1. ⇒ (MHT CET 2023 13th May Morning Shift )

The temperature gradient in a rod of length $75 cm$ is $40^{\circ} C / m$ . If the temperature of cooler end of the rod is $10^{\circ} C$ , then the temperature of hotter end is

A. $50^{\circ} C$

B. $40^{\circ} C$

C. $35^{\circ} C$

D. $25^{\circ} C$

Correct Option is (B)

We know

$\begin{aligned} T_{g} = \frac{T_{1} - T_{2}}{x} \\ \Rightarrow \frac{T_{1} - 10}{0.75} = 40 \end{aligned}$

$∴$ Temperature of the hotter end is:

$\begin{aligned} T_{1} = (40 \times 0.75) + 10 \\ T_{1} = 40^{\circ} C \end{aligned}$

2. ⇒ (MHT CET 2023 12th May Morning Shift )

A composite slab consists of two materials having coefficient of thermal conductivity $K$ and $2 K$ , thickness $x$ and $4 x$ respectively. The temperature of the two outer surfaces of a composite slab are $T_{2}$ and $T_{1} (T_{2} > T_{1})$ . The rate of heat transfer through the slab in a steady state is $[\frac{A (T_{2} - T_{1}) K}{x}] \cdot f$ where ' $f$ ' is equal to

A. 1

B. $\frac{2}{3}$

C. $\frac{1}{2}$

D. $\frac{1}{3}$

Correct Option is (D)

$\begin{array}{ll} R_{eq} = R_{1} + R_{2} \\ ∵ & R_{1} = \frac{x}{KA}, R_{2} = \frac{4 x}{2 KA} \\ ∴ & R_{eq} = \frac{x}{KA} + \frac{2 x}{KA} = \frac{3 x}{KA} \end{array}$

Rate of heat transfer of composite slab is given by,

$\frac{dQ}{dt} = \frac{T_{2} - T_{1}}{R_{eq}} = \frac{KA (T_{2} - T_{1})}{3 x}$

$∴ f = \frac{1}{3}$

3. ⇒ (MHT CET 2023 11th May Morning Shift )

The rate of flow of heat through a metal rod with temperature difference $40^{\circ} C$ is $1600 cal / s$ . The thermal resistance of metal rod in $^{\circ} C s / cal$ is

A. 0.025

B. 0.25

C. 2.5

D. 40

Correct Option is (A)

Given: rate of flow of heat (conduction rate) $P_{cond} = 1600 cal / s$

Thermal resistance,

$\begin{aligned} R_{T} = \frac{Δ T}{P_{cond}} \\ R_{T} = \frac{40}{1600} \\ R_{T} = {0.025}^{\circ} Cs / cal \end{aligned}$

4. ⇒ (MHT CET 2021 21th September Evening Shift )

Two rods of same length and material are joined end to end. They transfer heat in 8 second. When they are joined in parallel they transfer same amount of heat in same conditions in time

A. 3 s

B. 2 s

C. 1 s

D. 4 s

Correct Option is (B)

Let $R$ be the thermal resistance of each rod when connected in series, the total resistance will be

$R_{s} = 2 R$

Whey they are connected in parallel, the effective resistance will be

$R_{p} = \frac{R}{2}$

$∴ \frac{R_{p}}{R_{s}} = \frac{1}{4}$

In parallel combination, the resistance becomes one-fourth and hence time will also become one-fourth. $∴$ Time taken $\frac{8}{4} = 2 s$

5. ⇒ (MHT CET 2021 21th September Morning Shift )

The temperature difference bewtween two sides of metal plate, $3 cm$ thick is $15^{\circ} C$ . Heat is transmitted through plate at the rate of $900 kcal$ per minute per $m^{2}$ at steady state. The thermal conductivity of metal is

A. $1.8 \times 10^{- 2} \frac{kcal}{{ms}^{\circ} C}$

B. $4.5 \times 10^{- 2} \frac{kcal}{{ms}^{\circ} C}$

C. $3 \times 10^{- 2} \frac{kcal}{{ms}^{\circ} C}$

D. $6 \times 10^{- 2} \frac{kcal}{{ms}^{\circ} C}$

Correct Option is (C)

$\begin{aligned} \frac{Q}{tA} = \frac{k Δ θ}{d} \\ ∴ k = \frac{Q}{tA} \cdot \frac{d}{Δ θ} \\ \frac{Q}{tA} = 900 kcal per minute per m^{2} = \frac{900}{60} = 15 kcal / {sm}^{2} \\ d = 3 cm = 3 \times 10^{- 2} m, Δ θ = 15^{\circ} C \\ ∴ k = \frac{15 \times 3 \times 10^{- 2}}{15} = 3 \times 10^{- 2} kcal / {ms}^{\circ} C \end{aligned}$

⇒Thermal Property of Matter