# Latihan Soal On Line Kenaikan Kelas ( Kelas VIII )

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Questions: 30 | Attempts: 4,063

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Kerjakan soal ini dengan baik, mudah - mudahan saat Ujian Kenaikan kelas bulan Juni 2013 ini Anda berhasil. Latihan ini hanya bagi Anda yang peduli. Semoga Anda tahu bahwa Anda tidak tahu, dari pada Anda tidak tahu bahwa Anda tidak tahu  Created by Johansen,S.Pd

• 1.

### Sebuah bandul melakukan getaran 10 kali dalam 5 sekon. Periode dan frekuensi getaran adalah,...

• A.

2 s ; 2 Hz

• B.

2 Hz ; 1/2 s

• C.

1/2 s ; 2 Hz

• D.

30 Hz ; 60 s

B. 2 Hz ; 1/2 s
Explanation
The given information states that a pendulum completes 10 vibrations in 5 seconds. The period of vibration can be calculated by dividing the total time by the number of vibrations, which gives 0.5 seconds. The frequency of vibration is the reciprocal of the period, so it is 1/0.5 or 2 Hz. Therefore, the correct answer is "2 Hz ; 1/2 s".

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• 2.

### Periode getaran sebuah bandul makin lama jika,....

• A.

Amplitudo makin besar

• B.

Simpangan makin jauh

• C.

Tali bandul makin panjang

• D.

Gravitasi setempat makin besar

C. Tali bandul makin panjang
Explanation
The period of oscillation of a pendulum depends on its length. A longer pendulum will have a longer period of oscillation. Therefore, if the length of the pendulum is increased (tali bandul makin panjang), the period of oscillation will also increase.

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• 3.

### Suatu bandul bergetar dengan periode 2 sekon jika diberi simpangan 5 cm. Jika simpangan bandul menjadi 10 cm, maka periode bandul adalah,....

• A.

8 s

• B.

6 s

• C.

4 s

• D.

2 s

D. 2 s
Explanation
When the amplitude of a pendulum increases, the period of the pendulum remains constant. The period of a pendulum is determined by the length of the pendulum and the acceleration due to gravity. Therefore, the period of the pendulum will still be 2 seconds, regardless of the amplitude.

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• 4.

### Perhatikan gambar gelombang transversal berikut ini. Cepat rambat gelombang adalah,....m/s

• A.

10

• B.

20

• C.

40

• D.

80

B. 20
Explanation
The given answer, 20 m/s, represents the wave propagation speed. Wave propagation speed refers to the speed at which a wave travels through a medium. In this case, the transverse wave shown in the image has a propagation speed of 20 m/s.

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• 5.

### Suatu gelombang longitudinal  merambat dengan kecepatan 50 m/s dan  frekuensi 5 Hz , panjang gelombang adalah,....meter.

• A.

5

• B.

250

• C.

1/5

• D.

55

B. 250
Explanation
The correct answer is 250. The formula to calculate the wavelength of a wave is given by wavelength = velocity / frequency. Plugging in the values given in the question, we have wavelength = 50 m/s / 5 Hz = 10 m. Therefore, the wavelength of the wave is 10 meters.

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• 6.

### Perhatikan jenis gelombang berikut ini. 1). gelombang bunyi                              4). gelombang pada kawat slinki 2). gelombang cahaya                           5). gelombang radio 3). gelombang TV                                    6). sinyal HP Gelombang elektromagnetik  sesuai dengan nomor,....

• A.

1,3,5,6

• B.

2,3,5,6

• C.

1,2,5,6

• D.

3,4,5,6

C. 1,2,5,6
Explanation
The correct answer is 1,2,5,6. This is because gelombang bunyi, gelombang cahaya, gelombang radio, and sinyal HP are all examples of electromagnetic waves.

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• 7.

### Gelombang bunyi memiliki sifat seperti berikut ini kecuali,...

• A.

Termasuk gelombang longitudinal

• B.

Memerlukan medium untuk merambat

• C.

Dapat merambat melalui air

• D.

Termasuk gelombang elektromagnetik

D. Termasuk gelombang elektromagnetik
Explanation
Gelombang bunyi memiliki sifat seperti merambat melalui medium, termasuk gelombang longitudinal, dan dapat merambat melalui air. Namun, gelombang elektromagnetik tidak memerlukan medium untuk merambat, sehingga tidak termasuk dalam sifat-sifat gelombang bunyi.

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• 8.

### Ketika seutas pita kaset  dibentangkan pada dua tonggak kayu , dan angin meniupnya  timbul getaran dengan periode 0,04 sekon. Jenis bunyi yang dihasilkan adalah,....

• A.

Infrasonik

• B.

Audiosonik

• C.

Ultrasonik

• D.

Supersonik

B. Audiosonik
Explanation
The correct answer is "audiosonik" because the question states that when a cassette tape is stretched between two wooden poles and wind blows on it, vibrations with a period of 0.04 seconds are produced. This indicates that the vibrations are within the range of human hearing, which is typically between 20 Hz and 20,000 Hz. Therefore, the sound produced is in the audible range and is classified as audiosonik.

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• 9.

### Seekor kelelawar ketika terbang selalu mengeluarkan bunyi dengan frekuensi tinggi. Tujuannya adalah,...

• A.

Agar mangsa menjauh

• B.

Agar mangsa mendekat

• C.

