A cylindrical tub of radius 5 cm and length 9.8 cm is full of water. A solid is in the form of a right circular cone mounted on a hemisphere is immersed in a tub. If the radius of the hemisphere is 3.5cm height of the cone outside the hemisphere is 5cm find the volume left in the tub.
Answers
Answered by
258
The tub is half of a cylinder
∴ volume of water = 1/2 x 22/7 x 5² x 9.8 = 385 cm³
Volume of the solid = volume of cone + volume of hemisphere
= 1/3 x 22/7 x 3.5² x 5 + 2/3 x 22/7 x 3.5³
= 64 1/6 + 89 5/6
= 154 cm³
Remaining volume = 385 - 154 = 231 cm³
∴ volume of water = 1/2 x 22/7 x 5² x 9.8 = 385 cm³
Volume of the solid = volume of cone + volume of hemisphere
= 1/3 x 22/7 x 3.5² x 5 + 2/3 x 22/7 x 3.5³
= 64 1/6 + 89 5/6
= 154 cm³
Remaining volume = 385 - 154 = 231 cm³
Answered by
391
volume of water in cylindrical tub=volume of the hemisphere+volume of the cone
={[(2×22×7×7×7)/(3×7×2×2×2)]+[(1×22×7×7×5)/(3×7×2×2)]}
=(539/6 +385/6)
=924/6
=154cm³
volume of tub=(22×5×5×9.8)/7
=770 cm³
volume of water left in the tub=volume of tub -volume of solid immersed
=770-154
=616cm³
={[(2×22×7×7×7)/(3×7×2×2×2)]+[(1×22×7×7×5)/(3×7×2×2)]}
=(539/6 +385/6)
=924/6
=154cm³
volume of tub=(22×5×5×9.8)/7
=770 cm³
volume of water left in the tub=volume of tub -volume of solid immersed
=770-154
=616cm³
Similar questions