Two blocks of masses 3 kg and 5 kg are connected by a metal wire going over a smooth pulley. The breaking stress of the metal is $\frac{24}{\pi} \times 10^{2} \mathrm{Nm}^{-2}$. What is the minimum radius of the wire? (take $\mathrm{g}=10 \mathrm{~ms}^{-2}$ )

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Laws of Motion

Two blocks of masses 3 kg and 5 kg are connected by a metal wire going over a smooth pulley. The breaking stress of the metal is

\frac{24}{π} \times 1 0^{2} Nm^{- 2}

. What is the minimum radius of the wire? (take

g = 10 ms^{- 2}

)

(A)1250 cm
(B)125 cm
(C)12.5 cm✓ Correct
(D)1.25 cm

Answer: (C)

Solution

For two masses over a light smooth pulley the tension in the connecting wire is

T = \frac{2 m _{1} m _{2} g}{m _{1} + m _{2}} = \frac{2 ( 3 ) ( 5 ) ( 10 )}{3 + 5} = 37.5 N

The wire is on the point of breaking when the stress equals the breaking stress:

\frac{T}{π r ^{2}} = \frac{24}{π} \times 1 0^{2} \Rightarrow r^{2} = \frac{37.5}{24 \times 1 0 ^{2}} = 1.5625 \times 1 0^{- 2} m^{2}

r = 0.125

m, that is a minimum radius of

12.5

cm.

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Single correct option

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