Chapter # 23
Heat and Temperature
SOLVED EXAMPLES
1.
The pressure of air in the bulb of a constant volume gas thermometer is 73 cm of mercury at 0°C, 100.3 cm of mercury at 100°C and 77.8 cm of mercury at room temperature. Find the room temperature in centigrades.
Sol.
We have t =
77.8 73 p p0 × 100°C = 17°C 100C = 100 .3 73 p100 p 0
QUESTIONS
FOR
SHORT
ANSWER
1.
If two bodies are in thermal equilibrium in one frame will they be in thermal equilibrium in all frames?
2.
Does the temperature of a body depend on the frame from which it is observed?
3.
It is heard sometimes that mercury is used in defining the temperature scale because it expands uniformly with the temperature. If the temperature scale is not yet defined, is it logical to say that a substance expands uniformly with the temperature?
4.
In defining the ideal gas temperature scale, it is assumed that the pressure of the gas at constant volume is proportional to the temperature T. How can we verify whether this is true or not? Are we using the kinetic theory of gases? Are we using the experimental result that the pressure is proportional to temperature?
5.
Can the bulb of a thermometer be made of an adiabatic wall?
6.
Why do marine animals live deep inside a lake when the surface of the lake freezes?
7.
The length of a brass rod is found to be smaller on a hot summer day than on a cold winter day as measured by the same aluminium scale. Do we conclude that brass shrinks on heating?
8.
If mercury and glass has equal coefficient of volume expansion, could we make a mercury thermometer in a glass tube?
9.
The density of water at 4°C is supposed to be 1000 kg/m3 . Is it same at the sea level and a high altitude?
10.
A tightly closed metal lid of a glass bottle can be opened more easily if it is put in hot water for some time. Explain.
11.
If an automobile engine is overheated, it is cooled by putting water on it. It is advised that the water should be put slowly with engine running. Explain the reason.
12. 13.
Is it possible for two bodies to be in thermal equilibrium if they are not in contact? A spherical shell is heated. The volume changes according to the equation V = V0 (1 + ). Does the volume refer to the volume enclosed by the shell or the volume of the material making up the shell?
Objective - I 1.
A system X is neither in thermal equilibrium with Y nor with Z. The system Y and Z (A) must be in thermal equilibrium (B) cannot be in thermal equilibrium (C*) may be in thermal equilibrium (D) None of these ,d fudk; X u rks Y ds lkFk vkSj u gh Z. ds lkFk Å"eh; lkE;koLFkk esa gSA fudk; Y rFkk fudk; Z (A) Å"eh; lkE;koLFkk esa gksxaAs (B) Å"eh; lkE;koLFkk esa ugh gks ldrs gSA (C*) Å"eh; lkE;koLFkk esa gks ldrs gSA (D) d
2.
Which of the curves in figure represents the relation between Celsius and Fahrenheit temperature?
fuEu esa ls dkSulk oØ lsfYl;l rFkk QkjsugkbV rkiksa esa lEcU/k O;Dr djrk gS&
manishkumarphysics.in
Page # 1
Chapter # 23 Heat and Temperature 3. Which of the following pairs may give equal numerical values of the temperature of a body? (A*) Fahrenheit and kelvin (B) Celsius and kelvin (C) kelvin and platinum
fuEu esa ls dkSulk ;qXe oLrq ds rki dk leku vkafdd eku n'kkZ ldrk gSA (A*) QkjsugkbV rFkk dsfYou (B) lsfYl;l rFkk dsfYou (C) dsfYou rFkk IysfVue 4.
For a constant volume gas thermometer, one should fill the gas at (A) low temperature and low pressure (B) low temperature and high pressure (C*) high temperature and low pressure (D) high temperature and high pressure.
fu;r vk;ru xSl rkiekih esa] xSl Hkjuh pkfg;sa& (A) de rki ,oa de nkc ij (C*) mPpre ,oa de nkc ij 5.
Consider the following statements. (A) The coefficient of liner expansion has dimension K-1 (B) The coefficient of volume expansion has dimension K-1 (A*) A and B are both correct. (B) A is correct but B is wrong. (C) B is correct but A is wrong. (D) A and B are both wrong.
fuEufyf[kr dFkuksa ij fopkj dhft;s % (A) jsf[kd izlkj xq.kkad dh foek K-1 gSA (B) vk;ru izlkj xq.kkad dh foek K-1 gSA (A*) A rFkk B nksukas lR; gSa (C) B lR; gS fdUrq A vlR; gSA 6.
(B) de rki ,oa mPp nkc ij (D) mPp rki ,oa mPp nkc ij
(B) A lr; gS fdUrq B vlR; (D) A ,oa B nksuksa vlR; gSA
A metal sheet with a circular hole is heated. The hole (A*) gets large (B) gets smaller (C) remains of the size (D) gets deformed
,d o`Ùkkdkj fNnz ;qDr /kkrq dh pknj xeZ dh tkrh gSA fNnz (A*) cM+k gks tk;sxkA (B) NksVk gks tk;sxk 7.
gS
(C) mlh
vkdkj dk jgsxkA (D) vi:fir tk;sxkA
Two identical rectangular strips, one of copper and the other of steel, are rivetted together of form a bimetallic strip (copper>steel). On heating, this strip will (A) remain straight (B*) bend with copper on convex side (C) bend with steel on convex side (D) get twisted
rkacs rFkk LVhy dh nks ,d tSlh vk;rkdkj ifV~V;k¡ fjfcVksa ls tksMdj ,d f}&/kkrqd iV~Vh¼ckbZesVfs yd fLVªe½ cuk;h x;h gSA (copper>steel). xeZ djus ij ;g iV~Vh& (A) lh/kh jgsxhA (B*) bl izdkj eqM+sxh fd rkack mÙkj vksj jgsA (C) bl izdkj eqMs+xh fd LVhy mÙky vksj jgsA (D) bl izdkj eqMsxh 8.
If the temperature of a uniform rod is slightly increased by t, its moment of inertia about a perpendicular bisector increases by (A) zero (B) t (C*) 2t (D) 3t. ;fn ,d le:i NM ds rki esa FkksM+k lh o`f) t dh tk;s] yEc v/kZd ds ikfjr% NM+ ds tM+Ro vk?kw.kZ esa o`f) gksxhA (A) 'kwU; (B) t (C*) 2t (D) 3t.
9.
