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  1. Yes, the law of conservation of energy is obeyed in the interference of light. While energy is redistributed between constructive and destructive interference regions, the total energy across the entire interference pattern remains constant. No energy is created or destroyed in the process. For moreRead more

    Yes, the law of conservation of energy is obeyed in the interference of light. While energy is redistributed between constructive and destructive interference regions, the total energy across the entire interference pattern remains constant. No energy is created or destroyed in the process.

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  2. In the case of collisions, momentum and kinetic energy behave differently. Momentum is a fundamental property of motion, which is always conserved in all types of collisions provided no external forces act on the system. This universal principle applies to both elastic and inelastic collisions. KineRead more

    In the case of collisions, momentum and kinetic energy behave differently. Momentum is a fundamental property of motion, which is always conserved in all types of collisions provided no external forces act on the system. This universal principle applies to both elastic and inelastic collisions.

    Kinetic energy, however, is not conserved. It is conserved only in elastic collisions, where there is no loss of energy to heat, sound, or deformation. In such cases, the total kinetic energy of the system before and after the collision remains the same. Elastic collisions typically occur at a microscopic level, such as between gas particles, where energy is perfectly transferred between colliding objects.

    In inelastic collisions, kinetic energy is not conserved. A part of it is converted into other forms of energy, such as heat, sound, or potential energy due to deformation of the colliding objects. Inelastic collisions are very common in everyday life, like a car crash, where deformation of the vehicles and heat generation result in loss of kinetic energy.

    Thus, momentum conservation is a universal law of all collisions, whereas the former depends on the nature of the collision and points out to be an important distinction between the two.

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  3. Angular momentum 𝐿 = 𝐼 𝜔, so it depends directly on the moment of inertia 𝐼 and angular velocity ω. This question related to Chapter 6 physics Class 11th NCERT. From the Chapter 6 System of Particles and Rotational Motion. Give answer according to your understanding. For more please visit here: httpRead more

    Angular momentum 𝐿 = 𝐼 𝜔, so it depends directly on the moment of inertia 𝐼 and angular velocity ω. This question related to Chapter 6 physics Class 11th NCERT. From the Chapter 6 System of Particles and Rotational Motion. Give answer according to your understanding.

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    https://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-6/

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  4. To solve the problem of calculating work done in stretching a spring, we must refer to the property of the spring as described in Hooke's Law. In this law, it is defined that the amount of force applied to stretch the spring is proportional to the stretching from its original length. The force increRead more

    To solve the problem of calculating work done in stretching a spring, we must refer to the property of the spring as described in Hooke’s Law. In this law, it is defined that the amount of force applied to stretch the spring is proportional to the stretching from its original length. The force increases linearly with the degree of stretching the spring. The work done in stretching the spring is equivalent to the energy stored in it, often referred to as elastic potential energy.

    The work done is visualized to be the area under a graph of force extension. Since this relationship between the force and extension is linear, the graph plots as a triangle. The extension of the spring is represented by the base of this triangle, while the height would represent the maximum force required. Therefore, work done is directly proportional to the square of extension.

    In this given case, the spring would need a force of 10 N for every millimeter of extension, and it is stretched 40 mm. When we use the formula for work done in a spring, substituting the given values enables us to calculate how much energy is in the spring due to this extension. When using the result of this computation, we obtain a total work done of 8 J, meaning it amounts to the amount of energy required to extend the spring by 40 mm.

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  5. Rotational kinetic energy is 𝐾𝐸 = 1/2𝐼𝜔². If 𝜔 is halved, 𝜔² becomes 1/4. Thus, the energy is reduced to one-fourth. This question related to Chapter 6 physics Class 11th NCERT. From the Chapter 6 System of Particles and Rotational Motion. Give answer according to your understanding. For more pleaseRead more

    Rotational kinetic energy is 𝐾𝐸 = 1/2𝐼𝜔². If 𝜔 is halved, 𝜔² becomes 1/4. Thus, the energy is reduced to one-fourth. This question related to Chapter 6 physics Class 11th NCERT. From the Chapter 6 System of Particles and Rotational Motion. Give answer according to your understanding.

    For more please visit here:
    https://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-6/

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