Concrete mixture is made by mixing cement, stone and sand in a rotating cylindrical drum. If the drum rotates too fast, the ingredients remain stuck to the wall of the drum and proper mixing of ingredients does not take place. The maximum rotational speed of the drum in revolutions per minute (rpm) to ensure proper mixing is close to:
(Take the radius of the drum to be 1.25 m and its axle to be horizontal):

To solve this problem, we need to find the maximum rotational speed of a cylindrical drum in revolutions per minute (rpm) that ensures proper mixing of cement, stone, and sand. The drum has a radius of 1.25 m and rotates horizontally. If the drum rotates too fast, the ingredients stick to the wall due to centrifugal force, preventing proper mixing. The key is to determine the speed at which the ingredients just begin to lose contact with the wall at the highest point of rotation, allowing them to fall and mix.

Consider a small particle of the mixture stuck to the inner wall of the drum. When the drum rotates, this particle experiences two main forces: gravitational force pulling it downward and centrifugal force pushing it outward against the wall. At the highest point of rotation, the gravitational force acts toward the center of the drum, while the centrifugal force acts outward. For the particle to lose contact with the wall and fall (enabling mixing), the centrifugal force must exactly balance the gravitational force. At this point, the normal force exerted by the wall on the particle becomes zero.

The gravitational force is given by $$ mg $$, where $$ m $$ is the mass of the particle and $$ g $$ is the acceleration due to gravity (approximately $$ 9.8 \, \text{m/s}^2 $$). The centrifugal force is $$ m \omega^2 r $$, where $$ \omega $$ is the angular velocity in radians per second and $$ r $$ is the radius of the drum (1.25 m). At the highest point, setting these equal gives:

$$ mg = m \omega^2 r $$

We can divide both sides by $$ m $$ (since mass is not zero) to simplify:

$$ g = \omega^2 r $$

Solving for $$ \omega^2 $$:

$$ \omega^2 = \frac{g}{r} $$

Taking the square root of both sides to find $$ \omega $$:

$$ \omega = \sqrt{\frac{g}{r}} $$

Substituting the known values $$ g = 9.8 \, \text{m/s}^2 $$ and $$ r = 1.25 \, \text{m} $$:

$$ \omega = \sqrt{\frac{9.8}{1.25}} $$

First, compute the fraction inside the square root:

$$ \frac{9.8}{1.25} = \frac{9.8 \times 100}{1.25 \times 100} = \frac{980}{125} $$

Simplify $$ \frac{980}{125} $$ by dividing numerator and denominator by 5:

$$ \frac{980 \div 5}{125 \div 5} = \frac{196}{25} = 7.84 $$

So:

$$ \omega = \sqrt{7.84} $$

Since $$ 2.8 \times 2.8 = 7.84 $$, we have:

$$ \omega = 2.8 \, \text{rad/s} $$

Now, convert angular velocity from radians per second to revolutions per minute (rpm). One revolution corresponds to $$ 2\pi $$ radians. First, find the rotational frequency in revolutions per second (rps):

$$ f = \frac{\omega}{2\pi} $$

Substituting $$ \omega = 2.8 \, \text{rad/s} $$ and $$ \pi \approx 3.1416 $$:

$$ f = \frac{2.8}{2 \times 3.1416} = \frac{2.8}{6.2832} \approx 0.4456 \, \text{rps} $$

To convert rps to rpm, multiply by 60 (since there are 60 seconds in a minute):

$$ \text{rpm} = 0.4456 \times 60 = 26.736 $$

Rounding 26.736 to the nearest tenth gives approximately 26.7 rpm, which is closest to 27.0 rpm among the given options.

Alternatively, using exact fractions for verification:

$$ \omega = \sqrt{\frac{9.8}{1.25}} = \sqrt{\frac{98/10}{125/100}} = \sqrt{\frac{98}{10} \times \frac{100}{125}} = \sqrt{\frac{98 \times 10}{125}} = \sqrt{\frac{980}{125}} = \sqrt{\frac{196}{25}} = \frac{14}{5} = 2.8 \, \text{rad/s} $$

Then:

$$ f = \frac{2.8}{2\pi} = \frac{7}{5\pi} \, \text{rps} $$

Convert to rpm:

$$ \text{rpm} = \frac{7}{5\pi} \times 60 = \frac{7 \times 60}{5\pi} = \frac{420}{5\pi} = \frac{84}{\pi} $$

Using $$ \pi \approx 3.1416 $$:

$$ \frac{84}{3.1416} \approx 26.75 $$

Which rounds to 27.0 rpm.

This speed is the maximum because if the drum rotates faster, $$ \omega $$ increases, making centrifugal force greater than gravity. This causes the ingredients to stick to the wall, preventing mixing. At or below 27.0 rpm, ingredients will fall and mix properly.

Hence, the correct answer is Option A.