If the temperature of a uniform rod is slightly increased by t, its moment of inertia about a line parallel to itself will increases by (A) zero (B) t (C*) 2t (D) 3t. ;fn ,d le:i NM+ ds rki esa vlR; o`f) tdh tk;s] yEc v/kZd ds ikfjr% NM+ ds tM+Ro vk?kw.kZ esa o`f) gksxh& (A) 'kwU; (B) t (C*) 2t (D) 3t.
10.
The temperature of water at the surface of a deep lake is 2oC. The temperature expected at the bottom is ,d xgjh >hy dh lrg ij ikuh dk rki 2oC gSA blds iSns ij ikuh dk lEHkkfor rki gks ldrk gSA (A) 0oC (B) 2oC (C*) 4oC (D) 6oC
11.
An aluminium sphere is dipped into water at 100 C. If the temperature is increased, the force of buoyancy (A) will increase (B*) will decrease (C) will remain constant (D) may increase of decrease depending on the radius of the sphere. ,Y;wfefu;e dk ,d xksyk 100 C rki okys ikuh esa Mqck;k x;k gSaA ;fn rki esa o`f) dh tk;s rks mRIykou cy& (A) c<+ tk;sxkA (B*) de gks tk;sxk (C) fu;r jgsxkA (D) c<+ Hkh ldrk gS o de Hkh gks ldrk gSA ;g xksys dh f=kT;k ij fuHkZj djsxkA manishkumarphysics.in
Page # 2
Chapter # 23
1.
Heat and Temperature
Objective - II
A spinning wheel is brought in contact with an identical wheel spinning at identical speed. The wheels slow down under the action of friction. Which of the following energies of the first wheel decrease? (A*) kinetic (B) total (C*) mechanical (D) internal
,d /kwerk gqvk ifg;k leku pky ls ?kwe jgs] blds leku ifg;s ds lEidZ esa yk;k tkrk gSA ?k"kZ.k dh fØ;k ds dkj.k ifg;s /khes gks tkrs gSA izFke ifg;s dh fuEu esa ls dksulh ÅtkZ,¡ de gks tk;sxh& (A*) xfrt (B) dqy (C*) ;kaf=kd (D) vkarfjd 2.
A spinning wheel A is brought in contact with another wheel B initially at rest. Because of the friction at contact, the second wheel also starts spinning. Which of the following energies of the wheel B increase (A*) kinetic (B*) total (C*) mechanical (D*) internal ,d ?kwerk gqvk ifg;k A ,d vU; fLFkj ifg;s B ds lEidZ esa yk;k tkrk gSA lEidZ LFkku ij ?k"kZd ds dkj.k nwljk ifg;k ?kweuk izkjHEk dj nsrk gSA ifg;s B dh fuEu esa ls dkSulh ÅtkZ,¡ (A*) xfrt (B*) dqy (C*) ;kaf=kd (D*) vkarfjd
3.
A body A is placed on a railway platform and an identical body B in moving train. Which of the following energies of B are greater than those of A as seen from the ground ? (A*) kinetic (B*) total (C*) mechanical (D) internal ,d oLrq A jsYos IysVQkeZ ij j[kh gqbZ gSA rFkk blh ds leku ,d oLrq B xfrf'ky Vsªu esa gSA tehu ls ns[kus ij fuEu esa ls B dh dkSulh ÅtkZ, A dh ÅtkZvksa ls vf/kd gksxh& (A*) xfrt (B*) dqy (C*) ;kaf=kkd (D) vkarfjd
4.
In which of the following pairs of temperature scales, the size of a degree is identical ? (A) mercury scale and ideal gas scale (B) Celsius scale and mercury scale (C*) Celsius scale and ideal gas scale (D*) ideal gas scale and absolute scale
rki iSekuksa ds fuEu esa ls dkSuls tksMs esa fMxzh dk vkdkj ,d leku gS& (A) ikjn~ iSekuk o vkn'kZ xSl iSekuk (B) lsfYl;k iSekuk o ikjn~ iSekuk (C*) lsfYl;e iSekuk o vkn'kZ xSl iSekuk (D*) vkn'kZ xSl iSekuk o ijerki iSekuk 5.
A solid object is placed in water contained in an adiabatic container for some time. The temperature of water falls during the period and there is no appreciable change in the shape of the object. The temperature of the solid object. (A*) must have increased (B) must have decreased (C) may have increased (D) may have remained constant.
,d :)ks"e ik=k esa Hkjs gq, ikuh esa dqN le; ds fy;s ,d Bksl oLrq j[kh tkrh gSA bl dky esa ikuh dk rki de gks tkrk gS rFkk oLrq dh vkd`fr esa dksbZ ekius ;ksX; ifjorZu ugh gksrk gSA Bksl oLrq dk rki& (A*) fuf'pr :i ls c<+sxkA (B) fuf'pr :i ls de gksxkA (C) c<+ ldrk gSA (D) fu;r jg ldrk gSA 6.
As the temperature is increased, the time period of a pendulum (A) increases proportionately with temperature (B*) increase (C) decrease (D) remains constant
tSls&tSls rki c<+rk gS] yksyd dk vkorZdky& (A) rki ds lekuqikrh :i ls c<+rk gSA (C) de gksrk gSA
(B*) c<+rk gSA (D) fu;r jgrk
gSA
WORKED OUT EXAMPLES
1. Sol.
The pressure of the gas in a constant volume gas thermometer at steam point (373.15 K) is 1.50 × 104 Pa. What will be the pressure at the triple point of water? The temperature in kelvin is defined as
p T = p × 273.16 K tr Thus, 373.15 =
1.50 10 4 Pa × 273.16 Pt r manishkumarphysics.in
Page # 3
Chapter # 23
Heat and Temperature Pt r = 1.50 × 104 Pa ×
or, 2. Sol.
273.16 = 1.10 × 104 Pa. 373.15
The pressure of air in the bulb of a constant volume gas thermometer at 0°C and 100°C are 73.00 cm and 100 cm of mercury respectively. Calculate the pressure at the room temperature 20°C. The room temperature on the scale measured by the thermometer is
p t p0 t= p × 100°C. 100 p 0 Thus,
p t 73.00 cm of Hg 20°C = 100 cm of Hg 73.00 cm of Hg × 100°C or, 3.
Sol.
pt = 78.4 cm of mercury.
