Answer:
Side of cube \[(l)=5\text{ }cm=\frac{5}{100}m=\frac{1}{20}m\] Volume of cube (V) \[={{\ell }^{3}}={{\left( \frac{1}{20} \right)}^{3}}=\frac{1}{8,000}{{m}^{3}}\]
Density of block \[(d)=5\text{ }g\text{ }c{{m}^{3}}\] \[=5\times 1000\text{ }kg{{m}^{3}}=5000\text{ }kg{{m}^{3}}\] \[[\because 1\text{ }gc{{m}^{3}}=1000\text{ }kg\text{ }{{m}^{3}}]\] Thrust (T) = ? We know, \[T=mg\] \[T=V\times d\times g\] \[\left( \because d=\frac{m}{V}\Rightarrow m=V\times d \right)\] \[T=\times d\times g\] \[(\because {{V}_{cube}}={{l}^{3}})\] \[\because \]\[T=\frac{1}{8,000}\times 5,000\times 10=6.25N\] Therefore, the thrust acting on the table is 6.25 N.
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