የ12ኛ ክፍል ፊዚክስ መግቢያ ፈተና፣ 2017 ዓ.ም. (2025)፣ እትም 1

Vectors add head to tail. Once B starts where A ends and C starts where B ends, the single vector that replaces all three runs from the tail of the very first one to the tip of the very last one. The first is A and the last is C, so the sum is drawn from the tail of A to the tip of C. Option A is a real resultant, but only of $A + B$. It stops before C is counted.

The question asks which statement is wrong. Displacement is the straight line from the starting point to the finishing point, while distance is the whole path walked. A straight line between two points cannot be longer than any other route between them, so the magnitude of displacement is always less than or equal to the distance. They are equal only when the motion is in one straight line with no turning back. Statement D claims displacement can be more than the distance, so D is the wrong statement and the answer. Statement B sounds loose, but calling displacement the shortest path between the two positions is the accepted description, so it is not the one at fault.

The block presses down with its weight, so the normal force is $N = mg = 25 \times 10 = 250$ N. The 75 N is the force needed to break it loose, which measures static friction: $\mu_s = 75/250 = 0.3$. The 60 N keeps it sliding at constant speed, which measures kinetic friction: $\mu_k = 60/250 = 0.24$, and the paper rounds this to 0.2. So the pair is 0.3 and 0.2. Option C has the same two numbers in the opposite order. Kinetic friction is always the smaller of the two, so the bigger coefficient has to be the static one.

This asks for the statement that is not true. Newton's third law says the action and the reaction act on two different bodies. Two forces can only cancel when they pull on the same body, so an action and reaction pair never cancels. That makes A the false statement and the answer. Option B is tempting to tick because it also sounds like something you could argue with, but it is true, and it is the very reason A is false.

Use $F = ma$ twice with the same mass. At first $F = ma$. With the force tripled, $3F = ma'$, so $a' = 3a$. The question says the acceleration went up by 6, so $a' = a + 6$. Setting the two equal: $3a = a + 6$, which gives $a = 3$ m/s². The new acceleration is $a' = 3 \times 3 = 9$ m/s². Option A, 6 m/s², is the amount the acceleration increased by, not the new value itself.

Vectors that act parallel to each other add like ordinary numbers. Let B be the larger one. Then $A = B/2$, and $A + B = 90$ gives $\tfrac{B}{2} + B = 90$, so $1.5B = 90$ and $B = 60$ units. Then $A = 30$ units. Option A carries the same two numbers but attaches them the wrong way round. The question says A is the half, so A has to be the smaller value.