23: Heat And Temperature : Neet-xii-physics

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NEET-XII-Physics

23: Heat and Temperature

with Solutions - page 4

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Qstn #11
A metre scale made of steel is calibrated at 20°C to give correct reading. Find the distance between the 50 cm mark and the 51 cm mark if the scale is used at 10°C. Coefficient of linear expansion of steel is 1.1 × 10^-5 °C^-1.
Ans : Given:
Temperature at which the steel metre scale is calibrated, t₁ = 20^oC
Temperature at which the scale is used, t₂ = 10^oC
So, the change in temperature, `` \Delta ``t = (20^o`` -``10^o) C
The distance to be measured by the metre scale, L_o = (51`` -``50) = 1 cm = 0.01 m
Coefficient of linear expansion of steel, `` {\alpha }_{steel}`` = 1.1 × 10^-5 °C^-1
Let the new length measured by the scale due to expansion of steel be L₂, Change in length is given by,
`` ∆L={L}_{1}{\alpha }_{steel}\Delta t``
`` \Rightarrow ∆L=1\times 1.1\times {10}^{-5}\times 10``
`` \Rightarrow ∆L=0.00011\,\mathrm{\,cm\,}``
`` ``
As the temperature is decreasing, therefore length will decrease by `` ∆L``.
Therefore, the new length measured by the scale due to expansion of steel (L₂) will be,
L₂ = 1 cm `` -`` 0.00011 cm = 0.99989 cm
Page No 13:

Qstn #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 rises to 48°C? Coefficient of linear expansion of iron = 11 × 10^-6 °C^-1.
Ans : Given:
Length of the iron sections when there's no effect of temperature on them, L_o = 12.0 m
Temperature at which the iron track is laid in winter, t_w = 18^oC
Maximum temperature during summers, t_s = 48^oC
Coefficient of linear expansion of iron, `` \alpha `` = 11 × 10^-6 °C^-1
Let the new lengths attained by each section due to expansion of iron in winter and summer be L_wand L_s_,respectively, which can be calculated as follows:
`` {L}_{w}={L}_{0}\left(1+\alpha {t}_{w}\right)``
`` \Rightarrow {L}_{w}=12\left(1+11\times {10}^{-6}\times 18\right)``
`` \Rightarrow {L}_{w}=12.00237\,\mathrm{\,m\,}``
`` {L}_{s}={L}_{0}\left(1+\alpha {t}_{s}\right)``
`` \Rightarrow {L}_{s}=12\left(1+11\times {10}^{-6}\times 48\right)``
`` \Rightarrow {L}_{s}=12.006336\,\mathrm{\,m\,}``
`` \therefore ∆L={L}_{s}-{L}_{w}``
`` \Rightarrow \Delta L=12.006336-12.002376``
`` \Rightarrow \Delta L=0.00396\,\mathrm{\,m\,}``
`` \Rightarrow \Delta L\approx 0.4\,\mathrm{\,cm\,}``
Therefore, the gap (`` \Delta ``L) that should be left between two iron sections, so that there is no compression during summer, is 0.4 cm.
Page No 13:

Qstn #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? α for aluminium = 2.3 × 10^-5 °C^-1.
Ans : Given:
Diameter of a circular hole in an aluminium plate at 0°C, d₁ = 2 cm = 2 × 10^-2 m
Initial temperature, t₁ = 0 °C
Final temperature, t₂ = 100 °C
So, the change in temperature, (`` \Delta ``t) = 100°C - 0°C = 100°C
The linear expansion coefficient of aluminium, α_al = 2.3 × 10^-5 °C^-1
Let the diameter of the circular hole in the plate at 100^oC be d₂, which can be written as:
`` {d}_{2}={d}_{1}\left(1+\alpha \Delta t\right)``
`` \Rightarrow {d}_{2}``= 2 × 10^-2(1 + 2.3 × 10^-5× 10²)
`` \Rightarrow {d}_{2}``= 2 × 10^-2 (1 + 2.3 × 10^-3)
`` \Rightarrow {d}_{2}``= 2 × 10^-2 + 2.3 × 2 × 10^-5
`` \Rightarrow {d}_{2}`` = 0.02 + 0.000046
`` \Rightarrow {d}_{2}``= 0.020046 m
`` \Rightarrow {d}_{2}``≈ 2.0046 cm
Therefore, the diameter of the circular hole in the aluminium plate at 100^oC is 2.0046 cm.
Page No 13:

Qstn #14
Two metre scales, one of steel and the other of aluminium, agree at 20°C. Calculate the ratio aluminium-centimetre/steel-centimetre at ( α for steel = 1.1 × 10^-5 °C^-1 and for aluminium = 2.3 × 10^-5°C^-1.)
Ans : Given:
At 20°C, length of the metre scale made up of steel, L_st= length of the metre scale made up of aluminium, L_al
Coefficient of linear expansion for aluminium, α_al = 2.3 × 10^-5°C^-1
Coefficient of linear expansion for steel, α_st = 1.1 × 10^-5°C^-1
Let the length of the aluminium scale at 0°C, 40°C and 100°C be L_0al,L₄_0aland L₁₀_0al.
And let the length of the steel scale at 0°C, 40°C and 100°C be L_0st,L₄_0stand L₁₀_0st.

