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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).

Correct Answer: B — √(3gh)

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Q. A solid sphere rolls down a frictionless incline. If it starts from rest, what is its final velocity at the bottom of the incline of height h?

A. √(gh)
B. √(5gh/7)
C. √(2gh)
D. √(3gh)

Solution

Using conservation of energy, the final velocity of the solid sphere at the bottom is v = √(5gh/7).

Correct Answer: D — √(3gh)

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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).

Correct Answer: A — √(2gh)

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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.

Correct Answer: B — 2:3

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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.

Correct Answer: B — 2:3

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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.

Correct Answer: B — 2:3

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Q. A solution contains 15% salt. If 200 grams of the solution is taken, how much salt is present?

A. 30 grams
B. 25 grams
C. 20 grams
D. 15 grams

Solution

15% of 200 grams = 0.15 * 200 = 30 grams of salt.

Correct Answer: A — 30 grams

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Q. A solution contains 20% (w/w) of glucose. If the total mass of the solution is 200 g, what is the mass of glucose in the solution?

A. 20 g
B. 40 g
C. 60 g
D. 80 g

Solution

Mass of glucose = (20/100) x 200 g = 40 g.

Correct Answer: B — 40 g

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Q. A solution contains 20% sugar. If 5 liters of the solution is taken, how much sugar is in that solution?

A. 1 liter
B. 0.5 liters
C. 2 liters
D. 1.5 liters

Solution

Sugar in 5 liters = 20% of 5 = 0.2 * 5 = 1 liter.

Correct Answer: A — 1 liter

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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%.

Correct Answer: A — 10%

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Q. A solution contains 25% alcohol. If 200 ml of the solution is taken, how much alcohol is present in it?

A. 50 ml
B. 25 ml
C. 75 ml
D. 100 ml

Solution

25% of 200 ml = 0.25 * 200 = 50 ml of alcohol.

Correct Answer: A — 50 ml

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Q. A solution contains 25% sugar. If 10 liters of the solution is taken, how much sugar is in that solution?

A. 2.5 liters
B. 3 liters
C. 4 liters
D. 5 liters

Solution

25% of 10 liters = 0.25 * 10 = 2.5 liters of sugar.

Correct Answer: A — 2.5 liters

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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%.

Correct Answer: B — 20%

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Q. A solution contains 25% sugar. If 200 grams of sugar is added to 800 grams of the solution, what is the new percentage of sugar in the solution?

A. 30%
B. 25%
C. 35%
D. 40%

Solution

Initial sugar = 0.25 * 800 = 200 grams. New sugar = 200 + 200 = 400 grams. Total solution = 800 + 200 = 1000 grams. New percentage = (400/1000) * 100 = 40%.

Correct Answer: A — 30%

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Q. A solution contains 25% sugar. If 200 grams of the solution is taken, how much sugar is in it?

A. 25 grams
B. 50 grams
C. 75 grams
D. 100 grams

Solution

25% of 200 grams = 0.25 * 200 = 50 grams of sugar.

Correct Answer: A — 25 grams

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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?

A. 20%
B. 25%
C. 30%
D. 15%

Solution

Initial sugar = 0.25 * 8 = 2 liters. New volume = 8 + 4 = 12 liters. New concentration = (2/12) * 100 = 16.67%.

Correct Answer: A — 20%

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Q. A solution has a concentration of 0.1 M NaCl. How many grams of NaCl are present in 1 liter of this solution? (Molar mass of NaCl = 58.5 g/mol)

A. 5.85 g
B. 58.5 g
C. 0.1 g
D. 0.585 g

Solution

Mass = moles x molar mass = 0.1 moles x 58.5 g/mol = 5.85 g.

Correct Answer: B — 58.5 g

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Q. A solution has a concentration of 0.2 M. How many moles of solute are present in 1.5 L of this solution?

A. 0.3 moles
B. 0.5 moles
C. 0.2 moles
D. 0.15 moles

Solution

Moles of solute = Molarity × Volume = 0.2 M × 1.5 L = 0.3 moles.

Correct Answer: B — 0.5 moles

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Q. A solution has a density of 1.2 g/mL and contains 10% (w/v) NaCl. What is the mass of NaCl in 1 liter of this solution?

A. 100 g
B. 120 g
C. 80 g
D. 60 g

Solution

Mass of NaCl = (10/100) x 1000 mL = 100 g.

Correct Answer: B — 120 g

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Q. A solution has a density of 1.2 g/mL and contains 10% (w/v) NaOH. What is the mass of NaOH in 1 L of this solution?

A. 100 g
B. 120 g
C. 80 g
D. 60 g

Solution

Mass of NaOH = (10/100) x 1000 mL = 100 g.

Correct Answer: B — 120 g

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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.

Correct Answer: B — 1 M

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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?

A. 15%
B. 16%
C. 17%
D. 18%

Solution

Total salt = (3 * 0.2) + (5 * 0.1) = 0.6 + 0.5 = 1.1. Total mixture = 3 + 5 = 8. Percentage = (1.1/8) * 100 = 13.75%.

Correct Answer: A — 15%

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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?

A. 24%
B. 26%
C. 28%
D. 30%

Solution

Total acid = (0.20 * 4) + (0.30 * 6) = 0.8 + 1.8 = 2.6 liters. Total volume = 4 + 6 = 10 liters. Concentration = (2.6/10) * 100 = 26%.

Correct Answer: B — 26%

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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?

A. 15%
B. 16%
C. 17%
D. 18%

Solution

Total acid = (0.1 * 5) + (0.2 * 15) = 0.5 + 3 = 3.5 liters. Total volume = 5 + 15 = 20 liters. Concentration = (3.5/20) * 100 = 17.5%.

Correct Answer: B — 16%

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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%.

Correct Answer: A — 20%

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Q. A solution is prepared by dissolving 58.5 g of NaCl in 1 L of water. 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 = moles of solute / liters of solution = 1 mole / 1 L = 1 M.

Correct Answer: A — 1 M

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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.

Correct Answer: A — 1 M

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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.

Correct Answer: A — 1 M

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Q. A solution of 0.1 molal urea in water has a freezing point depression of how much? (K_f for water = 1.86 °C kg/mol)

A. 0.186 °C
B. 0.372 °C
C. 0.186 K
D. 0.372 K

Solution

Freezing point depression = K_f * m = 1.86 * 0.1 = 0.186 °C

Correct Answer: A — 0.186 °C

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Q. A sound wave has a frequency of 440 Hz. What is its period?

A. 0.00227 s
B. 0.00455 s
C. 0.01 s
D. 0.1 s

Solution

The period (T) is the inverse of frequency (f), T = 1/f = 1/440 ≈ 0.00227 s.

Correct Answer: B — 0.00455 s

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Major Competitive Exams

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