
A wire is elongated by $2 \mathrm{mm}$ when a brick is suspended from it. When the brick is immersed in water, the wire contracts by $0.6 \mathrm{mm}$. What is the density of brick?
(A) $3333 \mathrm{kg} / \mathrm{m}^{3}$
(B) $4210 \mathrm{kg} / \mathrm{m}^{3}$
(C) $5000 \mathrm{kg} / \mathrm{m}^{3}$
(D) $2000 \mathrm{kg} / \mathrm{m}^{3}$
Answer
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Hint: We know that elasticity is an economic concept used to measure the change in the aggregate quantity demanded for a good or service in relation to price movements of that good or service. A product is considered to be elastic if the quantity demand of the product changes drastically when its price increases or decreases. It is also the ability of a deformed material body to return to its original shape and size when the forces causing the deformation are removed. A body with this ability is said to behave (or respond) elastically.
Complete step by step answer
$\text{Y=}\dfrac{\left( \dfrac{\text{F}}{\text{A}} \right)}{\left( \dfrac{\text{dL}}{\text{L}} \right)}$
Y=YOUNG'S MODULUS, F= FORCE, A= AREA, dL=CHANGE IN LENGTH, L= LENGTH
$\mathrm{F}=\mathrm{Y} \times \mathrm{dL} \times \mathrm{A} / \mathrm{L}$
In First Case $\mathrm{F}=\mathrm{Mg}$ or $\text{D}\times \text{V}\times \text{g}$
D = Density of Solid,
V = Volume
In Second Case,
$\mathrm{F}=\mathrm{Mg}$ - Upthrust $=\text{D}\times \text{V}\times \text{g}-\text{d}\times \text{V}\times \text{g}$
D is Density of Water
Put It in Formula and divide all terms will be cancelled left terms will be
$\mathrm{D} / \mathrm{D}-\mathrm{d}=2 / 1.4$
$\mathrm{d}=1000 \mathrm{kg} / \mathrm{m}^{3}$
hence,
$0.7 \mathrm{D}=\mathrm{D}-1000$
$0.3 \mathrm{D}=1000$
$\mathrm{D}=3333 \mathrm{Kg} / \mathrm{m}^{3}$
Therefore, the correct answer is Option A
Note: We can conclude that in the science of physics, elasticity is the ability of a deformable body (e.g., steel, aluminium, rubber, wood, crystals, etc.) to resist a distorting effect and to return to its original size and shape when that influence or force is removed. Solid bodies will deform when satisfying forces are applied to them. The three types of elastic constants are: Modulus of elasticity or Young's modulus (E), Bulk modulus (K) and. Modulus of rigidity or shear modulus (M, C or G).
Complete step by step answer
$\text{Y=}\dfrac{\left( \dfrac{\text{F}}{\text{A}} \right)}{\left( \dfrac{\text{dL}}{\text{L}} \right)}$
Y=YOUNG'S MODULUS, F= FORCE, A= AREA, dL=CHANGE IN LENGTH, L= LENGTH
$\mathrm{F}=\mathrm{Y} \times \mathrm{dL} \times \mathrm{A} / \mathrm{L}$
In First Case $\mathrm{F}=\mathrm{Mg}$ or $\text{D}\times \text{V}\times \text{g}$
D = Density of Solid,
V = Volume
In Second Case,
$\mathrm{F}=\mathrm{Mg}$ - Upthrust $=\text{D}\times \text{V}\times \text{g}-\text{d}\times \text{V}\times \text{g}$
D is Density of Water
Put It in Formula and divide all terms will be cancelled left terms will be
$\mathrm{D} / \mathrm{D}-\mathrm{d}=2 / 1.4$
$\mathrm{d}=1000 \mathrm{kg} / \mathrm{m}^{3}$
hence,
$0.7 \mathrm{D}=\mathrm{D}-1000$
$0.3 \mathrm{D}=1000$
$\mathrm{D}=3333 \mathrm{Kg} / \mathrm{m}^{3}$
Therefore, the correct answer is Option A
Note: We can conclude that in the science of physics, elasticity is the ability of a deformable body (e.g., steel, aluminium, rubber, wood, crystals, etc.) to resist a distorting effect and to return to its original size and shape when that influence or force is removed. Solid bodies will deform when satisfying forces are applied to them. The three types of elastic constants are: Modulus of elasticity or Young's modulus (E), Bulk modulus (K) and. Modulus of rigidity or shear modulus (M, C or G).
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