Q.43. Two wires as shown in the figure below, made of steel and have breaking stress of . Area of cr

Q: What is the correct answer for this exam Physics Q43?

The correct answer is B. Sol. Let $m$ be the mass added to the pan. The mass of the lower block is 10 kg, and the upper block is 30 kg. Total mass supported by the lower wire is $(10+m)$ kg. Total mass supported by the upper wire is $(30+10+m) = (40+m)$ kg. Breaking stress = $\frac{\text{Breaking Force}}{\text{Area}}$. For the upper wire: Breaking stress = $\frac{(40+m)g}{0.008 cm^2} = \frac{(40+m) \times 10}{0.008 \times

Q: Q.43. Two wires as shown in the figure below, made of steel and have breaking stress of . Area of cross-section of upper wire is and of lower wire is . The maximum mass that can be added to pan…

The correct answer is B. Sol. Let $m$ be the mass added to the pan. The mass of the lower block is 10 kg, and the upper block is 30 kg. Total mass supported by the lower wire is $(10+m)$ kg. Total mass supported by the upper wire is $(30+10+m) = (40+m)$ kg. Breaking stress = $\frac{\text{Breaking Force}}{\text{Area}}$. For the upper wire: Breaking stress = $\frac{(40+m)g}{0.008 cm^2} = \frac{(40+m) \times 10}{0.008 \times

Q: What are the options for this this exam Physics question?

A. 56 · B. 38 · C. 96 · D. 6

Question

Q.43. Two wires as shown in the figure below, made of steel and have breaking stress of . Area of cross-section of upper wire is  and of lower wire is . The maximum mass that can be added to pan without breaking any wire is ______ kg. (take g = 10 m/s) [Figure: Two wires supporting masses]

FirstInTest · Accepted Answer

Sol. Let $m$ be the mass added to the pan. The mass of the lower block is 10 kg, and the upper block is 30 kg. Total mass supported by the lower wire is $(10+m)$ kg. Total mass supported by the upper wire is $(30+10+m) = (40+m)$ kg. Breaking stress = $\frac{	ext{Breaking Force}}{	ext{Area}}$. For the upper wire: Breaking stress = $\frac{(40+m)g}{0.008 cm^2} = \frac{(40+m) 	imes 10}{0.008 	imes 10^{-4} m^2} = \frac{(40+m) 	imes 10}{8 	imes 10^{-7}}$. This must be $\le 12 	imes 10^8 N/m^2$. So, $\frac{(40+m) 	imes 10}{8 	imes 10^{-7}} \le 12 	imes 10^8$. $(40+m) 	imes 10 = 12 	imes 10^8 	imes 8 	imes 10^{-7} = 960$. $40+m = 96$. $m = 56$ kg. For the lower wire: Breaking stress = $\frac{(10+m)g}{0.004 cm^2} = \frac{(10+m) 	imes 10}{0.004 	imes 10^{-4} m^2} = \frac{(10+m) 	imes 10}{4 	imes 10^{-7}}$. This must be $\le 12 	imes 10^8 N/m^2$. So, $\frac{(10+m) 	imes 10}{4 	imes 10^{-7}} \le 12 	imes 10^8$. $(10+m) 	imes 10 = 12 	imes 10^8 	imes 4 	imes 10^{-7} = 480$. $10+m = 48$. $m = 38$ kg. The maximum mass that can be added without breaking any wire is the minimum of these two values, which is 38 kg.

Answer

56

Answer

38

Answer

96

Answer

6

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