Questions and answers for exam Preparation

MCQs

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

Total Questions : 247 | Page 5 of 25 pages

Question 41. It was a bright and breezy day in New York, the air was a cool 20 degrees Celsius, when Tony Stark, a.k.a the Iron Man's day took a dramatic turn as he got news of Mandarin's attack in the windy city of Chicago, and decided to immediately fly over there, in his iron body-armor, whose cavity had enough space to fit a man of 4,984

c m^{3}

in volume, but not more.
Being a smart man, he did a quick check of two important things - his own body volume, which he found was currently 4.980

c m^{3}

, and the temperature in Chicago, which was a cold 8 degrees Celsius that day. Knowing that the coefficient of volume expansion for iron,

γ

, is

33.3 \times 10^{- 6}

/^{\circ} C

, he decided to go. Check if that was a smart decision, by finding out whether the new volume of the iron suit when he reaches Chicago will crush him or not.

4.982 cm3 (he is going to be alright)
4.979 cm3 (it's going to be an uncomfortable encounter)
4.938 cm3 (he will have to take his suit off and call the Hulk)
4.980 cm3 (perfect fit!).

Discuss Question

Answer: Option A. -> 4.982 cm3 (he is going to be alright)
:
A
Mr. Stark's suit is made of iron, which should contract upon cooling with a coefficient

γ

33.3 \times 10^{- 6}

/^{\circ} C

. As he flies from New York to Chicago, there occurs a temperature drop of

12^{\circ} C

∴

Δ T

T_{C h i c a g o}

T_{N Y}

= (

8^{\circ} C

20^{\circ} C

) =

- 12^{\circ} C

The volume,

V_{N Y}

, of the suit's cavity in New York =

4, 984 c m^{3}

∴

The volume of the cavity in Chicago will be -

V_{C h i c a g o}

V_{N Y} (1 + γ Δ T)

4.984 [1 + 33.3 \times 10^{- 6} \times (- 12)] c m^{3}

\approx

4.982 c m^{3}

.
Awesome! The suit will still fit Tony Stark comfortably in Chicago, so he can just concentrate on Mandarin's onslaught. I feel much less worried about Chicago now!

Question 42. A block of ice at -

10^{\circ} C

is slowly heated and converted to steam at

100^{\circ} C

. Which of the following curves represent the phenomenon qualitatively?

Discuss Question

Answer: Option A. -> 4.982 cm3 (he is going to be alright)
:
B

A Block Of Ice At - 10∘C Is Slowly Heated And Converted To...

The change of ice at

- 10^{\circ} C

into steam at

100^{\circ} C

occurs in four stages.

The temperature of ice changes from $- 10^{\circ} C t o 0^{\circ} C$ .

Ice at $0^{\circ} C$ changes to water at $0^{\circ} C$ . The state changes as heat is supplied.

Water at $0^{\circ} C$ changes into water at $100^{\circ} C$ .

Water at $100^{\circ} C$ change into steam at $100^{\circ} C$ .The state changes as heat is supplied.

Question 43. Column-I gives some devices and Column-II gives some processes on which the functioning of these devices depend. Match the devices in Column-I with the processes in Column-II and indicates your answer by darkening appropriate bubbles in the 4

\times

4 matrix given in the ORS.
Column-I
Column-II
(p) Bimetallic strip
(a) Radiation from a hot body
(q) Steam engine
(b) Energy conversion
(r) Incandescent lamp
(c) Melting
(s) Electric fuse
(d) Thermal expansion of solids
(IIT JEE 2007)

(p → d);(q → b);(r → a,b);(s → b,c)
(p → a);(q → b);(r → c);(s → b,c)
(p → a);(q → c);(r → b);(s → c)
(p → d);(q → c);(r → b);(s → b)

Discuss Question

Answer: Option A. -> (p → d);(q → b);(r → a,b);(s → b,c)
:
A
Bimetallic strip is based on thermal expansion of materials.
In stream engine, internal energy of fuel (say coal) is converted into mechanical work.
In incandescent lamp and fuse also energy is converted from electrical to heat.
Fuse is based on melting of fuse wire if suddenly current increases.