The pressure of the gas in a constant volume gas thermometer is 80 cm of mercury in melting ice at 1 atm. When the bulb is placed in a liquid, the pressure becomes 160 cm of mercury. Find the temperature of the liquid. For an ideal gas at constant volume,
T1 p1 = T2 p2 or, 4.
Sol.
T2 =
160 × 273.15 K = 546.30 K 80
In a constant volume gas thermometer, the pressure of the working gas in measured by the difference in the levels of mercury in the two arms of a U-tube connected to the gas at the end. When the bulb is placed at the room temperature 27.0°C, the mercury column in the arm open to atmosphere stands 5.00 cm above the level of mercury in the other arm. When the bulb is placed in a hot liquid, the difference of mercury levels becomes 45.0 cm. Calculate the temperature of the liquid. (Atmospheric pressure = 75.0 cm of mercury) The pressure of the gas = atmospheric pressure + the pressure due to the difference in mercury levels. At 27°C, the pressure is 75 cm + 5 cm = 80 cm of mercury. At the liquid temperature, the pressure is 75 cm
P2 + 45 cm = 120 cm of mercury. Using T2 = P T1, The temperature of the liquid is 1 T= 5.
Sol.
120 × (27.0 × 273.15) K = 450.22 K = 177.07°C 177°C 80
The resistance of a platinum resistance thermometer at the ice point, the steam point and the boiling point of sulphur are 2.50, 3.50 and 6.50 respectively. Find the boiling point of the sulphur on the platinum scale. The ice point and the steam point measure 0° and 100° respectively. The temperature on the platinum scale is defined as
R t R0 100 , The boiling point of sulphur on this scale is t= R 100 R 0 t= 6.
Sol.
6.50 2.50 × 100° 400°. 3.50 2.50
A platinum resistance thermometer reads 0° and 100° at the ice point and the boiling point of water respectively. The resistance of a platinum wire varies with Celsius temperature as Rt = R0 (1 + + 2), where = 3.8 × 10–3/°C and = – 5.6 × 10–7/(°C)2. What will be the reading of this termometer if it is placed in a liquid bath maintained at 50°C? The resistance of the wire in the thermometer at 100°C and 50°C are R100 = R0 [1 + × 100°C + × (100°C)2] and, R50 = R0 [1 + × 50°C + × (50°C)2]. The temperature t measured on the platinum thermometer is given by
R 50 R 0 t= R × 100° 100 R 0 manishkumarphysics.in
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Chapter # 23
Heat and Temperature =
7.
Sol.
50C (50C)2 100 C (100C)2
× 100° = 50.4°
A platinum resistance thermometer is constructed which reads 0° at ice point and 100° at steam point. Let tp denote the temperature on this scale and let t denote the temperature on a mercury thermometer scale. The resistance of the platinum coil varies with t as Rt = R0 (1 + t + t2). Derive an expression for the resistance as a function of tp. Let R t denote the resistance of the coil at the platinum scale temperature tp. Then p
tp =
or,
R tp R 0 R100 R 0
R tp = =
=
tp 100
tp 100
tp 100
× 100
(R100 – R0) + R0
[R0 {1 + × 100 + × (100)2} – R0] + R0 [ + 100 + × (100)2] R0 + R0
2 tp = R0 1 { 100 (100 ) } 100 = R0 [1 + tp + × (100) tp]. Only numerical value of and are to be used. 8.
Sol.
An iron rod of length 50 cm is joined at an end to an aluminium rod of length 100 cm. All measurements refer to 20°C. Find the length of the composite system at 100°C and its average coefficient of linear expansion. The coefficient of linear expansion of iron and aluminium are 12 × 10–8 /°C and 24 × 10–6 /°C respectively. The length of the iron rod at 100°C is 1 = (50 cm) [1 + (12 × 10–11/°C)(100°C - 20°C)] = 50.048 cm. The length of aluminium rod at 100°C is 2 = (100 cm) [1 + (24 × 10–6/°C) (100°C – 20°C)] = 100.192 cm The length of the composite system at 100°C is 50.048 cm + 100.192 cm = 150.24 cm. The length of the composite system at 20°C is 50 cm. So, The average coefficient of linear expansion of the composite rod is
0.24 cm = 150 cm (100C 20C) = 20 × 10–6 /°C 9.
Sol.
An iron ring measuring 15.00 cm in diameter is to be shrunk on a pulley which is 15.05 cm in diameter. All measurements refer to the room temperature 20°C. To what minimum temperature should the ring be heated to make the job possible? Calculate the strain developed in the ring when it comes to the room temperature. Coefficient of linear expansion of iron = 12 × 10–6/°C. The ring should be heated to increase its diameter from 15.00 cm to 15.05 cm. Using 2 = 1 (1 + ), 0.05 cm
=
15.00 cm 12 10 6 /º C
= 278°C
The temperature = 20°C + 278°C = 298°C. The strain developed = 10.
2 1 = 3.33 × 10–3 . 1
A pendulum clock consists of an iron rod connected to a small, heavy bob. If it is designed to keep correct time at 20°C, how fast or slow will it go in 24 hours at 40°C? Coefficient of linear expansion of iron = 1.2 × 10– 6 /°C. manishkumarphysics.in
Page # 5
Chapter # 23 Sol. The time period at temperature is T = 2
g = 2
Heat and Temperature
0 (1 ) g
= 2
0 1/2 g (1 + )
1 = T0 1 . 2
Thus,
1 T20 = T0 1 ( 20C) 2
and,
1 T40 = T0 1 ( 40C) 2
or,
T40 –1 T20 = [1 + (20°C)] [1 + (10°C)] = [1 + (20°C)] [1 – (10°C)]
or,
T40 T20 = (10°C) = 1.2 × 10–4 T20
...(1)
This is fractional loss of time. As the temperature increases, the time period also increases. Thus, the clock goes slow. The time lost in 24 hours is, by (i), t = (24 hours) (1.2 × 10–4) = 10.4 s. 11. Sol.
A pendulum clock having copper rod keeps correct time at 20°C. It gains 15 seconds per day if cooled to 0°C. Calculate the coefficient of linear expansion of copper. The time period at temperature is T = 2
Thus, or,
g
1 = T0 1 2 T20 = T0 [1 + (10°C)]
(T20 T0 ) = (10°C). .......(i) T0
T20 is the correct time period. The period at 0°C is smaller so that the clock runs fast. Equation (i) gives approximately the fractional gain in time. The time gained in 24 hours is T = (24 hours) [(10°C)] or, 15 s = (24 hours) [(10°C)] or, 12.
