A) \[\frac{1}{{{2}^{(m+n)}}}\]
B) \[(m+n)\]
C) \[(n-m)\]
D) \[{{2}^{(n-m)}}\]
Correct Answer: D
Solution :
\[R=K{{(A)}^{n}}{{(B)}^{m}}\] \[R'=K{{(2A)}^{n}}{{\left( \frac{B}{2} \right)}^{m}}\]\[=K{{(A)}^{n}}{{2}^{n}}{{(B)}^{m}}{{2}^{-m}}\] \[=K{{(A)}^{n}}{{(B)}^{m}}{{2}^{n-m}}\],\[\frac{R'}{R}=\frac{K{{(A)}^{n}}{{(B)}^{m}}{{2}^{n-m}}}{K{{(A)}^{n}}{{(B)}^{m}}}={{2}^{n-m}}\]
You need to login to perform this action.
You will be redirected in
3 sec
