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| New Integrated Govt. School Class 10 Physical Science Model Activity Task Solutions |
Today we will discuss about the Solutions of New Integrated Govt. School (wbbse) class 10 Physical Science Model Activity Task (Part 5)
MODEL ACTIVITY TASK
PHYSICAL SCIENCE
CLASS – X
1. Choose the correct answer :
1.1 The compound that is a weak electrolyte is —
(a) Sodium chloride (b) Ammonium sulfate (c) Sulfuric acid (d) Acetic acid .
Ans: (d) Acetic acid
1.2 The compound that is suitable for drying ammonia is —
(a) 
Ans: (d) CaO
1.3 The SI unit of thermal conductivity is -
(a) WmK (b) 
Ans:(c)
2. Determine whether the following statements are true or false :
2.1 During electrolysis, electrons carry electricity through the solution.
Ans: False
2.2 Reduction is essential in the extraction of a metal from its ore.
Ans: True
2.3 The unit of the coefficient of linear expansion does not contain the unit of length.
Ans: True
3. Answer in brief :
3.1 Determine the percentage by mass of calcium in calcium carbonate ( Ca = 40 ).
Ans: Molecular weight of calcium carbonate (`CaCO_3`)= [ 40 +12+ 16x3] = 100
Calcium (Ca) = 40
The percentage by mass of calcium in calcium carbonate = ( 40/100 ) x 100 %
= 40 %
3.2 Why are the catalysts used in industrial manufacture of ammonia, sulfur trioxide etc. kept in a finely divided state ?
Ans: Catalysts used in industrial production of ammonia, sulfur trioxide, etc. are kept in finely divided state because the surface area of the solid catalyst used in this case increases when it is crushed. As a result, more reactants come in contact with the catalyst. So the efficiency of the catalyst increases. As a result, more and more reactive substances are produced.
4. Answer the following question :
4.1 The EMF of a cell is 10 V and its internal resistance is 2 ohm. The cell is connected to a resistor of 8 ohm. Calculate the amount of heat in Joule that will be produced in the resistor in 60 seconds.
Ans: The EMF of a cell (E) = 10 V
Internal resistance (r) = 2 ohm External resistance (R) = 8 ohm
Current in the circuit (I) = E / ( R + r ) = 10 / ( 8 + 2 ) = 1 A
The amount of heat produced in the resistor in 60 seconds (H) =
=
= 480 J
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