Get Answers to all your Questions

header-bg qa

 A, B, C and D are four different physical quantities having different dimensions. None of them is dimensionless.  But we know that the equation AD = C ln(BD) holds true.  Then which of the combination is not a meaningful quantity ?

  • Option 1)    (A2-B2C2)
  • Option 2)    ({A-C}/D)
  • Option 3)  {(A/B)-C} 
  • Option 4)  {(C/BD)-(AD2/C)} 
 

Answers (1)

best_answer

A.D= C \ln \left ( B.D \right )

Hence \left [ B.D \right ]= are \: dimensionless

\left [ B \right ]= \left [ \frac{1}{D} \right ]...............(1)

&   \left [ A.D \right ]= \left [ C \right ]...............(2)

Relevent Concept here is :

Dimension -

The power to which fundamental quantities must be raised in order to express the given physical quantities.

-

 

 

 


Option 1)

A^{2}-B^{2}C^{2}

A^{2}-B^{2}\, C^{2}

\left [ A \right ]= \left [ \frac{C}{D} \right ]= \left [ C.B \right ]

\therefore Dimensions of A & B.C are same.

Option 2)

\frac{\left ( A-C \right )}{D}

\frac{A-C}{D}

Dimension of \left [ A \right ]and \left [ C \right ] are not same hence its meaningless.

Option 3)

\frac{A}{B}-C

\frac{\left [ A \right ]}{\left [ B \right ]}-\left [ C \right ]= \left [ AD \right ]-\left [ C \right ]

Since A D & C are of same dimension hence its meaningful.

Option 4)

\frac{C}{BD}-\frac{AD^{2}}{C}

\frac{C}{BD}-\frac{AD^{2}}{C}

Its meaningful.

Posted by

perimeter

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE