**Notes**
**Assignments**
**Vides**
**Revision Notes**

**Question 1** the triple points of neon and carbon dioxide are 24.57 k and 216.55K respectively. Express these temperatures on Celsius and Fahrenheit scales

Solution
Triple point of neon = 24.57 K

and, triple point of CO_{2} = 216.55K

these two temperatures are given on absolute or Kelvin temperature scale.

On Celsius Scale:-

T_{C} = T_{K} - 273

Where T_{C} is temperature on Celsius scale

T_{K} is temperature on Kelvin scale

Thus,

Triple point of neon = 24.57 - 273.15 = -248.58°C

Triple point of C0_{2} = 216.55 - 273.15 = -56.6°C

On Fahrenheit scale:-

T_{F} = (9T_{C}/ 5 )+ 32

T_{F} is temperature on Fahrenheit scale.

T_{C} is temperature on Celsius scale

Thus,

Triple point of neon = (9/5)(-248.48) + 32

= 415.44 °F

Triple point of CO_{2} = (9/5)(-56.6) + 32

= - 69.88 ° F

**Question 2** Two absolute scales A and B have triple points of water defined as 200A and 350A. What is the relation between T

_{A} and T

_{B}
Solution
Given that on absolute scale

Triple point of water on scale A = 200 A

Triple point of water on scale B = 350 B

Also, triple point of water on Kelvin scale = 273.16 K

Now temperature on scale A and on scale B is equivalent to 273.16 K on absolute temperature scale.

Thus, value of one degree on absolute scale A = (273.16/200) K

Or,

Value of temperature T_{A} on absolute scale A = (273.16XT_{A})/200

Similarly value of temperature T_{B} on absolute scale B = (273.16XT_{B})/350

Since T_{A} and T_{B} represent the same temperature

273.16×T_{A}/200 = 273.16×T_{B}/350

Or, T_{A} = 200T_{B}/350 = 4T_{B}/ 7

**Question 3**The pressure of the gas in constant volume gas thermometer are 80 cm, 90 cm and 100 cm of mercury at the ice point, the steam point and in a heated wax bath respectively. Find the temperature of the wax bath.

Solution
Given that,

Pressure at ice point P_{ice} = 80 cm Hg

Pressure at steam P_{steam} = 90 cm Hg

Pressure at heated wax bath P_{wax} = 100 cm Hg

The temperature of wax bath on the scale measured by thermometer is

$T_wax =\frac {P_wax - P_ice}{P_steam - P_ice) X 100_0 C$

$=\frac {100cm Hg - 80 cm Hg}{90cm Hg - 80 cm Hg} X100_0 C$

$=200_0 C$

**Question 4 **A resistance thermometer reads R= 20.0Ω, 27.5Ω and 50.0 Ω at the ice point (0°C), the steam point (100°C) and the zinc point 420°C respectively. Assuming that the resistance varies with temperature as R

_{θ}=R

_{0}(1+αθ+βθ

^{2}), find the values of R

_{0}, α and β. Here θ represents temperature on Celsius scale.

Solution
Given are the values of resistances,

at ice point 0°C, R_{0} = 20.0Ω

at steam point 100°C, R_{100} = 27.5Ω

at zinc point 420°C, R_{z} = 50.0Ω

Now resistance R_{θ} varies as

R_{θ} = R_{0} (1+αθ+βθ^{2})

For resistance at ice point

R_{ice}= R_{0} (1 +α°C +β(0°C)^{2})

R_{ice} = R_{0} = 20°C

For resistance at steam point

R_{steam}= R_{0} (1 +α(100°C) +β(100°C^{2})

27.5α / 20.0α = 1 + 100 (α+ 100β)

or,

α+100β= 3.75 × 10^{-3} .......(i)

for resistance at zinc point

R_{z} = R_{0} [1 +α(420°C) +β(420°C)^{2}]

(50Ω/ 20.0Ω)- 1 = 420 (α+420β)

or α+ 420β= 3.75 × 10^{-3} .......(ii)

Subtracting (i) and (ii) we find the value of β

α+ 420 β- α-100 β= 3.57 × 10^{-3}-3.75× 10^{-3}

or

320β= - 0.18× 10^{-3}

or,

β = - 5.62× 10^{-3} °C^{-1}

Put the value of β in eqn (i) for finding value of α

thus α= 3.75× 10^{-3}- 100 (-5.62× 10^{-7})

or,

α= 3.8× 10^{-3} °C^{-1}

**Question 5**A circular hole of diameter 2.00 cm is made in an aluminum plate at 0°C. What will be the diameter at 100°C. α for aluminum = 2.3 × 10

^{-3} °C

^{-1}.