#14-a
0°C,
Ans : So, L_0st(1 - α_st × 20) = L_0al(1 - α_al × 20)
`` \frac{{L}_{0st}}{{L}_{0al}}=\frac{\left(1-{\alpha }_{al}\times 20\right)}{\left(1-{\alpha }_{st}\times 20\right)}``
`` \Rightarrow \frac{{L}_{0st}}{{L}_{0al}}=\frac{1-2.3\times {{10}^{-}}^{5}\times 20}{1-1.1\times {10}^{-5}\times 20}``
`` \Rightarrow \frac{{L}_{0st}}{{L}_{0al}}=\frac{0.99954}{0.99978}``
`` \Rightarrow \frac{{L}_{0st}}{{L}_{0al}}=0.999759``

#14-b
40°C and
Ans : `` \frac{{L}_{40al}}{{L}_{40st}}=\frac{{L}_{0al}\left(1+\alpha {}_{al}\times 40\right)}{{L}_{0st}\left(1+{\alpha }_{st}\times 40\right)}``
`` \Rightarrow \frac{{L}_{40al}}{{L}_{40st}}=\frac{{L}_{0al}}{{L}_{0st}}\times \frac{\left(1+2.3\times {10}^{-5}\times 40\right)}{\left(1+1.1\times {10}^{-5}\times 40\right)}``
`` \Rightarrow \frac{{L}_{40al}}{{L}_{40st}}=\frac{0.99977\times 1.00092}{1.00044}``
`` \Rightarrow \frac{{L}_{40al}}{{L}_{40st}}=1.0002496``
`` ``

#14-c
100°C.
Ans : `` \frac{{L}_{100al}}{{L}_{100st}}=\frac{{L}_{0al}\left(1+\alpha {}_{al}\times 100\right)}{{L}_{0st}\left(1+{\alpha }_{st}\times 100\right)}``
`` \Rightarrow \frac{{L}_{100al}}{{L}_{100st}}=\frac{0.99977\times 1.0023}{1.0023}``
`` \Rightarrow \frac{{L}_{100al}}{{L}_{100st}}=1.00096``
`` ``
Page No 13:

Qstn #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

#15-a
on a summer day when the temperature is 46°C and
Ans : Let the correct length measured by a metre scale made up of steel 16 °C be L.
Initial temperature, t₁ = 16 °C
Temperature on a hot summer day, t₂ = 46 °C
So, change in temperature, Δθ = t₂`` -``t₁= 30 °C
Coefficient of linear expansion of steel, `` \alpha `` = 1.1 × 10^-5°C^-1
Therefore, change in length,
ΔL = L αΔθ = L × 1.1 × 10^-5 × 30
`` \%\,\mathrm{\,of\,}\,\mathrm{\,error\,}=\left(\frac{∆L}{L}\times 100\right)\%``
`` =\left(\frac{L\alpha \Delta \theta }{L}\times 100\right)\%``
`` =\left[1.1\times {10}^{-5}\times 30\times 100\right]\%``
`` =3.3\times {10}^{-2}\%``
`` ``

#15-b
on a winter day when the temperature is 6°C? Coefficient of linear expansion of steel = 11 × 10^-6 °C^-1.
Ans : Temperature on a winter day, t₂ = 6 °C
So, change in temperature, Δθ = t₁`` -`` t₂= 10 °C
ΔL = L₂`` -`` L₁= L αΔθ = L × 1.1 × 10^-5 × 10
`` \%\,\mathrm{\,of\,}\,\mathrm{\,error\,}=\left(\frac{∆L}{L}\times 100\right)\%``
`` =\left(\frac{L\alpha \Delta \theta }{L}\times 100\right)\%``
`` =1.1\times {10}^{-5}\times 10\times 100\%``
`` =1.1\times {10}^{-2}``
`` ``
Page No 13:

