Tag: anomalous expansion of water

Questions Related to anomalous expansion of water

Assertion(A): Real expansion of liquid does not depend up on material of container.

Reason (R): Liquids have no definite shape. They acquire the mouth of the containers in which they are taken.

physics heat and temperature thermal expansion in liquids anomalous expansion of water thermal expansion in solids, liquids and gases

  1. Both A and R are true and R is the correct explanation of A

  2. Both A & R are true but R is not the correct explanation of A.

  3. A is true but R is false

  4. A is false but R istrue

Show answer
Correct Option: B
Explanation:

Real expansion would not be affected by the size and shape of container. Because of liquid has no any definate shape.
Both assertion and reason are true and reason is correct explanation.

(1): when a liquid with a coefficient of $\gamma $ is heated in a vessel of a coefficient of linear expansion $\frac{\gamma }{3}$, the level of liquid in the vessel remains unchanged.

 (2):$\gamma _{a}=\gamma _{r}-\gamma _{g}=\gamma _{r}-3\alpha =0$

physics heat and temperature thermal expansion in liquids anomalous expansion of water thermal expansion in solids, liquids and gases

  1. Both 1 and 2 are true and 2 is the correct explanation of 1

  2. Both 1 & 2 are true but 2 is not the correct explanation of 1.

  3. 1 is true but 2 is false

  4. 1 is false but 2 is true

Show answer
Correct Option: A
Explanation:

Coefficient of volume expansion is thrice the coefficient of linear expansion. Hence the coefficient of expansion of both liquid and vessel is same. Hence they expand equally with rise in temperature. Therefore no rise in the level of liquid in vessel is observed.

A weight thermometer contains 52 gms of a liquid at 10$^{0}$C. When it is heated to 110$^{0}$C, 2 gm of the liquid is expelled. The coefficient of real expansion of the liquid is [$\alpha $ of glass is $9\times 10^{-6/0}C$]

physics heat and temperature thermal expansion in liquids anomalous expansion of water thermal expansion in solids, liquids and gases

  1. $27\times 10^{-6/0}C$

  2. $412\times 10^{-6/0}C$

  3. $373\times 10^{-6/0}C$

  4. 0

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Correct Option: B
Explanation:

$V _{expelled} = V _0(\gamma _l -\gamma _g)\Delta T$
$ \Rightarrow  2 = 52 \times ( \gamma _l -3 \times \alpha _g)\times(110-10)$
$ \Rightarrow 2 =52 \times (\gamma _l - 3 \times  9 \times 10^{-6} ) \times 100$
$ \Rightarrow \gamma _l = \frac{2}{52} \times 10^{-2} + 27 \times 10^{-6}=385 \times 10^{-6} + 27 \times 10^{-6}=412 \times 10^{-6}$

When a liquid in a glass vessel is heated, its apparent expansion is $10.30\times 10^{-4}/^o$ C. Same liquid when heated in a metallic vessel, its apparent expansion is $10.06\times 10^{-4}/^o$ C. The coefficient of linear expansion of metal is :

($\alpha   _{glass}=9\times 10^{-6}/^o$ C)

physics heat and temperature thermal expansion in liquids anomalous expansion of water thermal expansion in solids, liquids and gases

  1. $51\times 10^{-6}/^oC$

  2. $43\times 10^{-6}/^oC$

  3. $25\times 10^{-6}/^oC$

  4. $17\times 10^{-6}/^oC$

Show answer
Correct Option: D
Explanation:

${\gamma} _r={{\gamma} _{a1}} +{\gamma} _{glass}={{\gamma} _{a2}} +{\gamma} _{metal}$  ......(1)
where ${{\gamma} _{a1}}$ is the apparent  expansion coefficients when the the liquid is kept in the glass container and ${{\gamma} _{a2}}$is the apparent expansion coefficients when the the liquid is kept in the metal container
Given,
${{\gamma} _{a1}}=10.30 \times {10}^{-4}$
${{\gamma} _{a2}}=10.06 \times {10}^{-4}$
${\alpha} _{glass}=9 \times {10}^{-6}$
${\gamma} _{glass}=3{\alpha} _{glass}=27 \times {10}^{-6}$
${\gamma} _{metal}=3{\alpha} _{metal}$


Substituting the values in (1)
$10.30 \times {10}^{-4} + 27 \times {10}^{-6}=10.06 \times {10}^{-4} + 3{\alpha} _{metal}$
$ \Rightarrow 24 \times {10}^{-6}  + 27 \times{10}^{-6} = 3{\alpha} _{metal}$
$ \Rightarrow {\alpha} _{metal}=(\dfrac{51}{3}){10}^{-6}=17 \times {10}^{-6}$

The co-efficient of apparent expansion of a liquid in a vessel A is $18\times 10^{-4}/^oC$ and in vessel B is $21\times 10^{-4}/^oC$. The difference in the coefficients of linear expansions of A and B is :

physics heat and temperature thermal expansion in liquids anomalous expansion of water thermal expansion in solids, liquids and gases

