JEE Main 26-Aug-2021 Shift 2 Solved Paper

Given, breaking stress of wire, σ=

24

π

×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=

T

πr2

24

π

×102=

37.5

πr2

⇒ r2=

37.5

24×102

=

1

64

⇒ r=

1

8

m=

100

8

‌cm=12.5‌cm
Thus, the minimum radius of wire should be 12.5 cm.