The dimensional formula for angular momentum is …..
$\left( a \right)\ \left[ {{M}^{0}}{{L}^{2}}{{T}^{-2}} \right]$
$\left( b \right)\ \left[ M{{L}^{2}}{{T}^{-1}} \right]$
$\left( c \right)\ \left[ ML{{T}^{-1}} \right]$
$\left( d \right)\ \left[ M{{L}^{2}}{{T}^{-2}} \right]$
Answer
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Hint: We know that the dimensional formula of any term depends on the unit formula of that term in MKS dimensional unit system. Now, the formula of angular momentum can be given as: $l=mvr$. Now, by converting this formula into unit form we will derive the dimensional formula of angular momentum.
Complete step-by-step answer:
As given in the question we want to derive the MLT form of angular momentum. Where, M is mass, L is length and T is time.
Now, we know that the formula of angular momentum is $l=mvr$, where l is angular momentum, m is mass, v is velocity and r is radius.
Considering the MKS unit system, the units of m, v and r are given as:
$\text{mass m}=kg$………………………..(i)
$\text{velocity v}={m}/{s}\;$ …………………(ii)
$\text{radius r}=m$ ………………………..(iii)
Now, converting the above expressions into MLT form we get,
$m=kg={{M}^{1}}$ …………………(iv)
$v={m}/{s}\;=\dfrac{{{L}^{1}}}{{{T}^{1}}}$ ……………….(v)
$r=m={{L}^{1}}$ ……………………..(vi)
Now, substituting the values of m, v and r in main equation of angular momentum we get,
$L=\left( {{M}^{1}} \right)\left( \dfrac{{{L}^{1}}}{{{T}^{1}}} \right)\left( {{L}^{1}} \right)$
Now, adding the adding the powers of common entities and equating further we get,
$L=\left( {{M}^{1}} \right)\left( \dfrac{{{L}^{1+1}}}{{{T}^{1}}} \right)$
$\Rightarrow L={{M}^{1}}\dfrac{{{L}^{2}}}{{{T}^{1}}}$
$\Rightarrow L={{M}^{1}}{{L}^{2}}{{T}^{-1}}$
Or $L=M{{L}^{2}}{{T}^{-1}}$
Hence, the dimensional formula of angular momentum is $M{{L}^{2}}{{T}^{-1}}$.
Thus, option(d) is correct.
Note: In such types of questions students should not get confused with MKS and CGS unit systems. If the units in the question are given in centimeter, gram or millisecond, then the student must convert into meter, kilogram and second and then derive the dimensional formula otherwise the answer will be wrong. Students should also know the dimensional formula of basic quantities such as mass, length, time etcetera so that other entities can be derived or solved easily.
Complete step-by-step answer:
As given in the question we want to derive the MLT form of angular momentum. Where, M is mass, L is length and T is time.
Now, we know that the formula of angular momentum is $l=mvr$, where l is angular momentum, m is mass, v is velocity and r is radius.
Considering the MKS unit system, the units of m, v and r are given as:
$\text{mass m}=kg$………………………..(i)
$\text{velocity v}={m}/{s}\;$ …………………(ii)
$\text{radius r}=m$ ………………………..(iii)
Now, converting the above expressions into MLT form we get,
$m=kg={{M}^{1}}$ …………………(iv)
$v={m}/{s}\;=\dfrac{{{L}^{1}}}{{{T}^{1}}}$ ……………….(v)
$r=m={{L}^{1}}$ ……………………..(vi)
Now, substituting the values of m, v and r in main equation of angular momentum we get,
$L=\left( {{M}^{1}} \right)\left( \dfrac{{{L}^{1}}}{{{T}^{1}}} \right)\left( {{L}^{1}} \right)$
Now, adding the adding the powers of common entities and equating further we get,
$L=\left( {{M}^{1}} \right)\left( \dfrac{{{L}^{1+1}}}{{{T}^{1}}} \right)$
$\Rightarrow L={{M}^{1}}\dfrac{{{L}^{2}}}{{{T}^{1}}}$
$\Rightarrow L={{M}^{1}}{{L}^{2}}{{T}^{-1}}$
Or $L=M{{L}^{2}}{{T}^{-1}}$
Hence, the dimensional formula of angular momentum is $M{{L}^{2}}{{T}^{-1}}$.
Thus, option(d) is correct.
Note: In such types of questions students should not get confused with MKS and CGS unit systems. If the units in the question are given in centimeter, gram or millisecond, then the student must convert into meter, kilogram and second and then derive the dimensional formula otherwise the answer will be wrong. Students should also know the dimensional formula of basic quantities such as mass, length, time etcetera so that other entities can be derived or solved easily.
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