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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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See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-8/
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/
Why is the classical (Rutherford) model for an atom, of electron orbiting around the nucleus, not able to explain the atomic structure?
The classical Rutherford model fails because orbiting electrons, according to electromagnetic theory, should continuously emit radiation, lose energy, and spiral into the nucleus. This instability cannot explain atomic stability or discrete spectral lines observed in atomic emission, contradicting eRead more
The classical Rutherford model fails because orbiting electrons, according to electromagnetic theory, should continuously emit radiation, lose energy, and spiral into the nucleus. This instability cannot explain atomic stability or discrete spectral lines observed in atomic emission, contradicting experimental results.
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See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-12/
Explain the purpose and benefits of sectional breathing.
Sectional breathing is a Prānāyāma technique that consciously directs breath to specific lung sections: abdominal, thoracic, and clavicular. This practice enhances respiratory awareness, improves lung efficiency, and ensures optimal oxygenation of the blood. By encouraging deep, mindful breathing, iRead more
Sectional breathing is a Prānāyāma technique that consciously directs breath to specific lung sections: abdominal, thoracic, and clavicular. This practice enhances respiratory awareness, improves lung efficiency, and ensures optimal oxygenation of the blood. By encouraging deep, mindful breathing, it reduces anxiety, calms the mind, and supports overall lung health. Regular practice fosters better control over breathing patterns, contributing to emotional balance and improved physical vitality.
See lessWhich 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).