Question 44. A lead bullet strikes against a steel armor plate with the velocity of 300 meter/second. If the bullet, assuming that the impact is perfectly inelastic find the rise in temperature of the bullet. Assume that the heat produced is shared equally between the bullet and the target, specific heat of lead =

0.03 c a l / g /^{\circ} C

178.6∘C
78.6∘C
2780.6∘C
100.6∘C

Discuss Question

Answer: Option A. -> 178.6∘C
:
A
Suppose m be the mass of the bullet and T be the rise in temperature.
Heat produced in the bullet

H = m \times s \times T = m \times 0.03 \times T c a l

Kinetic energy of the bullet

= \frac{1}{2} m v^{2} = \frac{1}{2} m (300 \times 10^{2})^{2} e r g s

Half of this energy goes into the bullet on impact, therefore

W = \frac{1}{2} \times [\frac{1}{2} m (300 \times 10^{2})^{2}] = \frac{1}{4} m \times 9 \times 10^{8} e r g s U s i n g t h e f o r m u l a W = J H, w e h a v e \frac{1}{4} m \times 9 \times 10^{8} = (4.2 \times 10^{7}) (m \times 0.03 \times T) T = \frac{9 \times 10^{8}}{4 \times (4.2 \times 10^{7}) \times 0.03} = {178.6}^{\circ} C

Question 45. A lump of

0.10 k g

of ice at

- 10^{\circ} C

is put in

0.15 k g

of water at

20^{\circ} C

. How much water and ice will be found in the mixture when it has reached in thermal equilibrium.
(Specific heat of water

= 1 k c a l k g^{- 1}

, specific heat of ice

= 0.50 k c a l k g^{- 1}

while its latent heat

= 80 k c a l k g^{- 1}

)

Ice is 68.75gm, and water is 181.25gm at 0∘C.
Ice is 16.75gm, and water is 11.25gm at 0∘C.
Ice is 108.75gm, and water is 101.25gm at 10∘C.
Ice is 6.75gm, and water is 181.25gm at 100∘C.

Discuss Question

Answer: Option A. -> Ice is 68.75gm, and water is 181.25gm at 0∘C.
:
A
The heat, which

0.15 k g

of water can release when its temperature is changed from

20^{\circ} C

0^{\circ} C

Q_{1} = m_{W} s_{W} Δ T_{W}

where,

m_{W}

is the mass of the water,

s_{W}

is the specific heat of water and

Δ T_{W}

is the change in temperature of water.
Given,

m_{W} = 0.15 k g

s_{W} = 1 k c a l k g^{- 1}

Q_{1} = 0.15 \times (1 \times 10^{3}) \times (20 - 0)

= 3000 c a l

...

(1)

Now, heat absorbed by

0.10 k g

of ice at

- 10^{\circ} C

to increase its temperature to

0^{\circ} C

Q_{2} = m_{i} S_{i} Δ T_{i}

where

m_{i}

is the mass of the ice,

s_{i}

is the specific heat of ice and

Δ T_{i}

is the change in temperature of ice.
Given,

m_{i} = 0.10 k g

s_{i} = 0.50 k c a l k g^{- 1}

Q_{2} = 0.10 \times (0.5 \times 10^{3}) \times [0 - (- 10)]

= 500 c a l

So, remaining heat,

Q = Q_{1} - Q_{2} = 3000 - 500 = 2500 c a l

Now as latent heat of ice is

l = 80 k c a l k g^{- 1}

, the remaining heat will melt ice only

L = \frac{Q}{m} \Rightarrow m = \frac{Q}{m} = \frac{2500}{80} = 31.25 g

of ice.
Initial amount of ice

= 0.10 k g = 100 g

So, the remaining ice,

= 100 - 31.25 = 68.75 g

Initial amount of water

= 0.15 k g = 150 g

Therefore, total water

= 150 + 31.25 = 181.25 g

The temperature of the system will be

0^{\circ} C

Question 46. The value of transmissivity may vary from

Discuss Question

Answer: Option A. -> 0-1
Answer: (a).0-1

Question 47. A cavity with a small hole will always behave as a

White body
Transparent body
Black body
Opaque body

Discuss Question

Answer: Option C. -> Black body
Answer: (c).Black body

Question 48. What is the wavelength band of gamma rays?

1 * 10¯⁷ to 4 * 10¯⁴ micron meter
2 * 10¯⁷ to 2.4 * 10¯⁴ micron meter
3 * 10¯⁷ to 3.4 * 10¯⁴ micron meter
4 * 10¯⁷ to 1.4 * 10¯⁴ micron meter

Discuss Question

Answer: Option D. -> 4 * 10¯⁷ to 1.4 * 10¯⁴ micron meter
Answer: (d).4 * 10¯⁷ to 1.4 * 10¯⁴ micron meter

Question 49. Of the radiant energy 350W/m² incident upon a surface 250W/m² is absorbed, 60W/m² is reflected and the remainder is transmitted through the surface. Workout the value for absorptivity for the surface material