Directions: In the following question more than one of the answers given may be correct. Select the correct answer and mark it according to the code:
A body of mass M is attached to the lower end of a metal wire, whose upper end is fixed. The elongation of the wire is\[l\]. (1) Loss in gravitational potential energy of M is\[Mgl\] (2) The elastic potential energy stored in the wire is\[\frac{1}{2}Mgl\] (3) The elastic potential energy stored in the wire is\[Mgl\] (4) Heat produced is\[\frac{1}{2}Mgl\]A) 1 and 2 are correct
B) 2 and 3 are correct
C) 1 and 4 and correct
D) 1, 2 and 4 are correct
Correct Answer: D
Solution :
When a body of mass M is attached to the lower end of a metal wire whose upper end is fixed and the elongation of wire is\[l\], then loss in gravitational potential energy of mass\[m=MgL\]As elastic potential energy stored in stretched wire \[=\frac{Mgl}{2}\] Hence, heat produced in wire \[=Mfl-\frac{Mgl}{2}=\frac{Mgl}{2}\]
You need to login to perform this action.
You will be redirected in
3 sec
