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Q. A 1 kg block of metal at 100°C is placed in 2 kg of water at 20°C. If the final temperature is 30°C, what is the specific heat capacity of the metal? (2023)

A. 0.5 J/kg°C
B. 1 J/kg°C
C. 2 J/kg°C
D. 4 J/kg°C

Solution

Using conservation of energy: m1c1(T1 - Tf) = m2c2(Tf - T2), we find c1 = 2 J/kg°C.

Correct Answer: C — 2 J/kg°C

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Q. A 10 kg mass is lifted to a height of 5 m. What is the potential energy gained? (Take g = 10 m/s²)

A. 50 J
B. 100 J
C. 150 J
D. 200 J

Solution

Potential energy PE = mgh = 10 kg * 10 m/s² * 5 m = 500 J.

Correct Answer: B — 100 J

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Q. A 10 kg object is dropped from a height of 5 m. What is its speed just before it hits the ground? (Use g = 9.8 m/s²) (2021)

A. 10 m/s
B. 15 m/s
C. 20 m/s
D. 25 m/s

Solution

Using energy conservation, Potential Energy = Kinetic Energy at the ground. mgh = (1/2)mv². v = sqrt(2gh) = sqrt(2 × 9.8 m/s² × 5 m) ≈ 10 m/s.

Correct Answer: B — 15 m/s

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Q. A 10 kg object is lifted to a height of 5 m. What is the potential energy gained? (Take g = 9.8 m/s²) (2021)

A. 490 J
B. 980 J
C. 390 J
D. 590 J

Solution

Potential Energy (PE) = mgh = 10 kg × 9.8 m/s² × 5 m = 490 J.

Correct Answer: B — 980 J

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Q. A 10 kg object is lifted to a height of 5 m. What is the potential energy gained? (Take g = 10 m/s²)

A. 50 J
B. 100 J
C. 150 J
D. 200 J

Solution

Potential energy PE = mgh = 10 kg * 10 m/s² * 5 m = 500 J.

Correct Answer: D — 200 J

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Q. A 10 kg object is lifted to a height of 5 m. What is the potential energy gained by the object? (Take g = 10 m/s²)

A. 50 J
B. 100 J
C. 150 J
D. 200 J

Solution

Potential energy PE = mgh = 10 kg × 10 m/s² × 5 m = 500 J.

Correct Answer: B — 100 J

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Q. A 10 kg object is lifted to a height of 5 m. What is the work done against gravity? (Take g = 10 m/s²)

A. 50 J
B. 100 J
C. 150 J
D. 200 J

Solution

Work done W = mgh = 10 kg × 10 m/s² × 5 m = 500 J.

Correct Answer: B — 100 J

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Q. A 10 kg object is moving with a speed of 2 m/s. What is its momentum?

A. 20 kg·m/s
B. 10 kg·m/s
C. 5 kg·m/s
D. 15 kg·m/s

Solution

Momentum (p) = mass × velocity = 10 kg × 2 m/s = 20 kg·m/s.

Correct Answer: A — 20 kg·m/s

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Q. A 10 kg object is moving with a speed of 5 m/s. What is its momentum? (2022)

A. 50 kg m/s
B. 10 kg m/s
C. 25 kg m/s
D. 15 kg m/s

Solution

Momentum = mass × velocity = 10 kg × 5 m/s = 50 kg m/s.

Correct Answer: A — 50 kg m/s

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Q. A 10 kg object is moving with a velocity of 5 m/s. What is its kinetic energy?

A. 125 J
B. 50 J
C. 100 J
D. 75 J

Solution

Kinetic energy KE = (1/2)mv² = (1/2)(10)(5²) = 125 J.

Correct Answer: A — 125 J

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Q. A 100 g piece of metal at 150°C is placed in 200 g of water at 25°C. What is the final temperature of the system assuming no heat loss? (2019)

A. 30°C
B. 40°C
C. 50°C
D. 60°C

Solution

Using the principle of conservation of energy, we can calculate the final temperature to be 50°C.

Correct Answer: C — 50°C

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Q. A 2 kg object is dropped from a height of 10 m. What is its speed just before it hits the ground? (Use g = 9.8 m/s²) (2021)

A. 14 m/s
B. 20 m/s
C. 10 m/s
D. 15 m/s

Solution

Using energy conservation, PE = KE: mgh = 0.5mv². Solving gives v = √(2gh) = √(2 × 9.8 m/s² × 10 m) = 14 m/s.

Correct Answer: A — 14 m/s

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Q. A 2 kg object is heated from 20°C to 80°C. If the specific heat capacity is 0.5 J/g°C, how much heat is absorbed? (2019)

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

Solution

Heat absorbed (Q) = mcΔT = (2000 g)(0.5 J/g°C)(60°C) = 60,000 J = 60 kJ.

Correct Answer: B — 80 kJ

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Q. A 2 kg piece of copper (specific heat = 0.385 J/g°C) is heated from 25°C to 75°C. How much heat is absorbed? (2022)

A. 2000 J
B. 3000 J
C. 4000 J
D. 5000 J

Solution

Q = mcΔT = 2000 g * 0.385 J/g°C * (75°C - 25°C) = 38500 J.

Correct Answer: B — 3000 J

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Q. A 2 kg piece of iron at 150°C is placed in 1 kg of water at 20°C. What will be the final temperature of the system? (Specific heat of iron = 0.45 J/g°C, water = 4.18 J/g°C) (2023)

A. 25°C
B. 30°C
C. 35°C
D. 40°C

Solution

Using heat transfer equations, we find the final temperature to be approximately 35°C.

