A block of steel of size 5 cm × 5 cm × 5 cm is weighed in water. If the relative density of steel is 7, its apparent weight is

To find the apparent weight of the steel block when weighed in water, we can use the principles of buoyancy and relative density.

First, let's identify the given values:
- The dimensions of the steel block are $5\text{cm}\times 5\text{cm}\times 5\text{cm}$.
- Therefore, the volume of the block ($V$) is:
$V=5\times 5\times 5{\text{cm}}^3$.
- The relative density ($\text{RD}$) of steel is $7$.

The relative density of a substance is the ratio of its density to the density of water. In the CGS system, the density of water is $1{\text{g/cm}}^3$. Thus, the density of steel ($d$) is:
$d=7{\text{g/cm}}^3$.

The actual weight of the steel block in air ($W_{\text{air}}$) is given by:
$W_{\text{air}}=\text{Volume}\times \text{Density}=V\times d\text{gf}$
$W_{\text{air}}=(5\times 5\times 5)\times 7\text{gf}$.

When the block is completely immersed in water, it experiences an upward buoyant force (or upthrust, $U$) equal to the weight of the water displaced by it. Since the density of water is $1{\text{g/cm}}^3$, the upthrust is:
$U=\text{Volume of block}\times \text{Density of water}=V\times 1\text{gf}$
$U=(5\times 5\times 5)\times 1\text{gf}$.

The apparent weight of the block in water ($W_{\text{apparent}}$) is the actual weight in air minus the buoyant force:
$W_{\text{apparent}}=W_{\text{air}}-U$
$W_{\text{apparent}}=\left[\right(5\times 5\times 5)\times 7]-\left[\right(5\times 5\times 5)\times 1]\text{gf}$.

Factoring out the volume term $(5\times 5\times 5)$:
$W_{\text{apparent}}=(7-1)\times 5\times 5\times 5\text{gf}$
$W_{\text{apparent}}=6\times 5\times 5\times 5\text{gf}$.

Therefore, the apparent weight of the steel block in water is $6\times 5\times 5\times 5\text{gf}$.