The spring balance $A$ reads $2\,\text{kg}$ with a block $m$ suspended from it. $A$ balance $B$ reads $5\,\text{kg}$ 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:

Step 1: Analyze the initial state in air.

When the block is suspended in air, balance $A$ measures the tension in the string, which equals the weight of the block.

$$W = 2\,\text{kgf}$$

Balance $B$ measures the weight of the beaker and the liquid.

$$W_{\text{liq}} = 5\,\text{kgf}$$

Step 2: Analyze the forces on the immersed block.

When the block is lowered into the liquid, the liquid exerts an upward buoyant force $F_B$ on the block.

$$T + F_B = W$$ $$T = W - F_B$$ $$T = 2 - F_B < 2\,\text{kgf}$$

Since balance $A$ reads the tension $T$, its reading becomes less than $2\,\text{kg}$.

Step 3: Analyze the forces on the beaker and liquid.

By Newton's third law, the block exerts an equal and opposite downward force $F_B$ on the liquid.

$$N = W_{\text{liq}} + F_B$$ $$N = 5 + F_B > 5\,\text{kgf}$$

Since balance $B$ reads the normal force $N$, its reading becomes greater than $5\,\text{kg}$.

Step 4: Verify using the total system equilibrium.

For the combined system of the block, beaker, and liquid, the total upward force must balance the total downward weight.

$$T + N = W + W_{\text{liq}}$$ $$T + N = 2 + 5 = 7\,\text{kgf}$$

The sum of the readings of $A$ and $B$ remains constant at $7\,\text{kg}$ regardless of the immersion.

Answer: The reading of balance $A$ is less than $2\,\text{kg}$ and the reading of balance $B$ is greater than $5\,\text{kg}$.