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Question Number 29520 by bbbbbb last updated on 09/Feb/18

$$\mathrm{to}\mathrm{make}\mathrm{an}\mathrm{open}\mathrm{fish}\mathrm{tank}\mathrm{a}\mathrm{glass}\mathrm{sheet}\mathrm{of}$$$$2\mathrm{mm}\mathrm{gauge}\mathrm{is}\mathrm{used}.\mathrm{the}\mathrm{outer}\mathrm{length}$$$$,\mathrm{breadth}\mathrm{and}\mathrm{height}\mathrm{are}60.4,40.4,$$$$40.2\mathrm{respectively}.\mathrm{how}\mathrm{much}\mathrm{maximum}$$$$\mathrm{volume}\mathrm{of}\mathrm{water}\mathrm{will}\mathrm{be}\mathrm{contained}\mathrm{in}\mathrm{it}?$$$$$$

Answered by Rasheed.Sindhi last updated on 09/Feb/18

$$\mathrm{Net}\mathrm{length}=60.4\mathrm{m}-4\mathrm{mm}=(60.4-0.0004)\mathrm{m}$$$$\mathrm{Net}\mathrm{breadth}=40.4-4\mathrm{mm}=(40.4-0.0004)\mathrm{m}$$$$\mathrm{Net}\mathrm{height}=40.2-2\mathrm{mm}=(40.2-0.0002)\mathrm{m}$$$$\mathrm{V}=60.3996\times 40.3996\times 40.1996$$$$\approx 98091.835{\mathrm{m}}^3$$$$$$

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