A) \[{{10}^{4}}\,N{{m}^{-2}}\]
B) \[{{10}^{8}}\,N{{m}^{-2}}\]
C)
\[{{10}^{6}}\,N{{m}^{-2}}\] D)
\[{{10}^{3}}\,N{{m}^{-2}}\]
Correct Answer:
C Solution :
= Kinetic energy of the ball\[=\frac{1}{2}m{{\text{v}}^{2}}\] therefore, \[\frac{1}{2}\times {{\left( \frac{20}{42} \right)}^{2}}\times Y\times \pi \times {{3}^{2}}\times {{10}^{-6}}\times 42\times {{10}^{-2}}=\] \[\frac{1}{2}\times 2\times {{10}^{-2}}\times {{(20)}^{2}}\] \[Y\simeq 3\times {{10}^{6}}N{{m}^{-2}}\]
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