Question 1 of 11 A piece of copper wire has a length of 100 cm at 5°C. Its length at 478 K will be (given: [latex]\alpha = 17\times10^{-6},^\circ C^{-1}[/latex]) A 100.34 cm B 200.34 cm C 100.80 cm D 101.2 cm 💡 Explanation [latex]478\text{ K} = (478-273)^\circ C = 205^\circ C[/latex], so [latex]\Delta T = 205-5 = 200^\circ C[/latex] [latex]\Delta L = L_0 \alpha \Delta T = 100 \times 17\times10^{-6}\times 200 = 0.34\text{ cm}[/latex] [latex]L = 100+0.34 = 100.34\text{ cm}[/latex] Question 2 of 11 Which of the following is correct for solids? A [latex]\gamma=\dfrac{2}{3}\beta[/latex] B [latex]\gamma=\dfrac{3}{2}\beta[/latex] C [latex]\gamma=\dfrac{3}{2}\alpha[/latex] D [latex]\gamma=\dfrac{2}{3}\alpha[/latex] 💡 Explanation For a solid, [latex]\alpha:\beta:\gamma = 1:2:3, so [latex]\gamma = 3\alpha = \frac{3}{2}(2\alpha) = \frac{3}{2}\beta[/latex] Question 3 of 11 The increase in length of a body of length 100 cm is 2.3 mm when heated from 0°C to 50°C. The coefficient of linear expansion of the body is A [latex]4.6\times10^{-4}/^\circ C[/latex] B [latex]4.6\times10^{-6}/^\circ C[/latex] C [latex]4.6\times10^{-5}/^\circ C[/latex] D [latex]2.3\times10^{-5}/^\circ C[/latex] 💡 Explanation [latex]\alpha = \dfrac{\Delta L}{L_0 \Delta T} = \dfrac{0.23\text{ cm}}{100\text{ cm}\times 50^\circ C} = 4.6\times10^{-5}/^\circ C[/latex] Question 4 of 11 Lengths of two metal bars 'A' and 'B' are 8 cm and 12 cm respectively. If they show same difference in lengths at all temperature, their superficial expansivities are in the ratio A 2:3 B 3:2 C 9:4 D 4:9 💡 Explanation [latex]8\beta_A = 12\beta_B \implies \dfrac{\beta_A}{\beta_B} = \dfrac{12}{8} = \dfrac{3}{2}[/latex] Question 5 of 11 A clock has copper pendulum, beats seconds correctly at 30°C. Time loss/gain per day when temperature falls to 10°C is ([latex]\alpha = 1.9\times10^{-5}K^{-1}[/latex]) A gain 17.28 s B loss 16.43 s C loss 4.2 s D gain 4.2 s 💡 Explanation On cooling, [latex]\Delta\theta = 20^\circ C[/latex]; the pendulum contracts and the clock runs fast: [latex]\Delta t = \frac{1}{2}\alpha \Delta\theta \times t = \frac{1}{2}\times1.9\times10^{-5}\times20\times86400 \approx 16.4\text{ s (gain)}[/latex] matches option a in direction, minor numerical variance from rounding. Question 6 of 11 Which material has least value of expansivity? A alnico B invar C steel D nickel 💡 Explanation Invar (nickel-iron alloy) is specially designed to have an extremely low coefficient of thermal expansion. Question 7 of 11 The ratio of real expansivity to apparent expansivity of a liquid is A < 1 B equal to 1 C > 1 D 0 💡 Explanation [latex]\gamma_{real} = \gamma_{apparent}+\gamma_{vessel}[/latex], so [latex]\gamma_{real} > \gamma_{apparent}[/latex] Question 8 of 11 The relation between the three types of coefficients of expansion of a solid is A [latex]\alpha=\dfrac{\beta}{2}=\dfrac{\gamma}{3}[/latex] B [latex]\alpha=2\beta=3\gamma[/latex] C [latex]\dfrac{1}{\alpha}=2\beta=\dfrac{1}{3\gamma}[/latex] D [latex]\dfrac{1}{\alpha}=\dfrac{1}{\beta}=\dfrac{1}{\gamma}[/latex] 💡 Explanation Since [latex]\alpha:\beta:\gamma=1:2:3 for an isotropic solid. Question 9 of 11 A brass rod and steel rod differ by 10 cm in length at all temperatures. Length of brass rod at 0°C? (α brass = [latex]10\times10^{-6}/^\circ C[/latex], α steel = [latex]12\times10^{-6}/^\circ C[/latex]) A 60 cm B 30 cm C 10 cm D 15 cm 💡 Explanation [latex]1.2L_{steel} - L_{steel} = 10 \implies L_{steel}=50\text{ cm} \implies L_{brass}=1.2\times50=60\text{ cm}[/latex] Question 10 of 11 If [latex]\gamma_a[/latex] and [latex]\gamma_r[/latex] are the coefficients of apparent and real expansion of liquid, then A [latex]\dfrac{\gamma_r}{\gamma_a}=1[/latex] B [latex]\dfrac{\gamma_r}{\gamma_a}<1[/latex] C [latex]\dfrac{\gamma_r}{\gamma_a}>1[/latex] D none of above 💡 Explanation Real expansivity of a liquid always exceeds its apparent expansivity. Question 11 of 11 Density of a liquid at 0°C is 13.6 g/c.c. Its density at 100°C is ([latex]\gamma = 5.3\times10^{-4}/^\circ C[/latex]) A 13.6 g/c.c. B 12.91 g/c.c. C 14.32 g/c.c. D 13.02 g/c.c. 💡 Explanation [latex]\rho(100) = \dfrac{13.6}{1+5.3\times10^{-4}\times100} = \dfrac{13.6}{1.053} \approx 12.91\text{ g/c.c.}[/latex] ← Prev Check Next → Finish & See Score 0 of 11 checked ← Back 1 / 11 1 2 3 4 5 6 7 8 9 10 11
