A) \[\frac{8}{93}m\]
B) \[\frac{13}{93}m\]
C) \[\frac{16}{93}m\]
D) \[\frac{23}{93}m\]
Correct Answer: C
Solution :
Volume of mud dug out in two hemispherical pit holes = \[2\times \frac{2}{3}\pi {{r}^{3}}=2\times \frac{2}{3}\times \frac{22}{7}\times {{2}^{3}}\] = \[\frac{2\times 2\times 22\times 8}{21}=\frac{704}{21}{{m}^{2}}\] Area on which the mud is spread over = Area of field - Area of pit holes = \[l\times b-2\times \pi {{r}^{2}}\] = \[22\times 10-2\times \frac{22}{7}\times {{2}^{2}}=\frac{1364}{7}{{m}^{2}}\] Now, let the rise in level be h m, then Area of remaining field x h = Volume of mud dug out \[\Rightarrow \] \[\frac{1364}{7}\times h=\frac{704}{21}\] \[\therefore \] \[h=\frac{704\times 7}{1364\times 21}=\frac{\mathbf{16}}{\mathbf{93}}\mathbf{m}\]You need to login to perform this action.
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