Correct Answer: C — 35°C

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Q. A 5 kg block of metal is heated from 20°C to 100°C. If the specific heat capacity of the metal is 0.5 J/g°C, how much heat is required? (2023)

A. 2000 J
B. 2500 J
C. 3000 J
D. 3500 J

Solution

Heat (Q) = m * c * ΔT = 5000 g * 0.5 J/g°C * (100°C - 20°C) = 5000 * 0.5 * 80 = 200000 J = 2000 J.

Correct Answer: C — 3000 J

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Q. A 5 kg object is dropped from a height of 10 m. What is its potential energy at the top? (Take g = 9.8 m/s²) (2020)

A. 49 J
B. 98 J
C. 39 J
D. 59 J

Solution

Potential Energy (PE) = mgh = 5 kg × 9.8 m/s² × 10 m = 490 J.

Correct Answer: B — 98 J

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Q. A 5 kg object is dropped from a height of 10 m. What is its speed just before it hits the ground? (Use g = 9.8 m/s²) (2021)

A. 14 m/s
B. 20 m/s
C. 10 m/s
D. 15 m/s

Solution

Using energy conservation, Potential Energy = Kinetic Energy. mgh = 1/2 mv². v = √(2gh) = √(2 × 9.8 m/s² × 10 m) = 14 m/s.

Correct Answer: A — 14 m/s

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Q. A 5 kg object is dropped from a height of 20 m. What is its velocity just before it hits the ground? (Take g = 10 m/s²)

A. 10 m/s
B. 20 m/s
C. 15 m/s
D. 25 m/s

Solution

Using the formula v² = u² + 2gh, where u = 0, we have v = √(0 + 2*10*20) = √400 = 20 m/s.

Correct Answer: B — 20 m/s

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Q. A 5 kg object is lifted to a height of 10 m. What is the potential energy gained by the object? (Take g = 10 m/s²)

A. 50 J
B. 100 J
C. 150 J
D. 200 J

Solution

Potential energy PE = mgh = 5 kg * 10 m/s² * 10 m = 500 J.

Correct Answer: B — 100 J

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Q. A 50 kg object is moving with a velocity of 10 m/s. What is its kinetic energy? (2022)

A. 2500 J
B. 500 J
C. 1000 J
D. 2000 J

Solution

Kinetic energy KE = (1/2)mv² = (1/2)(50)(10²) = 2500 J.

Correct Answer: A — 2500 J

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Q. A ball is dropped from a height of 80 m. How long will it take to reach the ground? (Take g = 10 m/s²) (2021)

A. 4 s
B. 5 s
C. 6 s
D. 8 s

Solution

Using the formula h = (1/2)gt², we have 80 = (1/2)(10)t². Solving gives t² = 16, so t = 4 s.

Correct Answer: B — 5 s

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Q. A ball is thrown upwards with a speed of 15 m/s. How long will it take to reach the highest point? (Take g = 10 m/s²) (2023)

A. 1.5 s
B. 2 s
C. 3 s
D. 4 s

Solution

Time to reach highest point = u/g = 15/10 = 1.5 s.

Correct Answer: B — 2 s

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Q. A beam of light passes from air into a medium with a refractive index of 1.5. If the angle of incidence is 30 degrees, what is the angle of refraction? (2022)

A. 19.1 degrees
B. 20 degrees
C. 25 degrees
D. 30 degrees

Solution

Using Snell's law, n1 * sin(i) = n2 * sin(r). Thus, sin(r) = (1 * sin(30)) / 1.5 = 0.333, giving r ≈ 19.1 degrees.

Correct Answer: A — 19.1 degrees

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Q. A beam of light passes through a prism with an angle of 60 degrees. If the refractive index of the prism is 1.5, what is the angle of deviation? (2023)

A. 30 degrees
B. 60 degrees
C. 45 degrees
D. 15 degrees

Solution

Using the formula for angle of deviation: D = A(n-1), we find D = 60(1.5-1) = 30 degrees.

Correct Answer: A — 30 degrees

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Q. A block of mass 2 kg is placed on a frictionless surface and is pushed with a force of 10 N. What is the acceleration of the block?

A. 2 m/s²
B. 5 m/s²
C. 10 m/s²
D. 20 m/s²

Solution

Using Newton's second law, F = ma, we have a = F/m = 10 N / 2 kg = 5 m/s².

Correct Answer: B — 5 m/s²

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Q. A block of mass 5 kg is placed on a frictionless surface and is pushed with a force of 20 N. What is the acceleration of the block?

A. 2 m/s²
B. 4 m/s²
C. 5 m/s²
D. 10 m/s²

Solution

Using Newton's second law, F = ma, we have a = F/m = 20 N / 5 kg = 4 m/s².

Correct Answer: B — 4 m/s²

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Q. A boat can travel at a speed of 10 km/h in still water. If it goes downstream, its speed increases by 2 km/h. What is its speed downstream? (2021)

A. 10 km/h
B. 12 km/h
C. 14 km/h
D. 16 km/h

Solution

Speed downstream = Speed in still water + Speed increase = 10 km/h + 2 km/h = 12 km/h.

Correct Answer: B — 12 km/h

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Q. A body is thrown vertically upwards with a speed of 20 m/s. What is the maximum height it reaches? (g = 10 m/s²)

A. 20 m
B. 40 m
C. 30 m
D. 10 m

Solution

Using the formula h = (u^2)/(2g), where u = 20 m/s and g = 10 m/s², we get h = (20^2)/(2*10) = 20 m.

Correct Answer: B — 40 m

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Q. A body is thrown vertically upwards with a speed of 20 m/s. What is the maximum height reached by the body? (Take g = 10 m/s²)

A. 20 m
B. 40 m
C. 30 m
D. 50 m

Solution

Using the formula h = (u²)/(2g), where u = 20 m/s and g = 10 m/s², we get h = (20²)/(2*10) = 400/20 = 20 m.

Correct Answer: B — 40 m

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