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Mohit Soni

· started a discussion

· 1 Months ago

Wrong answer (Question:273472.0)

16/93 is the right answer

Question:

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

Options:
A)

893    m

B)

\(\frac{16}{93}\)  m

C)

\(\frac{352}{177}\)   m

D)

2393    m

Solution:

 

Knowledge Expert

· commented

· 1 Months ago

Remaining area=1188/7
Volume of mud spread over the remaining field=1188/7*h=4/3*pi*8
The correct h=704/3564
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