Q: (a) "Room temperature" is often taken to be 68∘F; what will it be on Celsius scale?(b) The temperature of the filament in a light bulb is about 1800∘C; what will it be on Fahrenheit scale?

(i) (a) We use TC=95(TF−32)=95(68−32)=20∘C(b) We useTF=59TC+32=59×1800+32=3272∘F

Q: If a temperature scale defined the freezing point of water as 100∘ and the boiling point as 0∘. What temperature on this scale corresponds to 25∘C ?

We use 0−100T−100=100−025−0⇒T−100=−25⇒T=75∘

Q: In an alcohol-in-glass thermometer, the alcohol column has length 12.45 cm at 0.0∘C and length 21.30 cm at 100.0∘C. What is the temperature if the column has length (a) 15.10 cm, and (b) 22.95 cm

(a) We use100−0T−0=21.30−12.4515.0−12.45=8.852.65⇒T=8.852.65×100=29.94∘C(b) We use 100−0T−0=21.30−12.4522.95−12.45=8.8510.5⇒T=8.8510.5×100=118.64∘C

Q: At what temperature will the Fahrenheit and Centigrade scales yield the same numerical value?

We use 212−32TC−32=100−0TC−0At TF=TC=T0 we useT0−32=1.8T0⇒−0.8T0=32T0=−40∘F

Q: A steel tape measures the length of a copper rod as 90.0 cm and when both are at 10∘C, which is the graduation temperature of the tape. What would the tape read for the length of the rod when both are at 30∘C. Given that coefficient of linear expansion of steel is 1.2×10−5C−1 and that of copper is 1.7×10−5∘C−1.

(i) At 30∘C length of copper rod islcu=90(1+1.7×10−5×20)=90.0306 cmLength of 1 cim of steeel tảp àt30∘=1(1+1.2×10−5×20)=1.00024 cmReading at30∘C=l1 cm of tap lcu=1.0002490.0306=90.01 cm

Question 1

(a) "Room temperature" is often taken to be 68∘F; what will it be on Celsius scale?(b) The temperature of the filament in a light bulb is about 1800∘C; what will it be on Fahrenheit scale?

Accepted Answer

(i) (a) We use TC​=95​(TF​−32)=95​(68−32)=20∘C(b) We useTF​=59​TC​+32=59​×1800+32=3272∘F

Question 2

If a temperature scale defined the freezing point of water as 100∘ and the boiling point as 0∘. What temperature on this scale corresponds to 25∘C ?

Accepted Answer

We use 0−100T−100​=100−025−0​⇒T−100=−25⇒T=75∘

Question 3

In an alcohol-in-glass thermometer, the alcohol column has length 12.45 cm at 0.0∘C and length 21.30 cm at 100.0∘C. What is the temperature if the column has length (a) 15.10 cm, and (b) 22.95 cm

Accepted Answer

(a) We use100−0T−0​=21.30−12.4515.0−12.45​=8.852.65​⇒T=8.852.65​×100=29.94∘C(b) We use 100−0T−0​=21.30−12.4522.95−12.45​=8.8510.5​⇒T=8.8510.5​×100=118.64∘C

Question 4

At what temperature will the Fahrenheit and Centigrade scales yield the same numerical value?

Accepted Answer

We use 212−32TC​−32​=100−0TC​−0​At TF​=TC​=T0​ we useT0​−32=1.8T0​⇒−0.8T0​=32T0​=−40∘F

Question 5

A steel tape measures the length of a copper rod as 90.0 cm and when both are at 10∘C, which is the graduation temperature of the tape. What would the tape read for the length of the rod when both are at 30∘C. Given that coefficient of linear expansion of steel is 1.2×10−5C−1 and that of copper is 1.7×10−5∘C−1.

Accepted Answer

(i) At 30∘C length of copper rod islcu​=90(1+1.7×10−5×20)=90.0306 cmLength of 1 cim of steeel tảp àt30∘=1(1+1.2×10−5×20)=1.00024 cmReading at30∘C=l1 cm​ of tap lcu​​=1.0002490.0306​=90.01 cm

Question 6

In the construction of a flyover, steel girders of length 12 m are used to place one after another. How much gap should be left between the two at their junction so that in summers when peak temperature is about 48∘C, there should not be any compression? Given that the construction is done in peak winters when temperature is 18∘C. Given that the coefficient of linear expansion of stecl is 1.1×10−5∘C−1.