Sol.
15 s = (24 hours )(10C) = 1.7 × 10–5/°C
A piece of metal weighs 46 g in air and 30 g in a liquid of density 1.24 × 103 kg/m3 kept at 27°C. When the temperature of the liquid is raised to 42°C, the metal piece weighs 30.5 g. The density of the liquid at 42°C is 1.20 × 103 kg/m3. Calculate the coefficient of linear expansion of the metal. Let the volume of the metal piece be V0 at 27°C and V at 42°C. The density of the liquid at 27°C is 0 = 1.24 × 103 kg/m3 and the density of liquid at 420C is 1.20 × 103 kg/m3 . The weight of the liquid displaced = apparent loss in the weight of the metal piece when dipped in the liquid. Thus, V0 0 = 46 g – 30 g = 16 g and, V = 46 g – 30.5 g = 15.5 g Thus,
or, or, or,
V 0 15.5 V0 × 16 1 + 3 =
1.24 10 3 15.5
1.20 103 16 1 + 3(42°C – 27°C) = 1.00104 = 2.3 × 10–5/°C manishkumarphysics.in
Page # 6
Chapter # 23 13.
Sol.
Heat and Temperature
A sphere of diameter 7.0 cm and mass 266.5 g floats in a bath of liquid. As the temperature is raised, the sphere begins to sink at a temperature of 35°C. If the density of the liquid is 1.527 g/cm3 at 0°C, find the coefficient of cubical expension of liquid neglect the expansion of sphere. It is given that the expansion of the sphere is negligible as compared to the expansion of the liquid. At 0°C, the density of the liquid is 0 = 1.527 g/cm3. At 35°C, the density of the liquid equals the density of the sphere. Thus, 266.5 g 35 = 4 = 1.484 g/cm3 3 (3.5 cm) 3
14.
Sol.
We have
V0 1 0 = V = (1 )
or,
=
Thus,
0 35 = (35C) 35
0 1 =
(1.527 1.484 ) g / cm3 (1.484 g / cm3 ) (35C)
= 8.28 × 10–4 /°C
An iron rod and a copper rod lie side by side. As the temperature is changed, the difference in the lengths of the rods remains constant at a value of 10 cm. Find the lengths at 0°C. Coefficients of linear expansion of iron and copper are 1.1 × 10–5 /°C and 1.7 × 10–5 /°C respectively. Suppose the length of the iron rod at 0°C is i0 and the length of the copper rod at 0°C is c0. The lengths at temperature are i = i0 (1 + i ) ...(i) and c = c0 (1 + c). ...(ii) Subtracting, i – c = (i0 – c0) + (i0 i – c0 c). Now, i – c = i0 – c0 (= 10 cm) Thus , from (iii), i0 i – c0 c or,
i0 c c0 = i
i0 c 17 1.7 10 5 / C = = = 5 i0 c 0 c i 6 0.6 10 / C This shows that i0 – c0 is positive. Its value is 10 cm as given in the question. or,
Hence,
i0 =
17 × (i0 – c0) 6
17 × 10 cm = 28.3 cm 6 c0 = i0 – 10 cm = 18.3 cm.
= and 15.
Sol.
A uniform steel wire of cross-sectional area 0.20 mm2 is held fixed by clamping its two ends. Find the extra force exerted by each clamp on the wire if the wire is cooled from 100°C to 0°C. Young’s modulus of steel = 2.0 × 1011 N/m2. Coefficient of linear expansion of steel = 2.0 × 10–5/°C. Let us assume that tension is zero at 100°C so that is the natural length of the wire at 100°C. As the wire cools down, its natural length decrease to 0. As the wire is fixed at the clamps, its length remains the same as the length at 100°C. Thus , the extension of the wire over its natural length at 0°C is – 0 = 0 (1 + ) – 0 = 0. The strain developed is
0 0 = = . 0
The stress developed = Y × strain = Y . The tension in the wire at 0°C is T = stress × area = Y t × 0.20 mm2 = (2.0 × 1011 N/m2) × (1.2 × 10–5/°C) × 100°C × (0.20 × 10–6 m2) = 48 N This is equal to the extra force exerted by each clamp. manishkumarphysics.in
Page # 7
Chapter # 23 16.
Sol.
17.
Heat and Temperature
A glass vessel of volume 100 cm 3 is filled completely with mercury and is heated from 25°C to 75°C. What volume of mercury will overflow? Coefficient of linear expansion of glass = 1.8 × 10–6/°C and coefficient of volume expansion of mercury is 1.8 × 10–4/°C. 100 cm 3 vk;ru okys ,d dk¡p ds ik=k dks ikjs ls iwjh rjg Hkj fn;k tkrk gS vkSj bls 25°C ls 75°C rd xeZ fd;k tkrk gSA ikjs dk fdruk vk;ru crZu ls ckgj cg tk,xk \ dk¡p dk js[kh; çlkj xq.kkad 1.8 × 10–6/°C rFkk ikjs dk vk;ru çlkj xq.kkad 1.8 × 10–4/°C gSA HCV_Ch-23_WOE_16 The volume of mercury at 25°C is V0 = 100 cm 3 . The coefficient of volume expansion of mercury L = 1.8 × 10–4/°C. The coefficient of volume expansion of glass’ S = 3 × 1.8 × 10–6/°C = 5.4 × 10–6/°C. Thus, the volume of mercury at 75°C us VL = V0 (1 + L ) The volume of mercury overflown = VL – VS = V0 (L – S) ........(i) = (100 cm 3) (1.8 × 10–4 – 5.4 × 10–6)/°C × (50°C) = 0.87 cm 3 . Note that a = (L – S) acts as the effective coefficient of expansion of the liquid with respect to the solid. The expansion of mercury ‘as seen from the glass’ can be written as V – V0 = V0 a or, V = V0 (1 + a). The constant a is called the ‘apparent coefficient of expansion’ of the liquid with respect to the solid. A barometer reads 75.0 cm on a steel scale. The room temperature is 30°C. The scale is correctly graduated for 0°C. The coefficient of linear expansion of steel is = 1.2 × 10–5/°C and the coefficient of volume expansion of mercury is = 1.8 × 10–4/°C. Find the correct atmospheric pressure. ,d csjksehVj LVhy iSekus ij 75.0 cm i<+rk gSA dejs dk rkieku 30°C gSA iSekuk 0°C ij lgh :i ls i<+us ds fy, cuk;k tkrk gSA LVhy ds fy, js[kh; çlkj xq.kkad = 1.2 × 10–5/°C gS rFkk ikjs ds fy, vk;ru çlkj xq.kkad = 1.8 × 10–4/°C
gSA lgh ok;qe.Myh; nkc dh x.kuk dhft,A Sol.
