A block of steel of size 5 text{ cm} times 5 | vedclass

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(A) The volume of the steel block is $V = 5 	ext{ cm} 	imes 5 	ext{ cm} 	imes 5 	ext{ cm} = 125 	ext{ cm}^3$.Given the relative density of steel

Vedclass · Accepted Answer

$6 	imes 5 	imes 5 	imes 5 	ext{ gf}$

Vedclass · Answer

(A) The volume of the steel block is $V = 5 	ext{ cm} 	imes 5 	ext{ cm} 	imes 5 	ext{ cm} = 125 	ext{ cm}^3$.Given the relative density of steel is $ho_r = 7$,the density of steel is $ho = 7 	ext{ g/cm}^3$.The density of water is $\sigma = 1 	ext{ g/cm}^3$.The apparent weight of an object submerged in a liquid is given by $W_{app} = V(ho - \sigma)g$.Since we are calculating the weight in $	ext{gf}$ (gram-force),we use the formula $W_{app} = V(ho - \sigma) 	ext{ gf/cm}^3$.Substituting the values: $W_{app} = (5 	imes 5 	imes 5) 	imes (7 - 1) 	ext{ gf}$.$W_{app} = 125 	imes 6 	ext{ gf} = 6 	imes 5 	imes 5 	imes 5 	ext{ gf}$.

$A$ block of steel of size $5 \text{ cm} \times 5 \text{ cm} \times 5 \text{ cm}$ is weighed in water. If the relative density of steel is $7$,its apparent weight is:

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