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Physics MCQs for NEET — Practice Questions with Answers

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Mud houses are cooler in summer and warmer in winter because

Mud is superconductor of heat

Mud is good conductor of heat

Mud is bad conductor of heat

None of these

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Explanation

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The temperature of hot and cold end of a 20 cm long rod in thermal steady state are at 100 $° C$ and 20 $° C$ respectively. Temperature at the centre of the rod is

50 $° C$

60 $° C$

40 $° C$

30 $° C$

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Explanation

(b) Temperature gradient = $\frac{100 - 20}{20} = 4 ° C / cm$
temperature at centre = 100-4 $\times$ 10 = 60°C

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A heat flux of 4000 J/s is to be passed through a copper rod of length 10 cm and area of cross-section 100 ${cm}^{2}$ . The thermal conductivity of copper is 400 W/m $° C$ . The two ends of this rod must be kept at a temperature difference of

1 $° C$

10 $° C$

100 $° C$

1000 $° C$

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Explanation

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At a common temperature, a block of wood and a block of metal feel equally cold or hot. The temperatures of block of wood and block of metal are

Equal to temperature of the body

Less than the temperature of the body

Greater than temperature of the body

Either (b) or (c)

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Explanation

(a) When the temperature of an object is equal to that of human body, no heat is transferred from the object to body and vice versa, Therefore block of wood and block of metal feel equally cold and hot if they have same temperature as human body.

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A slab consists of two parallel layers of two different materials of same thickness having thermal conductivities $K_{1}$ and $K_{2}$ . The equivalent conductivity of the combination is

$K_{1} + K_{2}$

$\frac{K_{1} + K_{2}}{2}$

$\frac{2 K_{1} K_{2}}{K_{1} + K_{2}}$

$\frac{K_{1} + K_{2}}{2 K_{1} K_{2}}$

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Explanation

(b) In parallel combination equivalent conductivity

$K = \frac{K_{1} A_{1} + K_{2} A_{2}}{A_{1} + A_{2}} = \frac{K_{1} + K_{2}}{2} (As A_{1} = A_{2})$

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There are two identical vessels filled with equal amounts of ice. The vessels are of different metals., If the ice melts in the two vessels in 20 and 35 minutes respectively, the ratio of the coefficients of thermal conductivity of the two metals is

4 : 7

7 : 4

16 :49

49 : 16

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Explanation

(b) $Q = \frac{KA (θ_{1} - θ_{2})}{l} t \Rightarrow K_{1} t_{1} = K_{2} t_{2} \Rightarrow \frac{K_{1}}{K_{2}} = \frac{t_{2}}{t_{1}} = \frac{35}{20} = \frac{7}{4} (As Q, l, A and (θ_{1} - θ_{2}) are same)$

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Consider a compound slab consisting of two different materials having an equal thickness and thermal conductivities K and 2K respectively. The equivalent thermal conductivity of the slab if the materials are joined by ends is:

$\sqrt{2 K}$

$\frac{4}{3} K$

$\frac{2}{3} K$

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Explanation

$K = \frac{2 K_{1} K_{2}}{K_{1} + K_{2}} = \frac{2 . K . 2 K}{K + 2 K} = \frac{4}{3} K$

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Two rods having thermal conductivity in the ratio of 5 : 3 having equal lengths and equal cross-sectional area are joined by face to face. If the temperature of the free end of the first rod is 100 $° C$ and free end of the second rod is 20 $° C$ . Then temperature of the junction is

70 $° C$

50 $° C$

90 $° C$

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Explanation

(a) Temperature of interface $θ = \frac{K_{1} θ_{1} + K_{2} θ_{2}}{K_{1} + K_{2}}$

It is given that $\frac{K_{1}}{K_{2}} = \frac{5}{3} \Rightarrow K_{1} = 5 K$ and $K_{2} = 3 K$

$θ = \frac{5 K \times 100 + 3 K \times 20}{5 K + 3 K} = \frac{560 K}{8 K} = 70 ° C$

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Woollen clothes are used in winter season because woollen clothes

Are good sources for producing heat

Absorb heat from surroundings

Are bad conductors of heat

Provide heat to body continuously

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Explanation

(c) In winter, the temperature of surrounding is low compared to the body temperature $(37.4 ° C)$ . Since woollen clothes are bad conductors of heat, so they keep the body warm.

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A cylindrical rod having temperature $T_{1}$ and $T_{2}$ at its ends. The rate of flow of heat is $Q_{1}$ cal/sec. If all the linear dimensions are doubled keeping temperature constant then rate of flow of heat $Q_{2}$ will be

$4 Q_{1}$

$2 Q_{1}$

$\frac{Q_{1}}{4}$

$\frac{Q_{1}}{2}$

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Explanation

(b) Rate of heat flow

$(\frac{Q}{t}) = \frac{{kπr}^{2} (θ_{1} - θ_{2})}{L} \propto \frac{r^{2}}{L} ∴ \frac{Q_{1}}{Q_{2}} = {(\frac{r_{1}}{r_{2}})}^{2} (\frac{l_{2}}{l_{1}}) = {(\frac{1}{2})}^{2} \times (\frac{2}{1}) = \frac{1}{2} \Rightarrow Q_{2} = 2 Q_{1}$

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