Heat Test

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Heat Test - 5

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Weekly Quiz Competition

Question 1
1 / -0

If the readings on Celsius and Fahrenheit scale of thermometers read the same numerical value when placed on a substance, then the temperature of the substance measured is

A
-40^oC

B
40^oC

C
0^oC

D
32^oC

Solution

The relation between Celsius and Fahrenheit scales is given by .
Let the value of temperature be X.
Since, the temperature readings are same on both the scales.

9X = 5X - 160
Solving for X, we get
X = -40
Question 2
1 / -0

In a thermometer the boiling and freezing points are marked as 125°C and 25°C. It reads 75°C. Then the corresponding Celsius scale is

A
25°C

B
40°C

C
45°C

D
50°C

Solution

The interval between the upper and lower points of the thermometer is 125 - 25 = 100°. The scale of the thermometer starts from 25°. Hence

Solving for T, we get T = 50°C.
Question 3
1 / -0

The string of a simple pendulum is of length 1 m at 25°C and made of steel. Its time period is given by the expression. If the temperature of the surroundings increases to 45°C, then the difference in number of oscillations made by it in one hour due to the increase in temperature is
(Coefficient of linear expansion of steel = 1.1 x 10^-6 per degree Celsius)

A
5

B
12

C
15

D
18

Solution

At 25°C, the time period is =1.98 seconds approximately. Number of oscillations made by the pendulum in one hour is 3600/1.98 = 1818 Oscillations.
When the temperature is increased by 20°C, the change in length is

Hence the corresponding time period is T` = 2 seconds. Number of oscillations in one hour = 3600/2 = 1800. Difference in number of oscillations = 1818 - 1800 = 18 Oscillations.
Question 4
1 / -0

The difference between the two temperatures in Celsius scale is 100°. Then the difference between their corresponding Fahrenheit scales is

A
100°F

B
200°F

C
180°F

D
150°F

Solution

Let the temperatures be X and Y. Then Y - X = 100°
Then their corresponding Fahrenheit readings are
.
Subtracting one from the other, we get.
Solving for Z- W, we get Z- W = 180.
Question 5
1 / -0

The difference between the Fahrenheit scale and Celsius scale of a temperature is 80. The temperature in Celsius scale is

A
70°C

B
30°C

C
60°C

D
80°C

Solution

Let the X and Y be Celsius and Fahrenheit scale readings of the temperature.
Then, Y- X= 80
Also,
9X + 160 = 5Y,
But Y = X + 80
Hence, 9X + 160 = 5X + 400
Solving for X, we get X = 60
Question 6
1 / -0

There are two rings one made up of Copper of radius 1.59 cm and the other of Nickel of radius 1.6cm. The temperature at which one will fit into the other is
(Coefficients of linear expansion of Copper and Nickel are 1.7 x 10^-5 /°C and 1.2 x 10^-5 /°C respectively)

A
1289°C

B
1177°C

C
1377°C

D
1277°C

Solution

The increased length due to the change in temperature is given by the expression: . As they need to fit into each other, the increased length should be equal. Hence
1.59[1 + 1.7 x 10^-5(X)] = 1.6[1 + 1.2 x 10^-5(X)], Solving for X we get X = 1277°C
Question 7
1 / -0

The length of a glass rod is found to be 1m on a meter scale made of Copper when both were at 50°C. The copper scale is correct at 0°C. If the coefficient of linear expansion of Copper and glass are 1.7 x 10^-5 /°C and 0.8 x 10^-5 /°C, then the length of the glass rod at 0°C is

A
1.045 m

B
1.0045 m

C
1.0045 cm

D
1.045 cm

Solution

The length of the scale at 50°C is 1(1 + 1.7 x 10^-5 x 50) and the length of the glass rod is L (1 + 0.8 x 10^-5 x 50). As both are equal, 1(1 + 1.7 x 10^-5 x 50) = L (1 + 0.8 x 10^-5 x 50), solving for L we get, L =1.0045 m
Question 8
1 / -0

