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Questions Related to physics

A vessel contains a mixture of $7g$ of nitrogen and $8g$ of oxygen at temperature $T=300K$. If the pressure of the mixture is $1atm$, its density is 
$\left[ R=\cfrac { 25 }{ 3 } J/mol\quad K \right] $

physics upthrust in fluids, archimedes' principle and floatation density of a fluid density of fluid density and relative density

  1. $0.6kg/{m}^{3}$

  2. $1.2kg/{m}^{3}$

  3. $1.5kg/{m}^{3}$

  4. $2kg/{m}^{3}$

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Correct Option: A
Explanation:

Molar mass of mixture $=\cfrac{total mass}{total mole}$

$M=\cfrac{15}{\cfrac{4}{14}+\cfrac{8}{16}}=15\M=15\ PM=\rho RT\ \rho=\cfrac{1(15)\times10^{-3}}{(\cfrac{25}{3}300)}\ \rho=\cfrac{15}{25}=\cfrac{3}{5}=0.6kg/m^3$

The mass and volume of a body are found to be 5.00 $\pm$ 0.05 kg and 1.00 $\pm$ 0.05 $m^3$  respectively. Then the maximum possible percentage error in its density is

physics upthrust in fluids, archimedes' principle and floatation density of a fluid density of fluid density and relative density

  1. 6%

  2. 3%

  3. 10%

  4. 5%

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Correct Option: A

The volume of a cube is $\displaystyle 2.5{ cm }^{ 3 }$ and its mass is 20g. Calculate the density of the cube in MKS and CGS systems.

physics upthrust in fluids, archimedes' principle and floatation density of a fluid density of fluid density and relative density

  1. $\displaystyle 0.008{ g }/{ { cm }^{ 3 } }and\quad 8{ kg }/{ { m }^{ 3 } }$

  2. $\displaystyle 8000{ g }/{ { cm }^{ 3 } }and\quad 8{ kg }/{ { m }^{ 3 } }$

  3. $\displaystyle 8{ g }/{ { cm }^{ 3 } }and\quad 8000{ kg }/{ { m }^{ 3 } }$

  4. $\displaystyle 0.8{ g }/{ { cm }^{ 3 } }and\quad 800{ kg }/{ { m }^{ 3 } }$

Show answer
Correct Option: C
Explanation:

Explanation: In MKS system i.e. the SI system the unit of density is $\displaystyle { kg }/{ { m }^{ 3 } }$ and in CGS system it is $\displaystyle { g }/{ { cm }^{ 3 } }$. Density = mass/volume = $\displaystyle 20/2.5=8{ g }/{ { cm }^{ 3 } }$ in CGS system. In MKS system density = $\displaystyle 8\times 1000=8000{ kg }/{ { m }^{ 3 } }$

A goldsmith desires to test the purity of a gold ornament suspected to the mixed with copper. The ornament weights $0.25\ kg$ in air and is observe to displace $0.015$ litre of water when immersed in it. Densities of gold and copper with respect to water are, respectively, $19.3$ and $8.9$. The approximate percentage of copper in the ornament is

physics upthrust in fluids, archimedes' principle and floatation density of a fluid density of fluid density and relative density

  1. $5\%$

  2. $10\%$

  3. $15\%$

  4. $25\%$

Show answer
Correct Option: D
Explanation:

Let volume of gold in ornament $V _1$ and that of copper $=V _2$ and density of gold $=\rho g$ and that of copper$=\rho _c$

$\Rightarrow \rho g V _1+\rho _cV-2=0.25\rightarrow (1)$
Volume of ornament $=$Volume of water displaced
$\Rightarrow V _1+V-2=0.15\times10^{-3}ms\Rightarrow V _2=(0.015\times10^{-3}-V _1)$
According to question-
$\cfrac{\rho _g}{\rho _w}=19.3$ and $\cfrac{\rho _c}{\rho _w}=8.9\ \rho _g=19.3\times10^3 kg/m^3$
$\rho _c=8.9\times10^3 kg/m^3$
Putting these value in equation $(1)$
$(19.3\times 10^{ 3 })V _{ 1 }+(1.9\times 10^{ 3 })(0.0015\times 10^{ -3 }-V _{ 1 })0.25\ [(19.3\times 10^{ 3 })-(1.9\times 10^{ 3 })]V _{ 1 }+(8.9\times 0.015)=0.25\ 10.4\times 10^{ 3 }V _{ 1 }=0.12\ V _{ 1 }=\cfrac { 0.12 }{ 10.4\times 10^{ 3 } } =0.0115\times 10^{ -3 }=1.15\times 10^{ -5 }m^{ 3 }$
$V _2=(0.015\times10^3-0.0115\times10^{-3})=0.0035\times10^{-3}\ \% \quad of\quad copper=(\cfrac{V _2}{V _1+V _2})\times100=[\cfrac{0.0035\times10^{-3}}{(0.0115+0.0035)\times10^{-3}}]\times100$
$\approx 23.34\%\ \approx 25\%$

A liquid mixture of volume $V$ has two liquids as its ingredients with densities $\alpha  \; and\; \beta $. If the density of the mixture is $\sigma $, then the mass of the first liquid in the mixture is :

physics upthrust in fluids, archimedes' principle and floatation density of a fluid density of fluid density and relative density