Qstn #16
A metre 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 temperature in which the experiment can be performed with this metre scale. Coefficient of linear expansion of steel = 11 × 10^-6 °C^-1.
Ans : Given:
Temperature at which a metre scale gives an accurate reading, T₁ = 20 °C
The value of variation admissible, ΔL = 0.055 mm = 0.055 × 10^-3 m, in the length, L₀ = 1 m
Coefficient of linear expansion of steel, α = 11 × 10^-6 °C^-1
Let the range of temperature in which the experiment can be performed be T₂.
We know: ΔL = L₀ αΔT
`` \Rightarrow 0.055\times {10}^{-3}=1\times 11\times {10}^{-6}\times \left({T}_{1}\pm {T}_{2}\right)``
`` \Rightarrow 5\times {10}^{-3}=\left(20\pm {T}_{2}\right)\times {10}^{-3}``
`` \Rightarrow 20\pm {T}_{2}=5``
`` \Rightarrow \text{E}\,\mathrm{\,ither\,}{T}_{2}=20+5=25°\,\mathrm{\,C\,}``
`` \,\mathrm{\,or\,}{T}_{2}=20-5=15°\,\mathrm{\,C\,}``
Hence, the experiment can be performed in the temperature range of 15 °C to 25 °C .
Page No 13:

Qstn #17
The density of water at 0°C is 0.998 g cm^-3and at 4°C is 1.000 g cm^-1. Calculate the average coefficient of volume expansion of water in the temperature range of 0 to 4°C.
Ans : Given:
Density of water at 0°C, ( f₀)= 0.998 g cm^-3
Density of water at 4°C, (f₄) = 1.000 g cm-3
Change in temperature, (Δt) = 4^oC
Let the average coefficient of volume expansion of water in the temperature range of 0 to 4°C be γ.
`` \,\mathrm{\,We\,}\,\mathrm{\,know\,}:{f}_{4}={f}_{0}\left(1+\gamma ∆t\right)``
`` \Rightarrow {f}_{0}=\frac{{f}_{4}}{1+\gamma ∆t}``
`` \Rightarrow 0.998=\frac{1}{1+\gamma .4}``
`` \Rightarrow 1+4\gamma =\frac{1}{0.998}``
`` \Rightarrow 4\gamma =\left(\frac{1}{0.998}\right)-1``
`` \Rightarrow \gamma =0.0005=5\times {10}^{-4}{}^{\,\mathrm{\,o\,}}{\,\mathrm{\,C\,}}^{-1}``
As the density decreases,
`` \gamma =-5\times {10}^{-4}{}^{\,\mathrm{\,o\,}}{\,\mathrm{\,C\,}}^{-1}``
Therefore,the average coefficient of volume expansion of water in the temperature range of 0 to 4°C will be `` \gamma =-5\times {10}^{-4}``^oC^-1.
Page No 13:

Qstn #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^-1 and 23 × 10^-6 °C^-1 respectively.
Ans : Let the original length of iron rod be L_Fe and L^'_Febe its length when temperature is increased by ΔT.
Let the original length of aluminium rod be L_Al and L^'_Albe its length when temperature is increased by ΔT.
Coefficient of linear expansion of iron, `` {\alpha }_{Fe}`` = 12 × 10^-6 °C^{`` -``}1
Coefficient of linear expansion of aluminium, α_Al = 23 × 10^-6 °C^{`` -``}1
Since the difference in length is independent of temperature, the difference is always constant.
`` L{\text{'}}_{Fe}={L}_{Fe}\left(1+{\alpha }_{Fe}\times ∆T\right)``
`` \,\mathrm{\,and\,}L{\text{'}}_{Al}={L}_{Al}\left(1+{\alpha }_{Al}\times ∆T\right)``
`` \Rightarrow \begin{array}{cc}L{\text{'}}_{Fe}-L{\text{'}}_{Al}={L}_{Fe}-{L}_{Al}+{L}_{Fe}\times {\alpha }_{Fe}∆T-{L}_{Al}\times {\alpha }_{Al}\times ∆T& -\left(1\right)\end{array}``
`` \,\mathrm{\,Given\,}:``
`` L{\text{'}}_{Fe}-L{\text{'}}_{Al}={L}_{Fe}-{L}_{Al}``
`` \,\mathrm{\,Hence\,},{L}_{Fe}{\alpha }_{Fe}={L}_{Al}{\alpha }_{Al}[\,\mathrm{\,using\,}(1\left)\right]``
`` \Rightarrow \frac{{L}_{Fe}}{{L}_{Al}}=\frac{23}{12}``
The ratio of the lengths of the iron to the aluminium rod is 23:12.
Page No 13:

