how many litres of kerosine can be held by a cylindrical 21cm deep?
Live your answer to 2 significant figure
(2).a conical vessel is 10cm deep.if it half filled with liquid,find the depth of the liquid
#thanks
working or hint plz
just find the volume in cm^3
Then recall that 1L = 1000cm^3
what about the second question
and how can i answer my quetion 1 without a radius
pi(r^2)21=1000
can i find my radius from here, and then find the volume.?
yes, with no radius, it's difficult to find the volume. The problem has been garbled, apparently. Both the cylinder and the cone are incompletely specified.
And that doesn't even get into the bad grammar...
To find the volume of the cylindrical vessel, you need to know its depth and its radius. However, since only the depth is given, we can assume that the vessel is a perfect cylinder.
The formula to calculate the volume of a cylinder is V = πr²h, where V is the volume, r is the radius, and h is the height (depth) of the cylinder.
In this case, the depth (h) of the cylindrical vessel is given as 21 cm. However, the radius is not provided. Therefore, without knowing the radius, it is not possible to directly calculate the volume.
For the second question about the conical vessel, the depth of the liquid can be calculated using similar principles. The formula to find the volume of a cone is V = (1/3)πr²h, where V is the volume, r is the radius, and h is the height (depth) of the cone.
In this case, the depth (h) of the conical vessel is given as 10 cm, and it is mentioned that the vessel is half-filled with liquid. To find the depth of the liquid, we need to calculate the total volume of the cone and divide it by 2. With that information, we can then solve for h.
Please note that in both of these calculations, the radius of the vessels is not given. So, without the radius value, it is not possible to determine the volume or the depth of the liquid accurately.