Solve this


Wires $\mathrm{W}_{1}$ and $\mathrm{W}_{2}$ are made of same material having the breaking stress of $1.25 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}$.

$\mathrm{W}_{1}$ and $\mathrm{W}_{2}$ have cross-sectional area of $8 \times 10^{-7} \mathrm{~m}^{2}$ and $4 \times 10^{-7} \mathrm{~m}^{2}$, respectively. Masses of $20 \mathrm{~kg}$ and $10 \mathrm{~kg}$ hang from them as shown in the figure. The maximum mass that can be placed in the pan without breaking the wires is $\mathrm{kg} _________.$ (Use $g=10 \mathrm{~m} / \mathrm{s}^{2}$ )


B.S $_{1}=\frac{\mathrm{T}_{1 \max }}{8 \times 10^{-7}} \Rightarrow \mathrm{T}_{1 \mathrm{max}}=8 \times 1.25 \times 100$

$=1000 \mathrm{~N}$

B.S ${ }_{2}=\frac{\mathrm{T}_{2 \max }}{4 \times 10^{-7}} \Rightarrow \mathrm{T}_{2 \max }=4 \times 1.25 \times 100$

$=500 \mathrm{~N}$

$\mathrm{m}=\frac{500-100}{10}=40 \mathrm{~kg}$

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now