A) \[2\sqrt{2}\,m\]
B) \[1\,m\]
C) 2 m
D) None of these
Correct Answer: C
Solution :
Applying Bernoulli's principle between the points 1 and 2. \[\rho g{{h}_{1}}+{{p}_{1}}+\frac{1}{2}\rho v_{1}^{2}={{p}_{2}}+\frac{1}{2}\rho v_{2}^{2}+\rho g{{h}_{2}}\] Substituting, \[{{h}_{1}}=h+h'=\,\left( \sqrt{2}+\frac{1}{\sqrt{2}} \right)m\] and since a << A \[\therefore \] \[{{v}_{1}}\simeq 0\] Putting, \[{{v}_{1}}\simeq 0,\,\,{{p}_{2}}={{p}_{0}}+\rho gh,\,\,{{p}_{1}}={{p}_{0,}}\] In Eq. (i), we get \[\rho gh=\frac{1}{2}\rho v_{2}^{2},\,\,{{v}_{2}}=\sqrt{2gh}=v\] Here, \[{{p}_{0}}\] is the atmospheric pressure. \[\therefore \] The range, \[R={{v}_{2}}\times t\] \[\Rightarrow \] \[R={{V}_{2}}\,\sqrt{\frac{2{{h}_{d}}}{g}},\] Putting \[{{v}_{2}}=\sqrt{2gh},\] we obtain \[R=2\,\sqrt{hh'}\] Putting, \[h'\,=\frac{1}{\sqrt{2}}m\] and \[h=\sqrt{2}\,m\]. We obtain \[R=2\,m\]You need to login to perform this action.
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