Given, breaking stress of wire,
σ=×102Nm−2 Free body diagram of 5kg block is given as
where, a is common acceleration.
Value of acceleration due to gravity,
g=10ms−2 From free body diagram of block of mass 5 kg
5g − T = 5a
⇒ 5 × 10 − T = 5a
⇒ 50 − T = 5a ...(i)
Free body diagram of 3 kg block is given as
From free body diagram of block of mass 3 kg,
T − 3g = 3a
⇒ T – 3 ×10 = 3a
⇒ T − 30 = 3a ...(ii)
Add Eqs. (i) and (ii), we get
50 − T + T − 30 = 5a + 3a
⇒ 20 = 8a
⇒
a=2.5ms−2 Substituting the value of a in Eq. (i), we get
50 − T = 5 × 2.5
⇒ T = 37.5 N
Let us assume the minimum radius of wire is r.
The breaking stress is expressed as
σ= ×102= ⇒
r2== ⇒
r=m=cm=12.5cm Thus, the minimum radius of wire should be 12.5 cm.