Q. A solid sphere of radius R rolls without slipping down an inclined plane of angle θ. What is the acceleration of the center of mass of the sphere?
A.g sin(θ)
B.g sin(θ)/2
C.g sin(θ)/3
D.g sin(θ)/4
Solution
The acceleration of the center of mass of a solid sphere rolling down an incline is given by a = g sin(θ) / (1 + (2/5)) = g sin(θ) / (7/5) = (5/7) g sin(θ).
Q. A solid sphere of radius R rolls without slipping down an inclined plane of height h. What is the speed of the center of mass at the bottom of the incline? (2021)
A.√(2gh)
B.√(3gh)
C.√(4gh)
D.√(5gh)
Solution
Using conservation of energy, potential energy at height h = kinetic energy at the bottom. For a solid sphere, v = √(3gh).
Q. A solid sphere rolls down a hill without slipping. If the height of the hill is h, what is the speed of the sphere at the bottom of the hill?
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 sphere, I = (2/5)mr^2 and ω = v/r. Solving gives v = √(2gh).
Q. A solid sphere rolls down an inclined plane without slipping. What is the ratio of its translational kinetic energy to its total kinetic energy at the bottom of the incline?
A.1:2
B.2:3
C.1:3
D.1:1
Solution
The total kinetic energy is the sum of translational and rotational kinetic energy. For a solid sphere, the ratio of translational to total kinetic energy is 2:5, which simplifies to 2:3.
Q. A solid sphere rolls down an inclined plane without slipping. What is the ratio of its translational kinetic energy to its total kinetic energy at the bottom?
A.1:2
B.2:3
C.1:3
D.1:1
Solution
The total kinetic energy is the sum of translational and rotational kinetic energy. For a solid sphere, the ratio of translational to total kinetic energy is 2:3.
Q. A solid sphere rolls without slipping down an incline. What is the ratio of its translational kinetic energy to its total kinetic energy at the bottom?
A.1:2
B.2:3
C.1:1
D.1:3
Solution
For a solid sphere, the ratio of translational kinetic energy to total kinetic energy is 2:3.
Q. A solution contains 20% sugar. If 5 liters of this solution is diluted with 10 liters of water, what is the new percentage of sugar in the solution?
A.10%
B.15%
C.20%
D.25%
Solution
Initial sugar = 20% of 5 liters = 1 liter. Total volume after dilution = 5 + 10 = 15 liters. New percentage = (1/15) * 100 = 6.67%.
Q. A solution contains 25% sugar. If 10 liters of water is added to 30 liters of this solution, what is the new concentration of sugar in the solution?
A.15%
B.20%
C.25%
D.30%
Solution
Initial sugar = 0.25 * 30 = 7.5 liters. New total volume = 30 + 10 = 40 liters. New concentration = (7.5/40) * 100 = 18.75%.
Q. A solution contains 25% sugar. If 8 liters of this solution is diluted with 4 liters of water, what is the new concentration of sugar in the solution?
Q. A solution has a density of 1.2 g/mL and contains 30 g of solute. What is the molarity if the molar mass of the solute is 60 g/mol?
A.0.5 M
B.1 M
C.2 M
D.1.5 M
Solution
Volume of solution = mass / density = 30 g / 1.2 g/mL = 25 mL = 0.025 L. Moles of solute = 30 g / 60 g/mol = 0.5 moles. Molarity = 0.5 moles / 0.025 L = 20 M.
Q. A solution is made by mixing 3 parts of solution A and 5 parts of solution B. If solution A contains 20% salt and solution B contains 10% salt, what is the percentage of salt in the final mixture?
Q. A solution is made by mixing 4 liters of a 20% acid solution with 6 liters of a 30% acid solution. What is the concentration of acid in the final mixture?
Q. A solution is made by mixing 5 liters of a 10% acid solution with 15 liters of a 20% acid solution. What is the concentration of acid in the final mixture?
Q. A solution is prepared by dissolving 50 g of glucose (C6H12O6) in 250 g of water. What is the mass percent of glucose in the solution? (Molar mass of glucose = 180 g/mol)
A.20%
B.15%
C.25%
D.10%
Solution
Mass percent = (mass of solute / (mass of solute + mass of solvent)) × 100 = (50 g / (50 g + 250 g)) × 100 = 20%.
Q. A solution is prepared by dissolving 58.5 g of NaCl in 1 L of water. What is the concentration in terms of molarity? (Molar mass of NaCl = 58.5 g/mol)
A.1 M
B.2 M
C.0.5 M
D.0.25 M
Solution
Moles of NaCl = 58.5 g / 58.5 g/mol = 1 mole. Molarity = 1 mole / 1 L = 1 M.
Q. A solution is prepared by dissolving 58.5 g of NaCl in enough water to make 1 L of solution. What is the molarity of the solution? (Molar mass of NaCl = 58.5 g/mol)
A.1 M
B.2 M
C.0.5 M
D.0.1 M
Solution
Moles of NaCl = 58.5 g / 58.5 g/mol = 1 mole. Molarity = 1 mole / 1 L = 1 M.
