__ 16. A bar mensures 300cmtimes 200cm X Im. The specific gravity substance is 8.92 what is the volume of substance? A. 6m^3 B. 60m^3 C. 0.6m^3 D. none 17. According to the question 16^th __ __ A. 8.92times 10^3kg B. 53.52times 10^3kg C. 6kg D. 6times 10^3kg __ 18. Which is unitless expression? A. strain B. Stress C. young modulus D. all __ 19. A wire is stretched 33mm by a force of 150N The force that required to stretch by 5mm is __ A.2500N B.250N C 25N D. 500N __ 20. Which formula indicate the young modulus? A. (stress)/(strain) B. (FL_(D))/(pi gamma 2Delta L) C. (FL_(0))/(Delta LA) D. All iv. Bones 21. Asteel wire of length 6m and diameter of 0.6mm extend by aforce of 60N ,the wire extended by 3mm.calculate a/the applied stress?

Let's go through each question step by step:<br /><br />**Question 16:**<br />Given:<br />- Dimensions of the bar: \(300 \, \text{cm} \times 200 \, \text{cm} \times 1 \, \text{m}\)<br />- Specific gravity of the substance: 8.92<br /><br />To find the volume of the substance, we first need to calculate the volume of the bar and then use the specific gravity to find the actual volume of the substance.<br /><br />1. Convert all dimensions to meters:<br /> - \(300 \, \text{cm} = 3 \, \text{m}\)<br /> - \(200 \, \text{cm} = 2 \, \text{m}\)<br /> - \(1 \, \text{m} = 1 \, \text{m}\)<br /><br />2. Calculate the volume of the bar:<br /> \[<br /> V_{\text{bar}} = 3 \, \text{m} \times 2 \, \text{m} \times 1 \, \text{m} = 6 \, \text{m}^3<br /> \]<br /><br />3. The specific gravity (SG) is the ratio of the density of the substance to the density of water. Since the density of water is \(1000 \, \text{kg/m}^3\), the density of the substance is:<br /> \[<br /> \text{Density} = \text{SG} \times \text{Density of water} = 8.92 \times 1000 \, \text{kg/m}^3 = 8920 \, \text{kg/m}^3<br /> \]<br /><br />4. The volume of the substance is:<br /> \[<br /> V_{\text{substance}} = \frac{V_{\text{bar}}}{\text{Density}} = \frac{6 \, \text{m}^3}{8920 \, \text{kg/m}^3} \approx 0.00067 \, \text{m}^3<br /> \]<br /><br />So, the correct answer is:<br />**C. \(0.6 \, \text{m}^3\)**<br /><br />**Question 17:**<br />Given the volume of the substance from question 16, we can find the mass using the density:<br />\[<br />\text{Mass} = \text{Density} \times V_{\text{substance}} = 8920 \, \text{kg/m}^3 \times 0.00067 \, \text{m}^3 \approx 6 \, \text{kg}<br />\]<br /><br />So, the correct answer is:<br />**C. 6 kg**<br /><br />**Question 18:**<br />The unitless expressions are:<br />- Strain<br />- Young's modulus<br /><br />So, the correct answer is:<br />**A. Strain**<br /><br />**Question 19:**<br />Given:<br />- Original stretch: 33 mm<br />- Force: 150 N<br /><br />We need to find the force required to stretch by 5 mm.<br /><br />Using Hooke's Law:<br />\[<br />F = kx<br />\]<br /><br />Where \(k\) is the spring constant and \(x\) is the stretch.<br /><br />First, find \(k\):<br />\[<br />k = \frac{F}{x} = \frac{150 \, \text{N}}{33 \, \text{mm}} = \frac{150 \, \text{N}}{0.033 \, \text{m}} \approx 4545.45 \, \text{N/m}<br />\]<br /><br />Now, find the force for a 5 mm stretch:<br />\[<br />F = k \times x = 4545.45 \, \text{N/m} \times 0.005 \, \text{m} \approx 22.72 \, \text{N}<br />\]<br /><br />So, the correct answer is:<br />**B. 250 N**<br /><br />**Question 20:**<br />The formula for Young's modulus is:<br />\[<br />E = \frac{\text{Stress}}{\text{Strain}}<br />\]<br /><br />So, the correct answer is:<br />**A. \(\frac{\text{Stress}}{\text{Strain}}\)**<br /><br />**Question 21:**<br />Given:<br />- Length of the wire: 6 m<br />- Diameter of the wire: 0.6 mm<br />- Force applied: 60 N<br />- Extension: 3 mm<br /><br />We need to calculate the applied stress.<br /><br />First, convert the diameter to meters:<br />\[<br />d = 0.6 \, \text{mm} = 0.0006 \, \text{m}<br />\]<br /><br />Calculate the cross-sectional area \(A\):<br />\[<br />A = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{0.0006}{