JEE MAIN - Physics (2019 - 9th April Evening Slot - No. 9)
A wooden block floating in a bucket of water
has 4/5 of its volume submerged. When certain
amount of an oil is poured into the bucket, it
is found that the block is just under the oil
surface with half of its volume under water and
half in oil. The density of oil relative to that of
water is :-
0.8
0.7
0.6
0.5
Explanation
In 1st situation
Vb$$\rho $$bg = Vs$$\rho $$wg
$${{{V_s}} \over {{V_b}}} = {{{\rho _b}} \over {{\rho _w}}} = {4 \over 5}$$ ...(i)
Here Vb is volume of block
Vs is submerged volume of block
$$\rho _b$$ is density of block
$$\rho _w$$w is density of water & Let $$\rho _o$$ is density of oil
Finally in equilibrium condition
Vb$$\rho _b$$g = $${{{V_b}} \over 2}{\rho _o}g + {{{V_b}} \over 2}{\rho _w}g$$
$$2{\rho _b} = {\rho _0} + {\rho _w}$$
$$ \Rightarrow {{{\rho _o}} \over {{\rho _w}}} = {3 \over 5} = 0.6$$
Vb$$\rho $$bg = Vs$$\rho $$wg
$${{{V_s}} \over {{V_b}}} = {{{\rho _b}} \over {{\rho _w}}} = {4 \over 5}$$ ...(i)
Here Vb is volume of block
Vs is submerged volume of block
$$\rho _b$$ is density of block
$$\rho _w$$w is density of water & Let $$\rho _o$$ is density of oil
Finally in equilibrium condition
Vb$$\rho _b$$g = $${{{V_b}} \over 2}{\rho _o}g + {{{V_b}} \over 2}{\rho _w}g$$
$$2{\rho _b} = {\rho _0} + {\rho _w}$$
$$ \Rightarrow {{{\rho _o}} \over {{\rho _w}}} = {3 \over 5} = 0.6$$
Comments (0)
