✌PLEASE ANSWER✌
✌ QUESTION OF DIMENSIONAL ANALYSIS✌
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P= ax^2-bt
where P is pressure,x is distance and t is time.
Dimension of Pressure = [M L^-1 T^ -2.].
P=ax^2 and P=bt [By dimensional analysis rule]
Substitute dimensions:
P=ax^2
ML^-1T^-2=a L^2
Dimension of a = [ML^-3T^-2].
P= bt
ML^-1T^-2=b T
Dimension of b = ML^-1T^-3.
Now ,
Dimension of a x b :
ML^-3T^-2 x Ml^-1T^-3
=M^2L^-4T^-5.
Dimension of a/p:
ML^-3T^-2/ML^-1T^-2
= L^-2.
Hope it helps u!!!!
where P is pressure,x is distance and t is time.
Dimension of Pressure = [M L^-1 T^ -2.].
P=ax^2 and P=bt [By dimensional analysis rule]
Substitute dimensions:
P=ax^2
ML^-1T^-2=a L^2
Dimension of a = [ML^-3T^-2].
P= bt
ML^-1T^-2=b T
Dimension of b = ML^-1T^-3.
Now ,
Dimension of a x b :
ML^-3T^-2 x Ml^-1T^-3
=M^2L^-4T^-5.
Dimension of a/p:
ML^-3T^-2/ML^-1T^-2
= L^-2.
Hope it helps u!!!!
Answered by
1
deminsion of a/p is L^-2
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