Uncategorized - NEET Physics Chapterwise MCQs & PYQs

NEET Uncategorized MCQs & PYQs

Practice NEET Uncategorized Questions

Question 111:

moderate

A Centigrade and a Fahrenheit thermometer are dipped in boiling water. The water temperature is lowered until the Fahrenheit thermometer registers $140^\circ\text{F}$. What is the fall in temperature as registered by the centigrade thermometer? (1990)

Initial temperature of boiling water is $212^\circ\text{F}$. Fall in Fahrenheit $= 212^\circ\text{F} - 140^\circ\text{F} = 72^\circ\text{F}$. Using $\Delta C = \frac{5}{9} \Delta F$, fall in Celsius $= \frac{5}{9} \times 72 = 40^\circ\text{C}$.

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Question 112:

A copper rod of $88\text{ cm}$ and an aluminium rod of unknown length have their increase in length independent of increase in temperature. The length of aluminium rod is : (2019) ( $\alpha_{\text{Cu}} = 1.7 \times 10^{-5}\text{ K}^{-1}$ and $\alpha_{\text{Al}} = 2.2 \times 10^{-5}\text{ K}^{-1}$ )

For the difference in length to be independent of temperature, their expansions must be equal: $L_{\text{Cu}} \alpha_{\text{Cu}} \Delta T = L_{\text{Al}} \alpha_{\text{Al}} \Delta T$. Thus, $88 \times 1.7 \times 10^{-5} = L_{\text{Al}} \times 2.2 \times 10^{-5} \Rightarrow L_{\text{Al}} = 68\text{ cm}$.

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Question 113:

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? (2016 – I)

If the difference in lengths remains the same, the change in their lengths for a given temperature change must be equal. Therefore, $\Delta l_1 = \Delta l_2 \Rightarrow l_1 \alpha_1 \Delta T = l_2 \alpha_2 \Delta T \Rightarrow \alpha_1 l_1 = \alpha_2 l_2$.

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Question 114:

moderate

The value of coefficient of volume expansion of glycerin is $5 \times 10^{-4}\text{ /K}$. The fractional change in the density of glycerin for a rise of $40^\circ\text{C}$ in its temperature, is: (2015 Re)

The fractional change in density is approximately given by $\frac{\Delta \rho}{\rho} = \gamma \Delta T$. Substituting the values: $\frac{\Delta \rho}{\rho} = 5 \times 10^{-4} \times 40 = 200 \times 10^{-4} = 0.020$.

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Question 115:

moderate

Two identical bodies are made of a material for which the heat capacity increases with temperature. One of these is at $100^\circ\text{C}$, while the other one is at $0^\circ\text{C}$. If the two bodies are brought into contact, then, assuming no heat loss, the final common temperature is:

(2016 – II)

By conservation of energy, $\int_{T_f}^{100} C(T)dT = \int_{0}^{T_f} C(T)dT$. Since $C(T)$ is greater at higher temperatures, the change in temperature for the hotter body will be less than that for the colder body. Thus, $T_f > 50^\circ\text{C}$.

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Question 116:

moderate

Steam at $100^\circ\text{C}$ is passed into $20\text{ g}$ of water at $10^\circ\text{C}$. When water acquires a temperature of $80^\circ\text{C}$, the mass of water present will be: [Take specific heat of water $= 1\text{ cal /g }^\circ\text{C}$ and latent heat of steam $= 540\text{ cal g}^{-1}$]: (2014)

Heat gained by water $= 20 \times 1 \times (80 - 10) = 1400\text{ cal}$. Heat lost by $m$ grams of steam $= m \times 540 + m \times 1 \times (100 - 80) = 560m$. Equating them: $560m = 1400 \Rightarrow m = 2.5\text{ g}$. Total mass $= 20 + 2.5 = 22.5\text{ g}$.

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Question 117:

moderate

When $1 text{ kg}$ of ice at $0^circ text{C}$ melts to water at $0^circ text{C}$, the resulting change in its entropy, taking latent heat of ice to be $80 text{ Cal}/^circ text{C}$, is:
(2011 Pre)

Entropy change $Delta S = frac{Delta Q}{T} = frac{m cdot L}{T} = frac{1000 cdot 80}{273} = 293 text{ cal/K}$

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Question 118:

moderate

Thermal capacity of $40 text{ g}$ of aluminum ($s = 0.2 text{ cal/g K}$) is:
(1990)

Thermal capacity $= ms = 40 cdot 0.2 = 8 text{ cal/K} = 8 cdot 4.2 text{ J/K} = 33.6 text{ J/K}$

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Question 119:

moderate

$10 text{ gm}$ of ice cubes at $0^circ text{C}$ are released in a tumbler (water equivalent $55 text{ g}$) at $40^circ text{C}$. Assuming that negligible heat is taken from the surroundings the temperature of water in the tumbler becomes nearly ($L = 80 text{ cal/g}$):
(1988)

Heat lost by tumbler = Heat gained by ice
$55 cdot (40 - T) = 10 cdot 80 + 10 cdot T$
$2200 - 55T = 800 + 10T Rightarrow 65T = 1400 Rightarrow T approx 21.5^circ text{C} approx 22^circ text{C}$

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Question 120:

moderate

The unit of thermal conductivity is :
(2019)

Thermal conductivity $K = frac{Delta Q cdot x}{A cdot Delta T cdot t}$
Unit $= frac{text{J} cdot text{m}}{text{m}^2 cdot text{K} cdot text{s}} = frac{text{W}}{text{m K}} = text{W m}^{-1} text{K}^{-1}$

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