Thermal Properties of Matter - Online Test

Q1. A washer consists of a 3.00 cm diameter circle of sheet metal with a 1.00 cm diameter circular hole in the middle. If the metal washer is heated until the diameter of the washer is 3.03 cm, then the diameter of the hole will be
Answer : Option D
Explaination / Solution:


diameter of hole = 1+0.01 = 1.01cm


Q2. Find the wrong statement about specific heat of a material
Answer : Option A
Explaination / Solution:

extensive properties (i.e. mass) are dependent upon the amound of a substance, while intensive properties (i.e. density) are independent of quantity.

Q3. Molar Specific Heat capacity has units of
Answer : Option B
Explaination / Solution:

Q=nCΔTC=QnΔT
Q4. The triple points of neon is 24.57 K. Express this temperatures on the Celsius and Fahrenheit scales
Answer : Option B
Explaination / Solution:


TK = 24.57K

after solving 

TC = –248.58 C

TF = – 415.44 F

 


Q5. The triple points of carbon dioxide is 216.55 K. Express this temperatures on the Celsius and Fahrenheit scales
Answer : Option B
Explaination / Solution:


TK = 216.55K

after solving 

TC = – 56.60 C

TF = – 69.88 F


Q6. Two absolut scales A and B have triple points of water defined to be 200 A and 350 B. What is the relation betweene  and ?
Answer : Option D
Explaination / Solution:

for any temperature scale 


Q7.

The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law : R =R[1 + (T –T )] The resistance is 101.6  at the triple-point of water 273.16 K, and 165.5  at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 ?


Answer : Option C
Explaination / Solution:

It is given that:
R0 [1 + 
  1.  (– T0)] … (i)
where,
R0 and T0 are the initial resistance and temperature respectively

R and T are the final resistance and temperature respectively 
 is a constant
At the triple point of water, T0 = 273.15 K
Resistance of lead, R0 = 101.6 
At normal melting point of lead, T = 600.5 K
Resistance of lead, R = 165.5 

Substituting these values in equation (i), we get:
Ro [1 +  (– To)]
165.5 = 101.6 [ 1 + (600.5 - 273.15) ] 
1.629 = 1 + (327.35) 
∴  = 0.629 / 327.35  =  1.92 × 10-3 K-1

For resistance, R1 = 123.4 
R1 R0 [1 + α (– T0)]
where,
T is the temperature when the resistance of lead is 123.4 
123.4 = 101.6 [ 1 + 1.92 × 10-3( T - 273.15) ]
Solving for T, we get
T = 384.8 K.

Q8.

A steel tape 1m long is correctly calibrated for a temperature of 27.0 C. The length of a steel rod measured by this tape is found to be 63.0 cm on a hot day when the temperature is 45.0 C. What is the actual length of the steel rod on that day? Coefficient of linear expansion of steel = 1.20   


Answer : Option B
Explaination / Solution:


final length =63.0136cm


Q9.

A large steel wheel is to be fitted on to a shaft of the same material. At 27 C, the outer diameter of the shaft is 8.70 cm and the diameter of the central hole in the wheel is 8.69 cm. The shaft is cooled using ‘dry ice’. At what temperature of the shaft does the wheel slip on the shaft? Assume coefficient of linear expansion of the steel to be constant over the required temperature range: αsteel = 1.20   .


Answer : Option A
Explaination / Solution:


final temp. = 


Q10. A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0 C. What is the change in the diameter of the hole when the sheet is heated to 227 C? Coefficient of linear expansion of copper = 1.70  K–1.
Answer : Option A
Explaination / Solution:

Change in diameter