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Q. An object is placed 10 cm in front of a convex mirror with a focal length of 5 cm. What is the nature of the image formed?

A. Real and inverted
B. Virtual and erect
C. Real and erect
D. Virtual and inverted

Solution

Convex mirrors always produce virtual and erect images.

Correct Answer: B — Virtual and erect

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Q. An object is placed 15 cm in front of a convex lens of focal length 10 cm. What is the position of the image formed? (2020)

A. 5 cm
B. 10 cm
C. 30 cm
D. 15 cm

Solution

Using the lens formula: 1/f = 1/v - 1/u, where f = 10 cm and u = -15 cm, we find v = 30 cm.

Correct Answer: C — 30 cm

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Q. An object is placed 25 cm from a convex lens of focal length 10 cm. Where is the image formed?

A. 10 cm
B. 15 cm
C. 20 cm
D. 30 cm

Solution

Using the lens formula 1/f = 1/v - 1/u, we find v = 20 cm.

Correct Answer: C — 20 cm

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Q. An object is placed 30 cm from a concave lens of focal length 15 cm. What is the nature of the image formed?

A. Real and inverted
B. Virtual and erect
C. Real and erect
D. Virtual and inverted

Solution

For a concave lens, the image formed is virtual and erect when the object is placed in front of it.

Correct Answer: B — Virtual and erect

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Q. An object is placed 30 cm in front of a convex lens of focal length 15 cm. What is the position of the image formed? (2020)

A. 10 cm
B. 15 cm
C. 20 cm
D. 25 cm

Solution

Using the lens formula 1/f = 1/v - 1/u, where u = -30 cm and f = 15 cm, we find v = 10 cm.

Correct Answer: D — 25 cm

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Q. An object is placed 30 cm in front of a convex lens of focal length 15 cm. What is the image distance?

A. 10 cm
B. 15 cm
C. 20 cm
D. 25 cm

Solution

Using the lens formula 1/f = 1/v - 1/u, where u = -30 cm and f = 15 cm, we find v = 10 cm.

Correct Answer: A — 10 cm

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Q. An object is placed 40 cm from a convex lens with a focal length of 10 cm. Where is the image formed?

A. At 10 cm
B. At 20 cm
C. At 30 cm
D. At 40 cm

Solution

Using the lens formula, the image is formed at 30 cm on the opposite side.

Correct Answer: C — At 30 cm

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Q. An object is placed 50 cm from a converging lens of focal length 25 cm. Where will the image be formed?

A. 16.67 cm
B. 33.33 cm
C. 25 cm
D. 20 cm

Solution

Using the lens formula 1/f = 1/v - 1/u, we find v = 33.33 cm.

Correct Answer: B — 33.33 cm

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Q. An object is placed at a distance of 15 cm from a convex lens of focal length 10 cm. Where is the image formed?

A. 5 cm
B. 10 cm
C. 15 cm
D. 20 cm

Solution

Using the lens formula, 1/f = 1/v - 1/u, we find v = 30 cm, meaning the image is formed 30 cm on the opposite side.

Correct Answer: D — 20 cm

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Q. An object is placed at a distance of 30 cm from a convex lens of focal length 15 cm. What is the distance of the image from the lens?

A. 10 cm
B. 15 cm
C. 20 cm
D. 30 cm

Solution

Using the lens formula, 1/f = 1/v - 1/u. Here, f = 15 cm, u = -30 cm. Solving gives v = 10 cm.

Correct Answer: C — 20 cm

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Q. An object is placed at a distance of 40 cm from a convex lens of focal length 20 cm. Where will the image be formed?

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

Solution

Using the lens formula 1/f = 1/v - 1/u, we find v = 10 cm.

Correct Answer: A — 10 cm

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Q. An object is placed at the focus of a concave lens. What type of image is formed?

A. Real and inverted
B. Virtual and erect
C. Real and erect
D. Virtual and inverted

Solution

A concave lens always forms a virtual and erect image regardless of the object position.

Correct Answer: B — Virtual and erect

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Q. An object is placed at the focus of a concave mirror. What type of image is formed?

A. Real and inverted
B. Virtual and upright
C. No image
D. Real and upright

Solution

When the object is at the focus of a concave mirror, the rays are parallel after reflection, resulting in no image being formed.

Correct Answer: C — No image

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Q. An object is projected at an angle of 60 degrees with an initial speed of 30 m/s. What is the horizontal component of its velocity?

A. 15 m/s
B. 25 m/s
C. 30 m/s
D. 20 m/s

Solution

Horizontal component (v_x) = u * cos(θ) = 30 * (1/2) = 15 m/s.

Correct Answer: B — 25 m/s

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Q. An object is projected at an angle of 60 degrees with an initial speed of 30 m/s. What is the vertical component of its velocity?

A. 15 m/s
B. 25 m/s
C. 30 m/s
D. 20 m/s

Solution

Vertical component (v_y) = u * sin(θ) = 30 * (√3/2) = 25.98 m/s.

Correct Answer: B — 25 m/s

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Q. An object is projected at an angle of 60 degrees with an initial velocity of 30 m/s. What is the horizontal component of its velocity?

A. 15 m/s
B. 25 m/s
C. 30 m/s
D. 20 m/s

Solution

Horizontal component Vx = u * cos(θ) = 30 * (1/2) = 15 m/s.

