Physics NCERT Exemplar Solutions Class 11th Chapter Eleven: Overview, Questions, Preparation

Explanation- As l_iron-l_brass =10cm= constant at all temperature

Let l_o be the length of temperature at 0⁰C  and l be the length after change in temperature

l_iron-l_brass =10cm

l_iron(1+ ${\alpha }_{iron}∆t$ )-l_brass(1+ ${\alpha }_{brass}∆t$ )=10cm

I_{iron $\alpha$ iron}= I_{brass $\alpha$ brass}

$\frac{l_{iron}}{l_{brass}}$ =1.8/1.2=3/2

$\frac{1}{2}{l}_{brass}=10cm$

L_brass=20cm and l_iron=30cm

Explanation- As difference in volume is constant

By considering the diagram

Let V_io , V_bo be the volume of iron and brass vessel at 0⁰C

V_i,V_b be the volume of iron and brass vessel at $∆\theta$ ⁰C

_{${\gamma }_i,{\gamma }_b$} be the coefficient of volume expansion of iron and brass.

V_io -V_bo= 100cc= V_i-V_b

V_i =V_io(1+ $\gamma$ _{i $∆\theta$} )

V_b =V_bo(1+ $\gamma$ _{b $∆\theta$} )

V_i-V_b = (V_io -V_bo)+ $∆\theta \left(V_{io}{\gamma }_i-V_{bo}{\gamma }_b\right)$

Since V_i-V_b= constant

V_{io $\gamma$ i}= V_{bo ${\gamma }_b$}

$\frac{V_{io}}{V_{b0}}=\frac{{\gamma }_b}{{\gamma }_i}=\frac{\frac{3}{2}{\beta }_b}{\frac{3}{2}{\beta }_i}=\frac{{\beta }_b}{{\beta }_i}=\frac{6\times {10}^{-5}}{3.55\times {10}^{-5}}=\frac{6}{3.55}$

Using above equations

$V_{io}=244.9cc$

V_bo = 144.9cc

11.3 Calculate the stress developed inside a tooth cavity filled with copper when hot tea at temperature of 57 ⁰C is drunk. You can take body (tooth) temperature to be 37⁰ C and

α = 1.7× 10^-5 / ⁰C , bulk modulus for copper = 140 × 10⁹ N/m².

Explanation- Decrease in temperature $∆t$ = 57-37= 20⁰C

Coefficient of linear expansion $\alpha$ = 1.7 $\times {10}^{-5}/$ ^oC

Bulk modulus for copper B = 140 $\times {10}^{9}N/m^2$

Coefficient of cubical expansion $\gamma$ = 3 $\alpha$ = 5.1 $\times {10}^{-5}/°C$

Let initial volume of the cavity be V and its volume increases by $∆V$  due to increase in temperature.

$∆V=\gamma V∆t$

$\frac{∆V}{V}=\gamma ∆t$

Thermal stress produced = B $\times volumetricstrain$

= B $\times \frac{∆V}{V}=B\gamma ∆t$

= 140 $\times {10}^9\times 5.1\times {10}^{-5}\times 20$

= 1428 $\times {10}^{8}N/m$ ²

Explanation- By applying Pythagoras theorem in given figure

^{${\left(\frac{L+∆L}{2}\right)}^2={\left(\frac{L}{2}\right)}^2+x$ 2}

x= $\sqrt[]{{\left(\frac{L+∆L}{2}\right)}^2-{\left(\frac{L}{2}\right)}^2}$

=1/2 $\sqrt[]{{(L+∆L)}^2-L^2}$

= ½ $\sqrt[]{L^2+{∆L}^2+2L∆L-L^2}$

=1/2 $\sqrt[]{\left({∆L}^2+2L∆L\right)}$

As Increase in $∆L$  is very small so we neglect it

=1/2 $\times \sqrt[]{2∆L}$

$∆L=L\alpha ∆t$

By using this value in above in equation

x=1/2 $\times \sqrt[]{2L\times L\alpha ∆T}$ = ½ L $\sqrt[]{2\alpha ∆t}$

= $\frac{10}{2}\times \sqrt[]{2\times 1.2\times {10}^{-5}\times 20}$

= 5 $\times \sqrt[]{4\times 1.2\times {10}^{-4}}$

= 5 $\times 2\times 1.1\times {10}^{-2}=0.11=11cm$