To solve this problem, we can use the principle of flotation, which states that when a body is floating in a fluid, the weight of the body is equal to the weight of the fluid displaced by the body.

Step 1: Identify the given values.
The density of the silver block (ρ1) = 10.5 g cm^-3
The density of the mercury (ρ2) = 13.6 g cm^-3

Step 2: Use the principle of flotation.
The fraction of the volume of the silver block (V1) that is submerged in the mercury is equal to the ratio of the density of the silver block to the density of the mercury.

V1/V = ρ1/ρ2

Step 3: Substitute the given values into the equation.
V1/V = 10.5 g cm^-3 / 13.6 g cm^-3

Step 4: Solve the equation.
V1/V = 0.77

So, approximately 77% of the silver block is submerged in the mercury.

Question

To solve this problem, we can use the principle of flotation, which states that when a body is floating in a fluid, the weight of the body is equal to the weight of the fluid displaced by the body.

Step 1: Identify the given values.
The density of the silver block (ρ1) = 10.5 g cm^-3
The density of the mercury (ρ2) = 13.6 g cm^-3

Step 2: Use the principle of flotation.
The fraction of the volume of the silver block (V1) that is submerged in the mercury is equal to the ratio of the density of the silver block to the density of the mercury.

V1/V = ρ1/ρ2

Step 3: Substitute the given values into the equation.
V1/V = 10.5 g cm^-3 / 13.6 g cm^-3

Step 4: Solve the equation.
V1/V = 0.77

So, approximately 77% of the silver block is submerged in the mercury.

Knowee AI · Accepted Answer

To solve this problem, we can use the principle of flotation, which states that when a body is floating in a fluid, the weight of the body is equal to the weight of the fluid displaced by the body.

Step 1: Identify the given values.
The density of the silver block (ρ1) = 10.5 g cm^-3
The density of the mercury (ρ2) = 13.6 g cm^-3

Step 2: Use the principle of flotation.
The fraction of the volume of the silver block (V1) that is submerged in the mercury is equal to the ratio of the density of the silver block to the density of the mercury.

V1/V = ρ1/ρ2

Step 3: Substitute the given values into the equation.
V1/V = 10.5 g cm^-3 / 13.6 g cm^-3

Step 4: Solve the equation.
V1/V = 0.77

So, approximately 77% of the silver block is submerged in the mercury.

The density of the silver block is 10.5 g cm ^- 3. and it is floating in mercury of density 13.6 g cm^- 3. What portion of the silver block is immersed in mercury?

Question

Solution

Similar Questions

Upgrade your grade with Knowee