Tag: floating bodies

Quizzes (12) Questions (71)

Questions Related to floating bodies

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

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.

  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
Reveal answer Fill a bubble to check yourself
C Correct answer
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

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

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

  1. Increases by $2\%$
  2. Decreases by $2\%$

  3. Remains unchanged

  4. Sometime increases and sometime decreases
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $2\%$.
Multiple choice physics floating bodies pressure exerted by a liquid column pressure at a certain depth in liquid variation of pressure with depth

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:

  1. $2400$
  2. $2.4 \times {10^9}$
  3. ${10^{ - 11}} \times 6 \times {10^{23}}$
  4. $2.68 \times {10^{19}} \times {10^{11}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using the ideal gas law PV = NkT, where P = 10^-11, T = 300K, and k = 1.38 * 10^-23. For a unit volume (1 m^3), N = P / kT = 10^-11 / (1.38 * 10^-23 * 300) approx 2.4 * 10^9.

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

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=$10 m/s^2]$

  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}\ $
Reveal answer Fill a bubble to check yourself
A Correct answer
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
Multiple choice physics floating bodies pressure exerted by a liquid column pressure at a certain depth in liquid variation of pressure with depth

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 $g=10ms^{-2}$)

  1. $1.6 \times 10^3Nm^{-2}$
  2. $3.2\times 10^{-3}$
  3. $1.6 \times 10^4Nm^{-2}$
  4. $3.2 \times 10^4Nm^{-2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The tank is 4m high and half filled with water, meaning 2m of water and 2m of the other liquid. The pressure at the bottom is the sum of the hydrostatic pressures due to both liquid columns, calculated as P = h1*d1*g + h2*d2*g. Substituting the given values yields 1.6 x 10^4 N/m^2.

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

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$^{2}$):

  1. 0.77 km

  2. 7.7 km

  3. 77 km

  4. 0.077 km
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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

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

  1. $\displaystyle \dfrac{h}{dg}$
  2. $hdg$
  3. $\displaystyle \dfrac{h}{d}$
  4. $hg$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.