Page 47 - Class-11-Physics-Part-1_Neat

UNITS AND MEASUREMENT 33

test of consistency of units, but has the (b) and (d); [M L T ] for (c). The quantity on the
–2
advantage that we need not commit ourselves right side of (e) has no proper dimensions since
to a particular choice of units, and we need not two quantities of different dimensions have been
worry about conversions among multiples and added. Since the kinetic energy K has the
sub-multiples of the units. It may be borne in dimensions of [M L T ], formulas (a), (c) and (e)
2
–2
mind that if an equation fails this consistency are ruled out. Note that dimensional arguments
test, it is proved wrong, but if it passes, it is cannot tell which of the two, (b) or (d), is the
not proved right. Thus, a dimensionally correct correct formula. For this, one must turn to the
equation need not be actually an exact actual definition of kinetic energy (see Chapter
(correct) equation, but a dimensionally wrong 6). The correct formula for kinetic energy is given
(incorrect) or inconsistent equation must be by (b). t
wrong.
2.10.2 Deducing Relation among the
Example 2.15 Let us consider an equation Physical Quantities
The method of dimensions can sometimes be
t
1
2
m v = m g h used to deduce relation among the physical
2 quantities. For this we should know the
where m is the mass of the body, v its dependence of the physical quantity on other
velocity, g is the acceleration due to quantities (upto three physical quantities or
gravity and h is the height. Check linearly independent variables) and consider it
whether this equation is dimensionally as a product type of the dependence. Let us take
correct. an example.

Answer The dimensions of LHS are t Example 2.17 Consider a simple
–1
–2
2
[M] [L T ] = [M] [ L T ] pendulum, having a bob attached to a
2
= [M L T ] string, that oscillates under the action of
–2
2
The dimensions of RHS are the force of gravity. Suppose that the period
[M][L T ] [L] = [M][L T ] of oscillation of the simple pendulum
2
–2
–2
= [M L T ] depends on its length (l), mass of the bob
–2
2
The dimensions of LHS and RHS are the same and (m) and acceleration due to gravity (g).
hence the equation is dimensionally correct. t Derive the expression for its time period
using method of dimensions.
Example 2.16 The SI unit of energy is
t
J = kg m s ; that of speed v is m s and Answer The dependence of time period T on
–1
2 –2
–2
of acceleration a is m s . Which of the the quantities l, g and m as a product may be
formulae for kinetic energy (K) given below written as :
y
x
can you rule out on the basis of T = k l g m z
dimensional arguments (m stands for the where k is dimensionless constant and x, y
mass of the body) : and z are the exponents.
2
(a) K = m v 3 By considering dimensions on both sides, we
(b) K = (1/2)mv 2 have
(c) K = ma [L M T ]=[L ] [L T ] [M ]
1
o
o
1
–2 y
1 z
1 x
(d) K = (3/16)mv 2
= L x+y T –2y M z
2
(e) K = (1/2)mv + ma
On equating the dimensions on both sides,
Answer Every correct formula or equation must we have
have the same dimensions on both sides of the x + y = 0; –2y = 1; and z = 0
equation. Also, only quantities with the same 1 1
physical dimensions can be added or So that x = ,y = – ,z = 0
subtracted. The dimensions of the quantity on 2 2
3
–3
2
–2
the right side are [M L T ] for (a); [M L T ] for Then, T = k l g –½
2
½
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