Sunday, November 3, 2019

H C Verma solutions, Heat and Temperature, Exercises, Q11_Q20, Chapter-23, Concepts of Physics, Part-II

Heat and Temperature

Exercises - Q11 to Q20




11. A meter scale made of steel is calibrated at 20 °C to give a correct reading. Find the distance between 50 cm mark and 51 cm mark if the scale is used at 10 °C. The coefficient of linear expansion of steel is 1.1x10⁻⁵/°C.  


Answer: Here the original length at 20°C, L = 51-50 = 1 cm. Temperature difference, θ = 20° - 10° =10 °C, α = 1.1x10⁻⁵/°C.

The change in length ΔL = αθL
→ΔL = 1.1x10⁻⁵*10*1 cm =11x10⁻⁵ cm
Now actual length of 1 cm at 10°C
=L-ΔL
=1.0 - 11x10⁻⁵ cm
=1.0-0.00011 cm
=0.99989 cm
Hence this scale will measure the mark between 50 cm and 51 cm as
1/0.99989 cm
=1.00011 cm


 

12. A railway track (made of iron) is laid in winter when the average temperature is 18 °C. The track consists of sections of 12.0 m placed one after the other. How much gap should be left between two such sections so that there is no compression during summer when the maximum temperature goes to 48 °C? Coefficient of linear expansion of iron = 11x10⁻⁶. 


Answer: Here the gap should be equal to the expansion (ΔL) due to the temperature difference. The temperature difference, θ = 48 - 18 =30 °C.

L = 12 m. 
α = 11x10⁻⁶/°C
Hence the expansion,
ΔL = αθL
     = 11x10⁻⁶*30*12 m
     = 3.96x10⁻³ m
     = 0.396 cm
     ≈ 0.4 cm



13. 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 = 2.3x10⁻⁵/°C.  


Answer: Here the hole in the metal will expand as if made of the same metal. Hence, L = 2.00 cm, θ = 100 °C, α = 2.3x10⁻⁵/°C.

Hence the diameter of the hole at 100 °C = L(1+αθ)

= 2.00(1 + 2.3x10⁻⁵*100) cm 
= 2.00(1+0.0023) cm
= 2.00*1.0023 cm
= 2.0046 cm.  

 


14. Two meter scales, one of steel and the other of aluminum, agree at 20 °C. Calculate the ratio aluminum-centimeter/steel-centimeter at (a) 0 °C, (b) 40°C and (c) 100 °C. α for steel = 1.1x10⁻⁵/°C and for aluminum = 2.3x10⁻⁵/°C.  


Answer: Let the length of a centimeter in both the scale at 20 °C = L. For steel α = 1.1x10⁻⁵/°C and for aluminum α' =2.3x10⁻⁵/°C. For a temperature difference of θ the length of a centimeter in steel scale

L' = L(1+αθ) and the length on the aluminum scale, L" = L(1+α'θ).

Hence the ratio, R = L"/L'

=(1+α'θ)/(1+αθ)

(a) at 0°C, θ = 0-20 =-20°C, Hence 
R = (1-20*2.3x10⁻⁵)/(1-20*1.1x10⁻⁵)
=0.99954/0.99978
=0.99976

(b) at 40 °C, θ = 40-20 = 20°C, Hence
R = (1+20*2.3x10⁻⁵)/(1+20*1.1x10⁻⁵)
=1.00046/1.00022
= 1.00024

(c) at 100 °C, θ=100-20=80 °C, Hence
R =(1+80*2.3x10⁻⁵)/(1+80*1.1x10⁻⁵)
=1.00184/1.00088
=1.00096 

15. A meter scale is made up of steel and measures the correct length at 16°C. What will be the percentage error if this scale is used (a) on a summer day when the temperature is 46 °C and (b) on a winter day when the temperature is 6 °C? Coefficient of linear expansion of steel =11x10⁻⁶  /°C. 

Answer: Let the correct length at 16 °C = L. 

(a) on a summer day at 46°C, θ=46-16 =30 °C. The difference in length will be 
ΔL = αθL =30*11x10⁻⁶*L =0.00033 L
Hence the percentage error in measurement = 100*0.00033L/L
= 0.033%

(b) On a winter day at 6°C, θ=6-16 =-10°C
The difference in length at this temperature
=αθL
= -10*11x10⁻⁶*L
= -0.00011L
  The percentage error in measurement 
= -100*0.00011L/L
= -0.011%


16. A meter scale made of steel reads accurately at 20 °C. In a sensitive experiment, distances accurate up to 0.055 mm in 1 m are required. Find the range of temperatures in which the experiment can be performed with this meter scale. Coefficient of linear expansion of steel =11x10⁻⁶/°C.  


Answer: For the expansion of 土0.055 mm this condition will be fulfilled. Let the corresponding  temperature be T.  

θ = (T - 20) °C

α = 11x10⁻⁶/°C

ΔL = 土0.055 mm

L = 1000 mm 

 But ΔL = αθL

→土0.055 = 11x10⁻⁶*(T-20)*1000
→T-20 = 土0.055/(1000*11x10⁻⁶)
→T-20 = 土0.055x10⁶/11000
→T-20 = 土55/11 = 土5
→T = 20土5 = 15 °C to 25 °C 
Hence the required range of temperatures are 15 °C to 25 °C.




