A) 1.04
B) 1.03
C) 1.26
D) 1.01
Correct Answer: D
Solution :
[d] Law of floatation \[\frac{{{V}_{i}}}{V}=\frac{{{\rho }_{body}}}{{{\rho }_{liquid}}}\] In given case \[\therefore \frac{{{h}_{1}}}{{{h}_{2}}}=\frac{\left( {{\rho }_{4{}^\circ c}} \right)}{\left( {{\rho }_{0{}^\circ c}} \right)}\left[ {{\rho }_{body}}=\text{constant} \right]\] \[\frac{{{p}_{4{}^\circ c}}}{{{p}_{0{}^\circ c}}}\left( \frac{100-20}{100-21} \right)=\frac{80}{79}=1.01\]You need to login to perform this action.
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