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State Hooke’s law. How can it be verified experimentally?
Hooke’s Law states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position, provided the elastic limit is not exceeded. Mathematically, it is expressed as: F = -kx where F is the restoring force exerted by the spring k is the spring constant andRead more
Hooke’s Law states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position, provided the elastic limit is not exceeded. Mathematically, it is expressed as:
F = -kx
where F is the restoring force exerted by the spring k is the spring constant and x is the displacement from the equilibrium position.
Experimental Verification:
To verify Hooke’s Law experimentally, the following steps can be followed:
1. Setup: Attach a spring vertically to a fixed support and hang weights from the free end of the spring. Use a ruler or measuring tape to measure the displacement.
2. Apply Weights: Gradually add known weights to the spring and measure the extension (displacement) of the spring each time a weight is added. Ensure that the weights are added incrementally and the spring is not overstretched.
3. Record Data: Record the weight applied (force) and the corresponding displacement of the spring.
4. Plotting Graph: Plot a graph of the applied force (y-axis) against the displacement (x-axis). According to Hooke’s Law, the graph should be a straight line passing through the origin, indicating that the force is directly proportional to the displacement.
5. Determine the Spring Constant: The slope of the linear graph gives the value of the spring constant (k), confirming the validity of Hooke’s Law.
By conducting this experiment, one can observe that the extension of the spring is proportional to the applied force, thereby verifying Hooke’s Law.
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A thin ring of radius 𝑅 and mass 𝑀 rotates about an axis passing through its diameter. What is its moment of inertia?
The moment of inertia of a thin ring about a diameter is 𝐼 = 1/2MR² , as derived using the perpendicular axis theorem. This question related to Chapter 6 physics Class 11th NCERT. From the Chapter 6. System of particle and Rotational motion. Give answer according to your understanding. For more pleaRead more
The moment of inertia of a thin ring about a diameter is 𝐼 = 1/2MR² , as derived using the perpendicular axis theorem.
This question related to Chapter 6 physics Class 11th NCERT. From the Chapter 6. System of particle and Rotational motion. Give answer according to your understanding.
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What is the angular velocity of a particle moving in a circular path of radius 𝑟 and speed 𝑣?
Angular velocity 𝜔 = v/r , where v is the tangential speed, and 𝑟 is the radius. This question related to Chapter 6 physics Class 11th NCERT. From the Chapter 6. System of particle and Rotational Motion. Give answer according to your understanding. For more please visit here: https://www.tiwariacadeRead more
Angular velocity 𝜔 = v/r , where v is the tangential speed, and 𝑟 is the radius. This question related to Chapter 6 physics Class 11th NCERT. From the Chapter 6. System of particle and Rotational Motion. Give answer according to your understanding.
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The center of mass of a two-particle system lies closer to:
The center of mass of a two-particle system lies closer to the particle with the greater mass. This is because the position of the center of mass (COM) is calculated as a weighted average of the positions of the two particles, with their respective masses serving as the weights. Mathematically, theRead more
The center of mass of a two-particle system lies closer to the particle with the greater mass. This is because the position of the center of mass (COM) is calculated as a weighted average of the positions of the two particles, with their respective masses serving as the weights.
Mathematically, the center of mass for a two-particle system can be expressed as:
\[ x_{COM} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \]
Here:
– \( m_1 \) and \( m_2 \) are the masses of the two particles.
– \( x_1 \) and \( x_2 \) are the positions of the two particles along a straight line.
From this formula, it is clear that:
– If \( m_1 > m_2 \), the center of mass will be closer to \( x_1 \).
– If \( m_2 > m_1 \), the center of mass will be closer to \( x_2 \).
This principle holds true regardless of the spatial dimension, with similar calculations extending to three-dimensional systems using the vector positions of the particles.
This question related to Chapter 6 physics Class 11th NCERT. From the Chapter 6. System of particle and Rotational Motion. Give answer according to your understanding.
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Which of the following is a condition for static equilibrium of a rigid body?
For static equilibrium, there should be no linear or rotational motion. Thus, both net force and net torque must be zero. This question related to Chapter 6 physics Class 11th NCERT. From the Chapter 6. System of Particle and Rotational Motion. Give answer according to your understanding. For more pRead more
For static equilibrium, there should be no linear or rotational motion. Thus, both net force and net torque must be zero. This question related to Chapter 6 physics Class 11th NCERT. From the Chapter 6. System of Particle and Rotational Motion. Give answer according to your understanding.
For more please visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-6/