30. which of the following characteristics of the image formed by a diverging lens are correct?
a. virtual and diminished
b. real and diminished
c. virtual and magnified
d. real and magnified

31. light from an illuminated object passes through a converging lens and is reflected along its original path to form an image at the same position as the object by a plane mirror placed just behind the lens. if the distance between the object and the plane mirror is 20 cm, determine the focal length of the lens.
a. 40 cm
b. 20 cm
c. 10 cm
d. 5 cm

32. the advantage of the galilean telescope over the astronomical telescope is that in the galilean telescope, the
a. final image is sharper.
b. angular magnification is greater.
c. field of view is wider.
d. final image is erect.

33. a singer emits a note of frequency 200 hz. calculate the wavelength of the note. speed of sound in air = 340 m s⁻¹
a. 17.0 m
b. 6.0 m
c. 1.7 m
d. 0.6 m

34. echo is a phenomenon that shows that sound waves
a. are mechanical.
b. are longitudinal.
c. can be polarized.
d. can be reflected.

35. which of the following statements about force fields is correct?
a. electrostatic, gravitational and magnetic forces are always attractive.
b. electric, gravitational and magnetic fields obey inverse square laws.
c. field lines are real but their corresponding fields are imaginary.
d. field lines and their corresponding fields are both real.

36. which of the following statements about a satellite is/are correct? a satellite
i. describes a circle about the earth if its escape velocity is 8 km s⁻¹.
ii. cannot escape from the earth’s surface with a velocity less than 8 km s⁻¹.
iii. describes an ellipse about the earth if its escape velocity is greater than 11 km s⁻¹.
a. i only
b. ii only
c. i and ii only
d. ii and iii only

Question

30. which of the following characteristics of the image formed by a diverging lens are correct?
a. virtual and diminished
b. real and diminished
c. virtual and magnified
d. real and magnified

31. light from an illuminated object passes through a converging lens and is reflected along its original path to form an image at the same position as the object by a plane mirror placed just behind the lens. if the distance between the object and the plane mirror is 20 cm, determine the focal length of the lens.
a. 40 cm
b. 20 cm
c. 10 cm
d. 5 cm

32. the advantage of the galilean telescope over the astronomical telescope is that in the galilean telescope, the
a. final image is sharper.
b. angular magnification is greater.
c. field of view is wider.
d. final image is erect.

33. a singer emits a note of frequency 200 hz. calculate the wavelength of the note. speed of sound in air = 340 m s⁻¹
a. 17.0 m
b. 6.0 m
c. 1.7 m
d. 0.6 m

34. echo is a phenomenon that shows that sound waves
a. are mechanical.
b. are longitudinal.
c. can be polarized.
d. can be reflected.

35. which of the following statements about force fields is correct?
a. electrostatic, gravitational and magnetic forces are always attractive.
b. electric, gravitational and magnetic fields obey inverse square laws.
c. field lines are real but their corresponding fields are imaginary.
d. field lines and their corresponding fields are both real.

36. which of the following statements about a satellite is/are correct? a satellite
i. describes a circle about the earth if its escape velocity is 8 km s⁻¹.
ii. cannot escape from the earth’s surface with a velocity less than 8 km s⁻¹.
iii. describes an ellipse about the earth if its escape velocity is greater than 11 km s⁻¹.
a. i only
b. ii only
c. i and ii only
d. ii and iii only

Accepted Answer

### Question 30
# Brief Explanations:
A diverging lens always forms a virtual, upright, and diminished image, regardless of the object's position. So option A (Virtual and diminished) is correct.
# Answer:
A. Virtual and diminished