Correct Answer: B — 25 m/s

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Q. An object is projected at an angle of 60 degrees with an initial velocity of 30 m/s. What is the time of flight?

A. 3 s
B. 5 s
C. 6 s
D. 10 s

Solution

Time of flight (T) = (2u * sin(θ)) / g = (2 * 30 * √3/2) / 9.8 = 5.18 s.

Correct Answer: C — 6 s

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Q. An object is thrown at an angle of 30 degrees with a speed of 40 m/s. What is the time of flight until it returns to the same vertical level?

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

Solution

Time of flight (T) = (2 * u * sin(θ)) / g = (2 * 40 * 0.5) / 9.8 ≈ 4.08 s.

Correct Answer: C — 6 s

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Q. An object is thrown horizontally from a height of 45 m. How long will it take to hit the ground? (Take g = 10 m/s²) (2022)

A. 3 s
B. 4 s
C. 5 s
D. 6 s

Solution

Using h = (1/2)gt², 45 = (1/2)(10)t². Solving gives t² = 9, so t = 3 s.

Correct Answer: B — 4 s

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Q. An object is thrown horizontally from the top of a 45 m high cliff. How far from the base of the cliff will it land if the initial speed is 10 m/s?

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

Solution

Time of flight (t) = √(2h/g) = √(2*45/9.8) = 3.03 s. Horizontal distance = speed * time = 10 * 3.03 = 30.3 m.

Correct Answer: B — 20 m

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Q. An object is thrown horizontally from the top of a 45 m high cliff. How far from the base of the cliff will it land if it is thrown with a speed of 10 m/s?

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

Solution

Time of flight (t) = √(2h/g) = √(2*45/9.8) = 3.03 s. Horizontal distance = speed * time = 10 * 3.03 = 30.3 m.

Correct Answer: B — 30 m

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Q. An object is thrown horizontally from the top of a cliff 80 m high. How long does it take to hit the ground?

A. 2 s
B. 4 s
C. 5 s
D. 8 s

Solution

Using h = (1/2)gt^2, we have 80 = (1/2)(9.8)t^2, solving gives t ≈ 4 s.

Correct Answer: B — 4 s

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Q. An object is thrown horizontally from the top of a cliff of height 80 m. How long does it take to hit the ground?

A. 2 s
B. 4 s
C. 5 s
D. 8 s

Solution

Using h = (1/2)gt^2, we have 80 = (1/2)(9.8)t^2, solving gives t ≈ 4 s.

Correct Answer: B — 4 s

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Q. An object is thrown horizontally from the top of a tower 80 m high. How long will it take to hit the ground?

A. 4 s
B. 5 s
C. 3 s
D. 2 s

Solution

Using the equation h = (1/2)gt², where h = 80 m and g = 9.8 m/s², we have 80 = (1/2)*9.8*t². Solving gives t² = 16.33, so t ≈ 4 s.

Correct Answer: B — 5 s

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Q. An object is thrown upwards with a speed of 20 m/s. How long will it take to reach the maximum height? (Assume g = 10 m/s²)

A. 1 s
B. 2 s
C. 3 s
D. 4 s

Solution

Time to reach maximum height = initial velocity / g = 20 m/s / 10 m/s² = 2 s.

Correct Answer: B — 2 s

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Q. An object is thrown upwards with a velocity of 20 m/s. How high will it go before coming to rest? (g = 10 m/s²)

A. 20 m
B. 40 m
C. 50 m
D. 80 m

Solution

Using the formula h = v²/(2g) = (20 m/s)² / (2 * 10 m/s²) = 20 m.

Correct Answer: B — 40 m

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Q. An object is thrown vertically upward with a speed of 20 m/s. How high will it rise before coming to rest?

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

Solution

Maximum height (H) = (u²)/(2g) = (20²)/(2*9.8) ≈ 20.4 m.

Correct Answer: C — 30 m

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Q. An object is thrown vertically upward with a speed of 30 m/s. How high will it rise before coming to rest?

A. 45 m
B. 90 m
C. 135 m
D. 180 m

Solution

Using the formula: h = (v² - u²) / (2g), where v = 0 m/s, u = 30 m/s, g = 9.8 m/s². h = (0 - 30²) / (2 * -9.8) = 45.92 m, approximately 45 m.

Correct Answer: B — 90 m

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Q. An object is thrown vertically upward with a speed of 30 m/s. How high will it rise before coming to a momentary stop?

A. 45 m
B. 60 m
C. 90 m
D. 135 m

Solution

Using the formula: h = (v² - u²) / (2g), where v = 0 m/s (at the highest point), u = 30 m/s, g = 9.8 m/s². h = (0 - 30²) / (2 * -9.8) = 45.92 m, approximately 45 m.

Correct Answer: B — 60 m

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Q. An object is thrown vertically upwards with a speed of 15 m/s. What is the maximum height it reaches? (g = 10 m/s²)

A. 11.25 m
B. 22.5 m
C. 15 m
D. 20 m

Solution

Maximum Height (h) = v^2 / (2g) = (15^2) / (2 * 10) = 11.25 m

Correct Answer: B — 22.5 m

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Major Competitive Exams

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