Thermal Properties of Matter Test

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Thermal Properties of Matter Test - 34

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Weekly Quiz Competition

Question 1
1 / -0

If rc and rs respectively represents cuticular and stomatal resistance, the total resistance (R) could be expressed as

A
$$R=rc+rs$$

B
$$R=r-rs$$

C
$$\frac{1}{R}=\frac{1}{rc}+\frac{1}{rs}$$

D
$$\frac{r}{R}=\frac{1}{rc}-\frac{1}{rc}$$

Solution

Resistance to the transport of water vapor in the leaves is referred to as leaf resistance This has both cuticular and stomatal components. Cuticular resistance is larger than stomatal resistance. Further, both the upper side of the lower side of the leaves have resistance associated with them. If Rc and Rs represents the cuticular and stomatal resistance, the total Resistance (R) can be expressed as 1/R= 1/Rc + 1/Rs
Therefore, the correct answer is option C.
Question 2
1 / -0

There are two lead spheres, the ratio c being $$1 : 2$$. If both are at the same tempe then ratio of heat contents is

A
$$1 : 1$$

B
$$1 : 2$$

C
$$1 : 4$$

D
$$1 : 8$$

Solution

We know that the heat content is proportional to the mass and specific heat capacity of the substance as well as temperature. As both the spheres are of the same material and are at the same temperature, heat content will be dependent on mass only.

$$ m = \rho \times v$$

where $$ \rho $$ is the density of the sphere and $$v$$ the volume.

And $$ v $$ = $$ \dfrac {4 \pi r^3}{3} $$

The given radius ratio is $$ 1:2 $$.

Hence heat contents ratio is in the ratio $$1:8$$
Question 3
1 / -0

Producers gas is a mixture of

A
Carbon monoxide and nitrogen gas

B
Carbon monoxide and hydrogen gas

C
Carbon monoxide and water vapour

D
Carbon monoxide and nitrous oxide

Solution
Question 4
1 / -0

In a room there are four objects a wooden dish, steel dish, glass dish and a copper dish. If a fire is lit in the room so that it burns at 300 degrees centigrade, and is equidistant from all the four dishes, then after a long time the dishes can be listed in the increasing order of temperature. Which is the correct order of temperature of dishes?

A
wooden dish > steel dish > glass dish > copper dish

B
steel dish > copper dish > glass dish > wooden dish

C
copper dish > steel dish > glass dish > wooden dish

D
none of the above
Question 5
1 / -0

At pressure P and absolute temperature T a mass M of an ideal gas fills a closed container of volume V. An additional mass 2M of the same gas is added into the container and the volume is then reduced to $$\dfrac{v}{3}$$ and the temperature to $$\dfrac{T}{3}.$$ The pressure of the gas will now be :

A
$$\dfrac{P}{3}$$

B
$$P$$

C
$$3 P$$

D
$$9 P$$

Solution

If $$M_{0}$$ is molecular mass of the gas then for initial condition PV
$$=\dfrac{M}{M_{0}}.RT ........(1)$$
After 2M mass has been added
$${P}'.\dfrac{V}{3}=\dfrac{3M}{M_{0}}.R.\frac{T}{3} .........(2)$$
By dividing (2) and (1)
$${P}'=3P$$
Question 6
1 / -0

The rate of heat conduction is proportional to the cross-sectional area and temperature
gradient (temperature difference per unit length). On a typical day during the World Cup
tournament in South Africa, the air in a room is heated to $$25^{o}C$$ while the air outside is
$$-2^{o}C$$. The area of the window of the room is
$$2 m^{2}$$ and it is made of crown glass with
thickness 2 mm and thermal conductivity
$$1.0 WK^{-1}m^{-1}$$. What is the heat power loss
through the window?

A
1.2 kW

B
2.7 kW

C
27 kW

D
50 kW
Question 7
1 / -0

A rigid container has a hole in its wall. When the container is evacuated, its weight is 100 gm. When someair is filled in it at 27C, its weight becomes 200 gm. Now the temperature of air inside is increased by $$\Delta$$ T, the weight becomes 150 gm. $$\Delta$$ T should be :

A
$$27^{\circ}$$

B
$$\dfrac{27^{\circ}} {4}$$ C

C
$$300^{\circ}$$

D
$$327^{\circ}$$

Solution

Initially mass of air is 200 - 100 = 100gm. finally mass of air is 150 - 100 = 50 gm. As there is a hole in the wall, pressure inside the container will remain constant = $$P_{0}$$
$$ PV = nRT \Rightarrow T \propto \dfrac{1} {n}$$
as number of moles of gas is halved, the temperature should be doubled (in K)
$$T_{i} = 300K$$ So, $$T_{f} = 600 K \Rightarrow \Delta T = 300 K = 300^{\circ}C$$
Question 8
1 / -0