### Question 31
# Explanation:
## Step1: Analyze the setup
The light passes through the converging lens, reflects off the plane mirror, and retraces its path. This means the light rays from the object, after passing through the lens, are incident normally on the plane mirror (so they retrace). For the lens, when the object is at 2F (twice the focal length), the image is also at 2F. But here, the distance between the object and the mirror is 20 cm, and the lens is just behind the mirror? Wait, no—wait, the light passes through the lens, then hits the mirror (placed just behind the lens), so the object distance from the lens is equal to the distance from the object to the mirror (since the lens is just behind the mirror). Wait, actually, when the light retraces, the image formed by the lens should be at the position of the object (because the mirror reflects it back). So the object is at a distance such that the image formed by the lens is at the same position (so the object is at 2F? No, wait, when the object is at the focal point, the rays are parallel, but here the rays retrace, so the object must be at the focal point? Wait, no—wait, the distance between the object and the mirror is 20 cm. The lens is just behind the mirror, so the object distance from the lens is 20 cm. For the light to retrace, the lens must form an image at the position of the object (so that the mirror reflects it back along the same path). Wait, actually, when the object is at the focal point of the converging lens, the rays become parallel after passing through the lens. Then, when they hit the plane mirror (which is perpendicular to the axis), they reflect back along the same path, passing through the lens again and forming an image at the object's position. Wait, no—if the object is at the focal point, the image is at infinity. But here, the image is at the same position as the object. So maybe the object is at 2F? Wait, no, let's think again. The distance between the object and the mirror is 20 cm. The lens is just behind the mirror, so the object distance $ u = 20 $ cm. For the light to retrace, the lens must form an image at the position of the object (so that the mirror reflects it back). So the image distance $ v = u $ (since the image is at the object's position). For a converging lens, when the object is at 2F, the image is also at 2F. So $ u = 2f $, so $ f = \frac{u}{2} $. Wait, but if the object is at the focal point, the rays are parallel, so after reflection, they would converge at the focal point. Wait, maybe I made a mistake. Let's recall: when a ray is incident normally on a plane mirror, it retraces its path. So for the light to retrace, the rays after passing through the lens must be parallel (so that they hit the mirror normally). Wait, no—if the rays are parallel, they hit the mirror normally and retrace. So the lens must form a parallel beam from the object. That happens when the object is at the focal point of the converging lens. So the object distance $ u = f $. But the distance between the object and the mirror is 20 cm, and the lens is just behind the mirror, so $ u = 20 $ cm? No, that can't be. Wait, maybe the lens is between the object and the mirror. Wait, the problem says: "Light from an illuminated object passes through a converging lens and is reflected along its original path to form an image at the same position as the object by a plane mirror placed just behind the lens." So the order is: object -> lens -> mirror (just behind lens). So the light goes through the lens, then hits the mirror (which is right behind the lens), then reflects back through the lens, forming an image at the object's position. For the light to retrace, the rays after passing through the lens must be parallel (so that when they hit the mirror, they reflect back along the same path). So the lens must form a parallel beam, which means the object is at the focal point of the lens. So the object distance from the lens is equal to the focal length $ f $. But the distance between the object and the mirror is 20 cm. The mirror is just behind the lens, so the distance from the object to the lens is 20 cm? Wait, no—if the mirror is just behind the lens, then the distance from the object to the lens is equal to the distance from the object to the mirror (since the lens is between the object and the mirror? Wait, no, the object is in front of the lens, the lens is in front of the mirror. So the distance from the object to the mirror is the distance from the object to the lens plus the distance from the lens to the mirror. But the mirror is "just behind the lens", so the distance from the lens to the mirror is negligible. So the object distance $ u \approx 20 $ cm. For the light to retrace, the lens must form an image at the object's position (so that the mirror reflects it back). Wait, when the object is at 2F, the image is at 2F. So $ u = v = 2f $. The distance between the object and the mirror is 20 cm, which is $ u + $ (lens to mirror distance). Since the mirror is just behind the lens, lens to mirror distance is 0, so $ u = 20 $ cm. Then $ 2f = 20 $ cm, so $ f = 10 $ cm. Wait, that makes sense. Because if the object is at 2F (20 cm from the lens), the image is at 2F (20 cm on the other side of the lens). But the mirror is just behind the lens, so the image is at the mirror's position. Then the mirror reflects the light back, so the light retraces, and the image formed by the lens (after reflection) is at the object's position. So the focal length is 10 cm.
## Step2: Calculate focal length
Given the distance between the object and the mirror is 20 cm, and the mirror is just behind the lens, so the object distance $ u = 20 $ cm. For the image to be formed at the same position as the object (after reflection), the object must be at $ 2f $ (since for a converging lens, when $ u = 2f $, $ v = 2f $). So $ 2f = 20 $ cm, so $ f = \frac{20}{2} = 10 $ cm.
# Answer:
C. 10 cm

