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Multiple choice physics the kinetic model of matter gases and the kinetic theory concept of ideal gas and state equation of ideal gas behaviour of perfect gas and kinetic theory of gases

An air bubble rises from the bottom of a deep lake the radius of the air bubble near the surface is 'r'. Choose the appropriate radius of the air bubble.

a) r/2 at depth 30m 

b) r/2 at depth 70m

c) r/3 at depth 140m 

d) r/3 at depth 260m

  1. a,c

  2. a,d

  3. b,c

  4. b,d
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

At a depth of h water pressure will be
${P} _{h} = {P} _{a} + \rho gh$                eq(1)
where 
$\rho  =1000 kg/{m}^{3}$
$g  = 9.81 m/{s}^{2} $
$h = \text{depth  of  water  bubble}$
${P} _{a} = {10}^{5}Pa$

initially bubble is below water at h, so
${P} _{1} = {P} _{h}$
finally it rises to surface of lake so
${P} _{2} = {P} _{a}$

we assume that temperature is constant when it is rising from bottom to surface
by ideal gas equation
${P} _{1}{V} _{1} = {P} _{2}{V} _{2}$

volume can be written as $V = \dfrac{4}{3}\pi {r}^{3}$
where r is radius.
${P} _{h} \times \dfrac{4}{3}\pi {r} _{h}^{3} = {P} _{a} \times \dfrac{4}{3}\pi {r} _{s}^{3}$
${r} _{h} = {(\dfrac{{P} _{a}}{{P} _{h}})}^{1/3} r$
${r} _{h} = \dfrac{r}{{(\frac{{P} _{h}}{{P} _{a}})}^{1/3}}$
by eq(1)
${r} _{h} = \dfrac{r}{{(\dfrac{{P} _{a} +  \rho gh}{{P} _{a}})}^{1/3}}$
${r} _{h} = \dfrac{r}{{(1 + 0.0981h)}^{1/3}}$              eq(2)

(a) at depth 30m
h = 30m
put h=30 in eq(2)
${r} _{h} = \dfrac{r}{1.5}$
false

(b) at depth 70m
h = 70m
put h=70 in eq(2)
${r} _{h} = \dfrac{r}{2}$
true

(c) at depth 140m
h = 140m
put h=140 in eq(2)
${r} _{h} = \dfrac{r}{2.4}$
false

(d) at depth 260m
h = 260m
put h=260 in eq(2)
${r} _{h} = \frac{r}{3}$
true

Answer is D.

Multiple choice physics the kinetic model of matter gases and the kinetic theory concept of ideal gas and state equation of ideal gas behaviour of perfect gas and kinetic theory of gases

A barometer reads 75 cm of mercury. When 2.0cm$^{3}$ of air at atmospheric pressure is introduced into space above the mercury level, the volume of the space becomes 50cm$^{3}$. The length by which the mercury column descends is

  1. 3 cm of Hg

  2. 6 cm of Hg

  3. 30 cm of Hg

  4. 10 cm of Hg
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the new pressure inside be $P$ after air introduced.

For the air,
$P _1V _1=P _2V _2$
$\implies P(50)=(75)(2)$
$\implies P=3\ cm\  of\ Hg$

Multiple choice physics the kinetic model of matter gases and the kinetic theory concept of ideal gas and state equation of ideal gas behaviour of perfect gas and kinetic theory of gases

Boyle's law is applicable when

a) temperature is constant

b) gas is at high temperature and low pressure

c) the vessel enclosing the gas is good conductor

d) the process is isothermal

  1. a & b

  2. b,c & d

  3. a,b & c

  4. a,b,c & d
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

(a) Temperature must be constant to apply boyle's law
(b)From vanderwaal's equation when non idealities of a gas is undertaken then we can apply boyle's only when temperture is high and pressure is low
(c)Vessel enclosing must be a good conductor so there is no any possibilities of adiabatic process
(d) Temperature must be constant therefore, process must be isothermal
Hence all are correct
Hence option(D)

Multiple choice physics oscillatory motion motion of a mass suspended by two springs example of simple harmonic motion oscillations due to a spring

A spring of spring constant $k$ is cut into $3$ equal part find $k$ of each

  1. $3k$
  2. $\dfrac{k}{3}$
  3. $k$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The spring constant k is inversely proportional to the length of the spring (k * L = constant). If a spring is cut into 3 equal parts, each part has a length of L/3, so its spring constant becomes 3k.

