Two springs have their force constants k 1 and k 2 . Both are stretched till their…
Physics · JEE Main · NTA Exams — Work, Energy and Power
Two springs have their force constants k1 and k2. Both are stretched till their elastic energies are equal. Then, ratio of stretching forces F1/F2 is equal to:
- k1: k2
- k2 : k1
- \(\sqrt{k_{1}}: \sqrt{k_{2}}\)
- \(k_{2}^{2}: k_{1}^{2}\)
Answer
(C) k_ 1 : k_ 2
Sign up free on Edukali to view the step-by-step worked solution and practice thousands of similar questions.
Related practice questions
- A body is moving under the action of a force. Suddenly, force is increased to such an extent that its kinetic…
- A particle of mass 0.l kg is subjected to a force which varies with distance as shown in figure. If it starts…
- A long spring, when stretched by a distance x, has potential energy U. On increasing the stretching to nx…
- An object of mass m is allowed to fall from rest along a rough inclined plane. The speed of the object on…
- A machine gun fires 360 bullets per minute, with a velocity of 600 m/s. If the power of the gun is 5.4 kW…
- The relationship between force and position is shown in figure (in one dimensional case). The work done by…
- A spring of spring constant 8 N/cm has an extension of 5 cm. The minimum work done in joule in increasing the…
- Atomic power station at Tarapur has a generating capacity of 200 MW. The energy generated in a day at this…
More Work, Energy and Power questions · Browse all practice questions