Tag: fundamental and derived units

Questions Related to fundamental and derived units

Maximum density of $H _2O$is at the temperature

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

  1. $32^oF$

  2. $39.2^oF$

  3. $42^oF$

  4. $4^oF$

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Correct Option: B
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,

$T(°F) = (4 × 1.8) + 32=39.2^oF$

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

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

  1. $M$

  2. $M\left( {1-\dfrac{d}{{{d _2}}}} \right)$

  3. $M\left( {1 - \dfrac{d}{{{d _1}}}} \right)$

  4. $\dfrac{{M\left( {1 - d/{d _1}} \right)}}{{\left( {1 - d/{d _2}} \right)}}$

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Correct Option: D

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 $2.50atm?$   

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

  1. $68^0C$

  2. $50^0C$

  3. $40^0C$

  4. $45^0C$

Show answer
Correct Option: A
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 $T _{2}=\dfrac{2.5\times 273}{2}$

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

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 ? 

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

  1. $1.1 \mu C$

  2. $4.2 \mu C$

  3. $2.4 \mu C$

  4. $3.7 \mu C$

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Correct Option: A

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) $

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

  1. $3.8 gm/{ cm }^{ 3 }$

  2. $7.6 gm/{ cm }^{ 3 }$

  3. $1.9 gm/{ cm }^{ 3 }$

  4. $5.0 gm/{ cm }^{ 3 }$

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Correct Option: A

 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 

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

  1. 1%

  2. 2%

  3. 11%

  4. 25%

Show answer
Correct Option: B
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.

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:

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

  1. zero

  2. 15 $c{m^3}$

  3. 5 $c{m^3}$

  4. 20 $c{m^3}$

Show answer
Correct Option: B
Explanation:

$V _s=5cc$ As given

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

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

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

  1. $13.6 g cm^{-3}$

  2. $1360 g cm^{-3}$

  3. $136 g cm^{-3}$

  4. $1.36 g cm^{-3}$

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Correct Option: A
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.

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:

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

  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. $\dfrac{3p _0R^3}{4\epsilon _0r^2}$

Show answer
Correct Option: B
Explanation:

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

A body of uniform cross-sectional area floats in a liquid of dentisty thrice its value. The portion of exposed height will be:

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

  1. 2/3

  2. 5/6

  3. 1/6

  4. 1/3

Show answer
Correct Option: A