A cuboid $$ABCDEFGH$$ is anisotropic with $$\alpha _{x}=1 \times 10^{-5} /^\circ C,\:\alpha _{y}=2 \times 10^{-5} /^\circ C,\:\alpha _{z}=3 \times 10^{-5} /^\circ C$$. Coefficient of superficial expansion of faces can be

A
$$\beta _{ABCD}=5 \times 10^{-5}/ ^\circ C$$

B
$$\beta _{BCGH}=4 \times 10^{-5}/ ^\circ C$$

C
$$\beta _{CDEH}=3 \times 10^{-5}/ ^\circ C$$

D
$$\beta _{EFGH}=2 \times 10^{-5}/ ^\circ C$$

Solution

Coefficient of superficial expansion of face ABCD=$$\alpha_x+\alpha_z=4\times 10^{-5}/^{\circ}C$$
Coefficient of superficial expansion of face BCGH=$$\alpha_y+\alpha_z=5\times 10^{-5}/^{\circ}C$$
Coefficient of superficial expansion of face CDEH=$$\alpha_x+\alpha_y=3\times 10^{-5}/^{\circ}C$$
Coefficient of superficial expansion of face EFGH=$$\alpha_x+\alpha_z=4\times 10^{-5}/^{\circ}C$$
Question 9
1 / -0

A vessel of volume 4 litres contains a mixture of 8 g of O$$_2$$, 14 g of N$$_2$$ and 22g of CO$$_2$$ at 27$$^o$$C. The pressure exerted by the mixture is

A
$$5 \times 10^6 N/m^2$$

B
$$6 \times 10^3 N/m^2$$

C
10 atmosphere

D
$$7.79 \times 10^5 N/m^2$$

Solution

Moles of $$O_2=\dfrac{8}{32}=\dfrac{1}{4}$$
Moles of $$N_2=\dfrac{14}{28}=\dfrac{1}{2}$$
Moles of $$CO_2=\dfrac{22}{44}=\dfrac{1}{2}$$
Thus total moles in the mixture=$$\dfrac{1}{4}+\dfrac{1}{2}+\dfrac{1}{2}=\dfrac{5}{4}$$
Using Ideal Gas Equation,
$$PV=nRT$$
$$\implies P=\dfrac{nRT}{V}=\dfrac{\dfrac{5}{4}\times 8.3\times 300}{4\times 10^{-3}}N/m^2\approx 7.79\times 10^{5}N/m^2$$
Question 10
1 / -0

A vessel contains $$1$$ mole of O$$_2$$ (molar mass $$ 32 gm$$) at a temperature $$T$$. The pressure is $$P$$. An identical vessel containing $$1$$ mole of He (molar mass $$4 gm$$) at a temperature $$2T$$ has a pressure :

A
P

B
$$\displaystyle \frac{P}{8}$$

C
2P

D
8P

Solution

Using $$PV=nRT$$
$$\therefore P_{He} = 2 P$$ as temperature of He is doubled.

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Thermal Properties of Matter Test - 34

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Thermal Properties of Matter Test - 34

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Question 1

$$R=rc+rs$$

$$R=r-rs$$

$$\frac{1}{R}=\frac{1}{rc}+\frac{1}{rs}$$

$$\frac{r}{R}=\frac{1}{rc}-\frac{1}{rc}$$

Solution

Question 2

$$1 : 1$$

$$1 : 2$$

$$1 : 4$$

$$1 : 8$$

Solution

Question 3

Carbon monoxide and nitrogen gas

Carbon monoxide and hydrogen gas

Carbon monoxide and water vapour

Carbon monoxide and nitrous oxide

Solution

Question 4

wooden dish > steel dish > glass dish > copper dish

steel dish > copper dish > glass dish > wooden dish

copper dish > steel dish > glass dish > wooden dish

none of the above

Question 5

$$\dfrac{P}{3}$$

$$P$$

$$3 P$$

$$9 P$$

Solution

Question 6

1.2 kW

2.7 kW

27 kW

50 kW

Question 7

$$27^{\circ}$$

$$\dfrac{27^{\circ}} {4}$$ C

$$300^{\circ}$$

$$327^{\circ}$$

Solution

Question 8

$$\beta _{ABCD}=5 \times 10^{-5}/ ^\circ C$$

$$\beta _{BCGH}=4 \times 10^{-5}/ ^\circ C$$

$$\beta _{CDEH}=3 \times 10^{-5}/ ^\circ C$$

$$\beta _{EFGH}=2 \times 10^{-5}/ ^\circ C$$

Solution

Question 9

$$5 \times 10^6 N/m^2$$

$$6 \times 10^3 N/m^2$$

10 atmosphere

$$7.79 \times 10^5 N/m^2$$

Solution

Question 10

P

$$\displaystyle \frac{P}{8}$$

2P

8P

Solution

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