Science M.C.Q. [I MARKS]

The road $Q$ provides the maximum friction, I since the stopping distance for surface $Q$ is the lowest.

Lubricants are thick liquids that make rough surfaces smooth and slippery. The application of lubricants reduces the friction between two surfaces moving relative to each other. Since the use of lubricants between two surfaces reduces the value of $X, X$ must be the force of friction.

Irregularities on surfaces offer resistance to the motion of an object over it. This resistance is called friction. The more the irregularities on the surface, the more is the friction. Hence, more force is required to move an object placed on an irregular surface. Since irregularities present on a grass field are more than those present on a marble floor, more force is required to move a box placed in the grass field. Boxes $I$ and $II$ are placed in a grass field. Therefore, boxes $I$ and $II$ require more forces to be moved than those required by boxes $III$ and $IV$. The opposing force of friction does not remain the same for heavier and lighter objects. Frictional force is greater for heavier objects and smaller for lighter objects. Since box $II$ is heavier than box $I$, box $II$ requires the maximum force to be moved.

The relationship between the three types of friction is:
The static friction > Sliding Friction > Rolling friction Thus the rolling friction will be less than $20\ N.$

In the absence of friction (no opposing motion), a moving object will keep on moving. In this case, the ball will oscillate between points'?' and $'Q'$ due to the action of gravitational force.

Sliding friction is less than the friction that builds up before sliding, i.e., static friction. Hence, sliding friction is less than $7\ N.$

The force of friction acts opposite to the direction of applied force. Since, friction is acting along $P$, so the force must have been applied in the direction of $R.$

The force of friction on a body sliding on a horizontal surface is, $F = m$ mg Given,
$\text{F}=30\text{N},\ \text{mg}=50\text{N}$
$\mu=\frac{\text{F}}{\text{mg}}=\frac{30}{50}=0.6$

The force of friction will be from left to right $(-4)$ because the friction force always acts in opposite direction to the motion.

Sliding friction is always greater than rolling friction. Box $II$ is sliding on the ground because of the force applied by Shyam. Box $I$ is rolling on wheels because of the force applied by Ram. Hence, Shyam has to apply a greater force to maintain a steady motion of his box.

Friction opposes the motion of an object. For surface $O$, force needed to push the box is least i.e., $320N$, so it has least friction.

As the distance travelled is the least in $4.8\ m$, thus friction would have been the largest.

Rolling friction is smaller than sliding friction. In situation $III$, the person is sliding the box on the dry road. Therefore, the magnitude of opposing friction exerted on it will be the maximum as compared to the other situations. In situation $IV$, the person is rolling a box on an oily road with the help of a roller. Since the application of lubricants reduces friction, the magnitude of friction in this case will be the least as compared to the other three cases. Hence, the maximum and the minimum friction acting on the boxes are for the cases $III$ and $IV$ respectively.