Solution
Diameter of circular hole in aluminum plate at 0°C = 2.0 cm

With increase in temperature from 0°C to 100°C diameter of ring increases.

Using,

l = l_{0}(1 +αΔθ)

Where,

l_{0} = 2.0 cm

α= 2.3 x 10_{-3} °C^{-1}

and Δθ= (100°C - 0°C) = 100°C

we can find diameter at 100°C

l= 2(1 + 2.3 x 10^{-5}X100°C)

= 2.0046 cm.

**Question 6**The density of water at 0°C is 0.998 gm/cm

^{3} and at 4°C is 1.000 gm/cm

^{3}. Calculate the average coefficient of volume expansion of water in temperature range 0 to 4°C.

Solution
Density is mass per unit volume given in the question is density of water at 0°C and 4°C

ρ_{0} = .998 gm/cm^{3} and ρ_{4}=1.00 gm/cm^{3} with increase in temperature volume of water changes using,

V = V_{0} (1 +βθ)

β is the coefficient of volume expansion.

Volume at 4°C equals

V_{4} = V_{0}(1 +β (4°C - 0°C))

or V_{4} = V_{0} (1 + 4β)

If m is the mass of water then density of water at 4°C is

m/V_{4} =m/V_{0} [(1 + 4β)]

or ρ_{4} = ρ_{0} / (1 +4β)

1.00 =.998/(1+4β)

1 +4β = .998

β = -5 X 10^{-4} °C^{-1}

**Question 7**A pendulum clock gives correct time at 20°C at a place where g=9.8m/ s

^{2}. The pendulum consists of a light steel rod connected to a heavy ball. It is taken to a different place where of = 9.788m/s

^{2}. At what temperature will it give correct time? Coefficient of linear expansion of steel = 12X10

^{-6}°C

^{-1}.

Solution
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.

**Question 8** A glass vessel measures exactly 10cm × 10cm × 10cm at 0°C. It is filled completely with mercury at this temperature. When the temperature rises to 10°C , 1.6 cm

^{3} of mercury overflows. Calculate the Coefficient of volume of mercury. Coefficient of linear expansion of glass = 6.5 × 10

^{-6}°C

^{-1}.

Solution
At temperature θ= 0°C Volume of glass vessel is

V = 1000 cm^{3}

It is then completely filled with mercury.

At temperature θ= 10°C, 1.6 cm^{3} of mercury overflows

we have to calculate coefficient of volume expansion of mercury γ_{m}

Coefficient of volume expansion of glass γ_{g}= 3α

γ_{g} = 3 × 6.5 × 10^{-6} °C^{-1}

= 1.95 × 10^{-5} °C^{-1}

Volume expansion of mercury at 10°C

V_{mθ}=V_{0} (1+γ_{m}Δθ)

and that of glass.

V_{gθ}=V_{0} (1+γ_{g}Δθ)

now volume of mercury overflown is

V_{mθ}-V_{gθ} = V_{0}(γ_{m}-γ_{g})Δθ

1.6 cm^{3} = 1000 × 10°C (γ_{m} - 1.95 × 10^{-5})

γ_{m} = 1.795 × 10^{-4}

or approximately

γ_{m} = 1.8 × 10^{-4} °C^{-1}.

**Question 9** A steel rod is clamped at its two ends and rests on a fixed horizontal base. The rod is unstrained at 20°C. Find the longitudinal strain developed in the rod if temperature rises to 50°C. Coefficient of linear expansion of steel is 1.2 × 10

^{-5}°C

^{-1}
Solution
Due to increase in temperature there would be an fractional increase in length of the rod. But since the rod is rigidly clamped. The tension must increase by sufficient amount to produce and equal and opposite change in length. Thus total change in length due to thermal expansion plus electric strain is zero

αΔθ+longitudinal strain = 0

electric strain = -αΔθ

= - 1.2 × 10^{-5} × (50°C - 20°C)

= - 3.6 × 10^{-4}

**Question 10**To ensure a light fit aluminum reverts used in airplane construction are made slightly larger than the revert holes and cooled by dry ice (solid CO

