Question

Upper wire has cross-section area \(0.008\,\text{cm}^2\) and lower wire has area \(0.004\,\text{cm}^2\). Find the maximum mass of the pan that can be connected without breaking any wire. Given breaking stress of wire is \(\sigma = 8\times10^8\,\text{N/m}^2\). 


 

• A. \(22\ \text{kg}\)
• B. \(24\ \text{kg}\)
• C. \(26\ \text{kg}\)
• D. \(28\ \text{kg}\)

Answer 1

The Correct Option is A

Concept: Breaking stress: \[ \sigma = \frac{F}{A} \] Maximum tension allowed in the wire: \[ F = \sigma A \] Step 1: Lower wire Area \[ A_2 = 0.004\ \text{cm}^2 \] \[ A_2 = 0.004\times10^{-4}\ \text{m}^2 \] Tension in lower wire \[ T_2 = (10 + m)g \] Stress condition \[ \frac{(10+m)g}{0.004\times10^{-4}} = 8\times10^8 \] \[ 10+m = 8\times10^7 \times 0.004\times10^{-4} \] \[ 10+m = 32 \] \[ m = 22\ \text{kg} \] Step 2: Upper wire Area \[ A_1 = 0.008\ \text{cm}^2 \] \[ A_1 = 0.008\times10^{-4}\ \text{m}^2 \] Tension \[ T_1 = (40+m)g \] Stress condition \[ \frac{(40+m)g}{0.008\times10^{-4}} = 8\times10^8 \] \[ 40+m = 64 \] \[ m = 24\ \text{kg} \] Step 3: Maximum safe mass The wire that reaches breaking condition first determines the limit. Lower wire gives \[ m = 22\ \text{kg} \] Upper wire allows \[ m = 24\ \text{kg} \] Thus maximum safe mass \[ \boxed{22\ \text{kg}} \]