Question

Two block of masses 1 kg and 4 kg are connected by a metal wire going over a smooth pulley as shown in the figure. The breaking stress of the metal is $3.18 \times 10 ^ { 10 } \mathrm {~N} / \mathrm { m } ^ { 2 }$. The minimum radius of the wire so it will not break is

• A.
• B.
• C.
• D.

Answer 1

Answer: (D)

Answer 2

Tension in the wire $T = \frac { 2 m _ { 1 } m _ { 2 } } { m _ { 1 } + m _ { 2 } } g$ ⇒ $T = \frac { 2 \times 1 \times 4 } { 1 + 4 } \times 10$ ⇒

Breaking force = Breaking stress × Area of cross-section

Tension in the wire = $3.18 \times 10 ^ { 10 } \times \pi r ^ { 2 }$

$r = \sqrt { \frac { 16 } { 3.18 \times 10 ^ { 10 } \times 3.14 } }$