- 1(current)
Q. No.The amount of elastic potential energy per unit volume (in SI unit) of a steel wire of length \(100~\text{cm}\) to stretch it by \(1~\text{mm}\) is:
(given: Young's modulus of the wire = \(Y=2.0\times 10^{11}~\text{N/m}^2\) )
1. \(10^{11}~\text{J/m}^3\)
2. \(10^{17}~\text{J/m}^3\)
3. \(10^{7}~\text{J/m}^3\)
4. \(10^{5}~\text{J/m}^3\)
(given: Young's modulus of the wire = \(Y=2.0\times 10^{11}~\text{N/m}^2\) )
1. \(10^{11}~\text{J/m}^3\)
2. \(10^{17}~\text{J/m}^3\)
3. \(10^{7}~\text{J/m}^3\)
4. \(10^{5}~\text{J/m}^3\)
Subtopic: Â Potential energy of wire |
 66%
Level 2: 60%+
NEET - 2023
Hints
When a block of mass \(M\) is suspended by a long wire of length \(L,\) the length of the wire becomes \((L+l).\) The elastic potential energy stored in the extended wire is:
1. \(\frac{1}{2}MgL\)
2. \(Mgl\)
3. \(MgL\)
4. \(\frac{1}{2}Mgl\)
Subtopic: Â Potential energy of wire |
 72%
Level 2: 60%+
NEET - 2019
Hints
Links
- 1(current)
Q. No.
© 2025 GoodEd Technologies Pvt. Ltd.