Physics Syllabus (JEE Main)

Current Electricity Electrostatics Gravitation Kinematics Kinetic Theory of Gases Laws of Motion Magnetism & EMI Modern Physics Optics Oscillations & Waves Properties of Matter Rotational Motion Thermodynamics Units & Measurement Work, Energy & Power

Q. A length is measured as 100.0 m with an uncertainty of ±0.5 m. What is the significant figure of the measurement?

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

Solution

The measurement has 4 significant figures (100.0).

Correct Answer: C — 4

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Q. A length is measured as 15.0 m with an uncertainty of ±0.2 m. What is the total uncertainty if this length is used in a calculation involving addition with another length of 10.0 m (±0.1 m)?

A. 0.3 m
B. 0.2 m
C. 0.1 m
D. 0.4 m

Solution

Total uncertainty = √((0.2)² + (0.1)²) = √(0.04 + 0.01) = √0.05 ≈ 0.224 m.

Correct Answer: A — 0.3 m

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Q. A length is measured as 15.0 m with an uncertainty of ±0.3 m. If this length is used to calculate the area of a rectangle, what is the maximum possible error in the area calculation?

A. 9.0 m²
B. 1.5 m²
C. 0.9 m²
D. 0.45 m²

Solution

Area = length², maximum error = 2 * length * uncertainty = 2 * 15.0 * 0.3 = 9.0 m².

Correct Answer: C — 0.9 m²

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Q. A length is measured as 15.0 m with an uncertainty of ±0.3 m. What is the fractional error in this measurement?

A. 0.02
B. 0.03
C. 0.01
D. 0.005

Solution

Fractional error = (absolute error / measured value) = 0.3 / 15.0 = 0.02.

Correct Answer: B — 0.03

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Q. A length is measured as 15.0 m with an uncertainty of ±0.3 m. What is the total length if two such lengths are added?

A. 29.4 m
B. 30.0 m
C. 30.6 m
D. 31.0 m

Solution

Total length = 15.0 + 15.0 = 30.0 m; Total uncertainty = ±0.3 + ±0.3 = ±0.6 m.

Correct Answer: C — 30.6 m

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Q. A length is measured as 15.0 m with an uncertainty of ±0.5 m. If this length is used to calculate the area of a rectangle, what is the maximum possible error in the area calculation?

A. 15 m²
B. 7.5 m²
C. 3.75 m²
D. 1.5 m²

Solution

Maximum error in area = 2 * length * uncertainty = 2 * 15.0 * 0.5 = 15 m².

Correct Answer: B — 7.5 m²

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Q. A length is measured as 15.0 m with an uncertainty of ±0.5 m. If this length is used to calculate the area of a square, what is the maximum possible error in the area?

A. 3.0 m²
B. 1.5 m²
C. 0.5 m²
D. 2.0 m²

Solution

Area = L², maximum error = 2 * L * ΔL = 2 * 15.0 * 0.5 = 15.0 m².

Correct Answer: A — 3.0 m²

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Q. A length is measured as 150 cm with an error of 3 cm. What is the lower limit of the measurement?

A. 147 cm
B. 150 cm
C. 153 cm
D. 148 cm

Solution

Lower limit = Measured value - Error = 150 - 3 = 147 cm

Correct Answer: A — 147 cm

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Q. A length is measured as 20 cm with an error of 0.5 cm. What is the percentage error?

A. 2.5%
B. 5%
C. 0.5%
D. 1%

Solution

Percentage error = (Absolute error / True value) * 100 = (0.5 / 20) * 100 = 2.5%.

Correct Answer: A — 2.5%

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Q. A length is measured as 50.1 cm with an error of 0.1 cm. What is the minimum possible true length?

A. 50 cm
B. 50.1 cm
C. 50.2 cm
D. 49.9 cm

Solution

Minimum true length = Measured value - Error = 50.1 cm - 0.1 cm = 50 cm.

Correct Answer: D — 49.9 cm

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Q. A lens forms a real image at a distance of 30 cm from the lens when the object is placed at 10 cm. What is the focal length of the lens?

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

Solution

Using the lens formula, we find f = 1/(1/v + 1/u) = 1/(1/30 - 1/10) = 5 cm.

Correct Answer: A — 5 cm

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Q. A lens forms a real image at a distance of 40 cm from the lens when the object is placed at 20 cm. What is the focal length of the lens?

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

Solution

Using the lens formula, 1/f = 1/v - 1/u. Here, v = 40 cm, u = -20 cm. Solving gives f = 10 cm.

Correct Answer: A — 10 cm

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Q. A lens forms a real image of a height 5 cm at a distance of 40 cm from the lens. If the object is placed at 20 cm from the lens, what is the height of the object?

A. 2.5 cm
B. 5 cm
C. 10 cm
D. 20 cm

Solution

Using the magnification formula, m = h'/h = -v/u. Here, h' = 5 cm, v = 40 cm, u = -20 cm. Thus, h = (h' * u) / v = (5 * -20) / 40 = 2.5 cm.