Accepted Answer

(ii) For the given caseΔT=30∘CExpansion of girders isΔl=l×ΔT=12×1.1×10−5×30=0.00396 m

Question 7

A steel rod is clamped between two rigid supports at 20∘C. If the temperature of rod increases to 50∘C, find the strain developed in the rod due to this. Given that the coefficient of linear expansion of steel is 1.2×10−50C−1.

Accepted Answer

Thermal expansion in rod isΔl−l×ΔTdue to clamps elastic strain in rod=lΔl​=−αΔT=−1.2×10−5×30=−3.6×10−4

Question 8

Shows a steel rod of cross sectional area 2×10−6 m2 is fixed between two vertical walls. Initially at 20∘C, there is no force between the ends of the rod and the walls. Find the force which the rod will exert on walls at 100∘C. Given that the coefficient of linear expansion of steel is 1.2×10−50C−1 and its Young's modulus is 2×1011 N/m2.

Accepted Answer

Force by rod on wall isF=YA×ΔT=2×1011×2×106×1.2×10−5×80=384 N

Question 9

A glass window is to be fitted in an aluminium frame. The temperature on the working day is 40∘C and the size of the glass piece is a rectangle of sides 20 cm and 30 cm. What should be the dimensions of the aluminium frame so that in winters glass does not experience any stress when temperature drops to 0∘C. Given that the coefficients of linear expansion for glass is 9×10−6∘C−1 and that for aluminium is 2.4×10−5∘C−1.

Accepted Answer

For ΔT=40∘C dimansions of glass in winters isl=30(1−9×10−6×40)=29.989a=20(1−9×10−6×40)=19993dimensions of aluminimum frame at 40∘C arel1​=29.989×(1+2.4×10−5×40)=30.018 cmω1​=19.993(1+2.4×10−5×40)=20.012 cm

Question 10

A cylindrical steel component machined on an engine lathe is heated to temperature of 80∘C. The diameter of the component should be 5 cm at temperature of 10∘C and the permissible error should not exceed 10 microns from the specified dimension. Do we need corrections for the thermal expansion component be introduced during the process of machining? If so, what diameter should be prepared? Take : αsteel ​=12×10−6∘C−1

Accepted Answer

Permissible error=10×10−6 m=10−3 cmrequired diameter at 80∘C isd=5.00(1+1.2×10−5×70)=5.0052 cm

Question 11

At a certain temperature the pendulum of a clock keeps correct time. The coefficient of linear expansion for the pendulum material is α=1.85×10−5 K−1. How much will the clock gain ơ lóse in 24 höurs if the anbient lémpérature is 10∘C higher?

Accepted Answer

Time lost by clock per second istΔt​=21​αΔT=21​×1.85×10−5×10=9.25×10−5 sTotal time lost in 24 hours is=9.25×10−5×24×3600=7.992 s

Question 12

A clock pendulum made of invar has a period of 0.5 s at 20∘C. If the clock is used in a climate where average temperature is 30∘C, what correction may be necessary at the end of 30 days. αinvar ​=7×10−7(∘C)−1

Accepted Answer

Time lost by clock per second istΔt​=21​αΔT=21​×7×10−7×10=3.5×10−6 stime lost by clock in 30 days is=3.5×10−6×30×86400=9.072 s

Question 13

The time period of a physical pendulum is given by T=2πmglI​​ where m= mass of the pendulum, I= moment of inertia about the axis of suspension, l= distance of centre of mass of bob from the centre of suspension. Calculate the change in time period when temperature changes by ΔT. The coefficient of linear expansion of the material of pendulum is

Accepted Answer

As temperature changes by ΔT, new length l′ and moment of inertial I′ is given as-l′=l(1+αΔT)I′=I(1+2αΔT)Thus new time period isT′=2πmgl′I′​​=2πmglI​​(1+αΔT)1/2(1+2αΔT)1/2​As αΔT<<1, using binomial affroximate we getT′=2πmglI​​(1+21​αΔT)change in period isT′−T=2πmglI​​(21​αΔT)=παΔTmglI​​

Question 14

Two rods of material X sandwich another rod of material Y as shown in figure. At temperature T, the three rods are in a state of zero strain and of length L and riveted to each other in this state. If the temperature of the system increases to T+ΔT, find the final length of the system of the three rods. Given that the coefficients of linear expansions of the rods and their Young's modulus for material X and Y are αx​,αy​,Yx​ and Yy​ respectively. Consider αy​>αx​.