HCV_Ch-23_WOE_17
The 75 cm length of steel at 0°C will become at 30°C where, = (75 cm) [1 + (30°C)]. ....(1) The length of mercury column at 30°C is . Suppose the length of the mercury column, if it were at 0°C, is 0 , Then, 1 = 0 1 (30C) . 3
....(ii)
By (i) and (ii), 1 0 1 (30C) = 75 cm [1 + (30°C)] 3
or,
0 = 75 cm
[1 (30C)] = 75 cm 1 1 ( 30 C ) 3
1 (30C) = 74.89 cm 3
EXERCISE 1.
The steam point and the ice point of a mercury thermometer are marked as 80° and 20°. What will be the temperature in centigrade mercury scale when this thermometer reads 32°? ikjn~ FkekZehVj ds eki Hkki fcUnq ,oa fgekad Øe'k% 80° rFkk 20° fpfUgr fd;s x;s gSA tc ;g FkekZehVj 32°ikB~;kad nsrk
gS] lsfUVxszM ikjn iSekus esa rki fdruk gksxk\ Ans. 200 C 2.
A constant volume thermometer registers a pressure of 1.500 × 104 Pa at the triple point of water and a pressure of 2.050 × 104 Pa at the normal boiling point. What is the temperature at the normal boiling point? ,d fu;r vkiru rkiekih ty ds f=kd~ fcUnq ij 1.500 × 104 ikWLdy nkc rFkk lkekU; DoFkuu fcUnq ij 2.050 × 104
ikWLdy nkc n'kkZrk gSA
manishkumarphysics.in
Page # 8
Chapter # 23 Ans. 373.3 K 3.
Heat and Temperature
A gas thermometer measures the temperature from the variation of pressure of a sample of gas. If the pressure measured at the melting point of lead is 2.20 times the pressure measured at the triple point of water, find the melting point of lead.
,d xSl FkekZehVj] xSl ds ,d uewus ds nkc esa ifjorZu ds vk/kkj ij rki dk ekiu djrk gSA ;fn lhls ds xyu fcUnq ij ekik x;k nkc] ty ds f=kd~ fcUnq ij ekis x;s nkc dk 2.20 xquk gS] lhls dk xyu fcUnq Kkr dhft;sA Ans. 601 K 4.
The pressure measured by a constant volume gas thermometer is 40 kPa at the triple point of water. What will be the pressure measured at the boiling point of water (100°C)? ty ds f=kd~ fcUnq fu;r vk;ru xSl rkiekih }kjk ekik x;k nkc 40 fdyks ikWLdy ( kPa) gSA ty ds DoFku fcUnq (100°C)ij
ekik x;k nkc fdruk gksxk \ Ans. 55 kPa 5.
The pressure of the gas in a constant volume gas thermometer is 70 kPa at the ice point. Find the pressure at the steam point. ,d fu;r vk;ru xSl rkiekih esa fgekad ij nkc 70 kPa gSA Hkki fcUnq ij nkc Kkr dhft;sA Ans. 96 kPa
6.
The pressure of the gas in a 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. ,d fu;r vk;ru xSl rkiekih esa fgekad] Hkkx fcUnq rFkk xeZ ekse dh ukn esa nkc Øe'k% 80 lseh, 90 lseh rFkk 100 lseh
ikjn~ LrEHk gSA ekse dh ukn dk rki Kkr dhft;sA Ans. 2000 C 7.
In a Callender’s compensated constant pressure air thermometer, the volume of the bulb is 1800 cc. When the bulb is kept immersed in a vessel, 200 cc of mercury has to be poured out. Calculate the temperature of the vessel. dSy.s Mj esa izfrdkfjr fLFkj nkc ok;q rkiekih (callender's compensated constant air pressure thermometer) esa cYc vk;ru 1800 ?ku lseh gSA tc cYc fdlh ik=k esa Mqck;k tkrk gSA 200 ?ku lseh ikjk ckj fudkyk tkrk gSA ik=k ds rki dh
x.kuk dhft;sA Ans. 307 K 8.
A platinum resistance thermometer reads 0° when its resistance is 80 and 100° when its resistance is 90 . Find the temperature at the platinum scale at which the resistance is 86 . IysfVue izfrjks/k rkiekih dk ikB~;kad 0° ij 80 rFkk 100° ij 90 gSA tc ;g 86 ikB~;kad nsrk gS] IysfVue iSekus
ij rki Kkr dhft;sA 8. 60° 9.
A resistance thermometer reads R = 20.0 , 27.5 , and 50.0 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 = R0 (1 + + 2 ), find the value of R0, and . Here represents the Celsius scale. ,d izfrjks/k rkiekih dk ikB~;kad R = 20.0 , 27.5 , ,oa 50.0 Øe'k% fgekad ij (0°C), Hkki fcUnq ij (100°C) rFkk ftad fcUnq ij (420°C) gSA eku yhft;s fd rki ds lkFk izfrjks/k esa ifjorZu R = R0 (1 + + 2 ) ds vuqlkj gksrk gS] R0, ,oa ds eku Kkr dhft;sA ;gk¡ lsfYl;l iSekus ij rki iznf'kZr djrk gSA 9. 20.0 , 3.8 × 10–3 /0C, 5.6 × 10–7 /0C2
10.
A concrete slab has a length of 10 m on a winter night when the temperature is 0°C. Find the length of the slab on a summer day when the temperature is 35°C. The coefficient of linear expansion of concrete is 1.0 × 10–5 /°C. lnhZ dh jkr esa tc rki 0ºC gS ,d dØhV iV~Vh dh yEckbZ10 eh- gSA bldh yEckbZ xehZ ds fnuksa esa Kkr dhft;s] tc rki 35°C gksrk gSA daØhV dk jSf[kd izlkj xq.kkad 1.0 × 10–5 /°C gSA 10. 10.0035 m
11.
