Question

The two wires shown in figure are made of the
Figure
same material which has a breaking stress of 8 × 108 N m−2. The area of cross section of the upper wire is 0.006 cm2 and that of the lower wire is 0.003 cm2. The mass m1 = 10 kg, m2 = 20 kg 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 if m1 = 10 kg and m2 = 36 kg.

Answer 1

Maximum Load is 20 N

Explanation:

Let $m_1$ be 10 kg and $m_2$ be 36 kg

Tension in Lower wire $T_1 = m_1g + w$

Also, g is the acceleration due to gravity and w is the load

So, Stress in Lower wire = $\frac {T_1}{A_1} = \frac {m_1g + w}{A_1} = 8 \times 10^5$

or w = 140 N

For Tension in the upper wire, $T_2 = m_1g + m_2g + w$

Stress in upper wire ,  

$\frac {T_u}{A_u} = \frac {m_1g + m_2g + w}{A_u} = 8 \times 10^5$

or w = 20 N

The Maximum load which can be put for same breaking stress is 20 N or 2 kg.

The upper wire will be breaking first if load is increased.

Answer 2

See the attachement.....

Maximum mass = 14 kg,

when load is increased, T1 increases hence, Lower wire will break.

Yrr second part khud kr lena

same aise hi hoga sachhi..