Physical Quantities [ Mechanics] Class 11 Physics solutions

Unit 1 : Mechanics 

1. Physical Quantities 

Dimensional formula for class 11

Short Answer Questions          [2]

1.[2075 Set A Q.No. 1a] Differentiate between accuracy and precession of measurement.

    Ans: Accurate measurement means capable of providing a correct reading of measurement. In physical science, it means correct. A measurement is accurate if it correctly reflects the size of the things being measured.

            Precise measurement means exact, as in performance, execution, or amount. In physical science it means repeatable, reliable, getting the same measurement each time.

2.[ 2075 Set B Q.No. 1g] In one of the printed documents the unit of universal gravitational constant is given as  NmKg-2. Check its correctness from dimensional analysis.

  Ans: The unit of universal gravitational constant on printed document is NmKg-2. Here N is the unit of force, m is the unit of distance and kg is unit of mass. So, dimension of force = [MLT-2],  dimension of distance =[L] and dimension of mass = [M], then 

       NmKg-2 = [MLT-2].[L].[M-2]

                      = [M-1L2T-2]

       again, 

                   F = Gm1m2/d2 

         or,.    G = Fd2 / m1.m2 

                       = [MLT-2]. [L2] / [M-2] 

                       = [M-1 L3 T-2] 

          Thus, the given unit is dimensionally incorrect.

3.[ 2074 Supp. Q.No. 1a] Is dimensionally correct equation necessarily be correct physical relation? What about dimensionally wrong equation?

  Ans: No, dimensionally correct equation may not be correct physical relation. Consider the following relation,.     v = u+ 2at

          Dimensional formula of v is [ LT -1] and that of (u+ 2at)  is also [ LT -1]. Hence, the above relation is dimensionally correct. But, it is established fact the above relation isn't correct in physical relation. 

          Dimensionally wrong equation can't be correct physical relation. For example, v = u + at2 is dimensionally wrong equation and it is also incorrect physical relation.

4.[ 2074 Set B Q.No. 1f ] What is the difference between accurate and precise measurements?

  Ans: Please refer to 2075 Set A. Q.No. 1a 

5.[ 2073 Set C Q.No. 1a] Check dimensionally the correctness of the Stock's formula, F= 6πnrv, where symbols have their usual meanings.

  Ans: The given Stock's formula is 

                  F= 6πnrv..............( i ) 

          The dimension of F = [ ML T-2 ]

          The dimension of 6π= 0 ( dimensionless quantity)

          The dimension of n = [ ML-1 T-1 ]

          The dimension of r = [ L]

          The dimension of v = [ LT-1]

     Putting the value of dimension in ( i ), we get,                  [MLT-2] = [ ML-1 T-1] [L] [LT-1]

                   = [ ML-1+1+1 T-1-1]

                   = [ MLT-2] 

      Therefore dimension on left hand side is equal to the dimensions on right and side. So the formula is dimensionally correct.

6.[ 2073 Supp Q.No. 1g] A student writes an expression for the momentum (p)a body of mass (m) with total energy (E) and considering the duration of time (t) as p = √2mE/t ( it's whole root). Check its correctness on dimensional analysis.

  Ans: The given formula is 

        p = √ 2mE/t, ( whole root)

        

      where, 

      p = momentum,.             m = mass

      E = Energy                        t  = time 

         The dimension of each term are,

          p= [ MLT-1] 

          m= [ M] 

          t = [T] 

          E = [ ML2 T-2 ]  

    Then, 

          [ MLT-1] = √M.ML2 T-2/ T.    ( whole root)

                           = √M2 L2 T-1          ( '" " """  """ )

                           = [MLT-1/2].     ( -1/2 is in the power of T)

    Dimension of LHS is not equals to Dimension of RHS. So, the formula is dimensionally incorrect.

7.[ 2072 Supp Q.No. 1a] check the correctness of the formula, PV = RT.

  Ans: Given formula is PV = RT 

     where,  P = pressure, V = volume 

                    R= gas constant T = temperature.

     The dimensions of each term,

         P= [ML-1 T-2]

         V = [ L3] 

         R = [ ML2 T-2 K-1] 

          T = [ K] 

  Now,  

       LHS, PV = [ ML-1 T-2 L3] 

                      = [ ML2 T-2 ] 

 Therefore, Dimension of LHS = Dimension of RHS.

      Hence, the formula is dimensionally correct.

8.[ 2072 Set C Q.No. 1a ] The diameter of a steel rod is given as 56.47 +- 0.002 mm. What does it mean? 

  Ans: the diameter of a steel rod is given as 56.47 +- 0.02 mm. This means that 56.47 millimetre is the actual diameter of the rod and plus minus 0.02 mm is instrumental error. The measuring instrument either measures 0.02 mm less or more than the actual diameter. i.e., 56. 45 or 56.49.

9. [ 2072 Set D Q.No. 1a] The length of rod is exactly 1 cm. An observer records the reading as 1.0 cm, 1.00 cm , and 1.000 cm , which is the most accurate measurement? 

  Ans: 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 figure and the instrument which measure this data has less least count than the other.

10.[2071 Supp Q.No 1a ] What do you mean by significant figure?

  Ans: The meaningful digits in a number are called significant figure. The number of digits in a number is called its number of significant figures. The number of significant figures in a measurement depends upon the least count of the measuring instrument.For example, the length of a scale is 2.17 CM when measured by Vernier caliper and 2.176 CM when measured by micro metre screw guaze. The significant figures are 3 and 4. The more significant figure gives more precious value of a measurement.

11.[ 2071 Set D Q.No. 1d] A student writes an expression of the force causing a body of mass ( m) to move in a circular motion with velocity (v) as F= mv2. Use the Dimensional metod to check its correctness.

  Ans: The given relation is F = mv2 

     where, F is force , m is mass and v is velocity

          The Dimensions of each terms are,

           F= [ MLT-2] 

           v = [ LT-1]

           m = [ M ] 

        Now,. 

           Dimension of LHS,

             F= [ MLT-2]

           Dimension of RHS, 

             mv2 = [ M] [ ML2 T-2] 

      Therefore, LHS is not equal to RHS. So, the formula is dimensionally wrong.

12.[ 2070 Set C Q.No. 1a] Name any two physical quantities which have same Dimensions. Can a quantity have unit but no Dimensions? Explain.

  Ans: Two physical quantities which have same Dimensions are work and torque. Yes, a physical quantity which has unit but no dimension is plane angle. It's SI unit is radian but it has no dimension.

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