implies T < 2 ext{ kg-wt}$$ Thus, **the reading of spring balance $A$ decreases and becomes less than $2 ext{ kg}$**. ### Newton's Third Law and the Reaction Force Now, let's turn our attention to the beaker and the liquid resting on pan balance $B$. Many students make the mistake of thinking that since the block is suspended from above, its weight has no effect on the bottom balance. But we must remember **Newton's Third Law of Motion**: *action and reaction are equal and opposite*. If the liquid exerts an upward buoyant force $F_B$ on the block, the block must exert an equal and opposite downward reaction force $F_B'$ on the liquid: $$F_B' = F_B \quad ext{(downwards)}$$ This reaction force acts directly downwards on the liquid-beaker system. Therefore, the total downward force pressing on the pan balance $B$ is now: $$F_{ ext{total}} = W_{ ext{beaker}} + F_B' = 5 ext{ kg-wt} + F_B$$ Since $F_B > 0$, the normal force $N$ that the pan balance must exert to support the beaker is: $$N = 5 ext{ kg-wt} + F_B > 5 ext{ kg-wt}$$ Consequently, **the reading of pan balance $B$ increases and becomes more than $5 ext{ kg}$**. ### Final Conclusion By combining our two insights, we find that: - Spring balance $A$ reads **less than $2 ext{ kg}$**. - Pan balance $B$ reads **more than $5 ext{ kg}$**. This perfectly matches options **(b)** and **(c)** of the question. Therefore, the correct options are **(b)** and **(c)**.

Select Answer:

Visualized Solution

Understanding the Initial State

Archimedes' Principle & Upthrust

Free Body Diagram of Block $m$

Analyzing Spring Balance $A$ 's Reading

Newton's Third Law & Reaction Force

Forces on the Beaker System

Analyzing Pan Balance $B$ 's Reading

Selecting the Correct Options

The Way Forward (Alternative Scenarios)

The Sigma Insight: Buoyancy and Archimedes' Principle

Analyzing the Setup

The Role of Archimedes' Principle

Newton's Third Law and the Reaction Force

Final Conclusion

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