Qstn #19
A pendulum clock shows 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^-1. At what temperature will the clock show correct time? Coefficient of linear expansion of steel = 12 × 10^-6 °C^-1.
Ans : Given:
The temperature at which the pendulum shows the correct time, T₁ = 20 °C
Coefficient of linear expansion of steel, `` \alpha `` = 12 × 10^-6 °C^-1
Let T₂ be the temperature at which the value of g is 9.788 ms^-2and `` \Delta ``T be the change in temperature.
So, the time periods of pendulum at different values of g will be t₁ and t₂ , such that
`` {t}_{1}=2\,\mathrm{\,\pi \,}\sqrt{\frac{{l}_{\mathit{1}}}{{g}_{1}}}``
`` {t}_{2}=2\,\mathrm{\,\pi \,}\sqrt{\frac{{l}_{2}}{{g}_{2}}}``
`` =\frac{2\,\mathrm{\,\pi \,}\sqrt{{l}_{1}\left(1+\alpha \Delta T\right)}}{\sqrt{{g}_{2}}}\left(\because {l}_{2}={l}_{1}\left(1+\alpha ∆T\right)\right)``
`` \,\mathrm{\,Given\,},{t}_{1}={t}_{2}``
`` \Rightarrow \frac{2\pi \sqrt{{l}_{1}}}{\sqrt{{g}_{1}}}=\frac{2\pi \sqrt{{l}_{1}\left(1+\alpha ∆T\right)}}{\sqrt{{g}_{2}}}``
`` \Rightarrow \sqrt{\left(\frac{{l}_{1}}{{g}_{1}}\right)}=\frac{\sqrt{{l}_{1}\left(1+\alpha ∆T\right)}}{\sqrt{{g}_{2}}}``
`` \Rightarrow \frac{1}{9.8}=\frac{1+12\times {10}^{-6}\times ∆T}{9.788}``
`` \Rightarrow \frac{9.788}{9.8}=1+12\times {10}^{-6}\times ∆T``
`` \Rightarrow \frac{9.788}{9.8}-1=12\times {10}^{-6}\times ∆T``
`` \Rightarrow ∆T=\frac{-0.00122}{12\times {10}^{-6}}``
`` \Rightarrow {T}_{2}-20=-102.4``
`` \Rightarrow {T}_{2}=-102.4+20``
`` =-82.4``
`` \Rightarrow {T}_{2}\approx -82°\,\mathrm{\,C\,}``
Therefore, for a pendulum clock to give correct time, the temperature at which the value of g is 9.788 ms^-2 should be `` -`` 82^oC.
Page No 13:

Qstn #20
An aluminium plate fixed in a horizontal position has a hole of diameter 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? Coefficient of linear expansion of aluminium is 23 × 10^-6 °C^-1 and that of steel is 11 × 10^-6 °C^-1.
Ans : Given:
Diameter of the steel sphere at temperature (T₁= 10 °C) , d_st = 2.005 cm
Diameter of the aluminium sphere, d_Al = 2.000 cm
Coefficient of linear expansion of steel, α_st = 11 × 10^{`` -6``} °C^{`` -1``}
Coefficient of linear expansion of aluminium, α_Al = 23 × 10^{`` -``6} °C^{`` -``1}
Let the temperature at which the ball will fall be T₂_,so that change in temperature be ΔT.
d'_st= 2.005(1 + α_st ΔT)
`` \Rightarrow d{\text{'}}_{\,\mathrm{\,st\,}}=2.005+2.005\times 11\times {10}^{-6}\times ∆T``
`` d{\text{'}}_{\,\mathrm{\,Al\,}}=2\left(1+{\alpha }_{\,\mathrm{\,Al\,}}\times ∆T\right)``
`` \Rightarrow d{\text{'}}_{\,\mathrm{\,Al\,}}=2+2\times 23\times {10}^{-6}\times ∆T``
The steel ball will fall when both the diameters become equal.
So, d'_st= d'_Al
`` \Rightarrow 2.005+2.005\times 11\times {10}^{-6}∆T=2+2\times 23\times {10}^{-6}∆T``
`` \Rightarrow \left(46-22.055\right)\times {10}^{-6}∆T=0.005``
`` \Rightarrow ∆T=\frac{0.005\times {10}^{6}}{23.945}=208.81``
`` \,\mathrm{\,Now\,},∆T={T}_{2}-{T}_{1}={T}_{2}-10°\,\mathrm{\,C\,}``
`` \Rightarrow {T}_{2}=∆T+{T}_{1}=208.81+10``
`` \Rightarrow {T}_{2}=218.8\cong 219°\,\mathrm{\,C\,}``
Therefore, the temperature at which the ball will fall is 219 °C.
Page No 13: