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Dimensional analysis worksheet




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Question 1 Show that expression v=v0+at is dimensionally correct, where v and v0 represent velocities and a is acceleration and t represents time.

Answer

Here in this question , we need to compare the dimension of the Left hand side and right hand side
LHS
v=[LT1]
RHS
v0+at=[LT1]+[LT2][T]=[LT1]+[LT1]=[LT1]
So, the dimensions on both the sides are same
So , we can say that expression v=v0+at is dimensionally correct


Question 2 Each of the following equations was given by a student during an examination.
Dimensional analysis worksheet
Do the dimensional analysis of each equation and explain why the equations cannot be correct.

Answer

a. The expression is 12mv2=12mv20+mgh
LHS
12mv2=M(LT1)2=ML2T2
RHS
12mv20+mgh=M(LT1)2+[ML2T2]1/2=ML2T2+M1/2LT1
Now this cannnot be correct as we cannot add quantities of two different dimension.
So this equation is incorrect

b.The expression is v=v0+at2
LHS
v=[LT1]
RHS
v0+at2=[LT1]+[LT2]T2=[LT1]+[L]
Now this cannnot be correct as we cannot add quantities of two different dimension.
So this equation is incorrect

c.The expression is ma=v2
LHS
ma=[M][LT2]=[MLT2]
RHS
v2=[L2T2]
Now dimension of LHS is not equal to dimension of RHS
So this equation is incorrect


Question 3 Suppose that the displacement of an object is related to time according to the expression x=Bt2 . What are the dimensions of B?

Answer

Here x=Bt2
or
B=xt2
Therefore dimension of B will be
=[L][T2]=[LT2]


Question 4 Find the SI unit of pressure.

Answer

SI unit of pressure is pascal


Question 5 (a) Is it possible for two quantities to have the same dimensions but different units.
(b) Is it possible for two quantities to have same units but different dimensions?

Answer

a. This is possible . Example Work and Torque have different unit Joule and Newton-meter but the dimension [ML2T1] is same
b. No it is not possible.


Question 6 Suppose we are told that the acceleration ‘a’ of a particle moving with uniform speed ‘v’ in a circle of radius ‘r’ is proportional to some power of r, say rn, and some power of vm. How can we determine the value of n and m.

Answer

We can write
aαvmrn
a=Kvmrn
Where K is the dimensionless constant
Now writing dimension of various quantities in the equation
[LT2]=[LT1]m[L]n
[LT2]=[Lm+nTm]
Comparing the dimensions of similar quantities on both the sides, we get
m+n=1
m=2 or m=2
Therefore 2+n=1 or n=-1
Hence the values of m and n are 2 and -1 respectively


Question 7 The mass of a solid cube is 856 g, and each edge has a length of 5.35 cm. Determine the density ρ of the cube in basic SI units.

Answer

Now ρ=MV
Here M= 856 g = .856 kg
V=(5.35)3cm3=(5.35)3(1/100)3m3
ρ=.856(5.35)3(1/100)3kg/m3=5590kg/m3


Question 8 The period ‘T’ of a simple pendulum is measured in time units and is described by
T=2πlg Where l is the length of the pendulum and ‘g’ is the free fall acceleration in units of length divided by square of time. Show that this equation is dimensionally correct.

Answer

T=2πlg
LHS
T=[T]
RHS
2πlg=([L][LT2])1/2=([T2])1/2=[T]
So, this equation is dimensionally consistent



Question 9 Estimate the average mass density of a sodium atom assuming its size to be about 2.5 Å. (Use the known values of Avogadro’s number and the atomic mass of sodium). Compare it with the density of sodium in its crystalline phase: 970 kg m-3. Are the two densities of the same order of magnitude? If so, why?
Question 10 In the following x is in meters, t in seconds, v in m/s and a is in m/s2. Find the SI unit of each combination.
a. v2xa
b.xa
c. 12at2

Answer

a. v2xa
=(m/s)2mm/s2=(m/s)2(m/s)2 =dimensionless

b.
mm/s2=s2=s

c.Here constant 1/2 does not have any units
(m/s2)s2=m


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