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A cylinderical tank of radiuus `20 cm` and height `50 cm` has water up to `30 cm` of height. What will be the rise in level of liquid at the periphery if the cylinder be givenan angular velocity of `10 rad s^(-1)`? Also determine the frequency of rotation when water just starts spilling over the sides of the vessel. |
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Answer» At an angular sped of `10 rads^(-1)`, let us assume that the water does not spill. `h_(2)=H+(omega^(2)r^(2))/(4g)impliesh_(2)-H=(omega^(2)r^(2))/(4g)` Therefore, rise in the liquid level at the periphery is `10 cm`. For water to just spill over the sides ,the maximum height `(h_("max"))` is `50 cm` Again `h_("max")=y_(0)+(omega^(2)R^(2))/(4g)` Here `h_("max")=50 cm, y_(0)=30 cm, R=20 cm, g=10ms^(-2)` `:. 50xx10^(-2)=30xx10^(-2)+(omega^(2)(20xx10^(-2))^(2))/(4xx10)` `implies omega^(2)xx10^(-3)=20xx10^(-2)` `omega^(2)=200implies omega=10sqrt(2)rads^(-1)` Hence, freqency of rotation is `f=omega/(2pi)=(5sqrt(2))/pis^(-1)` |
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