Formula: The potential energy per unit volume of a taut wire may be given as follows: Energy per unit volume = ½ × Stress × Strain For a wire: Tension = Stress Stress = Y × Strain Substituting Stress to above equation: Potential Energy per unit volume = 1/2 x (Yx X) × X= 0.5Yx² Click for more: hRead more
Formula:
The potential energy per unit volume of a taut wire may be given as follows:
Energy per unit volume = ½ × Stress × Strain
For a wire:
Tension = Stress
Stress = Y × Strain
Substituting Stress to above equation:
Potential Energy per unit volume = 1/2 x (Yx X) × X= 0.5Yx²
The extension due to the weight of the rope can be found using the following formula: ΔL = (F L) / (A Y) Where: - F = Force due to weight = mg - L = Length of the rope = 8 m - A = Cross-sectional area of the rope - Y = Young's modulus - m = mass of the rope = density × volume - Volume = A × L AfterRead more
The extension due to the weight of the rope can be found using the following formula:
ΔL = (F L) / (A Y)
Where:
– F = Force due to weight = mg
– L = Length of the rope = 8 m
– A = Cross-sectional area of the rope
– Y = Young’s modulus
– m = mass of the rope = density × volume
– Volume = A × L
After computation, we get that the stretch is approximately:
The work done (W) in elongating a rod is given by the formula: W = 1/2 × Stress × Strain × Volume Where: - Stress = Force / Area - Strain = ΔL / L (elongation per unit length) As elongation, ΔL, is proportional to the applied force and Young's modulus, work done is proportional to the square of theRead more
The work done (W) in elongating a rod is given by the formula:
W = 1/2 × Stress × Strain × Volume
Where:
– Stress = Force / Area
– Strain = ΔL / L (elongation per unit length)
As elongation, ΔL, is proportional to the applied force and Young’s modulus, work done is proportional to the square of the elongation.
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
The breaking tension of a wire is proportional to its cross-sectional area. It can be determined using the formula: A = (π × d²) / 4 Step 1: Determine the areas First wire (diameter = 1 mm): A₁ = (π × 1²) / 4 = π / 4 Second wire (diameter = 2 mm): A₂ = (π × 2²) / 4 = 4π / 4 = π The area of the seconRead more
The breaking tension of a wire is proportional to its cross-sectional area. It can be determined using the formula:
A = (π × d²) / 4
Step 1: Determine the areas
First wire (diameter = 1 mm):
A₁ = (π × 1²) / 4 = π / 4
Second wire (diameter = 2 mm):
A₂ = (π × 2²) / 4 = 4π / 4 = π
The area of the second wire is **4 times** the area of the first wire.
Step 2: Calculate the breaking tension
Breaking tension is proportional to the area.
For the first wire:
T₁ = 1000 N
For the second wire:
T₂ = 4 × T₁ = 4 × 1000 = 4000 N
If in a wire of Young’s modulus Y, longitudinal strain X is produced, then the value of potential energy stored in its unit volume will be
Formula: The potential energy per unit volume of a taut wire may be given as follows: Energy per unit volume = ½ × Stress × Strain For a wire: Tension = Stress Stress = Y × Strain Substituting Stress to above equation: Potential Energy per unit volume = 1/2 x (Yx X) × X= 0.5Yx² Click for more: hRead more
Formula:
The potential energy per unit volume of a taut wire may be given as follows:
Energy per unit volume = ½ × Stress × Strain
For a wire:
Tension = Stress
Stress = Y × Strain
Substituting Stress to above equation:
Potential Energy per unit volume = 1/2 x (Yx X) × X= 0.5Yx²
Click for more:
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A thick copper rope of density 1.5 x 10³ kgm⁻³ and Young’s modulus 5 x 10⁶ Nm⁻², 8 m in length, its length due to its own weight is
The extension due to the weight of the rope can be found using the following formula: ΔL = (F L) / (A Y) Where: - F = Force due to weight = mg - L = Length of the rope = 8 m - A = Cross-sectional area of the rope - Y = Young's modulus - m = mass of the rope = density × volume - Volume = A × L AfterRead more
The extension due to the weight of the rope can be found using the following formula:
ΔL = (F L) / (A Y)
Where:
– F = Force due to weight = mg
– L = Length of the rope = 8 m
– A = Cross-sectional area of the rope
– Y = Young’s modulus
– m = mass of the rope = density × volume
– Volume = A × L
After computation, we get that the stretch is approximately:
ΔL = 9.6 x 10⁻⁵ m
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A metallic rod of length l and cross – sectional area A is made of a material of Young’s modulus Y. If the rod is elongated by an amount y, then the work done is proportional to
The work done (W) in elongating a rod is given by the formula: W = 1/2 × Stress × Strain × Volume Where: - Stress = Force / Area - Strain = ΔL / L (elongation per unit length) As elongation, ΔL, is proportional to the applied force and Young's modulus, work done is proportional to the square of theRead more
The work done (W) in elongating a rod is given by the formula:
W = 1/2 × Stress × Strain × Volume
Where:
– Stress = Force / Area
– Strain = ΔL / L (elongation per unit length)
As elongation, ΔL, is proportional to the applied force and Young’s modulus, work done is proportional to the square of the elongation.
Therefore, work done is proportional to y².
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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.
Click here for more:
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:
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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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A wire of diameter 1 mm breaks under a tension 0f 1000 N. Another wire, of same material as that of the first one, but of diameter 2mm breaks under a tension of
The breaking tension of a wire is proportional to its cross-sectional area. It can be determined using the formula: A = (π × d²) / 4 Step 1: Determine the areas First wire (diameter = 1 mm): A₁ = (π × 1²) / 4 = π / 4 Second wire (diameter = 2 mm): A₂ = (π × 2²) / 4 = 4π / 4 = π The area of the seconRead more
The breaking tension of a wire is proportional to its cross-sectional area. It can be determined using the formula:
A = (π × d²) / 4
Step 1: Determine the areas
First wire (diameter = 1 mm):
A₁ = (π × 1²) / 4 = π / 4
Second wire (diameter = 2 mm):
A₂ = (π × 2²) / 4 = 4π / 4 = π
The area of the second wire is **4 times** the area of the first wire.
Step 2: Calculate the breaking tension
Breaking tension is proportional to the area.
For the first wire:
T₁ = 1000 N
For the second wire:
T₂ = 4 × T₁ = 4 × 1000 = 4000 N
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