Tag: floating bodies

Questions Related to floating bodies

We use a lactometer to test whether the milk is pure or not. True or false 

physics floating bodies hydrometer and lactometer comparing density - relative density principle of floatation and its applications

  1. True

  2. False

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Correct Option: A
Explanation:

Lactometer is a device use to test the purity of milk by the virtue of specific gravity of cow's milk according to standards.

Lactometer is graduated into how many parts

physics floating bodies hydrometer and lactometer comparing density - relative density principle of floatation and its applications

  1. Hundred

  2. thousand

  3. ten

  4. one

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Correct Option: A
Explanation:

Lactometer is graduated into hundred parts.


Hence, option A is correct.

Milk purity is measured using:

physics floating bodies hydrometer and lactometer comparing density - relative density principle of floatation and its applications

  1. Hydrometer

  2. Lactometer

  3. Polarimeter

  4. Spectrometer

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Correct Option: B
Explanation:

Lactometer is an instrument used for measuring the purity of milk.

When a lactometer is used in a given sample of milk 'A' it floats in the milk such that half of the length of its stem is immersed in the milk. When it  is used in another sample of milk 'B', the length of the stem immersed in the milk is one fourth of the total length. Then which of the following statement(s) is(are) true?
(A) Water content in A is more than that in B.
(B) Density of the sample A is less than the density of the sample B.

physics floating bodies hydrometer and lactometer comparing density - relative density principle of floatation and its applications

  1. A is true, but B is false

  2. A and B are false

  3. A and B are true

  4. A is false but B is true

Show answer
Correct Option: C
Explanation:

As lactometer sinks less in case of B , we can easily say that the density of sample B is more than that of sample A , hence water content in A would be more as milk is denser than water .


so both statements (A) and (B) are correct .
so option (C) is correct

If the value of $g$ at a place is decreased by $2\%$. The barometric height of the mercury 

physics floating bodies pressure exerted by a liquid column pressure at a certain depth in liquid variation of pressure with depth

  1. Increases by $2\%$

  2. Decreases by $2\%$

  3. Remains unchanged

  4. Sometime increases and sometime decreases

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Correct Option: A
Explanation:
We know that,

$P=h\rho g$

Then,

$h=\dfrac{P}{\rho g}$

So, $h\propto \dfrac 1g$

If the value of $g$ decreased by $2\%$ then $h$ will increase by $2\%$.

The residual pressure of a vessel at ${27^0}C$ is  $1 \times {10^{ - 11}}N/{m^2}$. The number of molecules in this vessel is nearly:

physics floating bodies pressure exerted by a liquid column pressure at a certain depth in liquid variation of pressure with depth

  1. $2400$

  2. $2.4 \times {10^9}$

  3. ${10^{ - 11}} \times 6 \times {10^{23}}$

  4. $2.68 \times {10^{19}} \times {10^{11}}$

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Correct Option: B

If pressure at the half depth of a lake is equal to $\dfrac{3}{4}$ times the pressure at its bottom, then find the depth of the lake . [Take g=$10 m/s^2]$

physics floating bodies pressure exerted by a liquid column pressure at a certain depth in liquid variation of pressure with depth

  1. $ \dfrac{P _{0}}{\rho g}\ $

  2. $ \dfrac{2P _{0}}{\rho g}\ $

  3. $ \dfrac{P _{0}}{2\rho g}\ $

  4. $ \dfrac{3P _{0}}{\rho g}\ $

Show answer
Correct Option: A
Explanation:
Let depth of the lake be $h$ and pressure at bottom $= P$
Then $P=P _{0}+\rho gh\rightarrow (1)$    $(P _{0}=$ atmospheric pressure, $\rho $ = density of water)
At half depth $(h/2)$ pressure is $\dfrac{3P}{4}$ then :
$\dfrac{3P}{4}=P _{0}+\rho g\dfrac{h}{2}\rightarrow (2)$
On subtracting equation 2 from 1 we get :
$\dfrac{P}{4}=\rho g\dfrac{h}{2}$
$\Rightarrow P=2\rho gh$, substituting this value of $P$ in equation 1:
$2\rho gh=P _{0}+\rho gh$
$\Rightarrow h=\dfrac{P _{0}}{\rho g}\rightarrow $ Depth of the lake

A tank $4m$ high is half filled with water then filled to the top with a liquid of density $0.60 g/cc$ what is the pressure at the bottom of the tank due to these liquids? (take $g=10ms^{-2}$)

physics floating bodies pressure exerted by a liquid column pressure at a certain depth in liquid variation of pressure with depth

  1. $1.6 \times 10^3Nm^{-2}$

  2. $3.2\times 10^{-3}$

  3. $1.6 \times 10^4Nm^{-2}$

  4. $3.2 \times 10^4Nm^{-2}$

Show answer
Correct Option: C

If the air density were uniform, then the height of the atmosphere above the sea level to produce a normal atmospheric pressure of 1.0 x 10$^{5}$ Pa is(density of air is 1.3 kg/m$^{3}$ , g $=$ 10m/s$^{2}$):

physics floating bodies pressure exerted by a liquid column pressure at a certain depth in liquid variation of pressure with depth

  1. 0.77 km

  2. 7.7 km

  3. 77 km

  4. 0.077 km

Show answer
Correct Option: B
Explanation:

$\rho g h = 1 \times 10^5$
$\Rightarrow h = \dfrac {10^5}{\rho g} = \dfrac {10^5}{1.3 \times 10} = 7.7 km$

The pressure exerted by a liquid at depth $h$ is given by:

physics floating bodies pressure exerted by a liquid column pressure at a certain depth in liquid variation of pressure with depth

  1. $\displaystyle \dfrac{h}{dg}$

  2. $hdg$

  3. $\displaystyle \dfrac{h}{d}$

  4. $hg$

Show answer
Correct Option: B
Explanation:

Answer is B.

The pressure exerted by a liquid at a point depends on its vertical depth and density of the liquid only. It is independent of the shape of the container. The pressure at the bottom of the three vessels of different shapes containing the same liquid acts equally in all directions.
Thus, the pressure exerted by the liquid of height $h$ is given as P=hdg, where $h$ is the height, $d$ is the density and $g$ is the acceleration due to gravity.