Two blocks of masses 1 , kg and 2 , kg are co | vedclass

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(A) The stress in the wire is given by $	ext{Stress} = \frac{	ext{Tension}}{	ext{Area of cross-section}}$.To avoid breaking, the stress must not ex

Vedclass · Accepted Answer

$4.6 	imes 10^{-5} \, m$

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(A) The stress in the wire is given by $	ext{Stress} = \frac{	ext{Tension}}{	ext{Area of cross-section}}$.To avoid breaking, the stress must not exceed the breaking stress.Let the tension in the wire be $T$ and the acceleration of the system be $a$.The equations of motion for the two blocks are:For the $1 \, kg$ block: $T - 1(10) = 1a \implies T - 10 = a$ (Equation $1$)For the $2 \, kg$ block: $2(10) - T = 2a \implies 20 - T = 2a$ (Equation $2$)Adding Equation $1$ and Equation $2$:$(T - 10) + (20 - T) = a + 2a$$10 = 3a \implies a = \frac{10}{3} \, m/s^2$Substituting $a$ into Equation $1$:$T = 10 + \frac{10}{3} = \frac{40}{3} \, N$The breaking stress is $\sigma_{max} = 2 	imes 10^9 \, N/m^2$. The area of cross-section is $A = \pi r^2$.Setting the stress equal to the breaking stress to find the minimum radius $r$:$\frac{T}{A} = \sigma_{max} \implies \frac{40/3}{\pi r^2} = 2 	imes 10^9$$r^2 = \frac{40}{3 	imes \pi 	imes 2 	imes 10^9} = \frac{20}{3 \pi 	imes 10^9} \approx \frac{20}{9.4247 	imes 10^9} \approx 2.122 	imes 10^{-9} \, m^2$$r = \sqrt{2.122 	imes 10^{-9}} \approx 4.6 	imes 10^{-5} \, m$.

Two blocks of masses $1 \, kg$ and $2 \, kg$ are connected by a metal wire going over a smooth pulley as shown in the figure. The breaking stress of the metal is $2 \times 10^9 \, N/m^2$. What should be the minimum radius of the wire used if it is not to break? Take $g = 10 \, m/s^2$.

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