## Question

### Solution

Correct option is

60 N

The extension produced in the string when the mass is suspended from it is Thus the equilibrium position of the mass corresponds to a position when the length of the spring = 50 + 5 = 55 cm. the lowermost position of the mass corresponds to a position when the length of the spring is 58 cm, i.e. when the spring is stretched by 58 – 50 = 8 cm. This position is (8 – 5) = 3 cm below the equilibrium position. At the equilibrium position, no net force acts on the body because the downward force mg is balanced by the restoring ky of the spring. Therefore, at the lowermost position, the force exerted by the spring on the mass is

F = force constant × extension below the equilibrium position #### SIMILAR QUESTIONS

Q1

A tuning fork produces 4 beta per second when sounded with sonometer of vibrating length 48cm. It produces 4 beats per second also when the vibrating length is 50cm. What is the frequency of the tuning fork?

Q2

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Q3

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Q4

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Q5

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Q6

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Q7

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Q8

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Q9

A spring of negligible mass having a force constant k extends by an amount y when a mass m is hung from it. The mass is pulled down a little and released. The system begins to execute simple harmonic motion of amplitude A and angular frequency ω. The total energy of the mass-spring system will be

Q10

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