A meter scale made of steel is calibrated at 20°C to give correct reading. Find the distance between 50 cm mark and 51 cm mark if the scale is used at 10°C. Coefficient of linear expansion of steel is 1.1 × 10–5/°C. LVhy dk cuk gqvk ,d ehVj iSekuk 20°C ij lgh ikB~;kad nsus ds fy;s va'kkfdr fd;k x;k gSA ;fn iSekuks dks 10Crki ij iz;D q r fd;k tk;s rks 50 lseh ,oa 51 lseh ds fpUgks ds chp dh nwjh Kkr dhft;sA LVhy dk jSf[kd izlkj xq.kkad1.1 × –5 10 /°C gSA manishkumarphysics.in Page # 9
Chapter # 23 11. 1.00011 cm 12.
Heat and Temperature
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 = 11 × 10–8/°C. jsy dh iVfj;k¡ (yksgs dh cuh gqb)Z lnhZ ds fnuksa esa fcNk;h x;h gS] tc vkSlr rki 18°C gksrk gSA iVfj;ks ds12.0 eh- yEcs
VqdM+s ,d ds ckn ,d fcNk;s x;s gSA bl izdkj ds nks VqdM+kas ds chp fdruk varjky ¼[kkyh LFkku½ j[kuk pkfg;s fd xehZ;ksa esa tc vf/kdre rkieku 48°C gks tkrk gS] dksbZ ladqpu u gks\ yksgs dk jSf[kd izlkj xq.kkad = 11 × 10–8/°C. 12. 0.4 cm 13.
A circular hole of diameter 2.00 cm is made in an aluminium plate at 0°C. What will be the diameter at 100°C? for aluminium = 2.3 × 10–5/°C. 0ºC rki okyh ,Y;wfefu;e dh ,d IysV esa 2.00 lseh O;kl dk ,d o`Ùkkdkj fNnz gSA 100°C rki ij bldk O;kl fdruk gksxk\ ,Y;wfefu;e ds fy;s = 2.3 × 10–5/°C. 13. 2.0046 cm
14.
Two metre scales, one of steel and the other of aluminium, agree at 20°C. Calculate the ratio aluminiumcentimetre/steel-centimetre at (a) 0°C, (b) 40°C and (c) 100°C. for steel = 1.1 × 10–5/°C and for aluminium = 2.3 × 10–5/°C. nks ehVj iSekus] ,d LVhy dk rFkk ,d ,Y;wfefu;e dk 20°C ij leku ikB~;kad n'kkZrs gSA vuqikr
,Y;wfefu;e&lsUVhehVj@LVhy&lsUVhehVj dh x.kuk dhft;sA (a) 0°C,rki ij (b) 40°C rki ij ,oa (c) 100°C rki ijA LVhy ds fy;s = 1.1 × 10–5/°C ,oa ,Y;wfefu;e ds fy;s = 2.3 × 10–5/°C 14. (a) 0.99977 (b) 1.00025
(c) 1.00096
15.
A metre scale is made up of steel and measures 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 = 11 × 10–6/°C. LVhy ls cuk gqvk ,d ehVj&iSekuk 16°C rki ij lgh yEckbZ ekirk gSA izfr'kr =kqfV dk eku fdruk gksxk] ;fn ;g iSekuk iz;D q r fd;k tk;sA (a) xehZ ds fnukas esa tc rki 46°C gksrk gS rFkk (b) lnhZ ds fnuksa esa tc rki 6°C? gksrk gS\ LVhy dk jSf[kd izlkj xq.kkad = 11 × 10–6/°C Ans. (a) 0.033% (b) – 0.011%
16.
A metre scale made of steel reads accurately at 20°C. In a sensitive experiment, distances accurate upto 0.055 mm in 1m are required. Find the range of temperature in which the experiment can be performed with this metre scale. Coefficient of linear expansion of steel = 11 × 10–6/°C. LVhy dk ,d ehVj&iSekuk 20°C rki ij lgh ikB~;kad nsrk gSA ,d vR;ar lqxzkgh iz;ksx esa 1 eh- nwjh esa 0.055 feeh dh
;FkkFkZrk vko';d gSA og rki ijkl Kkr dhft;s fy;s ;g iz;ksx bl ehVj iSekuk dh lgk;rk ls fd;k tl ldsA LVhy dk jSf[kd izlkj xq.kkad = 11 × 10–6/°C. Ans. 150C to 250C 17.
The density of water at 0°C is 0.998 g/cm 3 and at 4°C is 1.000 g/cm 3. Calculate the average coefficient of volume expansion of water in the temperature range 0 to 4°C. 0°C rki ij ikuh dk ?kuRo 0.998 xzke@lseh3 rFkk 4°C rki ij 1.000 xzke@lseh3 gksrk gSA 0ºC ls 4°C rki ijkl esa ikuh
dk vkSlr vkiru izlkj xq.kkad Kkr dhft;sA Ans. – 5 × 10–4 / 0C 18.
Find the ratio of the lengths of an iron rod and an aluminium rod for which the difference in the lengths is independent of temperature. Coefficients of linear expansion of iron and aluminium are 12 × 10 –6/°C and 23 × 10–6/°C respectively.
,d yksgs dh NM+ rFkk ,d ,Y;wfefu;e dh NM+ dh yEckbZ;ks dk og vuqiky Kkr dhft;sA ftlds fy;s] yEckbZ;ksa dk vUrj rki ij fuHkZj ugh djrk gksA yksgs rFkk ,Y;wfefu;e dks jSf[kd izlkj xq.kkad Øe'k% 12 × 10–6/°C ,oa 23 × 10–6/°C gSA Ans. 23 : 12 19.
A pendulum clock gives correct time at 20°C at a place where g = 9.800 m/s 2. 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/s 2. At what temperature will it give correct time? Coefficient of linear expansion of steel = 12 × 10–6/°C HCV_Ch-23_Ex_19
,d is.Mqye ?kM+h tgka g = 9.800 m/s gS] 20°C ij lgh le; nsrh gS iS.Mqye ,d gYdh LVhy NM+ ¼jksM+½ ls ,d Hkkjh xsan dks tksM+dj cuk gSA bls ,slh nwljh txg ij ys tk;k tkrk gS tgk¡ g = 9.788 m/s2 gSA fdl rkieku ij ;g lgh le; nsxhA LVhy ds fy, js[kh; izlkj xq.kkad = 12 × 10–6/°C gSA 2
Ans. – 820C manishkumarphysics.in
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Chapter # 23
Heat and Temperature
20.
