A cylindrical tub of radius 5 cm and length 9.8 cm is full of water. A solid in the form of right circular cone mounted on a hemisphere is immersed into the tub. The radius of the hemisphere is 3.5 cm and height of cone outside the hemisphere is 5 cm. Find the volume of water left in the tub
Answers
Answer:
616 cm³
Step-by-step explanation:
Length of cylindrical tub = 9.8 cm.
Radius of cylindrical tub = 5 cm.
Volume of cylindrical tub = πr²h
= (22/7) * (5)² * 9.8
= 770 cm³
Radius of hemisphere = 3.5 cm
Volume of hemisphere = (2/3) πr³
= (2/3) * (22/7) * (3.5)³
= (11 * 7 * 7)/(3 * 2)
= 539/6 cm³
Height of cone = 5 cm, Radius of cone = 3.5 cm
Volume of cone = (1/3) πr²h
= (1/3) * (22/7) * (3.5)² * 5
= 385/6 cm³
Volume of solid = Volume of hemisphere + Volume of the cone
= 539/6 + 385/6
= 154 cm³
∴ Volume of water left in the tub = Volume of cylinder - Volume of solid
= 770 - 154
= 616 cm³
Hope it helps!
Step-by-step explanation:
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