Answer
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Hint: Here, in this solution, we will start using the rules of dimensional formulas to determine the necessary values. Here, the dimensional formula is a kind of expression of the physical quantity in terms of basic units with proper dimensions. We’ll write each dimensional formulas in the given relation $x = k{a^m}{t^n}$ and solve further.
Complete answer:
We have given that the position of a body with acceleration in a t time is given by $x = k{a^m}{t^n}$
As we know that, the dimensional formula of the following are:
Position – $\left[ {{M^0}{L^1}{T^0}} \right]$
Acceleration – $\left[ {{M^0}{L^1}{T^{ - 2}}} \right]$
Time – $\left[ {{M^0}{L^0}{T^1}} \right]$
Now, given that
$x = k{a^m}{t^n}$
Putting all the above values in terms of dimensional formulas, It will be,
$\left[ {{M^0}{L^1}{T^0}} \right] = k{\left[ {{M^0}{L^1}{T^{ - 2}}} \right]^m}{\left[ {{M^0}{L^0}{T^1}} \right]^n}$
Multiplying the exponents with the terms inside the bracket,
$\left[ {{M^0}{L^1}{T^0}} \right] = k\left[ {{M^0}{L^m}{T^{ - 2m}}} \right]\left[ {{M^0}{L^0}{T^n}} \right]$
Now, on the right side, the same bases are in multiplication with different exponents. So, add the exponents of the exponential terms i.e.,
${x^p}{x^q} = {x^{p + q}}$
$\left[ {{M^0}{L^1}{T^0}} \right] = k\left[ {{M^0}{L^m}{T^{ - 2m + n}}} \right]$
We know that according to the principal homogeneity dimensions of the RHS should be equal to the dimensions LHS. Thus on comparing both sides of the above equation and the term k is scalar. Only compare similar terms.
It implies that, $m = 1$ And $ - 2m + n = 0$
Thus, we will get:
$n = 2$
Hence, option (2) is the correct answer i.e., $m = 1$,$n = 2$
Note: In such cases, we must understand how to apply the rules of dimensional formula analysis to determine the various dimensional formulas. We should also be familiar with the dimensional formulas for basic kinematic quantities such as velocity, acceleration, distance, and so on.
Complete answer:
We have given that the position of a body with acceleration in a t time is given by $x = k{a^m}{t^n}$
As we know that, the dimensional formula of the following are:
Position – $\left[ {{M^0}{L^1}{T^0}} \right]$
Acceleration – $\left[ {{M^0}{L^1}{T^{ - 2}}} \right]$
Time – $\left[ {{M^0}{L^0}{T^1}} \right]$
Now, given that
$x = k{a^m}{t^n}$
Putting all the above values in terms of dimensional formulas, It will be,
$\left[ {{M^0}{L^1}{T^0}} \right] = k{\left[ {{M^0}{L^1}{T^{ - 2}}} \right]^m}{\left[ {{M^0}{L^0}{T^1}} \right]^n}$
Multiplying the exponents with the terms inside the bracket,
$\left[ {{M^0}{L^1}{T^0}} \right] = k\left[ {{M^0}{L^m}{T^{ - 2m}}} \right]\left[ {{M^0}{L^0}{T^n}} \right]$
Now, on the right side, the same bases are in multiplication with different exponents. So, add the exponents of the exponential terms i.e.,
${x^p}{x^q} = {x^{p + q}}$
$\left[ {{M^0}{L^1}{T^0}} \right] = k\left[ {{M^0}{L^m}{T^{ - 2m + n}}} \right]$
We know that according to the principal homogeneity dimensions of the RHS should be equal to the dimensions LHS. Thus on comparing both sides of the above equation and the term k is scalar. Only compare similar terms.
It implies that, $m = 1$ And $ - 2m + n = 0$
Thus, we will get:
$n = 2$
Hence, option (2) is the correct answer i.e., $m = 1$,$n = 2$
Note: In such cases, we must understand how to apply the rules of dimensional formula analysis to determine the various dimensional formulas. We should also be familiar with the dimensional formulas for basic kinematic quantities such as velocity, acceleration, distance, and so on.
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