The horse power transmitted through a pipe is maximum when the ratio of loss of head due to friction and total head supplied is A. $$\frac{1}{3}$$ B. $$\frac{1}{4}$$ C. $$\frac{1}{2}$$ D. $$\frac{2}{3}$$

$$ rac{1}{3}$$
$$ rac{1}{4}$$
$$ rac{1}{2}$$
$$ rac{2}{3}$$

The correct answer is $\boxed{\frac{1}{2}}$.

The head loss due to friction is given by the following equation:

$$h_f = \frac{L}{D} \cdot \frac{v^2}{2g}$$

where:

  • $h_f$ is the head loss due to friction
  • $L$ is the length of the pipe
  • $D$ is the diameter of the pipe
  • $v$ is the velocity of the fluid
  • $g$ is the acceleration due to gravity

The total head supplied is given by the following equation:

$$h_t = h_f + h_g$$

where:

  • $h_t$ is the total head supplied
  • $h_f$ is the head loss due to friction
  • $h_g$ is the head due to gravity

The horsepower transmitted through a pipe is given by the following equation:

$$P = \frac{Q \cdot h_t}{\eta}$$

where:

  • $P$ is the horsepower transmitted through the pipe
  • $Q$ is the flow rate
  • $h_t$ is the total head supplied
  • $\eta$ is the efficiency of the pump

The efficiency of a pump is typically between 0.7 and 0.9.

The horsepower transmitted through a pipe is maximum when the ratio of loss of head due to friction and total head supplied is $\frac{1}{2}$. This is because the head loss due to friction is proportional to the square of the velocity, while the total head supplied is proportional to the velocity. Therefore, when the ratio of loss of head due to friction and total head supplied is $\frac{1}{2}$, the velocity is at a minimum, and the horsepower transmitted through the pipe is at a maximum.

The other options are incorrect because they do not result in the maximum horsepower transmitted through the pipe.