An aluminium plate fixed in a horizontal position has a hole of diameter 2.00 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? Coefficient of linear expansion of aluminium is 23 × 10–6/°C and that of steel is 11 × 10–6/°C. ,d ,Y;wfefu;e IysV {kSfrt fLFkr esa dldj j[kh gqbZ gS] blesa 2.00 lseh O;kl dk ,d fNnz gSA bl fNnz ij 2.005 lseh O;kl okyk LVhy dk xksyk j[kk gqvk gSa leLr yEckbZ;k¡ 10°C rki ij gSA lEiw.kZ fudk; dk rki /khjs&/khjs c<+k;k tkrk gSA fdl rki ij xksyk fxj tk;sxk\ ,Y;wfefu;e dk jSf[kd izlkj xq.kkad 23 × 10–6/°C rFkk LVhy dk 11 × 10–6/°C dk gSA Ans. 2190 C
21.
A glass window is to be fit in an aluminium frame. The temperature on the working day is 40°C and the glass window measures exactly 20 cm × 30 cm. What should be the size of the aluminium frame so that there is no stress on the glass in winter even if the temperature drops to 0°C? Coefficient of linear expansion for glass and aluminium are 9.0 × 10–6/°C and 24 × 10–6/°C respectively. ,Y;wfefu;e dh pkS[kV esa dkap dh f[kM+dh yxk;h x;h gSA fdlh fnol ij rki 40°C gS rFkk dkap dh f[kM+dh dk eki 20 × 30 lseh gSA ,Y;wfefu;e dh pkS[kV dk vkdkj D;k gksuk pkfg;s fd ;fn lfnZ;ksa esa rki 0°C jg tk;s rks Hkh dkap ij dksbZ izfrcy u yxs? dkap rFkk ,Y;wfefu;e ds jSf[kd izlkj Øe'k% 9.0 × 10–6/°C rFkk 24 × 10–6/°C
gSA
Ans. 20.012 cm × 30.018 cm 22.
The volume of glass vessel is 1000 cc at 20°C. What volume of mercury should be poured into it at this temperature so that the volume of the remaining space does not change with temperature? Coefficient of cubical expansion of mercury and glass are 1.8 × 10–4/°C and 9.0 × 10–6/°C respectively. dk¡p ds ,d ik=k dk 20°C rki ij vk;ru 1000 ?ku lseh gSA bl rki ij blesa ikjs ds fdruk vk;ru Hkjk tk;s
fd 'ks"k cps LFkku dk vk;ru rki ds lkFk ifjofrZr ugha gks\ ikjk rFkk dk¡p dk vk;ru izlkj xq.kkad Øe'k% 1.8 × 10–4/°C rFkk 9.0 × 10–6/°C gSA Ans. 50 cc 23.
An aluminium can of cylindrical shape contains 500 cm 3 of water. The area of the inner cross-section of the can is 125 cm 2. All measurements refer to 10°C. Find the rise in the water level if the temperature increases to 80°C. The coefficient of linear expansion of aluminium = 23 × 10–6/°C and the average coefficient of volume expansion of water = 3.2 × 10–4/°C respectively. ,Y;wfefu;e dh ,d csyukdkj dsu esa 500 lseh3 ikuh gSA dsu ds vkarfjd vuqiLz Fk dkV dk {ks=kQy 125 lseh2 gSA leLr eki 10°C rki ds fy;s gSA ;fn rki c<+dj 80°C gks tkrk gS rks ikuh ds ry esa o`f) Kkr dhft;sA ,Y;wfefu;e dk jSf[kd izlkj xq.kkad = 23 × 10–6/°C rFkk ikuh dk vkSlr vk;ru izlkj xq.kkad = 3.2 × 10–4/°C gSA Ans. 0.089 cm
24.
A glass vessel measures exactly 10 cm × 10 cm × 10 cm at 0°C. it is filled completely with mercury at this temperature. When the temperature is raised to 10°C, 1.6 cm3 of mercury overflows. Calculate the coefficient of volume expansion of mercury. Coefficient of linear expansion of glass = 6.5 × 10–6/°C dk¡p ds ,d ik=k dk 0°C rki ij eki 10 × 10 × 10 lseh gSA bldks bl rki ij ikjs ls iw.kZr;k Hkj fn;k x;k gSA tc rki dks 10°C rd c<+k;k tkrk gS] 1.6 lseh3 ikjk ckgj fudy tkrk gSA ikjs ds vk;ru izlkj xq.kkad dh x.kuk dhft;sA dk¡p dk jSf[kd izlkj xq.kkad = 6.5 × 10–6/°C A Ans. 1.8 × 10–4 /0C
25.
The densities of wood and benzene at 0°C are 880 kg/m3 and 900 kg/m3 respectively. The coefficients of volume expansion are 1.2 × 10–3/°C for wood and 1.5 × 10–3/°C for benzene. At what temperature will a piece of wood just sink in benzene? 0°C rki ij ydM+h rFkk cSathu ds ?kuRo Øe'k% 880 fdxzk@eh3 rFkk 900 fdxzk@eh3 gSA ydM+h ds fy;s vk;ru izlkj xq.kkad 1.2 × 10–3/°C ,oa cSathu dk 1.5 × 10–3/°C gSA fdl rki ij ydM+h dk VqdM+k cSathu esa l| Mwc tk;sxk\ Ans. 830 C
26.
A steel rod of length 1m rests on a smooth horizontal base. If it is heated from 0°C to 100°C, what is the longitudinal strain developed? 1 eh yEch LVhy dh NM+ ,d fpdus vk/kkj ij j[kh gqbZ gSA ;fn bldks 0°C ls 100°C, rd xeZ fd;k tk;s rks fdruh
vuqizLFk foÑfr mRiUu gksxh\ Ans. zero 27.
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 the temperature rises to 50°C. Coefficient of linear expansion of steel = 1.2 × 10–5/°C. manishkumarphysics.in
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Chapter # 23
Heat and Temperature
,d LVhy NM+ blds nksuksa fljksa ij LysEi ls dlh gqbZ gS rFkk fLFkj {kSfrt vk/kkj ij j[kh gqbZ gSA 20°C rki ij NM+ esa dksbZ foÑfr ugha gSA ;fn rki c<+dj 50°C gks tk;s rks NM+ esa mRiUu vuqnS/;Z foÑfr Kkr dhft;sA LVhy dk jSf[kd izlkj xq.kkad = 1.2 × 10–5/°C Ans. – 3.6 × 10–4 28.
