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
299.7k+ views
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).
Recently Updated Pages
Four persons A B C and D initially at the corners of class 11 physics JEE_Main

What is the difference between Conduction and conv class 11 physics JEE_Main

Moment of inertia of solid sphere about its diameter class 11 physics JEE_Main

If a piece of ice floating on the surface of water class 11 physics JEE_Main

At what temperature speed of sound in air will be doubled class 11 physics JEE_Main

A closed organ pipe and an open organ pipe are tuned class 11 physics JEE_Main

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Understanding the Electric Field of a Uniformly Charged Ring

Electron Gain Enthalpy and Electron Affinity Explained

Derivation of Equation of Trajectory Explained for Students

Understanding Atomic Structure for Beginners

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
CBSE Notes Class 11 Physics Chapter 1 - Units And Measurements - 2026-27 PDF Download (Login Required)

NCERT Solutions For Class 11 Physics Chapter 1 Units And Measurements - 2026-27 Free PDF Download (Login Required)

NCERT Solutions For Class 11 Physics Chapter 2 Motion In A Straight Line - 2026-27 Free PDF Download (Login Required)

Important Questions For Class 11 Physics Chapter 1 Units and Measurement - 2026-27 Free PDF Download (Sign-in Required)

CBSE Notes Class 11 Physics Chapter 2 - Motion in a Straight Line - 2026-27 PDF Download (Login Required)

NCERT Solutions For Class 11 Physics Chapter 3 Motion In A Plane - 2026-27 Free PDF Download (Sign-in Required)

