Question

A cube having mass 5kg is placed on a weighing scale inside water.Find the reading of weighing scale​

Answer 1

$\huge{\underline{\sf{\red{Answer\leadsto \dfrac{5}{6} kg}}}}$

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Given:

Mass of Block = 5kg

$\rho_s = Specific Gravity = 1.2$

Density of Water = $\rho_{water} = 10^3$

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Formula

• $\rho = \dfrac{Mass\:of\:solid }{Volume\:of\:Liquid }$
• Reading =$\dfrac{Normal\:Reaction}{g}$
• Buoyant Force = $V\rho_{water}g$

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Solution:

$\large\star$ A Buoyant Force and a Normal Force is acting in upwards Direction on the block from below

$\large\star$ Gravity=g acts on the block from above i.e mg

Making an FBD from the given Information_____In attachment

By this We get an Equation

$F_b + N=mg \\ \\ V\rho_{water}g + Normal\:Reaction=mg$

Diving the Equation by g

$\dfrac{V\rho_{water}\cancel{g}}{\cancel{g}} + \dfrac{N}{g} = \dfrac{m\cancel{g}}{\cancel{g}} \\ \\ V\rho_{water} + \dfrac{N}{g} = m \\ \\ \dfrac{N}{g} = m-V\rho_{water}\large{\star}$

From Formula above

$\rho = \dfrac{m}{V} \\ \\ V = \dfrac{m}{\rho} = \dfrac{5}{1.2 \times 10^3}$

$Reading \leadsto \dfrac{N}{g} = 5 - \dfrac{5}{1.2 \times \cancel{10^3}} \times\cancel{ 10^3} \\ \\ \implies 5\bigg(\dfrac{1.2-1}{1.2}\bigg) \\ \\ \implies 5\bigg(\dfrac{0.2}{1.2} \bigg) \\ \\ \implies 5\bigg(\dfrac{1}{6}\bigg) \\ \\ \huge{\leadsto}{\boxed{\dfrac{5}{6} kg}}$

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Answer 2

$\huge\bold\star\red{Answer}\star$

$\huge{\underline{\sf{\purple{Answer\leadsto \dfrac{5}{6} kg}}}}$

____________________________________

Given:

Mass of Block = 5kg

$\rho_s = Specific Gravity = 1.2$

Density of Water = $\rho_{water} = 10^3$

____________________________________

Formula

$\rho = \dfrac{Mass\:of\:solid }{Volume\:of\:Liquid }$

Reading =$\dfrac{Normal\:Reaction}{g}$

Buoyant Force = $V\rho_{water}g$

___________________________________

Solution:

$\large\star$ A Buoyant Force and a Normal Force is acting in upwards Direction on the block from below

$\large\star$ Gravity=g acts on the block from above i.e mg

Making an FBD from the given Information_____In attachment

By this We get an Equation

$F_b + N=mg \\ \\ V\rho_{water}g + Normal\:Reaction=mg$

Diving the Equation by g

$\dfrac{V\rho_{water}\cancel{g}}{\cancel{g}} + \dfrac{N}{g} = \dfrac{m\cancel{g}}{\cancel{g}} \\ \\ V\rho_{water} + \dfrac{N}{g} = m \\ \\ \dfrac{N}{g} = m-V\rho_{water}\large{\star}$

From Formula above

$\rho = \dfrac{m}{V} \\ \\ V = \dfrac{m}{\rho} = \dfrac{5}{1.2 \times 10^3}$

$Reading \leadsto \dfrac{N}{g} = 5 - \dfrac{5}{1.2 \times \cancel{10^3}} \times\cancel{ 10^3} \\ \\ \implies 5\bigg(\dfrac{1.2-1}{1.2}\bigg) \\ \\ \implies 5\bigg(\dfrac{0.2}{1.2} \bigg) \\ \\ \implies 5\bigg(\dfrac{1}{6}\bigg) \\ \\ \huge{\leadsto}{\boxed{\dfrac{5}{6} kg}}$

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