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Physics NEET Questions — Free MCQ Practice

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The dimensions of thermal resistance are

A.

$M^{- 1} L^{- 2} T^{3} K$

B.

${ML}^{2} T^{- 2} K^{- 1}$

C.

${ML}^{2} T^{- 3} K$

D.

${ML}^{2} T^{- 2} K^{- 2}$

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Explanation

(a) Thermal resistance

$= \frac{l}{KA} = [\frac{L}{{MLT}^{- 3} K^{- 1} \times L^{2}}] = [M^{- 1} L^{- 2} T^{3} K]$

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Two walls of thicknesses $d_{1}$ and $d_{2}$ and thermal conductivities $k_{1}$ and $k_{2}$ are in contact. In the steady state, if the temperatures at the outer surfaces are $T_{1}$ and $T_{2}$ , the temperature at the common wall is -

A.

$\frac{k_{1} T_{1} d_{2} + k_{2} T_{2} d_{1}}{k_{1} d_{2} + k_{2} d_{1}}$

B.

$\frac{k_{1} T_{1} + k_{2} d_{2}}{d_{1} + d_{2}}$

C.

$(\frac{k_{1} d_{1} + k_{2} d_{2}}{T_{1} + T_{2}}) T_{1} T_{2}$

D.

$\frac{k_{1} d_{1} T_{1} + k_{2} d_{2} T_{2}}{k_{1} d_{1} + k_{2} d_{2}}$

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A slab consists of two parallel layers of copper and brass of the same thickness and having thermal conductivities in the ratio 1 : 4. If the free face of brass is at 100°C and that of copper at 0°C, the temperature of interface is

A.

80°C

B.

20°C

C.

60°C

D.

40°C

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Explanation

(a) Temperature of interface

$θ = \frac{k_{1} θ_{1} + k_{2} θ_{2}}{k_{1} + k_{2}} (∵ \frac{k_{1}}{k_{2}} = \frac{1}{4} \Rightarrow If k_{1} = k then k_{2} = 4 k) \Rightarrow θ = \frac{k \times 0 + 4 k \times 100}{5 k} = 80 ° C$

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Two thin blankets keep more hotness than one blanket of thickness equal to these two. The reason is

A.

Their surface area increases

B.

A layer of air is formed between these two blankets, which is bad conductor

C.

These have more wool

D.

They absorb more heat from outside

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Explanation

(b) A layer of air, which is a bad conductor of heat, is formed between these two blankets.

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Ice formed over lakes has

A.

Very high thermal conductivity and helps in further ice formation

B.

Very low conductivity and retards further formation of ice

C.

It permits quick convection and retards further formation of ice

D.

It is very good radiator

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Explanation

(b) Ice has law thermal conductivity, it prevents further loss of heat from water to surroundings.

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Wires A and B have identical lengths and have circular cross-sections. The radius of A is twice the radius of B i.e. $r_{A} = 2 r_{B}$ . For a given temperature difference between the two ends, both wires conduct heat at the same rate. The relation between the thermal conductivities is given by

A.

$K_{A} = 4 K_{B}$

B.

$K_{A} = 2 K_{B}$

C.

$K_{A} = K_{B} / 2$

D.

$K_{A} = K_{B} / 4$

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Explanation

(d) $\frac{Q}{t} = \frac{KA ∆ θ}{l} \Rightarrow \frac{K_{A}}{K_{B}} = \frac{A_{B}}{A_{A}} = {(\frac{r_{B}}{r_{A}})}^{2} = \frac{1}{4} \Rightarrow K_{A} = \frac{K_{B}}{4}$

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Two identical plates of different metals are joined to form a single plate whose thickness is double the thickness of each plate. If the coefficients of conductivity of each plate are 2 and 3 respectively, then the conductivity of composite plate will be

A.

5

B.

2.4

C.

1.5

D.

1.2

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Explanation

(b) Thermal conductivity of composite plate

$K_{eq} = \frac{2 K_{1} K_{2}}{K_{1} + K_{2}} = \frac{2 \times 2 \times 3}{2 + 3} = \frac{12}{5} = 2.4$

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The coefficients of thermal conductivity of copper, mercury and glass are respectively $K_{c}$ , $K_{m}$ and $K_{g}$ such that $K_{c} > K_{m} > K_{g}$ . If the same quantity of heat is to flow per second per unit area of each and corresponding temperature gradients are $X_{c}$ , $X_{m}$ and $X_{g}$ , then

A.

$X_{c} = X_{m} = X_{g}$

B.

$X_{c} > X_{m} > X_{g}$

C.

$X_{c} < X_{m} < X_{g}$

D.

$X_{m} < X_{c} < X_{g}$

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Explanation

(c) $\frac{Q}{At} = K \frac{∆ θ}{l} \Rightarrow K \frac{∆ θ}{l} = constant \Rightarrow \frac{∆ θ}{l} \propto \frac{1}{K}$

Hence If $K_{c} > K_{m} > K_{g}$ , then

${(\frac{∆ θ}{l})}_{c} < {(\frac{∆ θ}{l})}_{m} < {(\frac{∆ θ}{l})}_{g} \Rightarrow X_{c} < X_{m} < X_{g}$

because higher K implies lower value of the temperature gradient.

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The quantity of heat which crosses unit area of a metal plate during conduction depends upon

A.

The density of the metal

B.

The temperature gradient perpendicular to the area

C.

The temperature to which the metal is heated

D.

The area of the metal plate

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Explanation

(b) $\frac{dQ}{dt} = KA \frac{dθ}{dl} \Rightarrow \frac{dQ}{dt} \propto \frac{dθ}{dl}$ (Temperature gradient)

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Two spheres of different materials one with double the radius and one-fourth wall thickness of the other, are filled with ice. If the time taken for complete melting ice in the large radius one is 25 minutes and that for smaller one is 16 minutes, the ratio of thermal conductivities of the materials of larger sphere to the smaller sphere is

A.

4 : 5

B.

5 : 4

C.

25 : 1

D.

8 : 25

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Explanation

(d)

$For small sphere : radius = r and thickness = t Q_{s} = \frac{K_{s} 4 {πr}^{2} (∆ θ) t_{s}}{l} = m_{s} L = \frac{4}{3} {πr}^{3} \times ρ \times L For large sphere : radius = r and thickness = t Q_{L} = \frac{K_{L} 4 π {(2 r)}^{2} (∆ θ) t_{L}}{\frac{l}{4}} = m_{L} L = \frac{4}{3} π {(2 r)}^{3} \times ρ \times L \frac{Q_{S}}{Q_{L}} = \frac{K_{s} t_{s}}{16 K_{L} t_{L}} = \frac{1}{8} \Rightarrow \frac{K_{s}}{K_{L}} = \frac{1}{8} \times 16 \times \frac{25}{16} = \frac{25}{8} \Rightarrow \frac{K_{L}}{K_{S}} = \frac{8}{25}$

Physics MCQs for NEET — Practice Questions with Answers

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