CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

If the work done in stretching a wire by 1mm is 2 J, the work necessary for stretching another wire of same material but with double radius of cross section and half the length by 1mm is : (in joules)             


A
16
loader
B
8
loader
C
4
loader
D
1/4
loader

Solution

The correct option is C 16
Work done $$=\dfrac{AY}{2L}l^{2}$$$$=\dfrac{\pi r^{2}Yl^{2}}{2L}$$
Work done $$\propto \dfrac{r^{2}}{L}$$
$$\dfrac{w_{1}}{w_{2}}=\dfrac{r{_{1}}^{2}}{L_{1}}\times \dfrac{1/2L_{1}}{(2r_{1})^{2}}$$
$$\dfrac{w_{1}}{w_{2}}=\dfrac{1}{8}$$
$$w_{2}=8w_{1}$$
$$=8(2)$$
$$=16J$$

Physics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image