  1. $1\times 10^{-3}/^oC$

  2. $1\times 10^{-6}/^oC$

  3. $1\times 10^{-5}/^oC$

  4. $1\times 10^{-4}/^oC$

Show answer
Correct Option: D
Explanation:

${\gamma} _r={\gamma} _a +{\gamma} _g={\gamma} _a +3{\alpha} _g$
${\gamma} _r={\gamma} _{a1} +3{\alpha} _1={\gamma} _{a2} +3{\alpha} _2$ for both vessels A and B
$ \therefore 18 \times 10^{-4} + 3{\alpha} _1= 21 \times 10^{-4}+ 3{\alpha} _2$
$ \therefore3( {\alpha} _1- {\alpha} _2) = 3 \times 10^{-4}$
$ \therefore {\alpha} _1- {\alpha} _2 = 1 \times 10^{-4}$/ $^{0}$C

A specific gravity bottle is filled up to the brim with mercury of 400g, at $0^{0}$C. When heated to $90^{0}$C, the mass of the mercury that overflows from the specific gravity bottle is : (Coefficient of apparent expansion of mercury in glass is $\frac{1}{6500}/^{0}C$

physics heat and temperature thermal expansion in liquids anomalous expansion of water thermal expansion in solids, liquids and gases

  1. 5.46g

  2. 6.54g

  3. 10.92g

  4. 13.08 g

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Correct Option: A
Explanation:

Coefficient of apparent expansion is the ratio of liquid spilled out to the liquid left in the bottle per unit degree change in temperature.

$\gamma _{a}=\dfrac{mass   spilled }{mass   left \times \Delta \theta  }$
Let the mass spilled be x.
$\gamma _{a}=\dfrac{x}{(400-x)(90)}$
$\implies x=5.46g$

The coefficient of apparent expansion of a liquid when determined using two different vessels A and B are $\gamma _{1}$ and $\gamma _{2}$ respectively. If the coefficient of linear expansion of the vessel A is $\alpha $, the coefficient of linear expansion of the vessel B is

physics heat and temperature thermal expansion in liquids anomalous expansion of water thermal expansion in solids, liquids and gases

  1. $\dfrac{\alpha \gamma _{1}\gamma _{2}}{\gamma _{1}+\gamma _{2}}$

  2. $\dfrac{\gamma _{1}-\gamma _{2}}{2\alpha }$

  3. $\dfrac{\gamma _{1}-\gamma _{2}+\alpha }{3}$

  4. $\dfrac{\gamma _{1}-\gamma _{2}}{3}+\alpha $

Show answer
Correct Option: D
Explanation:

$\gamma _r = \gamma _l + \gamma _g=\gamma _r + 3\alpha _g$
Given,
$\gamma _r = \gamma _1 + 3\alpha _1$ ....(1)
$\gamma _r = \gamma _2 + 3\alpha _2$ ....(2)
Equating (1) and (2),
$\gamma _1 + 3\alpha _1= \gamma _2 + 3\alpha _2$
$ \therefore \alpha _2 =\frac{ \gamma _1 - \gamma _2}{3} + \alpha _1$

The coefficient of apparent expansion of mercury in a glass vessel is $153 \times 10^{-6}/^{\circ}C$ and in a steel vessel is $144 \times 10^{-6} / ^{\circ}C$. If $\alpha$ for steel is then, that of glass is $12 \times 10^{-6} / ^{\circ}C$.then that of glass is

physics heat and temperature thermal expansion in liquids anomalous expansion of water thermal expansion in solids, liquids and gases

  1. $ 9\times 10^{-6} / ^{\circ}C$.

  2. $ 6\times 10^{-6} / ^{\circ}C$.

  3. $36 \times 10^{-6} / ^{\circ}C$.

  4. $27 \times 10^{-6} / ^{\circ}C$.

Show answer
Correct Option: A
Explanation:

$We\quad know\quad that\quad γ _{ real }=γ _{ apparent }+γ _{ vessel }\ So,(γ _{ app }+γ _{ vessel }) _{ glass }=(γ _{ app }+γ _{ vessel }) _{ steel }\ (∴γ _{ real }\quad is\quad same\quad in\quad both\quad cases)\quad or\quad \ 153×{ 10 }^{ -6 }+(γ _{ vessel }) _{ glass }=144×{ 10 }^{ -6 }+(γ\ _ { vessel} )\ _ { steel} \ Further(\gamma \ _ { vessel })\ _ { steel} =3α=3×(12×{ 10 }^{ -6 })=36×{ 10 }^{ -6 }/∘C\ ∴153×10−6+(γ _{ vessel }) _{ glass }=144×{ 10 }^{ -6 }+36×{ 10 }^{ -6 }\ Solving\quad we\quad get\quad (γ\ _ { vessel} )\ _ { glass} =27×10−6/∘C(γ\ _ { vessel} )\ _ { glass} \quad or\\quad α=\frac{ γ _{ glass } }{ 3 }=9×10−6/∘C$