Tag: physics

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.

$V _{expelled} = V _0(\gamma _l -\gamma _g)\Delta T$
$ \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}$

${\gamma} _r={{\gamma} _{a1}} +{\gamma} _{glass}={{\gamma} _{a2}} +{\gamma} _{metal}$  ......(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,
${{\gamma} _{a1}}=10.30 \times {10}^{-4}$
${{\gamma} _{a2}}=10.06 \times {10}^{-4}$
${\alpha} _{glass}=9 \times {10}^{-6}$
${\gamma} _{glass}=3{\alpha} _{glass}=27 \times {10}^{-6}$
${\gamma} _{metal}=3{\alpha} _{metal}$

${\gamma} _r={\gamma} _a +{\gamma} _g={\gamma} _a +3{\alpha} _g$
${\gamma} _r={\gamma} _{a1} +3{\alpha} _1={\gamma} _{a2} +3{\alpha} _2$ 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

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

$\gamma _r = \gamma _l + \gamma _g=\gamma _r + 3\alpha _g$
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$

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

It works on the principle of Pascals law.
Pascal's law: Pressure on a confined fluid is transmitted undiminished and acts with equal force on equal areas and at 90 degrees to the container wall.
Uses of hydraulic press - Hydraulic presses are commonly used for forging, clinching, moulding, blanking, punching, deep drawing, and metal forming operations.

Answer is B.
Hydraulics is the use of a liquid under pressure to transfer force or motion, or to increase an applied force. The pressure on a liquid is called HYDRAULIC PRESSURE. And the brakes which are operated by means of hydraulic pressure are called HYDRAULIC BRAKES. These brakes are based on the principle of Pascals law. 
The Pascal law states that 'The pressure exerted anywhere in a mass of confined liquid is transmitted undiminished in all directions throughout the liquid'.

The answer is A.

Pressure is the amount of force acting per unit area. That is, P=F/A.
where:
p is the pressure,
F is the normal force,
A is the area of the surface on contact. Let us consider A = $\pi { r }^{ 2 }$.
Therefore, $\dfrac { { F } _{ 1 } }{ { \pi { r } _{ 1 } }^{ 2 } } =\dfrac { { F } _{ 2 } }{ { \pi { r } _{ 2 } }^{ 2 } } $.