_{2}) before being driven. If diameter of a hole is 0.2500 in, what should be the diameter of revert at 20°C if is diameter is equal to that of hole when revert is cooled to -78°C the temperature of dry ice? Coefficient of Linear expansion of Aluminum =2.4X10

^{-5}°C

^{-1}
Solution
Given the diameter of hole = 0.2500 inch

at -78°C temperature of dry ice diameter of revert D_{r} = 0.2500 in

To find diameter of revert at 20°C using,

D_{r20} = D_{r} (1 + α_{al} Δθ)

= 0.2500 (1 + 2.4 × 10^{-5}(20-(-78))

= .2506 inch

**Question 11**A steel ring of 3.000 in inside diameter at 20°C is to be heated and slipped over a brass shaft measuring 3.002 inch in diameter at 20°C α

_{brass} = 2 × 10

^{-5} °C

^{-1} α

_{steel} =1.2 × 10

^{-5} °C

^{-1}
(a) To what temperature should the ring be heated?

(b) If the ring and shaft together are cooled by some means such as liquid air, at what temperature will the ring just slip off the shaft.

Solution

At 20°C

(a) Diameter of brass shaft is 3.002 inch and inner diameter of steel ring is 3.000 inch. Steel ring is to be slipped into the brass shaft and is required to be heated till it fits the shaft. Using equation,

D_{θ }= D_{20 }(1+ α_{S}Δθ)

3.002 = 3.000(1+1.2 × 10 ^{-5} (θ-20°C))

Calculating for value of θ we finally get

θ = 75.5°C

(b) Now we have to find the temp until which system is to be cooled so that ring slips off the shaft.

If D_{rθ} and D_{sθ} be the diameter of ring and shaft and are equal at temperature θ, then D_{rθ}= D_{sθ}=D

Using,

(i) for ring

D= 3[1+α_{s}(θ- 20)]

(ii) for shaft

D= 3.002[1+α_{b}(θ- 20)]

Subtracting (i) from (ii)

3.002+α_{b} × 3.002(θ- 20) - 3-3α_{s}(θ- 20) = 0

or ,

(θ- 20) =- .002/[(3.002 X 2 X 10^{-5}) - (3 × 1.2 × 10^{-5})]

θ= - 0.002/(2.404 × 105 + 20)

= - 63.19°C.

**Question 12**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.

Solution
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

substuting 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.

**Question 13**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

_{steel} = 1.2 × 10

^{-5} °C

^{-1}
Solution
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.

**Question 14**Two brass rods and one steel rod of equal length L

_{0} and equal crossectional are joined rigidly at their ends .All rods are in the state of zero tension at 0° C. Find the length of the system when the temperature is increased to θ.

Coefficient of linear expansion of steel=α

_{s}
Coefficient of linear expansion of brass=α

_{b}
Young modulus of steel=Y

_{s}
Young modulus of brass=Y

_{b}
Solution
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

**Question 15** What will the length of steel and copper rod .So that length of steel rod is 5cm longer then the copper rod at all the temperature (α

_{copper}=1.7X10

^{-5} /C,α

_{steel}=1.1X10

^{-5} /C)

Solution
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

**Question 16**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

_{c} and α

_{B}.On heating the strip through the temperature change ΔT,the strip bends into an arc of the circle. Find the radius of the curvature of the strip

Solution
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 bimetatic 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

**Question 17** Show that the moment of inertia of the solid body of any shape changes with the temperature as I=I

_{0}(1+2αθ) .Where I

_{0} is the moment of inertia at 0° C and α coefficient of linear expansion of the solid

Solution
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)

**Question 18** A aluminum container of mass .65 kg contains .35kg of water at 18° C.A block of iron of mass .2kg at 100° C is put into the water. Find the equilibrium temperature of the mixture. Specific heat capacities of aluminum ,iron and water are 910 J/Kg-K,470J/kg-K and 4200 J/kg-K resp.

