When a body of mass M is hung from a spring, the spring extends by 1 cm. If the body of mass 2 M be hung from the same spring, the extension of spring will be
A spring is a mechanical device that stores and releases energy. It follows Hooke’s law, where the force needed to compress or stretch it is proportional to the displacement. Springs are commonly used in various applications such as shock absorbers, watches, and machines for their elastic properties.
Chapter 8 of Class 11 Physics focuses on the mechanical properties of solids. It explains concepts like stress strain Hooke’s law Young’s modulus bulk modulus and shear modulus. The chapter also covers elastic potential energy Poisson’s ratio and applications of elasticity in real-world situations. Understanding these properties helps in analyzing the behavior of materials under various forces.
Share
According to Hooke’s Law, an applied force directly stretches a spring and is defined by the formula given below.
F = kx
Here:
– F refers to the force applied
– k refers to the spring constant
– x refers to the extension of the spring.
When the mass M is hung on the spring, the force applied is F1 = Mg where g is acceleration due to gravity and the extension is x1 = 1 cm.
Now, if the mass is doubled to 2M then the force applied will be F2 = 2Mg, and the extension will be x2.
By using the proportionality between force and extension:
x2 / x1 = F2 / F1
x2 / 1 cm = 2Mg / Mg = 2
Therefore, the extension x2 = 2 cm.
The correct answer is: 2 cm
Check this for more info:
https://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-10/