Tag: properties of substances

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Questions Related to properties of substances

Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

By sucking through a straw, a boy can reduce the pressure in his lungs to $750\ mm$ of $Hg$ (density$ = 13.6\,\,g/c{m^3}$). Using a straw, he can drink water from a maximum depth of

  1. $13.6\ cm$
  2. $1.36\ cm$
  3. $0.136\ cm$
  4. $10\ cm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given Pressure in lungs${P} _{lungs}=750$ mm of Hg

and atmospheric pressure${P} _{atm}=760$ mm of Hg
Now pressure difference $\Delta{P}=760-750=10$ mm of Hg=1 cm of Hg
Now
1 cm of Hg$={\rho} _{water}gh$
$1{\rho} _{Hg}g={\rho} _{water}gh$
$13.6g=gh$
$h=13.6 cm$

Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

Maximum density of $H _2O$is at the temperature

  1. $32^oF$
  2. $39.2^oF$
  3. $42^oF$
  4. $4^oF$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Maximum density of water is at $4^0C$


Here, the value in degree celsius is converting to degree Fahrenheit.

$T(°F) = (T(°C) × \dfrac 95 )+ 32$

or

$T(°F) = (T(°C) × 1.8) + 32$

We have,

$T(^0C)=4^0C$

Then,

$T(°F) = (4 × 1.8) + 32=39.2^oF$

Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

A body of density ${d _1}$ is counterpoised by $Mg$ of weights of density ${d _2}$ in air of density $d.$ Then the true mass of the body is

  1. $M$
  2. $M\left( {1-\dfrac{d}{{{d _2}}}} \right)$
  3. $M\left( {1 - \dfrac{d}{{{d _1}}}} \right)$
  4. $\dfrac{{M\left( {1 - d/{d _1}} \right)}}{{\left( {1 - d/{d _2}} \right)}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The true mass M_true is found by equating the forces: M_true * g - V * d * g = M * g - V * d2 * g, where V is volume. Substituting V = M_true / d1 leads to the correct buoyant force balance equation.

Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

A $10L$ container is filled with a gas to a pressure of $2atm$ at $0^0C.$ At what temperature will the pressure inside the container be $2.50atm?$   

  1. $68^0C$
  2. $50^0C$
  3. $40^0C$
  4. $45^0C$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since at both times, container is same . which means that volume remains constant. By applying ideal gas equation.

$PV=nRT$

or $\dfrac{P _{1}}{T _{1}}=\dfrac{P _{2}}{T _{2}}$

Substituting as per question,
$P _{1}=2 atm$ 
$p _{2}=2.5atm$
$T _{1}=0^\circ or \ 273k \ use \ S.I.unit $
$T _{2}=?$

$\dfrac{2}{273}=\dfrac{2.5}{T _{2}}$ or $T _{2}=\dfrac{2.5\times 273}{2}$

$=341.25K$           $(341.25-273=68.25)$
$=68.25^\circ C$

Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

A copper ball of density $8.6 g cm^{-3}$, 1 cm is diameter is immersed in oil of density $0.8cm^{-3}$ . if the ball remains suspended in oil in a uniform electric field of intensity $36000 NC^{-1} $acting in upward direction, what is the charge on the ball ? 

  1. $1.1 \mu C$
  2. $4.2 \mu C$
  3. $2.4 \mu C$
  4. $3.7 \mu C$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For the ball to be suspended: electric force = weight - buoyant force. qE = ρ_ball V g - ρ_oil V g. q = (ρ_ball - ρ_oil)Vg/E. Volume of sphere = (4/3)πr³ = (4/3)π(0.5 cm)³. After unit conversions and calculation, q ≈ 1.1 μC. Option A is correct.

Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

A block of mass 20 kg and volume $ { 10 }^{ 3 }{ cm }^{ 3 }$ is suspended vertically from a ceiling by  a wire . The liner mass density of the its length  is 50 cm. The wire is vibrating in its  fundamental mode and producing beats with a tuning fork f frequency block is just completely immersed in a liquid and vibrated in its fundamental mode , it produces the same number  of beats with  earlier . Density of the liquid is $\left( g={ 10 }{ m/s }^{ 2 } \right) $

  1. $3.8 gm/{ cm }^{ 3 }$
  2. $7.6 gm/{ cm }^{ 3 }$
  3. $1.9 gm/{ cm }^{ 3 }$
  4. $5.0 gm/{ cm }^{ 3 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When block is immersed in liquid, tension decreases by buoyant force. T' = T - ρ_liquid V g. Fundamental frequency f ∝ √T. Beats occur when frequencies differ from tuning fork. For equal beats, the tension ratio gives: (1 - ρ_liquid/20)/1 = 1/4. Solving gives ρ_liquid = 3.8 g/cm³. Option A is correct.

Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

 A body of mass 20.00 g has volume $5.0 cm^3$. The maximum possible error in the measurement of mass and volume respectively are 0.01 g and$0.1 cm^3$. The maximum percentage error in the density will be nearest to 

  1. 1%

  2. 2%

  3. 11%

  4. 25%
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$m = 20.00g$

$v = 5.0\,c{m^2}$
$d = \dfrac{m}{v}$

$ \Rightarrow \dfrac{{\Delta d}}{d} \times 100 = \dfrac{{\Delta m}}{m} \times 100 + \dfrac{{\Delta v}}{v} \times 100$

$ \Rightarrow \dfrac{{\Delta d}}{d} \times 100 = \dfrac{{0.01}}{{20}} \times 100 + \dfrac{{0.1}}{5} \times 100 = \dfrac{1}{{20}} + 2$

$ \Rightarrow \dfrac{{\Delta d}}{d} \times 100 = 2.005$

$\therefore \% $ error in density $ \approx 2\% $
Hence,
option $(B)$ is correct answer.

Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

A metallic sphere with an internal cavity weighs 40 g weight in air and 20 g weight in water. If the density of the material with  cavity be 8 g per $c{m^3}$ then the volume of cavity is:

  1. zero
  2. 15 $c{m^3}$
  3. 5 $c{m^3}$
  4. 20 $c{m^3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$V _s=5cc$ As given

When it is placed in water net force$=$ new downward force
$\Rightarrow Hog-(V+V _s)g=20g\Rightarrow V+V _s=20\ \therefore V=20-V _s cc\V=20-5cm^2=15cm^2$

Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

The density of mercury is $13600 kg m^{-3} $.Its  value in CGS system will be:

  1. $13.6 g cm^{-3}$
  2. $1360 g cm^{-3}$
  3. $136 g cm^{-3}$
  4. $1.36 g cm^{-3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The $C.G.S.$ unit of kg is Gm$.$ and that of meter is cm$.$ When we convert $13600\,kg{m^{ - 3}}$ then $13600 \times 1000\left( {100 \times 100 \times 100} \right) = 13.6\,gc{m^{ - 3}}$

Hence,
option $(A)$ is correct answer.

Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

A solid ball of radius R has a charge density p given by $p=p _0(1 -r/R)$ for $ 0 \leq r \leq R.$ The electric field outside the ball is:

  1. $\dfrac{p _0R^3}{\epsilon _0r^2}$
  2. $\dfrac{p _0R^3}{12\epsilon _0r^2}$
  3. $\dfrac{4p _0R^3}{3\epsilon _0r^2}$
  4. $\dfrac{3p _0R^3}{4\epsilon _0r^2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$According\, to\, question..................... \ q=\int  _{ 0 }^{ R }{ pdv }=\int  _{ 0 }^{ R }{ po }\left( { 1-\dfrac { r }{ R }  } \right) \, .\, 4\pi { r^{ 2 } }drr \ \Rightarrow q=po.4\pi \, \left[ { \int  _{ 0 }^{ R }{ { r^{ 2 } }dr-\int  _{ 0 }^{ R }{ \dfrac { { { r^{ 3 } } } }{ R } dr } } } \right]  \ \Rightarrow q=po.4\pi \left( { \dfrac { { { R^{ 3 } } } }{ 3 } -\dfrac { { { R^{ 3 } } } }{ 4 }  } \right) =po\left( { 4\pi  } \right) \dfrac { { { R^{ 3 } } } }{ { 12 } }  \ \, \, \, \, \, \, \therefore \, \, \, q=\dfrac { { po.\pi { R^{ 3 } } } }{ 3 }  \ E=\dfrac { 1 }{ { 4\pi { \in _{ 0 } } } } \times \, \dfrac { q }{ { { r^{ 2 } } } } =\dfrac { 1 }{ { 4\pi { \in _{ 0 } } } } \, \times \dfrac { { po\, .\pi { R^{ 3 } } } }{ { 3{ r^{ 2 } } } } =\dfrac { { po\, .{ R^{ 3 } } } }{ { 12{ \in _{ 0 } }{ r^{ 2 } } } }  \ \, \, \, \therefore \, \, \, \, E=\dfrac { { po\, .{ R^{ 3 } } } }{ { 12{ \in _{ 0 } }{ r^{ 2 } } } }  \ there\, for\, the\, correct\, option\, is\, B.$