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(A) Let the height of the liquid in the tank be $H = 5 \,m$. Let the height of the bottom of the tank from the ground be $h_0 = 5 \,m$. Let the hole b

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(A) Let the height of the liquid in the tank be $H = 5 \,m$. Let the height of the bottom of the tank from the ground be $h_0 = 5 \,m$. Let the hole be made at a depth $y$ from the free surface of the liquid. The height of the hole from the ground is $h = h_0 + (H - y) = 5 + 5 - y = 10 - y$. The velocity of efflux is $v = \sqrt{2gy}$. The time taken for the liquid to reach the ground is $t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2(10-y)}{g}}$. The horizontal range $R$ is given by $R = v \cdot t = \sqrt{2gy} \cdot \sqrt{\frac{2(10-y)}{g}} = 2\sqrt{y(10-y)}$. To maximize $R$, we maximize $f(y) = y(10-y) = 10y - y^2$. Taking the derivative with respect to $y$ and setting it to zero: $\frac{df}{dy} = 10 - 2y = 0$, which gives $y = 5 \,m$. Substituting $y = 5 \,m$ into the range formula: $R_{max} = 2\sqrt{5(10-5)} = 2\sqrt{25} = 2 	imes 5 = 10 \,m$.

$A$ liquid is filled in a tank up to a height of $5 \,m$ and the bottom of the tank is at $5 \,m$ height from the ground. $A$ hole is punched in the tank so that the liquid coming out touches the ground at maximum horizontal distance. Then, the maximum horizontal distance is (in $\,m$)

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