Tag: physics

Questions Related to physics

A piece of iron of density $7800kg/m^3$ and volume $100cm^3$ is completely immersed in water. Calculate the upthrust on the iron piece.

[Take, $g=10m/s^2$.]

physics upthrust in fluids, archimedes' principle and floatation upthrust is equal to the weight of displaced liquid pressure in fluids buoyancy

  1. $1N$

  2. $2.8N$

  3. $7.8N$

  4. $6.8N$

Show answer
Correct Option: A

if a block of wood is floating in a river water, then the apparent weight of the floating block is

physics upthrust in fluids, archimedes' principle and floatation upthrust is equal to the weight of displaced liquid pressure in fluids buoyancy

  1. equal to the weight of the displaced water

  2. zero.

  3. greater than the weight of the displaced water

  4. equal to the actual weight of the block

Show answer
Correct Option: B
Explanation:
Apparent weight in calculated as
Apparent weight$=$(weight of body)$-$(Buoyancy force)
Where
weight of body$=$(mass of body)$\times$(Gravitational force)
Buoyancy force$=$weight of fluid displaced
for the floating body:-
weight of body$=$Buoyancy force
$\therefore$ The body floating and
apparent weight equals to zero.
$\therefore$ Answer is B.

A body of mass 50 kg has a volume 0.049 $m^3$ and is kept on a table. The buoyant force on it due to air is.

physics upthrust in fluids, archimedes' principle and floatation upthrust is equal to the weight of displaced liquid pressure in fluids buoyancy

  1. 0.588 N

  2. 0.49 N

  3. 58.8 N

  4. 49 kgf

Show answer
Correct Option: D
Explanation:

since the mass is in equilibrium. So, the net downward force =buoyant force

buoyant force=50x9.8=49kgf

Calculate the upthrust acting on a ball of volume $\displaystyle 0.2{ m }^{ 3 }$ immersed in a liquid having density $\displaystyle { 1kg }/{ { m }^{ 3 } }$.

physics upthrust in fluids, archimedes' principle and floatation upthrust is equal to the weight of displaced liquid pressure in fluids buoyancy

  1. 0.2f

  2. 2f

  3. 0.2g f

  4. 2g f

Show answer
Correct Option: C
Explanation:

Upthrust = weight of the liquid displaced by the submerged part of the body Upthrust = mass of liquid displaced x Acceleration due to gravity Upthrust = volume of liquid displaced x density of liquid displaced x Acceleration due to gravity,Since volume of solid immersed is equal to the volume of the liquid displaced, Upthrust = volume of solid immersed x density of liquid displaced x Acceleration due to gravity Upthrust = 0.2 x 1 x g = 0.2g f.

A cube of wood floats in water, with $42$% of its volume is submerged, then the density of the wood is

physics upthrust in fluids, archimedes' principle and floatation upthrust is equal to the weight of displaced liquid pressure in fluids buoyancy

  1. $42\ g\ cm^{-3}$

  2. $0.42\ g\ cm^{-3}$

  3. $0.58\ kg\ cm^{-3}$

  4. $600\ g\ cm^{-3}$

Show answer
Correct Option: B
Explanation:

By Archimedes Principle, the buoyant force on a body partially or fully immersed in a fluid is given by the weight of the fluid displaced.


Let the volume of wood be $V$
Thus, volume of wood submerged is $0.42V$

Thus, the buoyant force acting on the wood is $B = 0.42\rho gV$
Weight of wood is $\rho _\textrm{wood}gV$

Thus, in equilibrium, $0.42\rho gV = \rho _\textrm{wood}gV \Rightarrow \rho _\textrm{wood} = 0.42\rho$

As Density of water is $\rho = 1\textrm{ g cm}^{-3}$, we have $\rho _\textrm{wood} = 0.42 \textrm{ g cm}^{-3}$

A: Diver dives in and reaches a depth of 100m.
B: Diver swims up from the depth of 100m to the surface.
Choose the correct alternative:

physics upthrust in fluids, archimedes' principle and floatation upthrust is equal to the weight of displaced liquid pressure in fluids buoyancy

  1. A is easier than B

  2. B is easier than A

  3. A and B are equally hard

  4. A is easier than B if speed of descent and ascent is same and more.

Show answer
Correct Option: A
Explanation:
Assume:
$F _d: \text{Force applied by diver during descent in downward direction}$
$F _u: \text{Force applied by diver during ascent in upward direction}$
$U: \text{Upthrust}$
$m: \text{Mass of the diver}$
$g: \text{Acceleration due to gravity}$
Uniform ascent and descent.

