Question 189: Units , Dimensions and Error Analysis

Question

If speed (V), acceleration (A) and force (F) are considered as fundamental units, the dimension of Young,s modulus will be:

Accepted Answer

We know,
Young's modulus (Y) = $${{{F \over A}} \over {{{\Delta l} \over l}}}$$
$$ 	herefore $$ [Y] = $${{\left[ {ML{T^{ - 2}}} ight]} \over {\left[ {{L^2}} ight]}}$$ = [ ML
-1
T
-2
]
Let [Y] = [V]
x
[A]
y
[F]
z
$$ 	herefore $$ [ ML
-1
T
-2
] =
[LT
-1
]
x
[LT
-2
]
y
[MLT
-2
]
z
$$ \Rightarrow $$ [ ML
-1
T
-2
] =
[ M
z
L
x + y + z
T
-x -2y - 2z
For dimensional balance, the dimension on both sides should be same.
So, z = 1
x + y + z = -1
$$ \Rightarrow $$ x + y = -2 ........(1)
and -x -2y - 2z = -2
$$ \Rightarrow $$ x + 2y = 0 ...........(2)
By solving those two equations we get,
x = -4 and y = 2
$$ 	herefore $$ [Y] = V
$$-$$4
A
2
F
1

Answer

V$$-$$2A2F2

Answer

V$$-$$4A$$-$$2F

Answer

V$$-$$4A2F

Answer

V$$-$$2A2F$$-$$2

Units , Dimensions and Error Analysis

If speed (V), acceleration (A) and force (F) are considered as fundamental units, the dimension of Young,s modulus will be:

Choose the correct answer: