question_answer

A rectangular field is 22 m long and 10 m wide. Two hemispherical pit holes of radius 2 m are dug from two places and the mud is spread over the remaining part of the field. The rise in the level of the field is -

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 \[\Rightarrow \]   \[\frac{1364}{7}\times h=\frac{704}{21}\] \[\therefore \]      \[h=\frac{704\times 7}{1364\times 21}=\frac{\mathbf{16}}{\mathbf{93}}\mathbf{m}\]