Diver has more density than water. Hence, weight of diver is more than the upthrust and without any effort, the diver sinks.
i.e. $mg>U..................(1)$

During upward motion, $F _u+U-mg=0............(2)$
During downward motion, $F _d-U+mg=0...............(3)$

From (1),(2) and (3), 
$F _u-F _d=2mg-2U$
$F _u-F _d>0$
$F _u>F _d$

Mathematical proof of upthrust is based on 

physics upthrust in fluids, archimedes' principle and floatation upthrust is equal to the weight of displaced liquid pressure in fluids buoyancy

  1. Definition of pressure

  2. Weight of object

  3. Pressure exerted by a column of fluid

  4. Viscosity

Show answer
Correct Option: C
Explanation:
Mathematical proof:
Consider a cylinder of cross section area $A$ and height $L$ completely submerged in water. Let depth of upper surface be $h$.
Using, pressure exerted by fluid column:
Force on the upper face of the cylinder = $hρgA$
Force on the lower face of the cylinder = $[h + L]ρgA$
Difference in force = $LAρg$

But $LA$ is the volume of liquid displaced by the cylinder, and $LrgA$ is the weight of the liquid displaced by the cylinder.

Therefore there is a net upward force on the cylinder equal to the weight of the fluid displaced by it.

150 g of ice is mixed with 100 g of water at temperature $80^oC$. The latent heat of ice is 80 ca/g and the specific heat of water is $1 cal/g-^oC$. Assuming no heat loss to the environment, the amount of ice which does not melt is

physics heat - measurement introduction to temperature application of various thermometric scales different types of thermometers

  1. 100 g

  2. 0 g

  3. 150 g

  4. 50 g

Show answer
Correct Option: D
Explanation:

Heat given by water $=100\times 1\times 800=8000 cal$
Heat taken by ice $=8000 cal=m\times 80$
$m=100 gm$
So amount of ice which does not melt $=150-100=50 gm$.

The following three objects (1) a metal tray, (2) a block of wood,and (3) a wooden cap are left in a closed room overnight. Next day the temperature of each is recorded as $T _1, T _2$ and $T _3$ respectively. The likely situation is 

physics heat - measurement introduction to temperature application of various thermometric scales different types of thermometers

  1. $ T _1 = T _2 = T _3 $

  2. $ T _3 > T _2 > T _1 $

  3. $ T _3 = T _2 > T _1 $

  4. $ T _3 > T _2 = T _1 $

Show answer
Correct Option: A
Explanation:

All three will be inthermal equilibrium with air of room. so temprature of the three will be same

A new temperature scale uses X as a unit of temperature, where the numerical value of the temperature t$ _x$ in this scale is related to the absolute temperature T by t$ _x$ = 3T + 300. If the specific heat of a material using this unit is 1400 J kg$^{-1}$ X${^-1}$ its specificne in the S.I. system of units is

physics heat - measurement introduction to temperature application of various thermometric scales different types of thermometers

  1. 4200 J kg$^{-1}$ K$^{-1}$

  2. 1400 J kg$^{-1}$ K$^{-1}$

  3. 466.7 J kg$^{-1}$ K$^{-1}$

  4. impossible to determine from the information provided

Show answer
Correct Option: A
Explanation:

$t _x = 3 T + 300                 Q = ms \Delta 0$
$\Delta t _x = 3 \Delta T$
$\displaystyle S = \frac{Q}{m \Delta \theta}                 S = \frac{Q}{m \Delta \theta}$
Since, unit of 8 is joule in both system
            X                                    T
$m = m _0 kg$                   $m _0 kg$
$Q = Q _0 J$                       $Q _0  J$
$\Delta t _x$                               $\Delta T$
$S _x = \displaystyle \frac{Q _0}{m _0 \Delta t _x} = 1400     S _r = \frac{Q _0}{m _0 \Delta T} = \frac{3 Q}{m _0 \Delta t _x}$
$S _r = 3 \times 1400 = 4200 J-kg^{-1} K^{-1}$