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Heat Test - 2

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Weekly Quiz Competition

Question 1
1 / -0

A steel sphere of diameter 50 cm is resting on a circular ring made up of Brass of radius 24.8 cm. Both the measurements are at room temperature of 20°C. If the coefficient of linear expansion of Steel and Brass are 1.2 x 10^-5/°C and 1.8 x 10^-5/°C, then the temperature at which the sphere will fall down is

A
1400 °C

B
1500 °C

C
1286°C

D
1386 °C

Solution

For the sphere to fall down the dimension of the diameter of the ring should either be greater than or equal to the diameter of the sphere. Hence,
50 [1 + (T - 20) 1.2 x 10^-5] = 49.6 [1 + (T - 20) 1.8 x 10^-5]
T - 20 = 0.4 x 10⁵/[49.6 x 1.8 – 50 x 1.2]= 1366.12
T= 1386 °C.
Question 2
1 / -0

If two identical spheres of same radii R out of which one is hollow containing air and the other is solid are heated through the same temperature range, then the change in dimensions of the spheres will be

A
Greater in hollow sphere

B
Greater in Solid sphere

C
Same in both spheres

D
Greater initially in hollow sphere and decreases after certain temperature

Solution

Air being a bad conductor of heat compared to the metallic sphere which is a good conductor, maximum heat is utilized in expanding the metallic part. In case of solid, the change in radius will be greater due to the metallic substance present in it.
Question 3
1 / -0

A glass vessel of volume 600 cu cm at 20°C contains certain amount of Mercury. The amount of Mercury that it should contain at 20°C such that the empty space in the container is same for at all temperatures is [Coefficient of linear expansion of glass is 0.8 x 10^-5/°C and the coefficient of cubical expansion of Mercury is 18 x 10^-5/°C]

A
60 cu m

B
70 cu cm

C
60 cu cm

D
80 cu cm

Solution

Let the volume of Mercury present in the container be Y. Then the change in due to the increase in temperature is 18 x 10^-5Y Δ T, where Δ T is the change in temperature. For glass the change in volume is 600 x 3 x 0.8 x 10^-5ΔT. As the empty space remains constant at all the temperatures, the change in volume should be same. Hence 18 x 10^-5Y Δ T = 600 x 3 x 0.8 x 10^-5 Δ T, solving for Y, we get Y = 80 cu cm.
Question 4
1 / -0

A cubic crystal of dimension 3 x 10^-12 m at 293 K, has coefficient of linear expansion along its length as 13 x 10^-7 /°C but along breadth and height it has the coefficient of linear expansion of 231 x 10^-7 /°C. If temperature of the crystal is increased to 393 K, then the change in volume is

A
1.2825 x 10^-37 cu cm

B
1.2825 x 10^-36 cu m

C
1.2825 x 10^-37 cu m

D
1.2285 x 10^-37 c

Solution

Volume of the cube at 293 K = 27 x 10^-36 cu m.
Coefficient of cubical expansion of the crystal = (2 x 231 + 13) 10^-7 /°C = 475 x10^-7 /°C.
Hence the change in volume = V x g x ΔT, where g is the cubical expansion coefficient.
The change in temperature = 100 K.
Hence the change in volume = 1.2825 x 10^-37 cu m
Question 5
1 / -0

The door of a running refrigerator inside a closed room was left open. Which of the following observations will be seen?

A
The room will be warmed up gradually.

B
The temperature of the room will remain unaffected.

C
The room will become as cold as the refrigerator.

D
The room will be cooled slightly.

Solution

Refrigerator is a device that transfers heat from inside a box to its surroundings.
If you leave the door open, heat is merely recycled from the room into the refrigerator, then back into the room. A net room temperature increase would result from the heat of the motor that would be constantly running to move energy around in a circle.
Question 6
1 / -0

Mercury is filled in a glass flask of volume 500 cu cm at 0°C. The coefficient of cubical expansion of mercury is 182 ´ l0^-6/°C and that of glass is 30 ´ l0^-6/°C. If the temperature of the flask is increased to 100°C, then the amount of Mercury that has over flown out is

A
30 cm³

B
18.2 cm³

C
15.2 cm³

D
7.6 cm³

Solution

The change in volume of the flask = 500 (1 + 30 ´ l0^-6 x 100) = 501.5 cu cm
The change in volume of the Mercury = 500 (1 + 182 ´ l0^-6 x 100) = 509.1 cu cm
The volume of the Mercury that has over flown is 509.1 - 501.2 = 7.6 cu cm

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Re-Attempt Weekly Quiz Competition

Heat Test - 2

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Heat Test - 2

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SHARING IS CARING

Question 1

1400 °C

1500 °C

1286°C

1386 °C

Solution

Question 2

Greater in hollow sphere

Greater in Solid sphere

Same in both spheres

Greater initially in hollow sphere and decreases after certain temperature

Solution

Question 3

60 cu m

70 cu cm

60 cu cm

80 cu cm

Solution

Question 4

1.2825 x 10-37 cu cm

1.2825 x 10-36 cu m

1.2825 x 10-37 cu m

1.2285 x 10-37 c

Solution

Question 5

The room will be warmed up gradually.

The temperature of the room will remain unaffected.

The room will become as cold as the refrigerator.

The room will be cooled slightly.

Solution

Question 6

30 cm3

18.2 cm3

15.2 cm3

7.6 cm3

Solution

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1.2825 x 10^-37 cu cm

1.2825 x 10^-36 cu m

1.2825 x 10^-37 cu m

1.2285 x 10^-37 c

30 cm³

18.2 cm³

15.2 cm³

7.6 cm³