17. The density of water at 0 °C is 0.998 g/cm³ and at 4 °C is 1.000 g/cm³. Calculate the average coefficient of volume expansion of water in the temperature range 0 to 4 °C.  


Answer: Here per gram volume at 0°C, V₀ = 1/0.998 cm³ =1.002 cm³ and at 4 °C, V₄ = 1 cm³. Here θ = 4°C. Let average coefficient of volume expansion = ɣ. So

V₄ = V₀(1+ɣθ)

→1 = 1.002*(1+ɣ*4)

→4ɣ = (1/1.002) - 1 = 0.998 - 1

→4ɣ = -0.002
→ɣ = -0.002/4 = -0.0005 /°C
→ɣ = -5x10⁻⁴/°C

 

18. Find the ratio of lengths of an iron rod and an aluminum rod for which the difference in the lengths is independent of temperature. The coefficient of linear expansion of iron and aluminum are 12x10⁻⁶/°C and 23x10⁻⁶/°C respectively.   


Answer: Let the length of iron rod = Lᵢ, and the length of the aluminum rod = Lₐ.

For iron, αᵢ = 12x10⁻⁶/°C, and
for aluminum, αₐ = 23x10⁻⁶/°C.
For a temperature difference of T°, length of iron rod, Lᵢ' = Lᵢ + αᵢTLᵢ, and length of aluminum rod, Lₐ' = Lₐ+αₐTLₐ
Hence, Lᵢ'-Lₐ'=Lᵢ-Lₐ+αᵢTLᵢ-αₐTLₐ ----- (i)
{since the difference in length is constant/same for all temperature, Lᵢ'-Lₐ'=Lᵢ-Lₐ}, hence (i) becomes
αᵢTLᵢ-αₐTLₐ = 0
→αᵢLᵢ-αₐLₐ = 0
→αᵢLᵢ = αₐLₐ
→Lᵢ/Lₐ = αₐ/αᵢ =23/12
Hence Lᵢ : Lₐ = 23 : 12


 

19. A pendulum clock gives the correct time at 20 °C at a place where g = 9.800 m/s². The pendulum consists of a light steel rod connected to a heavy ball. It is taken to a different place where g = 9.788 m/s². At what temperature will it give the correct time? Coefficient of linear expansion of steel = 12x10⁻⁶/°C.  


Answer: Let the length of the pendulum at 20°C = L. Its time period,

T = 2π√(L/g)
At the other place let the length = L'. Here also time period is the same, hence T=2π√(L'/g'). Equating,
L/g = L'/g'
→L' = Lg'/g
If the temperature difference between two places = θ
L' = L(1+αθ)
→Lg'/g = L(1+αθ)
→1+αθ = g'/g
→θ = (g'-g)/αg
Here g = 9.800 m/s², g' = 9.788 m/s²
α = 12x10⁻⁶/°C, so
θ=(9.788-9.800)/(9.800*12x10⁻⁶)
  =-102 °C
Hence the temperature at the other place = 20 °C + θ =20°C - 102°C 
=-82°C


 


20. An aluminum plate fixed in a horizontal position has a hole of a diameter of 2.000 cm. A steel sphere of diameter 2.005 cm rests on this hole. All the lengths refer to a temperature of 10 °C. The temperature of the entire system is slowly increased. At what temperature will the ball fall down? The coefficient of linear expansion of aluminum is 23x10⁻⁶/°C and that of steel is 11x10⁻⁶/°C.   


Answer: The ball will fall down at that temperature at which diameters of the hole and the ball are equal. Let that temperature be T and diameter D.
Diagram for Q-20

The temperature difference = T-10 °C

Original dia of the aluminum hole, d=2.000 cm, and α=23x10⁻⁶/°C. Hence

D = d{1+α(T-10)}

The original dia of the ball, d' = 2.005 cm and α' = 11x10⁻⁶/°C. Hence

D = d'{1+α'(T-10)}

Equating,

d{1+α(T-10)} = d'{1+α'(T-10)}

→2.000{1+23x10⁻⁶(T-10)} =2.005{1+11x10⁻⁶(T-10)}

→1+23x10⁻⁶(T-10) =1.0025{1+11x10⁻⁶(T-10)}
→(T-10)*(23-11*1.0025)*10⁻⁶ =0.0025
→T-10 =0.0025x10⁶/11.9725 ≈ 209
→T = 209+10 =219 °C

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Links to the Chapters




CHAPTER- 20 - Dispersion and Spectra


CHAPTER- 19 - Optical Instruments

CHAPTER- 18 - Geometrical Optics



CHAPTER- 17 - Light Waves




CHAPTER- 14 - Fluid Mechanics



CHAPTER- 13 - Fluid Mechanics


CHAPTER- 12 - Simple Harmonic Motion








CHAPTER- 11 - Gravitation




CHAPTER- 10 - Rotational Mechanics




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CHAPTER- 8 - Work and Energy

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