A cylindrical tub of radius 5 cm and length 9.8 cm is full of water. A...
**Given:**
- Radius of cylindrical tub = 5 cm
- Length of cylindrical tub = 9.8 cm
- Radius of hemisphere = 3.5 cm
- Height of cone outside the hemisphere = 5 cm
- Value of pi = 22/7
**To find:**
- Volume of solid immersed in the tub
**Solution:**
- Let's first find the volume of the solid.
- The solid is made up of a cone and a hemisphere.
- Volume of cone = 1/3 * pi * r^2 * h
- Here, radius of the cone (r) = 3.5 cm (same as the radius of the hemisphere)
- Height of the cone (h) = 5 cm (given)
- Volume of cone = 1/3 * 22/7 * 3.5^2 * 5 = 64.75 cm^3
- Volume of hemisphere = 2/3 * pi * r^3
- Here, radius of the hemisphere = 3.5 cm (given)
- Volume of hemisphere = 2/3 * 22/7 * 3.5^3 = 89.25 cm^3
- Total volume of the solid = Volume of cone + Volume of hemisphere = 64.75 + 89.25 = 154 cm^3
- Now, let's find the volume of water displaced by the solid.
- When the solid is immersed in the water in the tub, it displaces some water.
- The volume of water displaced is equal to the volume of the solid.
- Therefore, the volume of water displaced by the solid = 154 cm^3
- Hence, the volume of solid immersed in the tub is 154 cm^3.
- Therefore, the correct option is (B) 154 cm^3.
A cylindrical tub of radius 5 cm and length 9.8 cm is full of water. A...
Volume of water in the tub=πr²h=π×5²×9.8cm³
volume of hemisphere= 2/3×π×3.5³cm³
volume of cone = 1/3π×3.5²×5cm³
Volume of water left in the tub = Volume of water in tub-Volume of toy.....
[ 2/3×π×3.5³ +1/3π×3.5²×5 ]
V=π(245−49)= 22/7×196 =22×8 = 616 cm³
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