Physics - Unit & Dimension Question with Solution | TestHub

To determine the dimensions of $x+y-z$, we first need to find the dimensions of $a$, $b$, and $c$.

The given equation for momentum $P$ is:

$P=a{t}^2+bt+c$

According to the principle of homogeneity of dimensions, each term in an equation must have the same dimensions. The dimension of momentum $P$ is $[M^1{L}^1{T}^{-1}]$.

Therefore, we can equate the dimensions of each term to the dimension of momentum:

1. Dimension of $a{t}^2$:

$[a{t}^2]=[P]$

$[a][T^2]=[M^1{L}^1{T}^{-1}]$

$[a]=[M^1{L}^1{T}^{-1}][T^{-2}]$

$[a]=[M^1{L}^1{T}^{-3}]$

2. Dimension of $bt$:

$[bt]=[P]$

$[b][T^1]=[M^1{L}^1{T}^{-1}]$

$[b]=[M^1{L}^1{T}^{-1}][T^{-1}]$

$[b]=[M^1{L}^1{T}^{-2}]$

3. Dimension of $c$:

$[c]=[P]$

$[c]=[M^1{L}^1{T}^{-1}]$

Now, we need to find the dimension of the product $abc$:

$[abc]=[a][b][c]$

$[abc]=[M^1{L}^1{T}^{-3}]\times [M^1{L}^1{T}^{-2}]\times [M^1{L}^1{T}^{-1}]$

$[abc]=[M^{(1+1+1)}{L}^{(1+1+1)}{T}^{(-3-2-1)}]$

$[abc]=[M^3{L}^3{T}^{-6}]$

The problem states that the dimension of $abc$ is $M^x{L}^y{T}^z$.

By comparing the exponents, we get:

$x=3$

$y=3$

$z=-6$

Finally, we need to calculate $x+y-z$:

$x+y-z=3+3-(-6)$

$x+y-z=6+6$

$x+y-z=12$

The final answer is $\boxed{12}$.