Thermal Properties of Matter Test

Result

Thermal Properties of Matter Test - 64

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Question 1
1 / -0

The equation of state for 5 g of oxygen at a pressure P and temperature T, when occupying a volume V, will be

A
$$PV=\Big(\dfrac{5}{32}\Big)RT$$

B
$$PV = 5 RT$$

C
$$PV = \Big(\dfrac{5}{2}\Big)RT$$

D
$$PV = \Big(\dfrac{5}{16}\Big)RT$$

Solution

Given,
$$m=5g$$
$$M=32g$$ (molecular mass of the oxygen molecule)
Number of moles, $$n=\dfrac{m}{M}$$
$$n=\dfrac{5}{32}$$
From the equation of gas,
$$PV=nRT$$
$$PV=(\dfrac{5}{32})RT$$
Question 2
1 / -0

The movement of water being heated in a pot on a stove is an example of

A
conduction

B
convection

C
radiation

D
condensation

Solution
Question 3
1 / -0

A solid shere is rotating with an angular velocity $$ \omega $$. It is of moment is inertia T about its diametrical axis and is of linear expansion coefficient $$\alpha$$. When temperature is increased by $$\Delta T$$, the functional change in its angular velocity is

A
$$ 2\alpha \Delta T$$

B
$$\alpha \Delta T$$

C
$$\dfrac { \alpha \Delta T }{ 2 } $$

D
$$\dfrac { 2\alpha \Delta T }{ 5 } $$

Solution
Question 4
1 / -0

In the given figure conainer $$A$$ holds an ideal gas at a pressure of $$5.0 \times 10^{5}\ Pa$$ and a temperature of $$300\ K.$$ It is connected by the tin tube (and a closed valve) to container $$B$$, with four times the volume of $$A$$. Container $$B$$ hold same ideal gas at a pressure of $$1.0\ \times 10^{5}\ Pa$$ and a temperature of $$400\ K$$. The valve is opened to allow the pressure to equalize, but the temperature of each container kept constant at its initial value. The final pressure in the two containers will be close to :

A
$$2.0 \times 10^{5}\ Pa$$

B
$$2.5 \times 10^{5}\ Pa$$

C
$$3.0 \times 10^{5}\ Pa$$

D
$$3.5 \times 10^{5}\ Pa$$

Solution
Question 5
1 / -0

Temperature of a body increases from $$-73^{\circ}C$$ to $$127^{\circ}C$$. The radiant energy per second will become.

A
$$2\ times$$

B
$$256\ times$$

C
$$16\ times$$

D
$$32\ times$$

Solution
Question 6
1 / -0

A rod of length $$l$$ and cross section area A has a variable thermal conductivity given by $$k=\alpha T$$ where $$\alpha $$ is a positive constant and $$T$$ is temperature in Kelvin. Two ends of the rod are maintained at temperature $$T_1$$ and $$T_2$$ $$(T_1>T_2)$$. Heat current flowing through the rod will be

A
$$\dfrac{A \alpha (T_1^2-T_2^2)}{l}$$

B
$$\dfrac{A \alpha (T_1^2+T_2^2)}{l}$$

C
$$\dfrac{A \alpha (T_1^2+T_2^2)}{3l}$$

D
$$\dfrac{A \alpha (T_1^2-T_2^2)}{2l}$$

Solution
Question 7
1 / -0

A pond has an ice layer of thickness 3 cm. If K of ice is 0.005 CGS units, surface temperature of surroundings is $$-20^\circ C$$, density of ice is $$0.9 gm/cc$$, the time taken for the thickness to increase by 1 cm is

A
30 min

B
35 min

C
42 min

D
60 min

Solution
Question 8
1 / -0

A gas thermometer measure the temperature from, the variation of pressure of a sample of gas. If the pressure measured at the melting point of lead is 2.20 times the pressure measured at the triple point of water find the melting point of lead.

A
601 K

B
420 K

C
790 K

D
510 K

Solution

From charle's law
$$\dfrac{P_2}{P_1} = \dfrac{t_2}{t_1}$$
Hence
$$T_2 = T_1 \times \dfrac{P_2}{P_1}$$

$$ = 273 \times 2.2$$
= 600.6 k
= 601 k
hence (A) is correct answer
Question 9
1 / -0

Three rods of a same material same area of cross-section but different lengths 10 cm, 20 cm, and 30 cm are connected at a point as shown. What is a temperature of junction O?

A
19.2

B
16.4

C
11.5

D
22

Solution

Let the temperature at point 0 be $$T_0$$
so,
$$ \dfrac{T_A -T_O}{l_1} = \dfrac{T_O - T_B}{l_2} + \dfrac{T_O - T_C}{l_3}$$

$$\dfrac{30-T_O}{30} = \dfrac{T_O -10}{10} + \dfrac{T_O - 20}{20}$$

Solving this
$$T_O = 19.2$$
Question 10
1 / -0

Radiation emitted from a radioactive element, of half life period of $$30$$ min, are measured by a Geigar - Mullar counter. The count rate of the sample reduces to $$ 5$$ disintegrations /sec. in $$2hrs$$. then the initial count rate in disintegration/sec is :-

A
$$805$$

B
$$200$$

C
$$250$$

D
$$40$$

Solution

[A]
Half life = 30 mins
N = 5 / sec
Time = 120 min
$$\begin{array}{l} \frac { N }{ { { N_{ 0 } } } } ={ \left( { \frac { 1 }{ 2 } } \right) ^{ \frac { { Time } }{ { half\, \, life } } } }={ \left( { \frac { 1 }{ 2 } } \right) ^{ \frac { { 120 } }{ { 30 } } } }=\frac { 1 }{ { 16 } } \\ { N_{ 0 } }=16\times N={ 805^{ -1 } } \end{array}$$