$Y = \frac{{Fl}}{{\pi {r^2}\Delta l}}$
$Given,\,radius = 5mm,forceF = 50\pi KN,$
$\frac{\ell }{{\Delta \ell }} = 0.01\,mm$
$\therefore \,Y = \frac{F}{{\pi {r^2}}}\frac{\ell }{{\Delta \ell }} = 2 \times {10^{14}}N/{m^2}.$
| $(i)$ Suspension fibre of galvanometer | $(a)$ Linear |
| $(ii)$ Bending of beam | $(b)$ Shear |
| $(iii)$ cutting piece of paper | $(c)$ Bulk |
| $(iv)$ mechanical waves in fluid | $(d)$ Shear |

