Questions and answers for exam Preparation

MCQs

Heat Radiation, Thermal Properties Of Matter, Dual Nature Of Matter

Total Questions : 247 | Page 2 of 25 pages

Question 11. The graph AB shown in figure is a plot of temperature of a body in degree Celsius and degree Fahrenheit. Then

Slope of line AB is 95
Slope of line AB is 59
Slope of line AB is 19
Slope of line AB is 39

Discuss Question

Answer: Option B. -> Slope of line AB is 59
:
B
Relation between Celsius and Fahrenheit scale of temperature is

\frac{C}{5} = \frac{F - 32}{9}

By rearranging we get, C =

\frac{5}{9} F - \frac{160}{9}

By equating above equation with standard equation of line

y = m x + c

we get

m = \frac{5}{9}

and

c = \frac{- 160}{9}

i.e. Slope of the line AB is

\frac{5}{9}

Question 12. In a constant volume gas thermometer, the pressure of the working gas is measured by the difference in the levels of mercury in the two arms of a U-tube connected to the gas at one end. When the bulb is placed at the room temperature 27.0℃, the mercury column in the arm open to atmosphere stands 5.00 cms above the level of mercury in the other arm. When the bulb is placed in a hot liquid, the difference of mercury levels becomes 45.0 cms. Calculate the temperature of the liquid. (Atmospheric pressure = 75.0 cm of mercury.)

155℃
177℃
187℃
165℃.

Discuss Question

Answer: Option B. -> 177℃
:
B
The pressure of the gas = atmospheric pressure + the pressure due to the difference in mercury levels
At 27℃, the pressure is (75 cm + 5 cm) = 80 cm of mercury. At the liquid's temperature, the pressure is (75 cm + 45 cm) = 120 cm of mercury
Using

T_{2} = (\frac{P_{2}}{P_{1}}) T_{1},

the temperature of the liquid is -

T = [(\frac{120}{80}) \times (27.0 + 273.15)] K = 450.22 K = {177.07}^{\circ} C \approx 177^{\circ} C .

Question 13. The ideal gas law, which relates the P, V and temperature T of an ideal gas in a closed box, given as PV = nRT, can be manipulated to construct an effective thermometer. A "constant volume gas thermometer” uses a sample of gas (commonly, nitrogen or helium) in a closed chamber, thus fixing the volume V. The temperature is then measured by reading off the corresponding pressure P. At the freezing point of water, it is seen that a gas thermometer records a pressure of 0.9 x 10⁵ Pa; it also records a pressure of 1.2 x 10⁵ Pa at the boiling point of water. What pressure will you find in the gas chamber at a room temperature of 23⁰C?

0.81 x 105 Pa
1.34 x 105 Pa
0.64 x 105 Pa
0.97 x 105 Pa.

Discuss Question

Answer: Option D. -> 0.97 x 105 Pa.
:
D
For an ideal gas at constant V,
P∝ T

(for some proportionality constant 'C').
Let (T_F, P_F) and (T_B,P_B) be the (T,P) values for the freezing and the boiling points of water respectively. We can write -

(1)
Subtracting,

The Ideal Gas Law, Which Relates The P, V And Temperature T ...

(2)
From relation (1) and (2), we can write -

(3)
Comparing (2) and (3),

Question 14. Two rods, one of aluminium and the other made of steel, having initial length

l_{1}

and

l_{2}

are connected together to form a single rod of length

l_{1} + l_{2} .

The coefficients of linear expansion for aluminium and steel are

α_{a}

and

α_{a}

and respectively. If the length of each rod increases by the same amount when their temperature are raised by

t^{o} C

, then find the ratio of length of aluminium rod to the whold rod.
(IIT JEE 2003)

αs/αa
αa/αs
αs/(αa+αs)
αa/(αa+αs)

Discuss Question

Answer: Option C. -> αs/(αa+αs)
:
C
The length of each rod increases by the same amount

∴ △ l_{a} = △ l_{s} \Rightarrow △ I_{1} α_{a} t = l_{2} α_{s} t

\Rightarrow \frac{l_{2}}{l_{1}} = \frac{α_{a}}{α_{s}} \Rightarrow \frac{l_{2}}{l_{1}} + 1 = \frac{α_{a}}{α_{s}} + 1

\Rightarrow \frac{l_{2} + l_{1}}{l_{1}} = \frac{α_{a} + α_{s}}{α_{s}} \Rightarrow \frac{l_{1}}{l_{1} + l_{2}} = \frac{α_{s}}{α_{a} + α_{s}}

Question 15. In a closed container of constant volume, the pressure of water exhibits an interesting dependence on temperature -

The point D (273.16 K,0.006 atm) is called the triple point, Tp, of water, where all there phases coexist. What is the observed phase change when the temperature is increased from -10⁰C to +10⁰C, while maintaining a constant pressure of 0.006 atm?

Solid → Liquid
Liquid → Gas
Solid → Gas
Gas → Liquid

Discuss Question

Answer: Option C. -> Solid → Gas
:
C
Along the constant pressure line of P=0.006, we have ice (solid H₂O) on the left of the triple point, and steam (gaseous H₂O) on the right.
Thus, as we increase the temperature from below Tpto above Tp, we observe a solid → gas phase transition, i.e., we see ice directly turning into gas without going through the water phase. Magical!

Question 16. An electron, an