The force constant of a wire is k and that of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The work done in stretching a wire (or spring) by a distance $x$ is given by the formula $W = \frac{1}{2}kx^2$.Here,$k$ is the force constant and

Vedclass · Accepted Answer

$W_2 = 2W_1$

Vedclass · Answer

(B) The work done in stretching a wire (or spring) by a distance $x$ is given by the formula $W = \frac{1}{2}kx^2$.Here,$k$ is the force constant and $x$ is the extension.For the first wire,the work done is $W_1 = \frac{1}{2}k_1x^2$,where $k_1 = k$.For the second wire,the work done is $W_2 = \frac{1}{2}k_2x^2$,where $k_2 = 2k$.Since both wires are stretched by the same distance $x$,we can see that $W \propto k$.Therefore,the ratio of the work done is $\frac{W_2}{W_1} = \frac{k_2}{k_1} = \frac{2k}{k} = 2$.This implies $W_2 = 2W_1$.

The force constant of a wire is $k$ and that of another wire is $2k$. When both the wires are stretched through the same distance,then the work done is:

Similar Questions

Work done in stretching a wire through $1 \ mm$ is $2 \ J$. What amount of work will be done for elongating another wire of same material,with half the length and double the radius of cross-section,by $1 \ mm$ (in $J$)?

The work done per unit volume in stretching a wire is :-

$A$ wire suspended vertically from one end is stretched by attaching a weight $200 \,N$ to the lower end. The weight stretches the wire by $1 \,mm$. The elastic potential energy gained by the wire is ....... $J$

Which of the following is true for elastic potential energy density?

$A$ brass rod of cross-sectional area $1 \, cm^2$ and length $0.2 \, m$ is compressed lengthwise by a weight of $5 \, kg$. If Young's modulus of elasticity of brass is $1 \times 10^{11} \, N/m^2$ and $g = 10 \, m/s^2$,then the increase in the energy of the rod will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module