A pendulum clock gives correct time at 20°C at a place where g=9.8m/ s

Time period of pendulum clock at temperature θ is T=2π√(l_{θ}/g)

at temperature θ

l_{θ}= l_{0} (1+αθ)

T = 2π√[l_{0}(1+αθ)/g]

Now at temperature 20°C,

T_{20}=(2π√l_{0}) [(1 + 12 × 10^{-6}X20)/9.8]^{1/2}

= 2.007√l_{0}

Now at some other place where acceleration due to gravity is g = 9.788 time period at temperature θ is

T = (2π√l_{0}) [(1 + 12 × 10^{-6}Xθ)/9.788] ^{1/2}

but T = 2.007 as calculated earlier so

(2.007√l_{0}) = (2π√l_{0})[(1+12 × 10^{-6}Xθ)/9.788]^{1/2}

calculating for value of θ we get

θ=-82.06°C.

A metal rod 30.0 cm long expands by 0.075 cm when its temperature is raised from 0°C to 100°C.A rod of a different metal and of same length expands by 0.045 cm for the same raise in temperature. A third rod also 30.0 cm long is made up of the pieces of each of the above metals placed and to end. and expends 0.065 cm between 0°C and 100°C. Find the length of each portion of the composite bar.

we would first calculate the coefficient of linear expansion of metals 1 and 2.For first metal given l = 30cm at temp 0°C and change in length when temperature changes to 100°C

Δl = 0.075

using Δl/l = αΔθ

α_{m1} = (0.075/30 )×(1/100)

= 2.5 x 10^{-3} °C^{-1}

for second metal at 0°C l= 30cm

at 100ºC change in length due to thermal expansion Δl= 0.045.

α_{m2}= Δl/lΔθ

= (0.045/30) ×( 1/100)

= 1.5 × 10^{-5} °C^{-1}.

Composite rod made up of both the metals has l_{1} length of first metal and l_{2} of second metal such that at 0°C . l_{1}+l_{2} = 30 cm. (i)

In an increase in temperature from 0°C to 100°C change in length

Δl = Δl_{1}+Δl_{2}= 0.065 cm

but ,

Δl_{1}=α_{m1}l_{1}Δθ

Δl_{2}=α_{m2}l_{2}Δθ

thus,

α_{m1}l_{1}Δθ+α_{m2}l_{2}Δθ= 0.065

since, Δθ= 100°C

α_{m1}l_{1}+α_{m2}l_{2} = 0.065/100

substituting the values of α_{m1} and α_{m2} we get

2.5l_{1}+1.5l_{2} = 65 (ii)

using equations (i) and (ii) we can find the values of l_{1}=20cm and l_{2}=10cm.

Suppose that a steel hoop could be constructed around the earth's equator, just fitting it at a temperature 20°C. What would be the thickness of space between hoop and the earth if the temperature of the hoop were increased by 1 °C.

α

hoop made of steel is constructed around earth's equator. So at temperature 20°C

R_{Hoop} = radius of earth

= 6.38 × 10^{6} m

with one degree increase in temperature expansion in the hoop

ΔR = αR_{Hoop}Δθ

= 1.2 × 10-5 × 6.38 × 106 m

= 76.5m.

This would be thickness of space between hoop and earth.

Two brass rods and one steel rod of equal length L

Coefficient of linear expansion of steel=α

Coefficient of linear expansion of brass=α

Young modulus of steel=Y

Young modulus of brass=Y

Suppose α_{s} > α_{b}

It means steel rod will expand more that brass rod.

Since they are all joined at their ends. Steel rod will try to expand brass when heated to θ and it will resisted by the brass rod. It means a tensile force F will develop in brass and while a compressive force of 2F will develop in steel

So final strain in each case

For steel

Thermal strain-Longitudinal strain due to compression=Final strain

(α_{s}θ)-(2F/AY_{s})=ΔL/L_{0}---(1)

For brass

Thermal strain+Longitudinal strain=Final strain

(α_{b}θ)+(F/AY_{b})=ΔL/L_{0}---(2)

Multiplying Equation (2) by 2 and then Adding to Equation (1),it can be calculated that

$\Delta L= \frac {L_0( 2\alpha _b Y_b + \alpha _s Y_s) \theta}{2Y_b + Y_s}$

Now length of the system when temperature is changed to θ

L=L_{0}+ΔL where ΔL is calculated as above

Let α_{c} and α_{s} are the coefficient of linear expansion of copper and steel. Suppose the steel rod is of length L_{1} and copper rod of length L_{2},Such that al all the temperatures

L_{1}-L_{2}=5----(1)

This would happen when increase in length of steel rod ΔL_{1} is of equal to increase in length of copper rod ΔL_{2} at all change in temperature ΔT

or

L_{1}α_{s}ΔT=L_{2}α_{c}ΔT

or

L_{2}=L_{1}α_{s}/α_{c}

putting the values of α

L_{2}=11/17L_{1}

Putting this in equation ---(1)

L_{1}=14.17cm

and also L_{2}=9.17 cm

A bimetallic strip is formed out of two identical strips .One of the copper and other of brass. The coefficient of linear expansion of the two metals are α

Let d be the thickness of the strip and L_{0} be the length of the two identical strips of copper and brass. On heating the bimetallic strip through the temperature change

ΔT,length of copper strip becomes l_{c} and length of brass strip becomes l_{b} and as given in question, strip bends to a circular arc.

Let R be the mean radius of curvature of the bi-metallic strip.

