MEDIUMPearson | PTE
Trusted by 3,500+ universities and colleges globally | Accepted for migration visa applications to AUS, CAN, New Zealand , and the UK
Law Of Thermal Conductivity MCQ - Practice Questions with Answers
Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET
Quick Facts
-
Law of Thermal Conductivity is considered one of the most asked concept.
-
14 Questions around this concept.
Solve by difficulty
The two ends of a metal rod are maintained at temperatures 100°C and 110°C. The rate of heat flow in the rod is found to be 4.0 J/s. If the ends are maintained at temperatures 200°C and 210°C, the rate of heat flow will be:
The unit of thermal conductivity is :
A slab of stone of area 0.36 m2 and thickness 0.1 m is exposed on the lower surface to steam at 100 °C. A block of ice 0 °C rests on the upper surface of the slab. In one hour 4.8 kg of ice is melted. The thermal conductivity of slab is:
(Given latent heat of fusion of ice = 3.36 x 105 J kg-1)
Three rods of Copper, Brass and Steel are welded together to form a Y - shaped structure. Area of cross - section of each rod = 4 cm2. End of copper rod is maintained at 1000C where as ends of brass and steel are kept at 00C. Lengths of the copper, brass and steel rods are 46, 13 and 12 cms respectively. The rods are thermally insulated from surroundings except at ends. Thermal conductivities of copper, brass and steel are 0.92, 0.26 and 0.12 CGS units respectively. Rate of heat flow through copper rod is :
Two rods, one semi-circular of thermal conductivity $\mathrm{K}_1$ and the other straight of thermal conductivity $\mathrm{K}_2$ and of the same cross-sectional area, are joined as shown in the figure. The points $A$ and $B$ are maintained at the same temperature difference. If the rate of flow of heat is the same in two rods, then $\frac{k_1}{k_2}$ is.
Two rods A and B of different materials are welded together. Their thermal conductivities are $K_1$ and $K_1$. The thermal conductivity of the composite rod will be
Temperature difference of $120^{\circ} \mathrm{C}$ is maintained between two ends of a uniform rod AB of length 2 L. The $ \frac{3 L}{2}$ is connected across AB (See figure). In another bent rod $P Q$, of the same cross-section as $A B$ and length $\overline{2}$, is connected across $A B$ (See figure). In steady state, temperature(in ${ }^0 \mathrm{C}$ ) difference between P and Q will be close to :
A cylindrical metallic rod in thermal contact with two reservoirs of heat at its two ends conducts an amount of heat Q in time t. The metallic rod is melted and the material is formed into a rod of half the radius of the original rod. What is the amount of heat conducted by the new rod, when placed in thermal contact with the two reservoirs in time t?
Enlargement of cotyledons is an indication of
Most Scoring concepts for NEET
This ebook serves as a valuable study guide for NEET exams, specifically designed to assist students in light of recent changes and the removal of certain topics from the NEET exam.
Download EBookConcepts Covered - 1
Law of Thermal Conductivity
Law of Thermal Conductivity -
$$
Q=\frac{K A\left(\theta_1-\theta_2\right) t}{l}
$$
$Q=$ Amount of heat transfer
$t=$ Time of heat flow
$K=$ Thermal conductivity of the material
So, from the above equation we can calculate the- Rate of flow of heat i.e. heat current which can be written as -
$$
\frac{Q}{t}=H=\frac{K A\left(\theta_1-\theta_2\right)}{l}
$$
In the differential form, this heat current can also be written as -
$$
\frac{d Q}{d t}=-K A \frac{d \theta}{d x}
$$
In the case of a non-steady state or variable cross-section, this is the more general equation that can be used to solve problems.
Relation of thermal conductivity of some material -
$$
\begin{aligned}
& K_{A g}>K_{C u}>K_{A l} \\
& K_{\text {Solid }}>K_{\text {Liquid }}>K_{\text {Gas }} \\
& K_{\text {Metals }}>K_{\text {Non-metals }}
\end{aligned}
$$
Thermal resistance $\left(\mathbf{R}_{\mathbf{t h}}\right)$: The thermal resistance of a body is defined as the measure of its opposition to the flow of heat through it. It is defined as the ratio of temperature difference to the heat current $\left(\frac{Q}{t}\right)$.
$$
R_{t h}=\frac{\theta_1-\theta_2}{H}=\frac{\theta_1-\theta_2}{K A\left(\theta_1-\theta_2\right) / I}=\frac{l}{K A}
$$
Study it with Videos
Law of Thermal Conductivity"Stay in the loop. Receive exam news, study resources, and expert advice!"
Get Answer to all your questions
NEET Articles
Mar 13, 2025
Back to top