### Question 32
# Brief Explanations:
A Galilean telescope uses a converging objective lens and a diverging eyepiece, resulting in an erect final image. Astronomical telescopes use a converging eyepiece, giving an inverted image. The other options: A (sharper image) is not a specific advantage, B (greater angular magnification) is not true, C (wider field of view) is not an advantage of Galilean. So D (final image is erect) is correct.
# Answer:
D. final image is erect

### Question 33
# Explanation:
## Step1: Recall the wave speed formula
The formula relating speed ($ v $), frequency ($ f $), and wavelength ($ \lambda $) is $ v = f\lambda $.
## Step2: Solve for wavelength
We need to find $ \lambda $, so rearrange the formula: $ \lambda = \frac{v}{f} $. Given $ v = 340 \, \text{m/s} $ and $ f = 200 \, \text{Hz} $, substitute the values: $ \lambda = \frac{340}{200} = 1.7 \, \text{m} $.
# Answer:
C. 1.7 m

### Question 34
# Brief Explanations:
Echo is the reflection of sound waves (e.g., sound bouncing off a surface and returning). Option A: sound is mechanical, but echo doesn't show that. Option B: sound is longitudinal, but echo doesn't show that. Option C: sound can't be polarized (it's longitudinal). Option D: echo shows sound waves can be reflected. So D is correct.
# Answer:
D. can be reflected

### Question 35
# Brief Explanations:
- Option A: Electrostatic force can be repulsive (e.g., like charges), so not always attractive.
- Option B: Electric, gravitational, and magnetic fields obey the inverse square law (e.g., Coulomb's law, Newton's law of gravitation, magnetic field from a point source). This is correct.
- Option C: Field lines are a visual tool, fields are real; field lines aren't real.
- Option D: Field lines are not real, fields are real. So B is correct.
# Answer:
B. Electric, gravitational and magnetic fields obey inverse square laws.

### Question 36
# Brief Explanations:
- I: A satellite in circular orbit has orbital velocity, not escape velocity. Escape velocity is the minimum speed to escape Earth's gravity. So I is incorrect.
- II: Escape velocity from Earth is ~11.2 km/s. If a satellite's velocity is less than 8 km/s (wait, no, escape velocity is ~11.2, orbital velocity is ~7.9). Wait, the problem says "escape velocity is 8 km/s"—maybe a typo, but assuming: If a satellite has a velocity less than escape velocity (11.2 km/s), it can't escape. So II: "cannot escape from the earth’s surface with a velocity less than 8 km s⁻¹"—wait, escape velocity is ~11.2, so if velocity is less than 11.2, it can't escape. But the option says 8 km/s. Maybe the question has a typo, but let's check the options.
- III: If escape velocity is greater than 11 km/s (close to 11.2), a satellite with escape velocity >11 km/s would escape, not orbit in an ellipse. Elliptical orbits are for velocities between orbital (7.9) and escape (11.2). So III is incorrect.
Wait, maybe the question's escape velocity values are approximate. Let's re-express:
- I: A satellite describes a circle if its escape velocity is 8 km/s? No, circular orbit has orbital velocity (~7.9), escape velocity is higher. So I is wrong.
- II: "cannot escape... with a velocity less than 8 km s⁻¹"—if escape velocity is ~11.2, then any velocity less than that can't escape (if it's less than orbital, it falls back; between orbital and escape, it orbits elliptically). So II is correct (assuming 8 km/s is less than escape velocity).
- III: "describes an ellipse... if its escape velocity is greater than 11 km s⁻¹"—escape velocity is ~11.2, so greater than 11 km/s is close to escape. If velocity is greater than escape, it escapes, not orbits. So III is wrong.
Wait, the options:
A. I only – wrong.
B. II only – maybe correct.
C. I and II only – I is wrong.
D. II and III only – III is wrong.
So the correct answer is B.
# Answer:
B. II only

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QUESTION IMAGE

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Explanation:

Question 30

Step1: Analyze the setup

Answer:

Question 31