Directions : In the following question more than one of the answers given may be correct. Select the correct answer and mark it according to the code:
A steel cube weighs 1 kg in air and 0.88 kg in water. The density of the steel is\[7.71\times {{10}^{3}}\] \[kg/{{m}^{3}}\]and of water is\[{{10}^{3}}kg/{{m}^{3}}\]. The cube (1) must be solid (2) consists of impure steel (3) must be hollow (4) consists of pure steelA) 1, 2 and 3 are correct
B) 1 and 2 are correct
C) 2 and 4 are correct
D) 1 and 3 are correct
Correct Answer: B
Solution :
\[{{W}_{0}}=\]weight in air = 1 kg,\[{{W}_{W}}=\]weight in water =0.88 kg \[{{\rho }_{s}}=\]density of steel\[=7.71\times {{10}^{3}}kg/{{m}^{3}}\] \[{{\rho }_{w}}=\]density of water\[={{10}^{3}}kg/{{m}^{3}}\] Loss in weight\[=1-0.88=0.12\text{ }kg\] \[{{(RD)}_{s}}=\frac{1}{0.12}\] \[\rho _{s}^{'}=\]density of steel \[=\frac{1}{0.12}\times {{10}^{3}}\,kg/{{m}^{3}}\] \[=8.3\times {{10}^{3}}kg/{{m}^{3}}\] ie, \[p_{s}^{'}\ne {{p}_{s}}\] This shows that cube is made of impure steel. Now volume of body\[=\frac{0.12}{{{10}^{3}}}=0.12\times {{10}^{-3}}{{m}^{3}}\] Volume of material\[=\frac{1\,kg}{8.3\times {{10}^{3}}}\] \[=0.12\times {{10}^{-3}}{{m}^{3}}\] So, body must be solid.
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