Wires W1 and W2 are made of same material having the breaking stress of 1-25- 109 N-m2- W1 and W2 have cross-sectional area of 8- 10-7 m2 and 4- 10-7 m2- respectively- Masses of 20 kg and 10 kg hang from them as shown in the figure- The maximum mass that can be placed in the pan without breaking the wires is kg- Use g-10 m-s2

Let m be mass in the pan Load on wire W_2 is (m+10)g Cross section area of W_2 is 4× 10^ 7 m^2 Stress on wire W_2 is (m+10)g4× 10^ 7 For limiting case, nbs ...

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Byju's Answer

Standard XII

Physics

Longitudinal Stress

Wires W1 and ...

Question

Wires $W_{1}$ and $W_{2}$ are made of same material having the breaking stress of $1.25\times 10^{9}\text{ N/m}^{2}. W_{1}~\text{and}~W_{2}$ have cross-sectional area of $8\times 10^{-7}~\text m^{2}~\text{and}~ 4\times 10^{-7}~\text m^{2},$ respectively. Masses of $20~\text {kg}$ and $10~\text{kg}$ hang from them as shown in the figure. The maximum mass that can be placed in the pan without breaking the wires is

$\text{kg}$. (Use $g=10\text{ m/s}^{2}$)
$ w_{1} $ $ 20 kg $ $ 10 kg $ pan

40.0

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40.00

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Solution

Let $m$ be mass in the pan

Load on wire $W_{2}$ is $(m + 10) g$
Cross section area of $W_{2}$ is $4 \times 10^{- 7} m^{2}$

Stress on wire $W_{2}$ is $\frac{(m + 10) g}{4 \times 10^{- 7}}$
For limiting case,
Breaking stress= stress on the wire
$\frac{(m + 10) g}{4 \times 10^{- 7}} = 1.25 \times 10^{9}$
$m + 10 = 50$
$m = 40 kg$
$\Rightarrow$ If mass is greater than $40 kg,$ string $W_{2}$ breaks.
Stress on wire $W_{1}$ is $\frac{(m + 30) g}{8 \times 10^{- 7}}$
For limiting case,
Breaking stress= stress on the wire
$\frac{(m + 30) g}{8 \times 10^{- 7}} = 1.25 \times 10^{9}$
$m + 30 = 100$
$m = 70 kg$
$\Rightarrow$ If mass is greater than $70 kg,$ string $W_{1}$ breaks.

Hence, maximum mass that can be kept in the pan without breaking any of strings is $40 kg$

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