A body weights 50 N in air and 45 N when wholly immersed in water calculate (i) the loss in weight of the body in water (ii) the upthrust on the body. (iii) volume of the body.

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XxGamerPrinceXx XxGamerPrinceXx · Answer 1 · 2021-08-23T13:58:14+02:00

Answer:

[tex]difference \: in \: weight = 150n - 100n = 50n[/tex]

Now,buyantant force

[tex]difference \: in \: weight \: = volume(body) \times density \: of \: water \: \times g[/tex]

so;

[tex]50 = {v}^{b} \times 1 \times {10}^{3} \times 9.8m {s}^{2}[/tex]

[tex] {v}^{b} = \frac{50}{1000 } \times 9.8[/tex]

[tex] = \frac{50}{9800} [/tex]

[tex] = 0.0051[/tex]

Now,

[tex]mass \: in \: air \: = 150n = \frac{150}{9.8kg} [/tex]

[tex]density = \frac{weght}{volume} [/tex]

[tex] = \frac{150}{0.0051} \times 9.8 \\ x = 3000[/tex]

And now,

[tex]specific \: density \: = \frac{density of \: the \: body}{density \: of \: water} [/tex]

[tex] = \frac{3000}{1000} [/tex]

[tex] = 3[/tex]

Hence that,specific density of a given body is 3

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