Solution
Mass of aluminum container=.65 Kg

Mass of water=.35 Kg

Mass of iron block=.2 Kg

Temperature of iron block=100° C

Temperature of aluminum water system=18° C

Now heat gained by water

Q_{1}=.35*4200(T-18)

Heat gained by the aluminum

Q_{2}=.65*910(T-18)

Heat lost by iron block

Q_{3}=.2*4700(100-T)

Now

Q_{1}+Q_{2}=Q_{3}

Substituting all values and solving

T=21.58° C

**Question 19** An iron piece of mass 200 gm is kept inside a furnace for a considerable long time and then put in calorimeter of water equivalent 20 Gm containing 280 gm of water at 22° C. The mixture attains an equilibrium temperature of 80° C. Find the temperature of the furnace.

Specific heat capacities of iron 470 J/kg-° C.

Solution

Mass of iron piece= 200 × 10^{-3} Kg

mass of water and calorimeter= 300 × 10^{-3} Kg

Heat gained by calorimeter

Q=msΔθ

s=4186 J/kg-K is the specific heat capacity of water.

thus

Q_{1}=300 × 10^{-3} Kg × 4186 J/kg-K (80° C - 22° C)

or Q_{1}=1255.8 J/K × 58° C

now, heat lost by iron piece

Q_{2}=200 × 10^{-3} Kg × 470 J/kg-K (θ° C-80° C)

heat lost by iron piece=heat gained by calorimeter.

thus,

1255.8 J/K × 58° C=200 × 10^{-3} Kg × 470 J/kg-K (θ° C-80° C)

calculating for θ we get

θ=858.4° C

**Question 20** 1 kg of ice at 0 °C is mixed with 1 kg of steam at 100 °C. What will be the composition of the system when thermal equilibrium is reached. latent heat of fusion of ice = 3.36X10

^{5} J/kg and latent heat of vaporization of water = 2.26X10

^{6} J/kg.

Solution
Heat required to melt ice at 0 °C to water = 1 kg × 3.36 × 10^{5} J / K.

= 336000 J

heat required to take 1 kg of water to temperature 100 from 0 °C to convert into water

=4200 × 100J

Total Heat(Q_{1})=336000+420000=756000=7.56X10^{5}

heat lost by steam at 100°C to convert into water

Q_{2}=2.26 × 10^{6} J/kg.

=22.6 X10^{5}

since Q_{2} > Q_{1}. It means whole steam will not be cooled down .Some steam will cooled down to make ice reach the temperature of 100°C.it means thermal Equilibrium is reached at 100 °C and some mass of steam will be cooled down

Let m be the mass of steam cooled down

Heat lost=mX22.6 X10^{5}

From principal of calorimetry heat lost = heat gained

7.56X10^{5} = mX22.6 X10^{5}

or m = .334

= >.334 kg of heat gets converted into water this total amount of water is 1.335 kg and steam left is .665 kg.

**Question 21** A bullet of mass 30 gm. enters into a fixed wooden block with a speed 50 m / s and stops in it. Calculate the change in internal energy during this process.

Solution
given that velocity of bullet v = 50 m/s

and its mass is m = 30 gm

internal energy of the system is Kinetic Energy plus Potential Energy of the system. Initially all the energy of system is kinetic which become potential when bullet enters wooden block so change in internal energy

Δu = (1 / 2 )mu^{2}

= 1 / 2 (.30) × (50)^{2}

= 37.5J

**Question 22** An animal of 70 kg is running with a speed of 6 m/s. If all the KE of animal can be used in increasing water from 18 °C to 32 °C, how much water can be heated with this energy

Solution
KE of animal = 1/2 × 70 × (6)^{2}

= 12.60J

Q = 1260 J is the amount of heat being given to heat the water

since, Q = ms Δθ

m = Q/ sΔθ

= 12605 / 4186 × (32-18)

= 0.0215 kg = 21.5 gm

**Question 23** A copper cube of mass 300g slides down on a rough inclined plane of inclination 40 ° at a constant speed. Assume that any loss in mechanical energy goes into copper block as thermal energy find the increase in thermal energy of the block as it slides down through 50 cm. specific heat capacity of copper = 420J/kg-k.

Solution
work done in sliding 50 cm=Change in Potential energy=mgh

=mg(dsin40)

W = = .95 J.

If this mechanical work W produces same temperature change as heat Q then

W = J Q

where J = mechanical equivalent of heat and J = 1 if same unit of W and Q are used

Q= .95 J

since, Q= msΔθ

change in temperature Δθ of block

Δθ= Q/ ms

= .95 / .3 x 420

= 7.53 x 10^{-3} °C.

Class 11 Mathse
Class 11 Physics

### NCERT Solutions