Answer:
We know that, momentum \[P=m\times v\] …..(1) But \[m=d\times v\] …… (2) where d is the density of the sphere and V is the volume of a cube. since volume of a cube\[={{(side)}^{3}}\] \[\therefore \,\,m=d\,\times {{(l)}^{3}}\] ……… (3) Substituting (2) and (3) in (1), We have, \[P=[d\times {{(l)}^{3}}]\times v\] As the same material is taken, its density is same and also given that its velocity is same for the two silver cubes. \[\Rightarrow \,\,P\propto {{(l)}^{3}}\] ……… (4) \[\frac{{{P}_{2}}}{{{P}_{1}}}=\frac{l_{2}^{3}}{l_{1}^{3}}\Rightarrow \frac{{{P}_{2}}}{K}={{\left( \frac{20}{10} \right)}^{3}}=8\,\,\,\Rightarrow {{P}_{2}}=8k\]
Case-I
Case-II
\[{{l}_{1}}=10\,cm\]
\[{{l}_{2}}=20\,cm\]
\[{{P}_{1}}=k\,\,units\]
\[{{P}_{2}}=?\]
\[{{v}_{1}}=v\,\,units\]
\[{{v}_{2}}=v\,\,units\]
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