When a rubber is stretched in length, it experiences a deformation in which the material attempts to preserve its volume. Since the length increases, the cross-sectional area reduces, bringing down the thickness. In most cases, width can increase as rubber stretches in length to preserve the overallRead more
When a rubber is stretched in length, it experiences a deformation in which the material attempts to preserve its volume. Since the length increases, the cross-sectional area reduces, bringing down the thickness. In most cases, width can increase as rubber stretches in length to preserve the overall volume.
Young's modulus is defined as stress divided by strain, where: Stress = Force / Area (unit: Nm⁻² or Pascal) Strain is dimensionless, so Young's modulus has the same unit as stress, i.e., Nm⁻², Pascal (Pa), or derived units like megapascals (MPa). Now, let's analyze the options: - Nm⁻¹: This represenRead more
Young’s modulus is defined as stress divided by strain, where:
Stress = Force / Area (unit: Nm⁻² or Pascal)
Strain is dimensionless, so Young’s modulus has the same unit as stress, i.e., Nm⁻², Pascal (Pa), or derived units like megapascals (MPa).
Now, let’s analyze the options:
– Nm⁻¹: This represents force per unit length, not a valid unit for Young’s modulus.
– Nm⁻²: Correct unit of Young’s modulus.
– dyne cm⁻²: A valid unit (CGS system).
– mega pascal: A legitimate unit (1 MPa = 10⁶ Pa).
Liquid crystals are substances that exhibit properties intermediate between those of crystalline solids and amorphous liquids. They possess an ordered structure like solids but also flow like liquids.
Liquid crystals are substances that exhibit properties intermediate between those of crystalline solids and amorphous liquids. They possess an ordered structure like solids but also flow like liquids.
In solids, the interatomic forces are **both attractive and repulsive**. The atoms or molecules in a solid are held together by attractive forces such as covalent, ionic, or metallic bonds; however, as atoms approach closer, repulsive forces begin acting due to the electron cloud interactions. ClickRead more
In solids, the interatomic forces are **both attractive and repulsive**. The atoms or molecules in a solid are held together by attractive forces such as covalent, ionic, or metallic bonds; however, as atoms approach closer, repulsive forces begin acting due to the electron cloud interactions.
In formula format: ΔL = (F × L) / (A × Y) Where: - F= applied load which is the same for both the wires - L= length of wire which is the same for both - A= cross sectional area of the wire A = πd² / 4 - Y = Young's modulus (same material, so constant for both wires) The extension is **inversely propRead more
In formula format:
ΔL = (F × L) / (A × Y)
Where:
– F= applied load which is the same for both the wires
– L= length of wire which is the same for both
– A= cross sectional area of the wire A = πd² / 4
– Y = Young’s modulus (same material, so constant for both wires)
The extension is **inversely proportional to the cross-sectional area (A)
Step 1: Calculate the area ratio
For the first wire (diameter = d):
A₁ = (π × d²) / 4
For the second wire (diameter = 2d):
A₂ = (π × (2d)²) / 4 = (π × 4d²) / 4 = πd²
Area ratio is:
A₁ : A₂ = (πd² / 4) : πd² = 1 : 4
Step 2: Ratio of extensions
Since extension (ΔL) is inversely proportional to the area:
ΔL₁ : ΔL₂ = A₂ : A₁ = 4 : 1
A long piece of rubber is wider than it is thick. When it is stretched in length by some amount
When a rubber is stretched in length, it experiences a deformation in which the material attempts to preserve its volume. Since the length increases, the cross-sectional area reduces, bringing down the thickness. In most cases, width can increase as rubber stretches in length to preserve the overallRead more
When a rubber is stretched in length, it experiences a deformation in which the material attempts to preserve its volume. Since the length increases, the cross-sectional area reduces, bringing down the thickness. In most cases, width can increase as rubber stretches in length to preserve the overall volume.
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See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-8/
Which one of the following is not a unit of Young’s modulus?
Young's modulus is defined as stress divided by strain, where: Stress = Force / Area (unit: Nm⁻² or Pascal) Strain is dimensionless, so Young's modulus has the same unit as stress, i.e., Nm⁻², Pascal (Pa), or derived units like megapascals (MPa). Now, let's analyze the options: - Nm⁻¹: This represenRead more
Young’s modulus is defined as stress divided by strain, where:
Stress = Force / Area (unit: Nm⁻² or Pascal)
Strain is dimensionless, so Young’s modulus has the same unit as stress, i.e., Nm⁻², Pascal (Pa), or derived units like megapascals (MPa).
Now, let’s analyze the options:
See less– Nm⁻¹: This represents force per unit length, not a valid unit for Young’s modulus.
– Nm⁻²: Correct unit of Young’s modulus.
– dyne cm⁻²: A valid unit (CGS system).
– mega pascal: A legitimate unit (1 MPa = 10⁶ Pa).
The term liquid crystal refers to a state that is intermediate between
Liquid crystals are substances that exhibit properties intermediate between those of crystalline solids and amorphous liquids. They possess an ordered structure like solids but also flow like liquids.
Liquid crystals are substances that exhibit properties intermediate between those of crystalline solids and amorphous liquids. They possess an ordered structure like solids but also flow like liquids.
See lessIn solids interatomic forces are
In solids, the interatomic forces are **both attractive and repulsive**. The atoms or molecules in a solid are held together by attractive forces such as covalent, ionic, or metallic bonds; however, as atoms approach closer, repulsive forces begin acting due to the electron cloud interactions. ClickRead more
In solids, the interatomic forces are **both attractive and repulsive**. The atoms or molecules in a solid are held together by attractive forces such as covalent, ionic, or metallic bonds; however, as atoms approach closer, repulsive forces begin acting due to the electron cloud interactions.
Click for more:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-8/
There are two wires of same material and same length while the diameter of second wire is two times the diameter of first wire, then the ratio of extension produced in the wire by applying same load will be
In formula format: ΔL = (F × L) / (A × Y) Where: - F= applied load which is the same for both the wires - L= length of wire which is the same for both - A= cross sectional area of the wire A = πd² / 4 - Y = Young's modulus (same material, so constant for both wires) The extension is **inversely propRead more
In formula format:
ΔL = (F × L) / (A × Y)
Where:
– F= applied load which is the same for both the wires
– L= length of wire which is the same for both
– A= cross sectional area of the wire A = πd² / 4
– Y = Young’s modulus (same material, so constant for both wires)
The extension is **inversely proportional to the cross-sectional area (A)
Step 1: Calculate the area ratio
For the first wire (diameter = d):
A₁ = (π × d²) / 4
For the second wire (diameter = 2d):
A₂ = (π × (2d)²) / 4 = (π × 4d²) / 4 = πd²
Area ratio is:
A₁ : A₂ = (πd² / 4) : πd² = 1 : 4
Step 2: Ratio of extensions
Since extension (ΔL) is inversely proportional to the area:
ΔL₁ : ΔL₂ = A₂ : A₁ = 4 : 1
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See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-8/