Question

A wooden cylinder of diameter 4r, height H and density p/3 is kept on a hole of diameter 2r of a lank,filled with liquid of density p as shown in the figure.If level of the liquid starts decreasing slowly when the level of liquid is at a height hi above the cylinderthe block starts moving up. At what value of hi, will the block rise:[IIT-JEE 2006,5/184)- FB- upon ha presanp/3(4H(B) 6H(D) Romains samo​

Answer 1

Answer:

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Explanation:

Answer 2

The value of $h_{i}$ is $\dfrac{5H}{3}$.

Explanation:

Given that,

Diameter of wooden cylinder = 4r

Height = h

Density $\rho'=\dfrac{\rho}{3}$

Diameter of hole = 2r

Density of liquid =ρ

We know that,

The cross section area of cylinder

$A_{1}=\pi\times r^2$

Put the value of r in to the formula

$A_{1}=\pi\times (2r)^2$

$A_{1}=4\pi r^2$

The cross section area of base cylinder in air

$A_{2}=\pi\times r^2$

The cross section area of base cylinder in water

$A_{3}=4\pi\times r^2-\pi\times r^2$

$A_{3}=3\pi\times r^2$

We need  to calculate the value oh height $h_{i}$

Using formula of force acting on cylinder

$(P_{0}+h_{1}\rho g)A_{1}+\dfrac{\rho}{3}gHA_{1}=P_{0}A_{2}+[P_{0}+\rho gh_{1}+\rho g H]A_{3}$

$(P_{0}+h_{1}\rho g)4\pi r^2+\dfrac{\rho}{3}gH4\pi r^2=P_{0}\pi r^2+[P_{0}+\rho gh_{1}+\rho g H]3\pi r^2$

$P_{0}4\pi r^2+h_{1}\rho g4\pi r^2+\dfrac{\rho}{3}\times gH4\pi r^2=P_{0}\pi r^2+P_{0}3\pi r^2+\rho gh_{1}3\pi r^2+\rho g h3\pi r^2$

$\dfrac{\rho hH5\pi r^2}{3}=\rho g h_{1}\pir^2$

$h_{1}=\dfrac{5H}{3}$

Hence, The value of $h_{i}$ is $\dfrac{5H}{3}$.

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