The two wires shown in figure are made of the same material which has a breaking stress of `8xx10^8Nm^-2`. The area of cross section of the upper wire is `0.006 cm^2` and that of the lower wire is `0.003 cm^2`. The mass `m_1=10 kg`, `m_2=20kg` and the hanger is light. a. Find the maximum load that can be put on the hanger without breaking a wire. Which wire will break first if the load is increased? b. Repeat the above part `m_1=10 kg and m_2=36kg`.

Answer

A, B, D

Explanation

a. Given that
`Stress =8xx10^8 1/m^2`
`A_u=0.006cm^2`
`A_1=0.003cm^2`
`m_1=10 kg, m_2=20kg`
Therefore stress in lower rod
`=T_1/A_1=(m_1g+wg)/A_1`
(from free body diagram -2)

`rarr (m_1g+wg)/A_1=8xx10^8`
`rarr w=14kg`
Stess in upper rod
`=T_2/A_2=(m_2g+m_1g+wg)/A_2`
For FBD 2
`rarr (m_2g+m_1g+wg)/A_2=8xx10^8`
`rarr 2=18kg`
For same stress the maximum load that can be ut is 14 kg. If the load is increased the lower wire wil break first.
b. if `m_1=10 kg`
and `m_2=36kg`
`rarr T_1/A_1``=(m_1g+wg)/A_1=8xx10^5`
`rarr w=14kg`
`rarr T_2/A_2=(m_2g+m_1g+wg)/A_2`
The maximum load that can be put is 2 kg. upper wire wil break first if load is increased.