Multiple choice physics oscillatory motion motion of a mass suspended by two springs example of simple harmonic motion oscillations due to a spring

A block of mass m is suddenly released from the top of a string of stiffness constant k.
(i) The maximum compression in the spring will be
(ii) at equilibrium, the compression in the spring will be .......... 

  1. 2mg/k, mg/k

  2. mg/k, mg/k

  3. mg/k, 2mg/k

  4. 2mg/k, 2mg/k
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When a mass is released suddenly, the maximum compression is 2mg/k due to energy conservation (potential energy lost equals elastic potential energy gained). At equilibrium, the forces balance (mg = kx), resulting in a compression of mg/k.

Multiple choice physics oscillatory motion motion of a mass suspended by two springs example of simple harmonic motion oscillations due to a spring

A block of mass $m$ moving with speed v compresses a spring through distance $x$ before is halved. What is the value of spring constant?

  1. $\dfrac { 3 m v ^ { 2 } } { 4 x ^ { 2 } }$
  2. $\dfrac { m v ^ { 2 } } { 4 x ^ { 2 } }$
  3. $\dfrac { m v ^ { 2 } } { 2 x ^ { 2 } }$
  4. $\dfrac { 2 m v ^ { 2 } } { x ^ { 2 } }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the velocity at starting is $v$.

After compression change in velocity $ = \dfrac{v}{2}$
Here, Initial kinetic energy of a block $ = \left( {\dfrac{1}{2}} \right)m{v^2}$
After compression of spring,
Total energy at the point $x$= Kinetic energy of a block +Potential Energy which stored in the spring.
$\begin{array}{l} \left( { \dfrac { 1 }{ 2 }  } \right) m{ v^{ 2 } }=\dfrac { 1 }{ 2 } m{ \left( { \dfrac { v }{ 2 }  } \right) ^{ 2 } }+\dfrac { 1 }{ 2 } k{ v^{ 2 } } \ \dfrac { 1 }{ 2 } k{ v^{ 2 } }=\dfrac { 1 }{ 2 } m{ \left( { \dfrac { v }{ 2 }  } \right) ^{ 2 } }-\dfrac { 1 }{ 2 } \left( { m{ v^{ 2 } } } \right)  \ k{ x^{ 2 } }=m\left( { { v^{ 2 } }-\dfrac { { { v^{ 2 } } } }{ 4 }  } \right)  \ k{ x^{ 2 } }=\dfrac { { 3m{ v^{ ^{ 2 } } } } }{ 4 }  \ \therefore k=\dfrac { { 3m{ v^{ ^{ 2 } } } } }{ { 4{ x^{ 2 } } } }  \end{array}$

Multiple choice physics oscillatory motion motion of a mass suspended by two springs example of simple harmonic motion oscillations due to a spring

A hollow pipe of length $0.8\ m$ is closed at one end. At its open end, a $0.5\ m$ long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire is $50\ N$ and the speed of sound is $320\ ms^{-1}$, the mass of the string is

  1. $5\ grams$
  2. $10\ grams$
  3. $20\ grams$
  4. $40\ grams$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Velocity of sound $= c$

$\dfrac{c}{4L} = \dfrac{2v}{l}$

$\Rightarrow \dfrac{320}{4\times 0.8} = \dfrac{1}{5} \sqrt{\dfrac{T}{\mu}}$

$\Rightarrow \mu = 0.02\space kgm^{-1}$

$\Rightarrow m = \mu l = 10\space g $