Sol.
A steel wire of cross-sectional area 0.5 mm2 is held between two fixed supports. If the wire is just taut at 20°C, determine the tension when the temperature falls to 0°C. Coefficient of linear expansion of steel is 1.2 × 10–5/°C and its Young’s modulus is 2.0 × 1011 N/m2. [Q.28/HCV-2/Ch-23/Exercise] [2] 0.5 feeh2 vuqizLFk dkV {ks=kQy okyk LVhy dk ,d rkj nks fLFkj vk/kkjksa ds chp dlk x;k gSA ;fn 20°C rki ij rki l| ruk gqvk gS] tc rki 0°C rd fxjus ij ruko fu/kkZfjr dhft;sA LVhy dk jSf[kd izlkj xq.kkad 1.2 × 10–5/°C rFkk ;ax dk izR;kLFkrk 2.0 × 1011 U;wVu@eh2 gSA F = A Y t = 0.5 × 10–6 × 2 × 1011 × 1.2 × 10–5 × 20 = 24 N Ans. 24 N
29.
A steel rod is rigidly clamped at its ends. The rod in under zero tension at 20°C. If the temperature rises to 100°C, what force will the rod exert on one of the clamps. Area of cross-section of the rod = 2.00 mm2. Coefficient of linear expansion of steel = 12.0 × 10–6 /°C and Young’s modulus of steel = 2.00 × 1011 N/m2. ,d LVhy NM+ blds nksuksa fljksa ij n`<+rk iwod Z dlh gqbZ gSA 20°C ij NM+ esa ruko 'kwU; gSA ;fn rki c<+dj 100°C gks tkrk gS] rks NM+ fdlh ,d DysEi ij fdruk cy yxk;sxhA NM+ ds vuqizLFk dkV dk {ks=kQy = 2.00 eheh2. ,oa LVhy dk jSf[kd izlkj xq.kkad = 12.0 × 10–6 /°C rFkk LVhy ds fy;s ;ax dk izR;kLFkrk xq.kkad = 2.00 × 1011 U;wVu@eh2 Ans. 384 N
30.
Two steel rods and an aluminium rod of equal length 0 and equal cross-section are joined rigidly at their ends as shown in the figure below. All the rods are in a state of zero tension at 0°C. Find the length of the system when the temperature is raised to . Coefficient of linear expansion of aluminium and steel are a and s respectively. Young’s modulus of aluminium is Ya and of steel is Ys leku yEckbZ 0 rFkk leku vuqizLFk dkV okyh nks LVhy dh NM+s rFkk ,Y;wfefu;e dh NM+ fuEu fp=k esa n'kkZ;s vuqlkj buds fljksa ls tksM+h x;h gSA leLr NM+sa 0°C rki ij 'kwU; ruko dh fLFkfr esa gSA tc rki c<+kdj dj fn;k tkrk gSA bl fudk; dh yEckbZ Kkr dhft;sA ,Y;wfefu;e ,oa LVhy ds jSf[kd izlkj xq.kkad Øe'k% a rFkk s gSA ,Y;wfefu;e ds fy;s ;ax dk izR;kLFkrk xq.kkad Ya rFkk LVhy ds fy;s Ys gSA
.
a Ya 2 S YS Ans. 0 1 Ya 2YS
31.
A steel ball initially at a pressure of 1.0 × 105 Pa is heated from 20°C to 120°C keeping its volume constant. Find the pressure inside the ball. Coefficient of linear expansion of steel = 12 × 10–6/°C and bulk modulus of steel = 1.6 × 1011 N/m2. LVhy dh ,d xsan izkjfEHkd nkc 1.0 × 105 ikLdy ij 20°C ls 120°C rd vk;ru fu;r j[krs gq, xeZ dh tkrh gSA xsan ds vUnj nkc Kkr dhft;sA LVhy dk jSf[kd izlkj xq.kkad = 12 × 10–6/°C ,oa LVhy dk vkiru izR;kLFkrk xq.kkad = 1.6 × 1011 U;wVu@eh2 A Ans. 5.8 × 10–8 Pa
32.
Show that moment of inertia of a solid body of any shape changes with temperature as = 0 (1 + 2), where 0 is the moment if inertia at 0°C and is the coefficient of linear expansion of the solid. O;Dr dhft;s fd fdlh Hkh vkÑfr dh Bksl oLrq dk tM+Ro vk?kw.kZ rki ds lkFk = 0 (1 + 2), ds vuqlkj ifjofrZr gksrk gS] tgk¡ 0 0°C rki ij tM+Ro vk?kw.kZ gS rFkk Bksl dk jSf[kd izlkj xq.kkad gSA
33.
A torsional pendulum consists of a solid disc connected to a thin wire ( = 2.4 × 10–5/°C) at its centre. Find the percentage change in the period between peak winter (5°C) and peak summer (45°C). ,d ejksM+h nkstd dks ,d&,d irys rkj ( = 2.4 × 10–5/°C) dks ,d Bksl pdrh ds dsUnz ij tksM+dj cuk;k x;k gSA ?kksj lnhZ (5°C) ls ?kksj xehZ (45°C) rd blds vkorZdky esa izfr'kr ifjorZu Kkr dhft;sA Ans. 9.6 × 10–2 manishkumarphysics.in
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Chapter # 23 Heat and Temperature 34. A circular disc made of iron is rotated about its axis at a constant velocity . Calculate the percentage change in the linear speed of a particle of the rim as the disc is slowly heated from 20°C to 50°C keeping the angular velocity constant. Coefficient of linear expansion of iron = 1.2 × 10–5/°C. yksgs dh ,d o`Ùkkdkj pdrh bldh v{k ds ifjr% fu;r dks.kh; osx ls ?kqek;h tk jgh gSA tc pdrh dks bldk dks.kh; osx fu;r j[krs gq, 20°C ls 50°C rd /khjs&/khjs xeZ fd;k tkrk gS rks bldh usfe ij fLFkr d.k dh jSf[kd pky ls izfr'kr ifjorZu dh x.kuk dhft;s = 1.2 × 10–5/°C Ans. 3.6 × 10–2
manishkumarphysics.in
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