Two rods, one made of copper and other made of steel, of the same length and same cross sectional area are joined together. The thermal conductivity of copper and steel are $$385~\text{Js}^{-1}\text{K}^{-1}\text{m}^{-1}$$ and $$50~\text{Js}^{-1}\text{K}^{-1}\text{m}^{-1}$$ respectively. The free ends of copper and steel are held at $$100^\circ \text{C}$$ and $$0^\circ \text{C}$$ respectively. The temperature at the junction is, nearly:
1. $$12^\circ \text{C}$$
2. $$50^\circ \text{C}$$
3. $$73^\circ \text{C}$$
4. $$88.5^\circ \text{C}$$
Subtopic: Â Conduction |
Â 70%
From NCERT
NEET - 2022
To view explanation, please take trial in the course.
NEET 2023 - Target Batch - Aryan Raj Singh
Hints
To view explanation, please take trial in the course.
NEET 2023 - Target Batch - Aryan Raj Singh

A cup of coffee cools from $$90^{\circ}\mathrm{C}$$ $\mathrm{to}$ $$80~^{\circ}\mathrm{C}$$ in t minutes, when the room temperature is $$20^{\circ}\mathrm{C}$$$\mathrm{}$. The time taken by a similar cup of coffee to cool from $$80^{\circ}\mathrm{C}$$ $\mathrm{to}$ $$60^{\circ}\mathrm{C}$$ at room temperature same at $$20^{\circ}\mathrm{C}$$ is:
1. $$\frac{10}{13}t$$
2. $$\frac{5}{13}t$$
3. $$\frac{13}{10}t$$
4. $$\frac{13}{5}t$$

Subtopic: Â Newton's Law of Cooling |
Â 62%
From NCERT
NEET - 2021
To view explanation, please take trial in the course.
NEET 2023 - Target Batch - Aryan Raj Singh
Hints
To view explanation, please take trial in the course.
NEET 2023 - Target Batch - Aryan Raj Singh

The quantities of heat required to raise the temperature of two solid copper spheres of radii $$r_1$$ and $$r_2$$ $$(r_1=1.5~r_2)$$ through $$1~\text{K}$$ are in the ratio:
1. $$\frac{9}{4}$$
2. $$\frac{3}{2}$$
3. $$\frac{5}{3}$$
4. $$\frac{27}{8}$$

Subtopic: Â Conduction |
Â 51%
From NCERT
NEET - 2020
To view explanation, please take trial in the course.
NEET 2023 - Target Batch - Aryan Raj Singh
Hints
To view explanation, please take trial in the course.
NEET 2023 - Target Batch - Aryan Raj Singh

Three stars $$A,$$ $$B,$$ and $$C$$ have surface temperatures $$T_A,~T_B$$and $$T_C$$ respectively. Star $$A$$ appears bluish, star $$B$$ appears reddish and star $$C$$ yellowish. Hence,

1. $$T_A>T_B>T_C$$
2. $$T_B>T_C>T_A$$
3. $$T_C>T_B>T_A$$
4. $$T_A>T_C>T_B$$
Subtopic: Â Wien's Displacement Law |
Â 68%
From NCERT
NEET - 2020
To view explanation, please take trial in the course.
NEET 2023 - Target Batch - Aryan Raj Singh
Hints
To view explanation, please take trial in the course.
NEET 2023 - Target Batch - Aryan Raj Singh

A copper rod of $$88$$ cm and an aluminium rod of an unknown length have an equal increase in their lengths independent of an increase in temperature. The length of the aluminium rod is:
$$\left(\alpha_{Cu}= 1.7\times10^{-5}~\text{K}^{-1}~\text{and}~\alpha_{Al}= 2.2\times10^{-5}~\text{K}^{-1}\right)$$
1. $$68~\text{cm}$$
2. $$6.8~\text{cm}$$
3. $$113.9~\text{cm}$$
4. $$88~\text{cm}$$

Subtopic: Â Thermal Expansion |
Â 76%
From NCERT
NEET - 2019
To view explanation, please take trial in the course.
NEET 2023 - Target Batch - Aryan Raj Singh
Hints
To view explanation, please take trial in the course.
NEET 2023 - Target Batch - Aryan Raj Singh

An object kept in a large room having an air temperature of $$25^\circ \text{C}$$ takes $$12$$ minutes to cool from $$80^\circ \text{C}$$ to $$70^\circ \text{C}.$$ The time taken to cool for the same object from $$70^\circ \text{C}$$ to $$60^\circ \text{C}$$ would be nearly:
1. $$10$$ min
2. $$12$$ min
3. $$20$$ min
4. $$15$$ min

Subtopic: Â Newton's Law of Cooling |
Â 77%
From NCERT
NEET - 2019
To view explanation, please take trial in the course.
NEET 2023 - Target Batch - Aryan Raj Singh
Hints
To view explanation, please take trial in the course.
NEET 2023 - Target Batch - Aryan Raj Singh

A deep rectangular pond of surface area A, containing water (density = $$\rho,$$ specific heat capacity = $$s$$), is located in a region where the outside air temperature is at a steady value of $$-26^{\circ}\mathrm{C}$$. The thickness of the ice layer in this pond at a certain instant is $$x$$. Taking the thermal conductivity of ice as $$k$$, and its specific latent heat of fusion as $$L$$, the rate of increase of the thickness of the ice layer, at this instant, would be given by:
1. $$\frac{26k}{x\rho L-4s}$$
2. $$\frac{26k}{x^2\rho L}$$
3. $$\frac{26k}{x\rho L}$$
4. $$\frac{26k}{x\rho L+4s}$$

Subtopic: Â Conduction |
Â 52%
From NCERT
NEET - 2019
To view explanation, please take trial in the course.
NEET 2023 - Target Batch - Aryan Raj Singh
Hints