Accepted Answer

Natural elongation in x and y rods arelx​=L(1+αx​ΔT) ly​=L(1+αy​ΔT)Due to pivots if final length is Lf​ we use strain in x and y rods as( strain )x​=lx​Lf​−lx​​≃LLf​−lx​​( strain )y​=ly​ly​−Lf​​≃Lly​−Lf​​There we can use lx​≃ly​≃L as numerator is very small difference Now stress on y is twice that of x so we useYy​(strain)y​=2Yx​(strain)x​ ⇒Yy​(L(1+αy​ΔT)−Lf​)=2Yx​(Lf​−L(1+αx​ΔT)) Yy​L+αy​Yy​LΔT−Lf​Yy​=2Yx​Lf​−2Yx​L−2Yx​αx​LΔT⇒Lf​=L[1+(Yy​+2Yx​αy​Yy​+2αx​Yx​​)ΔT]

Question 15

In a mercury in glass thermometer the cross-section of the capillary is A0​ and volume of the bulb is V0​ at 0∘C. If mercury just fills the bulb at 0∘C, find that the length of mercury in the capillary at temperature t∘C. Given that β= coefficient of cubical expansion of mercury and α= coefficient of linear expansion of glass.

Accepted Answer

Volume expansion of mercury with respect to bulb isΔV=V0​(β−3α)tFind area of cross section of capillary isAf​=A0​(1+2αt)height of mercury diased in capillaryh=Af​ΔV​=A0​(1+2αt)V0​(β−3α)t​

Question 16

A one-litre flask contains some mercury. It is found that at different temperatures the volume of air inside the flask remains the same. What is the volume of mercury in flask ? Given that the coefficient of linear expansion of glass =9×10−6∘C−1 and coefficient of volume expansion of mercury =1.8×10−4∘C−1.

Accepted Answer

If x is the volume of mercury we use1−x=1(1+3×9×10−6×ΔT)−x(1+1.8×10−4ΔT)⇒27×10−6=x×1.8×10−4⇒x=18027​=1.15ltr=150 cm3

Question 17

Some steel balls are fixed at the bottom of a silica bulb of negligible expansivity. The bulb holds 340 gm of mercury at 0∘C when filled, in the absence of the steel balls and 255gm of mercury, when the steel balls are inside. On heating the bulb and its contents (steel balls and mercury) to 100∘C,4.8gm of mercury overflows. Find the coefficient of linear expansion of steel. γHg​=180×10−6∘C−1.

Accepted Answer

Let volume of silica balls is V0​Thus at 0∘C340=V0​ρHg​ at 0∘Cif V1​ is the volume of steel balls we use255=(V0​−V1​)ρHg​ at 0∘C⇒85=V1​ρHg​ at 0∘C at 100∘C 250.2=[V0​−V1​(1+γs​(100))]1+γHg​(100)ρHg​​ at 0∘C250.2[1+γHg​(100)]=255−85γs​×100⇒254.7=255−8500γs​⇒γs​=35.29×10−6∘C−1⇒αs​=3γs​​=11.76×10−6∘C−1

Question 18

If a liquid is contained in a long narrow rigid vessel so it an expand in essentially one direction only, show that the effective coefficient of linear expansion α of liquid is approximately equal to the coefficient of volume expansion β of liquid.

Accepted Answer

If volume expansion of a liquid is ΔV, we useΔV=VγΔtheight raised by liquid in vessel ish=AΔV​[A→ constant ]h=AVγΔt​=hi​γΔt[hi​→ initial leved of liquid ]⇒γ→ coefficient of linear expansion of liquid

Question 19

In observing the real expansion of a liquid a certain part of glass vessel is to be filled with a liquid of volume expansion γ=18×10−5 in order to exclude the effect of the change in remaining volume of the vessel during heating. What fraction of the vessel should be filled with the liquid ? The coefficient of linear expansion of glass is α=9.0×10−6.

Accepted Answer

If vessel volume is V and its f fraction is filled with liquid, we useV−fV=V(1+27×10−6Δt)−fV(1+18×10−5Δt)⇒f=18×10−527×10−6​=203​

Question 20

A clock with a metallic pendulum gains 6 seconds each day when the temperature is 20∘C and loses 6 second when the temperature is 40∘C. Find the coefficient of linear expansion of the metal.

Accepted Answer

For simple pendulum T=2πgl​​⇒T∝l1/2⇒TΔT​=21​lΔl​=21​αΔθAssuming clock gives correct time at temperature θ0​⇒24×36006​=21​α(θ0​−20)24×36006​=21​α(40−θ0​)⇒θ0​=30∘C⇒α=1.4×10−5∘C−1

Filters

Filters

book

topic

difficulty

Solutions for Physics Galaxy Thermodynamics, Oscillations and Waves Vol 2 by Ashish Arora

Found the solution, but did not understand the concept?

Ask my first question