Thermal Expansion MCQ Questions & Answers in Heat and Thermodynamics | Physics

Young’s modulus $$Y = \frac{{{\text{stress}}}}{{{\text{strain}}}}$$
$$ = \frac{{\frac{F}{A}}}{{\left( {\frac{{\Delta \ell }}{\ell }} \right)}}$$
Using, co - efficient of linear expansion,
$$\eqalign{ & \frac{{\Delta \ell }}{\ell } = \alpha \,\Delta T \cr & \therefore \,\,Y = \frac{F}{{A\left( {\alpha \,\Delta T} \right)}} \cr} $$

$$\eqalign{ & Y = \frac{{\frac{F}{S}}}{{\frac{{\Delta L}}{L}}} \cr & \Rightarrow \Delta L = \frac{{FL}}{{SY}}\, \cr & \therefore \,\,L\alpha \,\Delta T = \frac{{FL}}{{SY}}\,\,\left[ {\because \,\Delta L = L\alpha \,\Delta T} \right] \cr & \therefore \,\,F = SY\alpha \,\Delta T \cr} $$
$$\therefore $$ The ring is pressing the wheel from both sides,
$$\therefore \,\,{F_{{\text{net}}}} = 2\,F = 2\,YS\alpha \,\Delta T$$

Mercury thermometer is a liquid thermometer and it is based upon the uniform variation in volume of a liquid with temperature. Mercury is opaque and bright and therefore can be easily seen in the glass tube and it is good conductor of heat and attains the temperature of the hot bath quickly. A mercury thermometer can be used to measure temperature upto $${300^ \circ }C$$  or so, as before boiling at $${376^ \circ }C,$$  the vaporisation of mercury will start.
NOTE
By filling nitrogen gas at high pressure above the mercury column, a mercury thermometer may be used to measure temperature upto $${500^ \circ }C.$$

$$\eqalign{ & \frac{{\Delta \ell }}{\ell } = \frac{{\pi \times 0.05}}{{\pi \times 40}} = \frac{1}{{800}} \cr & T = Y\frac{{\Delta \ell }}{\ell } \times A \cr & = 200 \times {10^9} \times \frac{1}{{800}} \times 1 \times {10^{ - 6}} = 250\,N \cr} $$

Young's modulus $$Y = \frac{{{\text{stress}}}}{{{\text{strain}}}}$$
stress $$ = Y \times \,\,{\text{strain}}$$
Stress in steel wire = Applied pressure
Pressure = stress $$ = Y \times \,\,{\text{strain}}$$
Strain $${\text{ = }}\frac{{\Delta L}}{L} = \alpha \,\Delta T$$    (As length is constant)
$$\eqalign{ & = 2 \times {10^{11}} \times 1.1 \times {10^{ - 5}} \times 100 \cr & = 2.2 \times {10^8}\,Pa \cr} $$

Young's modulus $$ = \frac{{{\text{Thermal stress}}}}{{{\text{Strain}}}} = \frac{{\frac{F}{A}}}{{\frac{{\Delta L}}{L}}}$$
$$Y = \frac{F}{{A \cdot \alpha \cdot \Delta \theta }}\left( {\because \frac{{\Delta L}}{L} = \alpha \Delta \theta } \right)$$
Force developed in the rail $$F = YA\alpha \Delta t$$
$$\eqalign{ & = 2 \times {10^{11}} \times 40 \times {10^{ - 4}} \times 1.2 \times {10^{ - 5}} \times 10 \cr & = 9.6 \times {10^4} = 1 \times {10^5}\,N \cr} $$

$$\eqalign{ & \ell = 5\;m,{t_1} = {0^ \circ }C \cr & {\ell _2} = 5.01\;m,{t_2} = {100^ \circ }C \cr & \alpha = \frac{{{\ell _2} - {\ell _1}}}{{{\ell _1}\left( {{t_2} - {t_1}} \right)}} = \frac{{5.01 - 5}}{{5 \times 100}} = 2 \times {10^{ - 5}}{/^ \circ }C \cr} $$

$$\eqalign{ & L = {L_0}\left( {1 + \alpha \Delta \theta } \right) \Rightarrow \frac{{{L_1}}}{{{L_2}}} = \frac{{1 + \alpha {{\left( {\Delta \theta } \right)}_1}}}{{1 + \alpha {{\left( {\Delta \theta } \right)}_2}}} \cr & \Rightarrow \frac{{10}}{{{L_2}}} = \frac{{1 + 11 \times {{10}^{ - 6}} \times 20}}{{1 + 11 \times {{10}^{ - 6}} \times 19}} \Rightarrow {L_2} = 9.99989 \cr} $$
⇒ Length is shorten by $$10 - 9.99989 = 0.00011 = 11 \times {10^{ - 5}}\,cm.$$

(The relation $$R = {R_0}\left( {1 + \alpha \,\Delta t} \right)$$    is valid for small values of $${\Delta t}$$ and $${R_0}$$ is resistance at 0°$$C$$ and also $$\left( {R - {R_0}} \right)$$  should be much smaller than $${R_0}.$$  So, statement (1) is wrong but statement (2) is correct.

From question, $$\left( {{\ell _2} - {\ell _1}} \right)$$  is maintained same at all temperatures hence change in length for both rods should be same i.e., $$\Delta {\ell _1} = \Delta {\ell _2}$$
As we know, coefficient of linear expansion, $$\alpha = \frac{{\Delta \ell }}{{{\ell _0}\Delta T}}$$
$${\ell _1}{\alpha _1}\Delta T = {\ell _2}{\alpha _2}\Delta T \Rightarrow {\ell _1}{\alpha _1} = {\ell _2}{\alpha _2}$$