question_answer

A small hole is made at a height of \[h'\left( =\frac{1}{\sqrt{2}} \right)m,\] from the bottom of a cylindrical water tank and at a depth of \[h=\sqrt{2}\,m,\]from the upper level of water in the tank. The distance where the water emerging from the hole strikes the ground is

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 \[R=2\,m\]