  1. $\dfrac{\alpha V[\sigma \beta +1]}{\beta [\alpha +\sigma ]}$

  2. $\dfrac{\alpha V[\sigma -\beta ]}{ [\sigma +\beta]}$

  3. $\dfrac{\alpha V[\beta-\sigma ]}{ \beta-\alpha }$

  4. $\dfrac{\alpha V[1-\sigma\alpha ]}{ \beta[\alpha-\sigma ] }$

Show answer
Correct Option: C
Explanation:
Let mass of liquid with density $\alpha =M _1$
mass of liquid with density $\beta =M _2$
Total volume$=V$
Net density of mixture$=\sigma$
Total mass$=M _1+M _2$
$\Rightarrow V\sigma =M _1+M _2$
$\Rightarrow M _2=V\sigma -M _1$ ......$(1)$

$\left[\because \dfrac{Total \, Mass}{v}=\sigma\right]$

$T=\dfrac{Total \,mass}{Total \, volume}=\dfrac{M _1+M _2}{\dfrac{M _1}{\alpha}+\dfrac{M _2}{\beta}}$ .......$(2)$
sub $(1)$ in $(2)$

$\Rightarrow \sigma =\dfrac{M _1+(v\sigma -M _1)}{\dfrac{M _1}{\alpha}+\left(\dfrac{v\sigma -M _1}{\beta}\right)}$

$\Rightarrow M _1=\dfrac{\alpha V(\beta -\sigma)}{\beta -\alpha}$.

A long straight cable of length l is placed symmetrically along z-axis and has radius $a(<<l)$. the cable consists of a thin wire and co-axial conducting tube. An alternating current I(t) = ${ I } _{ 0 }\  \sin { \  (2\pi \nu t) }$flows down the central thin wire and returns along the co-axial conducting tube. The induced electric field at a distance s from the wire inside the cable is E(s, t) = ${ \mu  } _{ 0 }{ I } _{ 0 }\cos { (2\pi \nu t)\  ln \left( \dfrac { s }{ a }  \right) \hat{ k } }$. The displacement current density inside the cable is

physics upthrust in fluids, archimedes' principle and floatation density of a fluid density of fluid density and relative density

  1. $\dfrac { 2\pi }{ { \lambda }^{ 2 } } { I } _{ 0 }ln \left( \dfrac { a }{ s } \right) \sin { (2\pi \nu t) } \hat{ k }$

  2. $\dfrac { 1 }{ { \lambda }^{ 2 } } { I } _{ 0 }ln \left( \dfrac { a }{ s } \right) \sin { (2\pi \nu t) } \hat{ k }$

  3. $\dfrac { \pi }{ { \lambda }^{ 2 } } { I } _{ 0 }ln \left( \dfrac { a }{ s } \right) \sin { (2\pi \nu t) } \hat{ l }$

  4. Zero

Show answer
Correct Option: A
Explanation:

Given :  The induced electric field at a distance from wire is $E(s,t)=\mu _0I _0v\,\cos(2\pi v t)ln(\dfrac{s}{a})\hat{k}$

Displacement current density is given by :
$\vec{J} _d=\epsilon _0\dfrac{dE}{dt}$

$\Rightarrow \epsilon _0\mu _0I _0v \begin{matrix}\partial  \\partial(t)  \end{matrix}(\cos 2\pi vt)(\begin{matrix}  \dfrac{s}{a} \end{matrix})\hat{k}$

Substitute for $\dfrac{1}{\sqrt{\mu _0\epsilon _0}}$

$\Rightarrow \dfrac{1}{c^2}I _0 2\pi v^2(-\sin(2\pi vt))ln(\begin{matrix}  \dfrac{s}{a} \end{matrix})\hat{k}$

$\Rightarrow (\begin{matrix} \dfrac{v} {c}\end{matrix})^2  2\pi I _0 \sin(2\pi vt))ln(\begin{matrix}  \dfrac{a}{s}\end{matrix})\hat{k}$

$\Rightarrow \dfrac{2\pi}{\lambda^2}I _0 ln (\begin{matrix} \dfrac{a}{ s} \end{matrix})\sin (2\pi vt)\hat{k}$

Option (A) is correct.

In India, a nuclear power plant is located in 

physics towards green energy electricity from nuclear energy nuclear power nuclear energy

  1. Rampur

  2. Tarapur

  3. Sahranpur

  4. Kanpur

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Correct Option: B
Explanation:
Nuclear power plants in India are: 
1. Rawatbhata, Rajasthan
2. Tarapur, Maharashtra
3. Kaiga, Karnataka
4. Kudankulam, Tamil Nadu
5. Kalpakkam, Tamil Nadu
6. Kakrapar, Gujarat
7. Narora, Uttar Pradesh

The heat energy produced in the nuclear fission reactions can be used to produce

physics towards green energy electricity from nuclear energy nuclear power nuclear energy

  1. Solar energy

  2. Electricity

  3. Nuclear energy

  4. None of these

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Correct Option: B

Which is used as 'coolants' in a nuclear power station?

physics towards green energy electricity from nuclear energy nuclear power nuclear energy

  1. water and nitrogen

  2. liquid sodium and carbon dioxide

  3. liquid ammonia and carbon monoxide

  4. heavy water and uranium oxide

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Correct Option: B
Explanation:

Coolant is used to remove the heat produced from the reactor core. This heat is then transferred to the electrical generators or the environment. Water, liquid sodium, helium and carbon dioxide are most commonly used coolants.

Which of the following will produce more energy in a short time?

physics towards green energy electricity from nuclear energy nuclear power nuclear energy

  1. Nuclear fission

  2. Nuclear fusion

  3. Both will produce the same amount of energy

  4. Nothing can be decided

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Correct Option: B