R_{B} --> radius of curvature of the brass strip

R_{C} --> radius of curvature of the copper strip

Then R_{B}-R_{C}=d the thickness of the bimetallic strip

Now

l_{c}=l_{0}(1+α_{c}ΔT)

l_{b}=l_{0}(1+α_{b}ΔT)

then

l_{b}-l_{c}=l_{0}(α_{b}-α_{c})ΔT---(1)

If θ is the angle subtended by the bend strip at the center of curvature O,

then

l_{b}=θR_{B}

l_{c}=θR_{C}

or

l_{b}-l_{c}=θ(R_{B}-R_{C})

=θd--(2)

from equation 1 and 2

θd=l_{0}(α_{b}-α_{c})ΔT ---(3)

The mean radius of curvature of bimetallic strip is approximately given by

Rθ=l_{0} ---(4)

So from equation (3) and (4)

(l_{0}d/R)=l_{0}(α_{b}-α_{c})ΔT

Thus R=d/(α_{b}-α_{c})ΔT

Show that the moment of inertia of the solid body of any shape changes with the temperature as I=I

At 0° C, the Moment of Inertia
of the solid is given by

I_{0}=mR_{0}^{2}

m --mass of the body

R_{0}--- radius at 0° C

At temperature θ° C

ΔR=R_{0}αθ

Radius of the body now becomes

R=R_{0}(1+αθ)

Now I=mR^{2}

=mR_{0}^{2}(1+2αθ+α^{2}θ^{2})

=I_{0}(1+2αθ+α^{2}θ^{2})

As α^{2}θ^{2} is negligible as comparison to αθ

Thus ,at temperature moment of inertia of body is I=I_{0}(1+2αθ)

A bimetallic strip is made of aluminium and steel ($\alpha _{Al} > \alpha _{Steel}$) on heating, the strip will

(a) remain straight

(b) get twisted

(c) will bend with aluminium on concave side

(d) will bend with steel on concave side.

Both strips of Al and steel are fixed together initially in bimetallic strip. When both are heated then expansion in steel will be smaller than aluminium. So Al
strip will be convex side and steel on concave side verifies the option (d).

A uniform metallic rod rotates about its perpendicular bisector with constant angular speed. If it is heated uniformly to raise its temperature slightly

(a) its speed of rotation increases.

(b) its speed of rotation decreases.

(c) its speed of rotation remains same.

(d) its speed increases because its moment of inertia increases

(b): On heating a uniform metallic rod its length will increase so Moment of Inertia of rod increased from $I_1$ to $I_2$

Now from law of conservation of angular momentum

$I_1 \omega_1 = I_2 \omega_2$

As $I_1 < I_2$ , $\omega _1 > \omega _2$, so Angular Speed decreased.

Verifies option (b)

Two rods one of aluminium and other made of steel having initial length $L_1$ and $L_2$ are connected together to form a single rod of length $L_1 + L_2$ . The coefficient of linear expansion for aluminium and steel are $\alpha _a$ and $\alpha _S$ respectively. If the length of each road increases by the same amount when their temperature are raised by T degree Celsius, the what is the ratio between $ \frac {L_1}{L_1 + L_2}$ ?

a.$ \frac {\alpha _s}{\alpha _a}$

b.$ \frac {\alpha _a}{\alpha _s}$

c.$ \frac {\alpha _s}{\alpha _a + \alpha _s}$

d.$ \frac {\alpha _a}{\alpha _a + \alpha _s}$

Now

$L_1 \alpha _a t=L_2 \alpha _s t$

$L_1 \alpha _a =L_2 \alpha _s $

$\frac {L_1}{L_2} = {\alpha _s}{\alpha _a}$

$\frac {L_1}{L_1 + L_2}= \frac {\alpha _s}{\alpha _a + \alpha _s}$

A pendulum clock loses 12s a day if the temperature is 40 °C and gains 4s a day if the temperature is 20°C. The temperature at which the clock will show correct time, and the co-efficient of linear expansion ($\alpha$) of the metal of the pendulum shaft are respectively :

(a)25°C ; $\alpha =1.85 \times 10^{-5} $/°C

(b)60°C ;$\alpha =1.85 \times 10^{-4}$ /°C

(c)30°C ;$\alpha=1.85 \times 10^{-3}$/°C

(d)55°C ;$\alpha=1.85 \times 10^{-2}$/°C

Time loss or gain per sec is given by

$ \Delta t=\frac {1}{2} \alpha ? \Delta T$

$12=12 \alpha (40-T) \times 86400$ ----(1) ( as 1 day=84400 sec)

$4=12 \alpha (T-20) \times 86400$ ----(2)

From (1) and (2)

$3=\frac {40-T}{T-20}$

Solving T=25°C and putting in (2)

$\alpha =\frac {8}{5 \times 86400}$

$=1.85 \times 10^{-5}$ /°C

A cubical block is floating inside a bath. The temperature of system is increased by small temperature $\Delta T$. It was found that the depth of submerged portion of cube does not change. Find the relation between coefficient of linear expansion ($\alpha$) of the cube and volume expansion of liquid ($\gamma$).

a. $\gamma =2\alpha$

b. $\gamma =\alpha$

c. $\gamma =3\alpha$

d. $\gamma =\frac {1}{2} \alpha$

Let x be the depth submerged.At initial temperature for the equilibrium of the block

$AL \rho _b g = Ax \rho _l g$

$L\rho_b = x \rho _l$ . . . (i)

At final temperature

$A^{'} = A(1 + 2 \alpha \Delta T)$

$\rho _{l}^{'}= \rho _l (1 - \gamma \Delta T)$

For the equilibrium of the block

$A(1 + 2 \alpha \Delta T)(x) (1 - \gamma \Delta T) = AL \rho _b = A x \rho _l$

$ 1 + 2\alpha \Delta T - \gamma \Delta T= 1$

$\gamma =2\alpha$

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