Class 11th

11.8 Initial temperature, $T_1$ = 27.0 °C, Initial diameter of the hole, $d_1$ = 4.24 cm

Final temperature, $T_2$ = 227.0 °C, Final diameter of the hole $=d_2$

Coefficient of linear expansion of copper, ${\alpha }_{copper}$ = 1.70 $*{10}^{-5}$ K–1

We know

$\frac{Changeinarea(?A)}{Originalarea\left(A\right)}$ = $\beta ?$ T where $\beta$ is the coefficient of superficial expansion, $\beta =2{\alpha }_{copper}$

$\frac{(\mathit{\pi }\frac{{\mathit{d}}_2^2}{4}-\mathit{}\mathit{\pi }\frac{{\mathit{d}}_1^2}{4})}{\mathit{\pi }\frac{{\mathit{d}}_1^2}{4}}$ = $2{\alpha }_{copper}?$ T

$\frac{{\mathit{d}}_2^2-\mathit{}{\mathit{d}}_1^2}{{\mathit{d}}_1^2}$ = $2{\alpha }_{copper}?$ T

$\frac{{\mathit{d}}_2^2}{{\mathit{d}}_1^2}$ - 1= $2{\alpha }_{copper}\left(T_2-T\right)=$ 2 $*$ 1.70 $*{10}^{-5}$ (227-27) = 6.8 $*{10}^{-3}$

$d_2^2=1.0068*4.24*4.24$

$d_2$ = 4.2544 cm

So change in diameter = 4.2544 – 4.24 = 0.01439 cm

Diameter increase by 1.44 $*{10}^{-2}$ cm

11.7 Given, temperature $T_1$ = 27 °C = 27 + 273.16 K = 300.16 K

Outer dia of the shaft at temp $T_1$ ,  $d_1$ = 8.7 cm

Diameter of the central hole of the wheel,  $d_2$ = 8.69 cm

The change in diameter, Δd= $d_2-d_1=$ 8.69 – 8.7 = $-$ 0.01 cm

After the shaft is cooled in dry ice, its temperature becomes $T_2$ . It can be calculated from the relation

Δd= $d_1*{\alpha }_{steel}$  ( $T_2-T_1)$

-0.01 = 8.7 $*1.20\mathrm{}*\mathrm{}{10}^{-5}(T_2-300)$

$T_2-300$ = -95.78

$T_2$ = 204.22 K = -68.94 $?$$$

11.6 Length of the steel tape, l = 1 m = 100 cm, At temperature T = 27 $°$ C

Coefficient of linear expansion of steel $\alpha$ = 1.2 $*{10}^{-5}$ / K

Let $l_1=63cm$ be the length of the steel rod at temperature $T_1$ = 45.0 °C and

$l_2$ be the length of the steel rod and l' be the length of the steel tape at 45.0 °C

We have l' = l + $\alpha l(T_1-T)$ = 100 + 1.2 $*{10}^{-5}*100*$  (45-27) = 100.0216 cm

$l_2$ can be calculated as $l_2$ = $\frac{l\text{'}}{l}*63$ = 63.0136 cm

11.3 It is given that R = $R_0$ [1 + α(T – $T_0$ )] ………(i)

Where Ro and To are the initial resistance and temperature and R and T are the final resistance and temperature.

At the triple point of water, To = 273.15 K, $R_0$ = 101.6 Ω

At normal melting point of lead, T = 600.5 K, R = 165.5 Ω

Substituting these values in equation (i), we get

165.5 = $101.6$ [1 + α(600.5 – $273.15$ )]

$\frac{165.5}{101.6}$ = 1 + 327.35 α, or α = $\frac{0.629}{327.35}$ = 1.92 $*{10}^{-3}$ $K^{-1}$

When R = 123.4 Ω, T can be calculated as

123.4 = $101.6$ [1 + 1.92 $*{10}^{-3}*$ (T – $273.15$ )]

1.214 = 1 + T1.92 $*{10}^{-3}$ - 1.92 $*{10}^{-3}*$ 273.15

T = 384.6 K

11.2 Triple point of water on absolute scale A,  $T_1$ = 200 A

Triple point of water on absolute scale B,  $T_2$ = 350 B

Triple point of water on absolute Kelvin scale,  $T_k$ = 273.15 K

The temperature 273.15 K on Kelvin scale is equivalent to 200 on absolute scale A

$T_1=T_k$

200 A = 273.15 K, Therefore A = $\frac{273.15}{200}$

Similarly B = $\frac{273.15}{350}$

If $T_A$ is the triple point of water on scale A and $T_B$ is the triple point of water on scale B, we have

$\frac{273.15}{200}$ $*T_A$ = $\frac{273.15}{350}*$ $T_B$

$T_A=$ $\frac{200}{350}*$ $T_B$ = $\frac{4}{7}*$ $T_B$

11.1 Kelvin and Celsius scales are related as

$T_c$ = $T_k$ - 273.15 ….(i),

Where $T_c=$ temperature in Celsius scale and $T_k$ = temperature in Kelvin scale

Celsius and Fahrenheit scales are related as

$\frac{T_c\mathrm{}}{5}$ = $\frac{T_f-32}{9}$ , where $T_f$ = temperature in Fahrenheit scale

(a) For Neon, $T_k$ = 24.57. Hence $T_c$ = 24.57 – 273.15 = -248.58 degree Celsius

$T_f=\frac{9}{5}*T_c+32$ = $\frac{9}{5}*(-248.58)+32$ = -415.44 degree Fahrenheit

(b) For Carbon dioxide, $T_k$ = 216.55. Hence $T_c$ = 216.55 – 273.15 = -56.6 degree Celsius

$T_f=\frac{9}{5}*T_c+32$ = $\frac{9}{5}*(-56.6)+32$ = -69.88 degree Fahrenheit

Let $\theta$ be the angle made by the radius vector joining the bead and the centre of the wire with the downward direction. Let, N be the normal reaction.

mg = N $\cos ?\theta$ …….(1)

mr ${\omega }^2$ = N $\sin ?\theta$ ……(2)

m(R $\sin ?\theta$ ) ${\omega }^2$ = N $\sin ?\theta$

Hence N = m(R) ${\omega }^2$

Substituting the value on N in eqn (1)

mg = mR ${\omega }^2\cos ?\theta$

or $\cos ?\theta$ = g/ R ${\omega }^2$ ………(3)

As $\left|\cos ?\theta \right|$ $\le$ 1, the bead will remain at the lowermost point

g/ R ${\omega }^2$ $\le 1or\omega \le \sqrt{g}/R$

For $\omega$ = $\frac{\sqrt{2g}}{R},equation\left(3\right)$ becomes

$\cos ?\theta$ = g/ R ${\omega }^2$

$\cos ?\theta$ =(g/R)(R/2g) = ½

$\theta =60°$