When a block of mass M is suspended by a long wire of length L, the le

1.	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:(A) \(\frac{1}{2}\)Mgl(B) \(\frac{1}{2}\)MgL(C) Mgl(D) MgL
Answer» (A) \(\frac{1}{2}\)Mgl elastic potential energy: \(\frac{1}{2}\) × F × L = \(\frac{1}{2}\)Mgl

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:(A) \(\frac{1}{2}\)Mgl(B) \(\frac{1}{2}\)MgL(C) Mgl(D) MgL

Answer»

(A) \(\frac{1}{2}\)Mgl

elastic potential energy: \(\frac{1}{2}\) × F × L = \(\frac{1}{2}\)Mgl