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(D) The tension $T$ in the wire provides the necessary centripetal force for circular motion: $T = m \omega^{2} L$.The breaking stress is defined as t

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$4$

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(D) The tension $T$ in the wire provides the necessary centripetal force for circular motion: $T = m \omega^{2} L$.The breaking stress is defined as the maximum force per unit area: $	ext{Breaking Stress} = \frac{T}{A}$.Substituting the expression for tension: $	ext{Breaking Stress} = \frac{m \omega^{2} L}{A}$.Given: $m = 10 \; kg$,$L = 0.3 \; m$,$	ext{Breaking Stress} = 4.8 	imes 10^{7} \; N m^{-2}$,and $A = 10^{-2} \; cm^{2} = 10^{-2} 	imes 10^{-4} \; m^{2} = 10^{-6} \; m^{2}$.Rearranging for $\omega^{2}$: $\omega^{2} = \frac{(	ext{Breaking Stress}) 	imes A}{m 	imes L}$.$\omega^{2} = \frac{4.8 	imes 10^{7} 	imes 10^{-6}}{10 	imes 0.3} = \frac{48}{3} = 16$.Therefore,$\omega = \sqrt{16} = 4 \; rad \; s^{-1}$.

Column-$I$	Column-$II$
$(A)$ Shear modulus	$(I)$ Resistance to change in volume
$(B)$ Shearing stress	$(II)$ Proportionality constant
$(C)$ Elastic fatigue	$(III)$ Tangential stress
$(D)$ Modulus of elasticity	$(IV)$ Temporary loss of elastic property
	$(V)$ Resistance to change against deformation force

System	Type of Elasticity
$(i)$ 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

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