Railway tracks are usually laid during winter times. The rails are of length 10 m. The difference in peak temperatures in India is 50°C. Then the length of the gap between the two rails that should accommodate the expansion of rails due to the change in temperature is
(Coefficient of Linear expansion of Iron = 1.2 x 10^-5/°C)

A
10 mm

B
6 mm

C
8 mm

D
6 cm

Solution

Assuming that one end of the Iron rail is fixed, then the linear expansion of the rails due to the temperature difference is. The difference in length is 10.006 -10 = 0.006 m = 6 mm.
Question 9
1 / -0

The lengths of the rods at two different temperatures T₁ and T₂ are found to be X and Y. Then the length of the rod at another temperature T₃is

A

B

C

D

Solution

The dependence of length on temperature is given by the expression. If L₀ is the length of the material at 0°C, then
Solving, we get. If L₃ is the length at temperature T₃, then

Substituting the values of L₀ and, we get
Question 10
1 / -0

The coefficients of linear expansion of Copper and Nickel are 1.7 x 10^-5 /°C and 1.2 x 10^-5 /°C respectively. The lengths of the copper strip such that the difference in length between it and the Nickel strip is always 20 cm at all temperatures is

A
68 cm

B
48 cm

C
40 cm

D
38 cm

Solution

The dependence of length on temperature is given by the expression,
. At any temperature T, let the lengths of the strips be X and Y respectively and also let their initial length be U and V.
Then X - Y = U - V = 20
. Subtracting one from another, we get
. Substituting in U - V = 20 and solving for U we get, U = 68 cm
Question 11
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 12
1 / -0

The volume of a metallic sphere increases by X % on increasing the temperature by T°C. Then the coefficient of linear expansion of the metal is

A

B

C

D

Solution

But V = V₀(1 + 3T), where V₀ is the volume at 273 K. Then the change in Volume = V₀(3T). Hence
Question 13
1 / -0

A container having certain amount of liquid expels 1/200^th of its mass on heating through a temperature range of 200°C. Then the apparent coefficient of expansion of the liquid is

A
3.5 x 10^-5 /°C

B
2.25 x 10^-5 /°C

C
2.75 x 10^-5 /°C

D
2.5 x 10^-5 /°C

Solution

If the original mass of the liquid present in the solid is M, and the apparent change in Mass is M₁ then the relation between them is given by
M₁= M (1 + apparent expansion coefficient of liquid x change in temp)
Apparent coefficient of expansion of liquid
=
Question 14
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 15
1 / -0

The change in temperature required to increase the length of a rod by 0.12% of its original lenght L is_____ (Given that the coefficient of linear expansion is 1.2 x 10^-5 /°C)

A
50°C

B
100°C

C
75°C

D
80°C

Solution

The temperature dependence of length on the temperature is given by.
Change in length = 0.12% of the original length
Question 16
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 T, where T is the change in temperature. For glass the change in volume is 600 x 3 x 0.8 x 10^-5T. As the empty space remains constant at all the temperatures, the change in volume should be same. Hence 18 x 10^-5Y T = 600 x 3 x 0.8 x 10^-5T, solving for Y, we get Y = 80 cu cm.
Question 17
1 / -0

The change in time period of a simple pendulum is dependent on the temperature according the relation , where t is the time period of the pendulum, T is the change in temperature and is the coefficient of linear expansion of the suspended wire. If a simple pendulum is fast by 0.5 seconds at 10°C and is slow by 1.5 seconds at 40°C, then the temperature at which it will oscillate with correct time period is

A
20°C

B
25°C

C
30°C

D
35°C

Solution

Let the temperature at which it oscillates with correct time period be T. Then
0.5 = t (X - 10)/2 and - 1.5 = t (X - 40)/2. Dividing one by the other, we get
- 3 = (X - 40)/(X - 20). Solving for X, we get X = 25
Question 18
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 cu m

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 T, 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 19
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 20
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