Correct Answer: C — 10 cm

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Q. A lens forms a real image of an object placed 60 cm away from it. If the image distance is 20 cm, what is the focal length of 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, where u = -60 cm and v = 20 cm, we find f = 30 cm.

Correct Answer: D — 30 cm

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Q. A lens forms a real image of an object placed at a distance of 60 cm from it. If the image distance is 15 cm, what is the focal length of the lens?

A. 10 cm
B. 12 cm
C. 20 cm
D. 25 cm

Solution

Using the lens formula 1/f = 1/v - 1/u, we have 1/f = 1/15 - 1/60. Solving gives f = 10 cm.

Correct Answer: A — 10 cm

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Q. A lens forms a real image that is three times the size of the object. If the object is placed 20 cm from the lens, what is the focal length of the lens?

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

Solution

Using magnification m = -v/u = 3, we find v = -60 cm and then use the lens formula to find f = 15 cm.

Correct Answer: B — 15 cm

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Q. A lens forms a real image that is twice the size of the object. If the object is placed 10 cm from the lens, what is the focal length of the lens?

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

Solution

Using the lens formula and magnification, we find the focal length to be 15 cm.

Correct Answer: C — 15 cm

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Q. A lens forms a virtual image at a distance of 10 cm when the object is placed at 5 cm. What type of lens is it?

A. Convex lens
B. Concave lens
C. Bifocal lens
D. Plano-convex lens

Solution

A virtual image is formed by a concave lens when the object is placed within its focal length.

Correct Answer: B — Concave lens

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Q. A lens forms a virtual image at a distance of 12 cm when the object is placed at 8 cm. What is the focal length of the lens?

A. 4 cm
B. 6 cm
C. 8 cm
D. 10 cm

Solution

Using the lens formula, 1/f = 1/v - 1/u, we find f = 4 cm.

Correct Answer: A — 4 cm

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Q. A lens forms a virtual image at a distance of 20 cm from the lens when the object is placed at 10 cm. What is the focal length of the lens?

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

Solution

Using the lens formula 1/f = 1/v - 1/u, where v = -20 cm and u = -10 cm, we find f = 5 cm.

Correct Answer: A — 5 cm

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Q. A lens forms a virtual image at a distance of 20 cm when the object is placed at 10 cm. What is the type of lens?

A. Convex lens
B. Concave lens
C. Bifocal lens
D. Plano-convex lens

Solution

A virtual image formed by a lens indicates that it is a concave lens.

Correct Answer: B — Concave lens

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Q. A lens has a focal length of +20 cm. What type of lens is it?

A. Concave lens
B. Convex lens
C. Bifocal lens
D. Cylindrical lens

Solution

A positive focal length indicates that the lens is a convex lens, which converges light rays.

Correct Answer: B — Convex lens

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Q. A lens has a focal length of -10 cm. What type of lens is it?

A. Convex lens
B. Concave lens
C. Biconvex lens
D. Biconcave lens

Solution

A negative focal length indicates that the lens is a concave lens.

Correct Answer: B — Concave lens

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Q. A lens has a focal length of 40 cm. If an object is placed 80 cm from the lens, what is the image distance?

A. 40 cm
B. 60 cm
C. 80 cm
D. 100 cm

Solution

Using the lens formula, we find the image distance to be 40 cm.

Correct Answer: A — 40 cm

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Q. A lens has a focal length of 50 cm. If an object is placed at 100 cm, what type of image is formed?

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

Solution

Since the object distance is greater than the focal length, a real and inverted image is formed.

Correct Answer: A — Real and inverted

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Q. A lens has a power of +2 diopters. What is its focal length?

A. 0.5 m
B. 1 m
C. 2 m
D. 3 m

Solution

Power (P) is given by P = 1/f (in meters). Thus, f = 1/P = 1/2 = 0.5 m.

Correct Answer: B — 1 m

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Q. A lens has a power of +2.0 D. What is its focal length?

A. 50 cm
B. 25 cm
C. 20 cm
D. 10 cm

Solution

Power (P) = 1/f (in meters), so f = 1/2.0 = 0.5 m = 50 cm.

Correct Answer: B — 25 cm

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Q. A lens has a power of +2.5 D. What is its focal length?

A. 40 cm
B. 25 cm
C. 50 cm
D. 20 cm

Solution

Power (P) = 1/f (in meters). Therefore, f = 1/2.5 = 0.4 m = 40 cm.

Correct Answer: B — 25 cm

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Q. A lens has a power of +2.5 D. What is the focal length of the lens in meters?

A. 0.4 m
B. 0.5 m
C. 0.6 m
D. 0.7 m

Solution

Power (P) = 1/f, so f = 1/P = 1/2.5 = 0.4 m.

Correct Answer: B — 0.5 m

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Q. A lens has a power of +5 diopters. What is its focal length?

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

Solution

Power (P) = 1/f (in meters). Therefore, f = 1/P = 1/5 = 0.2 m = 20 cm.

Correct Answer: A — 20 cm

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Physics Syllabus (JEE Main)

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