Q. A cyclist is negotiating a circular track of radius 30 m. If the cyclist's speed is 15 m/s, what is the net force acting on the cyclist if the mass of the cyclist is 60 kg?
A.180 N
B.120 N
C.90 N
D.60 N
Solution
Centripetal force F_c = mv²/r = 60 kg * (15 m/s)² / 30 m = 180 N.
Q. A cyclist is negotiating a circular track of radius 30 m. If the cyclist's speed is 15 m/s, what is the net force acting on the cyclist towards the center of the track?
A.50 N
B.75 N
C.100 N
D.125 N
Solution
Centripetal force (F_c) = mv²/r. Assuming mass m = 100 kg, F_c = (100 kg)(15 m/s)² / (30 m) = 75 N.
Q. A cyclist is negotiating a circular turn of radius 30 m at a speed of 15 m/s. What is the minimum coefficient of friction required to prevent slipping?
A.0.25
B.0.5
C.0.75
D.1
Solution
Frictional force = m * a_c; μmg = mv²/r; μ = v²/(rg) = (15²)/(30*9.8) = 0.25.
Q. A cyclist is pedaling at a constant speed and exerts a power of 200 W. If the cyclist increases their power output to 400 W, what happens to their speed assuming no other forces act?
A.Speed remains the same
B.Speed doubles
C.Speed increases by 41%
D.Speed increases by 100%
Solution
Power is proportional to the cube of the speed in cycling. If power doubles, speed increases by a factor of (2)^(1/3) which is approximately 1.26, or about a 41% increase.
Q. A cylinder rolls down a hill of height h. What is the speed of the center of mass when it reaches the bottom?
A.√(2gh)
B.√(3gh)
C.√(4gh)
D.√(5gh)
Solution
Using conservation of energy, potential energy at the top (mgh) converts to kinetic energy (1/2 mv^2 + 1/2 Iω^2). For a solid cylinder, I = (1/2)mR^2 and ω = v/R. Solving gives v = √(3gh).
Q. A cylinder rolls down a hill. If it has a radius R and rolls without slipping, what is the relationship between its linear velocity v and its angular velocity ω?
A.v = Rω
B.v = 2Rω
C.v = ω/R
D.v = R^2ω
Solution
For rolling without slipping, the relationship is v = Rω.
Q. A cylinder rolls down a hill. If the height of the hill is h, what is the speed of the center of mass of the cylinder at the bottom of the hill?
A.√(gh)
B.√(2gh)
C.√(3gh)
D.√(4gh)
Solution
Using conservation of energy, potential energy at the top (mgh) converts to kinetic energy (1/2 mv^2 + 1/2 Iω^2). For a solid cylinder, I = 1/2 mr^2, leading to v = √(2gh).
Q. A cylindrical conductor of radius R carries a uniform charge per unit length λ. What is the electric field at a distance r from the axis of the cylinder (r > R)?
A.0
B.λ/(2πε₀r)
C.λ/(2πε₀R)
D.λ/(4πε₀r²)
Solution
For a point outside the cylinder, the electric field is given by E = λ/(2πε₀r).
Q. A cylindrical Gaussian surface encloses a charge Q. If the height of the cylinder is doubled while keeping the radius constant, what happens to the electric flux through the curved surface?
A.It doubles
B.It halves
C.It remains the same
D.It becomes zero
Solution
The electric flux through the curved surface is proportional to the charge enclosed, which remains constant, so the flux through the curved surface doubles if the height is doubled.
Q. A cylindrical Gaussian surface encloses a charge Q. If the radius of the cylinder is r and its height is h, what is the electric flux through the curved surface?
A.Q/ε₀
B.Q/(2ε₀)
C.Q/(4ε₀)
D.0
Solution
The electric flux through the curved surface of a cylinder is given by Φ = Q_enc/ε₀, where Q_enc = Q.
Q. A cylindrical Gaussian surface of length L and radius R encloses a charge Q uniformly distributed along its length. What is the electric field at a distance R from the axis of the cylinder?
A.Q/(2πε₀R)
B.Q/(4πε₀R²)
C.0
D.Q/(ε₀L)
Solution
Using Gauss's law, the electric field outside the cylinder is E = Q/(2πε₀R).
Q. A cylindrical Gaussian surface of length L and radius R encloses a charge Q. What is the electric field E at a distance R from the axis of the cylinder?
A.Q/(2πε₀R)
B.Q/(4πε₀R²)
C.Q/(ε₀L)
D.0
Solution
Using Gauss's law, the electric field E at a distance R from the axis of a long charged cylinder is E = Q/(2πε₀L) for points outside the cylinder.
Q. A cylindrical rod is subjected to a tensile force. If the diameter of the rod is doubled while keeping the length constant, what happens to the stress in the rod?
A.Increases
B.Decreases
C.Remains the same
D.Becomes zero
Solution
Stress is defined as force per unit area. Doubling the diameter increases the area by a factor of four, thus reducing the stress.
Q. A cylindrical rod is subjected to a tensile force. If the radius of the rod is halved while keeping the length constant, how does the tensile stress change?
A.It doubles
B.It halves
C.It quadruples
D.It remains the same
Solution
Tensile stress is given by force/area. Halving the radius reduces the area by a factor of four, thus the stress quadruples for the same force.
