A) \[14\,\,\text{cm}\]
B) \[28\,\,\text{cm}\]
C) \[\frac{14}{3}\,\,\text{cm}\]
D) \[\frac{28}{3}\,\,\text{cm}\]
Correct Answer: B
Solution :
Radius of the iron ball \[=\frac{14}{2}=7\,\,\text{cm}\] Volume of the ball \[=\frac{4}{3}\pi \times {{(7)}^{3}}\,\,\text{c}{{\text{m}}^{3}}\] Let the radius of cylinder = r cm \[\therefore \] Volume of cylinder \[=\pi {{r}^{2}}\times \frac{7}{3}\,\,\text{c}{{\text{m}}^{\text{3}}}\] Clearly, \[\pi {{r}^{2}}\times \frac{7}{3}=\frac{4}{3}\pi \times {{(7)}^{3}}{{r}^{2}}=\frac{4\times 7\times 7\times 7\times 3}{7\times 3}\] \[r=\sqrt{4\times 7\times 7}=14\,\,\text{cm}\] Diameter \[=2\times 14=28\,\,\text{cm}\]
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