Thermometry and Thermal Expansion — NEET Physics PYQ MCQs with Solutions
Free NEET Physics PYQ Thermometry and Thermal Expansion MCQs with step-by-step solutions (7 questions). Part of Thermal Properties of Matter. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Thermometry and Thermal Expansion · medium · numerical
A copper rod of 88 cm and an aluminium rod of unknown length have their increase in length independent of increase in temperature. The length of aluminium rod is
A. 113.9 cm
B. 88 cm
C. 68 cm ✓ Correct
D. 6.8 cm
Solution: For the increase in length to be independent of temperature, $L\alpha\Delta T$ must be the same for both rods, i.e. $L_{Cu}\alpha_{Cu}=L_{Al}\alpha_{Al}$. With $\alpha_{Cu}=1.7\times10^{-5}$ K$^{-1}$, $\alpha_{Al}=2.2\times10^{-5}$ K$^{-1}$ and $L_{Cu}=88$ cm, $L_{Al}=\dfrac{L_{Cu}\alpha_{Cu}}{\alpha_{Al}}=\dfrac{1.7\times10^{-5}\times88}{2.2\times10^{-5}}\approx68$ cm.
Q2 — Thermometry and Thermal Expansion · medium · numerical
Coefficient of linear expansion of brass and steel rods are $\alpha_1$ and $\alpha_2$. Lengths of brass and steel rods are $l_1$ and $l_2$ respectively. If $(l_2-l_1)$ is maintained same at all temperatures, which one of the following relations holds good?
A. $\alpha_1 l_2^2=\alpha_2 l_1^2$
B. $\alpha_1^2 l_2=\alpha_2^2 l_1$
C. $\alpha_1 l_1=\alpha_2 l_2$ ✓ Correct
D. $\alpha_1 l_2=\alpha_2 l_1$
Solution: For the difference $(l_2-l_1)$ to stay the same, the increases must be equal: $l_2'-l_1'=l_2-l_1$, so $l_2(1+\alpha_2\Delta t)-l_1(1+\alpha_1\Delta t)=l_2-l_1$, giving $l_2\alpha_2=l_1\alpha_1$, i.e. $\alpha_1 l_1=\alpha_2 l_2$.
Q3 — Thermometry and Thermal Expansion · medium · numerical
The value of coefficient of volume expansion of glycerin is $5\times10^{-4}$ K$^{-1}$. The fractional change in the density of glycerin for a rise of 40°C in its temperature is
A. 0.015
B. 0.020 ✓ Correct
C. 0.025
D. 0.010
Solution: Since $\rho=\rho_0(1+\gamma\Delta T)$, the fractional change in density is $\dfrac{\rho-\rho_0}{\rho_0}=\gamma\Delta T=5\times10^{-4}\times40=200\times10^{-4}=0.020$.
Q4 — Thermometry and Thermal Expansion · medium · numerical
On a new scale of temperature (which is linear) and called the W scale, the freezing and boiling points of water are 39°W and 239°W respectively. What will be the temperature on the new scale, corresponding to a temperature of 39°C on the celsius scale?
A. 78°W
B. 117°W ✓ Correct
C. 200°W
D. 139°W
Solution: Equating the fractional rise on both scales, $\dfrac{39°C-0°C}{100°C-0°C}=\dfrac{t-39°W}{239°W-39°W}$, so $\dfrac{39}{100}=\dfrac{t-39}{200}\Rightarrow t=117°$W.
Q5 — Thermometry and Thermal Expansion · medium · numerical
The coefficients of linear expansions of brass and steel are $\alpha_1$ and $\alpha_2$ respectively. When we take a brass rod of length $l_1$ and a steel rod of length $l_2$ at 0°C, then the difference in their lengths $(l_2-l_1)$ will remain the same at all temperatures, if
A. $\alpha_1 l_1=\alpha_2 l_2$ ✓ Correct
B. $\alpha_1 l_2=\alpha_2 l_1$
C. $\alpha_1^2 l_2=\alpha_2^2 l_1$
D. $\alpha_1 l_2^2=\alpha_2 l_1^2$
Solution: For brass, $\Delta l_1=l_1\alpha_1 t$; for steel, $\Delta l_2=l_2\alpha_2 t$. Since $l_2-l_1$ is constant, $\Delta l_2=\Delta l_1$, so $l_2\alpha_2 t=l_1\alpha_1 t$. As $t\neq0$, $l_2\alpha_2=l_1\alpha_1$, i.e. $\alpha_1 l_1=\alpha_2 l_2$.
Q6 — Thermometry and Thermal Expansion · medium · theory
Mercury thermometer can be used to measure temperature upto
A. 260°C
B. 100°C
C. 360°C ✓ Correct
D. 500°C
Solution: A mercury thermometer is a liquid thermometer based on the uniform expansion of mercury. It can be used up to about 300°C or so; mercury begins to vaporise near 367°C. Hence the highest usable value among the options is 360°C.
Q7 — Thermometry and Thermal Expansion · medium · numerical
A Centigrade and a Fahrenheit thermometer are dipped in boiling water. The water temperature is lowered until the Fahrenheit thermometer registers 140°. What is the fall in temperature as registered by the Centigrade thermometer?
A. 80°
B. 60°
C. 40° ✓ Correct
D. 30°
Solution: Using $\dfrac{C}{100}=\dfrac{F-32}{180}$ with $F=140°$, $\dfrac{C}{100}=\dfrac{140-32}{180}=0.6\Rightarrow C=60°$. Fall in temperature $=100°C-60°C=40°$C.