Short Answer Questions of Physical Quantities Mechanics Grade 11

Last Updated On:
Short Answer Questions of Physical Quantities Mechanics Grade 11
Short Answer Questions of Physical Quantities Mechanics Grade 11

Short Answer Questions of Physical Quantities (Mechanics)

  • Question. State principle of Homogeneity
Answer. It states that for a corect physical equation, the dimension on the left hand side must be equal to the dimension on the right hand side.
If the equation is in the form of Z = X + Y
According to this principle, Dimension of Z = Dimension of Y = Dimension of X
If Z = XY
Dimension of Z = Dimension of (XY)
  • Question. Write some uses of dimension.
Answer. The uses of dimensions are as follows.
A.      To check the correctness of physical relation.
B.      To derive the relation between different physical quantities.
C.      To convert the value of a physical quantity from one system of unit to another system of unit.
  • Question. Write some limitations of dimension analysis.
Answer.  The limitations of dimensional analysis are as follows.
A.      The dimensional analysis can’t be used to derive the physical relation, if the physical quantity depends on more than three fundamental quantities i.e. mass, length and time.
B.      This analysis can’t be used to derive the relation between logarithm functions, trigonometric functions, exponential functions, etc.
C.      This analysis doesn’t give any information about the physical quantity whether it is a vector or scalar.
  • Question. Define the following terms.
Answer.  Significant Figure: The digits which are accurately measured by an instrument in the particular measurement is called significant measurement. The numbers of accurately measured digits are called number of significant figure. It depends on the least count of the measuring instrument.
Accuracy: The degree of closeness between the mean value of experimental readings and standard value is called accuracy.
Precision: The degree of closeness among the experimental readings is called precision.
  • Question. Name any two physical quantities which have the same dimensions. Can a quantity have unit but no dimension? Explain.
Answer. Two physical quantities which have the same dimensions are work and torque. Yes, a physical quantity which has unit but no dimension is plane angle. Its SI unit is radian but it has no dimension.
  • Question. The diameter of a steel rod is given as 56.47 ± 0.02 mm. What does it mean?
Answer. The diameter of a steel rod is given as 56.47 ± 0.02 mm. This means that 56.47 mm is the actual diameter of the rod and + 0.02 mm is instrumental error. The measuring instrument either
measures 0.02 mm less or more than the actual diameter.
  • Question. The length of rod is exactly 1 cm. An observer records the readings as 1.0 cm, 1.00 cm, and 1.000 cm, which is the most accurate measurement?
Answer. If the length of rod is exactly 1 cm. When an observer records the readings as 1.0 cm, 1.00 cm and 1.000 cm, then the measurement which measures 1.000 cm is most accurate because it has more number of significant figures and the instrument which measures this data has less least count than the other.
  • Question. Is dimensionally correct equation necessarily to be a correct physical relation? What about dimensionally wrong equation?
Answer.  No, dimensionally correct equation may not be a correct physical relation. Consider the following relation,
v = u +2 at
Dimensional formula of v is [LT-1] and that of (u + 2 at) İs also [LT-1]. Hence, the above relation is dimensionally correct. But, it is established fact that the above relation isn’t correct physical relation.
Dimensionally wrong equation can’t be correct physical relation. For example, v = u+ at^2   is dimensionally wrong equation and it is also incorrect physical relation.
  • Question. Check the correctness of the formula, PV = RT.
Answer. 
Given,
PV= RT, where, P = pressure,
V =  volume, R =  gas constant and T = temperature.
The dimension of pressure, P = [ML-1T-2]
The dimension of volume, V = [L3]
The dimension of gas constant, R [ML2 T-2 K-1]
The dimension of temperature, T [K]
Now,
Dimensions of left hand side,
PV = [ML-1T-2L3]
     =[ML2T-2]
Dimensions of right hand side,
RT= [ML2T-2K-1K]
    = [ML2 T-2 ]
Dimension on left hand side is equal to dimension on right hand side.
Hence, the formula is dimensionally correct.

LEAVE A REPLY

Please enter your comment!
Please enter your name here