Tag: physics

Quizzes (598) Questions (13441)

Questions Related to physics

Multiple choice physics forces and matter pressure in fluids pressure in liquids introduction to pressure

How much is the hydrostatic pressure exerted by water at the bottom of a beaker? Take the depth of water as 45 cm. (density of water $10^3 kg m^{-3})$.

  1. <span>2410 Pa</span>

  2. <span>3410 Pa</span>

  3. <span>4410 Pa</span>

  4. <span>5410 Pa</span>
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$Height=45 cm=0.45 m$,
$gravity=9.8 ms^{-2}$
$Density=1000 kg m^{-3}$
$Pressure=hdg$
$=0.45\times 9.8\times 1000Pa=4410 Pa$.

Multiple choice physics forces and matter pressure in fluids pressure in liquids introduction to pressure

The pressure in a water pipe on the second floor of a building is 60,000 Pa, and on the third floor it is 30,000 Pa. Find the height of the second floor. (Density of water $=1000 kg m^{-3}, g=10 m s^{-2})$.

  1. <span>3 m</span>

  2. <span>4 m</span>

  3. <span>5 m</span>

  4. <span>6 m</span>
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Second floor :
$P _1=60,000 Pa, g=10 ms^{-2}$
$P _1=h _1dg$


$60,000=h _1\times 1000\times 10$

$h _1=\dfrac {60,000}{1000\times 10}=6 m$

[where $h _1=$ height of water tank above second floor]

Third floor :
$P _2=30,000 Pa, g=10 m s^{-2}$,


$\therefore 30,000=h _2\times 1000\times 19$

$\Rightarrow h _2=\dfrac {30000}{1000\times 10}=3 m$

[where $h _2=$ height of water tank above first floor]

$\therefore $ height of the second floor
$=h _1-h _2=6m-3m=3m$

Multiple choice physics forces and matter pressure in fluids pressure in liquids introduction to pressure

The pressure in water pipe at the ground floor of a building is 120000 Pa, where as the pressure on a third floor is 30000 Pa. What is the height of third floor?
[Take $g=10 m s^{-2}$, density of water $=1000 kg m^{-3}]$.

  1. <span>9 m</span>

  2. <span>10 m</span>

  3. <span>11 m</span>

  4. <span>12 m</span>
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Difference in pressure of water at ground floor and third floor
$=(120000-300000)=900000 Pa$
Density of water $=1000 kg m^{-3}$
Let 'h' be the height third floor.
$P=hdg$
$h=\dfrac {p}{dg}=\dfrac {90000}{1000\times 10}=9m$.

Multiple choice physics forces and matter pressure in fluids pressure in liquids introduction to pressure

The pressure of water on the ground floor is 50000 Pa and at the first floor is 20000 Pa. Find the height of the first floor. Take density of water is $10^3 kg m^{-3}$ and $g=10 ms^{-2}$

  1. <span>2 m</span>

  2. <span>3 m</span>

  3. <span>4 m</span>

  4. <span>5 m</span>
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Pressure on the ground floor $=$ Pressure on the first floor + hdg
$\Rightarrow 50000=20000+hdg$
$\Rightarrow 50000=20000+h\times 10^3\times 10$
$\Rightarrow h\times 10^4=50000-20000$
$\Rightarrow h=\frac {30000}{10000}m$
$\Rightarrow height=3 m.$

Multiple choice physics forces and matter pressure in fluids pressure in liquids introduction to pressure

Calculate the pressure exerted by 0.8 m vertical length of alcohol of density $0.8 g cm^{-3}$. (Acceleration due to gravity $(g)=10 m s^{-2})$.

  1. <span>3200 Pa</span>

  2. <span>6400 Pa</span>

  3. <span>800 Pa</span>

  4. <span>5000 Pa</span>
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Vertical length of the alcohol column
$(h)=0.8 m$
Density of alcohol $(d)=0.8 g cm^{-3}$
$=0.8\times 1000=800 kg m^{-3}$
$[\therefore 1 g cm^{-3}=1000 kg m^{-3}]$
$Pressure = hdg$
$=0.8\times 800\times 10$
$=6400 Pa$
Therefore, pressure exerted by the alcohol column is 6400 Pa.

Multiple choice physics forces and matter pressure in fluids pressure in liquids introduction to pressure

Calculate the pressure exerted by water at the bottom of a lake of depth 6 m. (Density of water $=1000 kg m^{-3}, g=10 ms^{-2})$.

  1. <span>$2\times 10^4 Pa$</span>

  2. <span>$4\times 10^4 Pa$</span>

  3. <span>$6\times 10^4 Pa$</span>

  4. <span>$8\times 10^4 Pa$</span>
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Height $(h)=6 m$,

$density=1000 kg m^{-3}$,
$g=10 m s^{-2}, Pressure = ?$
$P=hdg$
$=6\times 1000\times 10 Pa$
$=60000 Pa$
$=6\times 10^4 Pa.$

Multiple choice physics forces and matter pressure in fluids pressure in liquids introduction to pressure

Calculate the vertical height of a mercury column which exerts a pressure of 81600 Pa. (Density of mercury is $13.6 g cm^{-3}$ and $g=10 ms^{-2})$.

  1. <span>0.2 m</span>

  2. <span>0.4 m</span>

  3. <span>0.6 m</span>

  4. <span>0.8 m</span>
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$Pressure (P)=81600 Pa$
$Density (d)=13.6 g cm^{-3}$
$=13.6\times 1000 kg m^{-3}$
$(\therefore 1 g cm^{-3}=1000 kg m^{-3})$

Acceleration due to gravity $(g)=10 ms^{-2}$
$P=hdg$
$\Rightarrow h=\dfrac {P}{dg}=\dfrac {81600}{13.6\times 1000\times 10}$
$\Rightarrow h=0.6 m$
Therefore, vertical height of mercury column is 0.6 m.

Multiple choice physics forces and matter pressure in fluids pressure in liquids introduction to pressure

A liquid of density $12 kg m^{-3}$ exerts a pressure of 600 Pa at a point inside a liquid. What is height of liquid column above that point? $[g=10 ms^{-2}]$

  1. 4 m

  2. 5 m

  3. 6 m

  4. 7 m
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$Denstiy = 12 kg m^{-3}, Pressure=600 Pa$
$Pressure=hdg$
$\Rightarrow hdg=Pressure\Rightarrow height=\dfrac {P}{dg}$
$=\dfrac {600}{12\times 10}m=5m.$

Multiple choice physics forces and matter pressure in fluids pressure in liquids introduction to pressure

A metal cube is placed in an empty vessel. When water is filled in the vessel so that  cube is completely immersed in the water, the force on the bottom of the vessel in contact with the cube-

  1. Will increase

  2. Will decrease

  3. Will remain the same

  4. Will become zero
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

According to Archimedes' principle, the buoyant force acting on the cube is equal to the weight of the water displaced. Since the cube is resting on the bottom, the total downward force exerted on the bottom of the vessel remains the same because the weight of the water added is balanced by the buoyant force pushing up on the cube.