A) \[92\frac{2}{3}%\]%
B) \[46\frac{1}{3}%\]%
C) \[53\frac{2}{3}%\]%
D) \[7\frac{1}{3}%\]%
Correct Answer: A
Solution :
\[\therefore \] Height of cone \[=7\,cm\] and radius of base \[=1\,cm\] \[\therefore \] Volume of cone \[=\frac{1}{3}\times 7\times 1\times 1\times \frac{22}{7}=\frac{22}{7}c{{m}^{3}}\] And volume of conical block \[=10\times 5\times 2=100\,c{{m}^{2}}\] \[\therefore \] Wastage \[=100-\frac{22}{3}=\frac{278}{3}c{{m}^{3}}\] \[\therefore \] Percentage of wastage \[=\frac{278}{3}\times \frac{100}{100}%=92\frac{2}{3}%\]
You need to login to perform this action.
You will be redirected in
3 sec
