Q. A box contains 5 red, 3 blue, and 2 green balls. If one ball is drawn at random, what is the probability that it is blue given that it is not red?
A.1/2
B.1/4
C.1/3
D.1/5
Solution
The total number of balls that are not red is 5 (3 blue + 2 green). The probability that the ball is blue given it is not red is P(Blue | Not Red) = 3/5.
Q. A box contains 5 red, 3 green, and 2 blue marbles. If a marble is drawn and it is known to be red, what is the probability that it is the first marble drawn?
A.1/5
B.1/3
C.1/2
D.1/10
Solution
The probability of drawing a red marble is independent of the order. Therefore, P(First | Red) = 1/5.
Q. A box contains 5 red, 3 green, and 2 blue marbles. If a marble is drawn at random, what is the probability that it is green given that it is not red?
A.1/2
B.1/3
C.1/4
D.1/5
Solution
The total number of non-red marbles is 5 (3 green + 2 blue). Therefore, P(Green | Not Red) = 3/5.
Q. A box contains 5 red, 3 green, and 2 blue marbles. If one marble is drawn at random, what is the probability that it is green given that it is not red?
A.1/2
B.1/3
C.1/4
D.1/5
Solution
The total number of non-red marbles is 5 (3 green + 2 blue). The probability that the marble is green given that it is not red is P(Green | Not Red) = 3/5.
Q. A box is pushed across a floor with a force of 50 N. If the coefficient of kinetic friction is 0.4, what is the net force acting on the box if the normal force is 100 N?
A.10 N
B.20 N
C.30 N
D.40 N
Solution
Frictional force = μk * N = 0.4 * 100 N = 40 N. Net force = applied force - frictional force = 50 N - 40 N = 10 N.
Q. A box is pushed with a force of 50 N on a surface with a coefficient of kinetic friction of 0.3. If the normal force is 100 N, what is the net force acting on the box?
A.20 N
B.30 N
C.50 N
D.70 N
Solution
Frictional force = μk * N = 0.3 * 100 N = 30 N. Net force = applied force - frictional force = 50 N - 30 N = 20 N.
Q. A box is pushed with a force of 50 N on a surface with a coefficient of kinetic friction of 0.4. What is the acceleration of the box if its mass is 10 kg?
A.1 m/s²
B.2 m/s²
C.3 m/s²
D.4 m/s²
Solution
Net force = applied force - frictional force. Frictional force = μ_k * N = 0.4 * 10 kg * 9.8 m/s² = 39.2 N. Net force = 50 N - 39.2 N = 10.8 N. Acceleration = F/m = 10.8 N / 10 kg = 1.08 m/s², approximately 1 m/s².
Q. A building is 40 m high. From a point on the ground, the angle of elevation to the top of the building is 60 degrees. What is the distance from the point to the base of the building?
A.20√3 m
B.40 m
C.30 m
D.10√3 m
Solution
Using tan(60°) = height/distance, we have distance = height/tan(60°) = 40/√3 = 20√3 m.
Q. A capacitor in an AC circuit has a capacitive reactance of 50 ohms. If the frequency of the AC source is increased, what happens to the capacitive reactance?
A.Increases
B.Decreases
C.Remains the same
D.Becomes infinite
Solution
Capacitive reactance (X_C) is given by X_C = 1/(2πfC). If the frequency (f) increases, X_C decreases.
Q. A capacitor in an AC circuit has a capacitive reactance of 50 ohms. What is the frequency if the capacitance is 10 microfarads?
A.1 kHz
B.10 kHz
C.100 Hz
D.1000 Hz
Solution
Capacitive reactance (X_C) is given by X_C = 1 / (2πfC). Rearranging gives f = 1 / (2πX_CC). Substituting X_C = 50 ohms and C = 10 x 10^-6 F gives f = 318.31 Hz, approximately 1 kHz.