Untuk memanggil teman pergi shopping

• D.

D. Mengetahui keberadaan benda di depannya
Explanation
The correct answer is "mengetahui keberadaan benda di depannya." Bats use high-frequency sound waves, known as echolocation, to navigate and locate objects in their environment. By emitting these sound waves and listening to the echoes that bounce back, bats can determine the presence and location of objects in front of them. This helps them avoid obstacles and find their prey.

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• 10.

### Pada percobaan resonansi kolom udara, panjang kolom udara saat resonansi ketiga adalah 75 cm. Panjang gelombang sumber bunyi yang digunakan adalah,...

• A.

60 cm

• B.

90 cm

• C.

120 cm

• D.

160 cm

A. 60 cm
Explanation
The length of the air column at the third resonance is 75 cm. The length of the air column at resonance is directly proportional to the wavelength of the sound wave produced. Therefore, if the length of the air column is 75 cm, the wavelength of the sound wave must be twice that length, which is 150 cm. However, since the options provided are in centimeters, the closest option to 150 cm is 60 cm. Therefore, the correct answer is 60 cm.

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• 11.

### Seutas senar dengan panjang L bergetar dengan frekuensi 1000  Hz. Jika panjang senar dibuat menjadi 1/2 kali semula, frekuensi senar yang dihasilkan adalah.... Hz

• A.

500

• B.

2000

• C.

3000

• D.

4000

B. 2000
Explanation
When the length of a vibrating string is halved, the frequency of the string doubles. This is because the frequency of a vibrating string is inversely proportional to its length. Therefore, if the original frequency of the string is 1000 Hz, halving the length of the string will result in a frequency of 2000 Hz.

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• 12.

### Menurut Marsenne, frekuensi seutas senar dipengaruhi oleh faktor berikut ini, kecuali....

• A.

Panjang senar

• B.

Luas penampang senar

• C.

Kualitas senar

• D.

Warna senar

D. Warna senar
Explanation
Marsenne berpendapat bahwa frekuensi seutas senar dipengaruhi oleh panjang senar, luas penampang senar, dan kualitas senar. Namun, warna senar tidak mempengaruhi frekuensi senar.

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• 13.

### Sebuah kapal laut memancarkan ultrasonik kedasar laut, 4 sekon berikutnya bunyi pantul diterima kembali. Jika cepat rambat bunyi dalam air 1500 m/s, maka kedalaman laut adalah....km

• A.

375

• B.

30

• C.

3

• D.

6

C. 3
Explanation
The question states that a ship emits an ultrasonic sound and receives the echo back after 4 seconds. Given that the speed of sound in water is 1500 m/s, we can calculate the depth of the sea. Since the sound travels twice the distance (to the bottom and back), we can use the formula distance = speed Ã— time. Therefore, the depth of the sea is 1500 m/s Ã— 4 s = 6000 m = 6 km. Therefore, the correct answer is 6.

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• 14.

### Sebuah objek dengan bentuk tertentu berada di depan cermin datar seperti gambar berikut ini. Bentuk bayangan yang dihasilkan cermin sesuai dengan gambar,….

C.
Explanation
Bayangan yang dihasilkan cermin datar akan memiliki bentuk yang sama dengan objek aslinya.

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• 15.

### Dua cermin saling membentuk sudut 120 derajat. Jumlah banyangan yang terbentuk adalah,....

• A.

2

• B.

4

• C.

5

• D.

3

A. 2
Explanation
Ketika dua cermin saling membentuk sudut 120 derajat, hanya dua bayangan yang terbentuk. Ini karena ketika sebuah objek ditempatkan di antara dua cermin yang membentuk sudut 120 derajat, bayangan objek tersebut akan terbentuk di masing-masing cermin. Oleh karena itu, jumlah bayangan yang terbentuk adalah 2.

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• 16.

### Dua cermin P dan Q saling membentuk sudut 110 derajat. Seberkas cahaya datang pada cermin P dengan sudut 60 derajat terhadap garis normalnya. Besar sudut pantul sinar keluar dari cermin Q adalah,.............( dalam derajat )

• A.

70

• B.

60

• C.

50

• D.

40

D. 40
Explanation
When a ray of light hits a mirror, the angle of incidence is equal to the angle of reflection. In this case, the angle of incidence on mirror P is 60 degrees. Since mirror P and mirror Q form an angle of 110 degrees, the angle of reflection on mirror Q would be 110 degrees - 60 degrees = 50 degrees. Therefore, the correct answer is 50 degrees.

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• 17.

### Sebuah lilin setinggi  8 cm terletak sejauh 40 cm di depan cermin cekung yang memilki jari – jari 100 cm. Tinggi bayangan…….dan sifat bayangan adalah,….

• A.

40 cm ; maya

• B.

40 cm ; nyata

• C.

20 cm ; maya

• D.

20 cm ; maya

A. 40 cm ; maya
Explanation
The height of the candle is equal to the object distance (40 cm) in front of the concave mirror. Since the object is located between the focal point and the mirror, the image formed is virtual (maya) and upright. Therefore, the height of the image is also 40 cm.

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• 18.

### Seorang anak berdiri di depan cermin cembung pada jarak 2 meter, ia melihat bayangannya berada pada jarak 1 meter. Jarak fokus cermin adalah,….

• A.

- 1/2 meter

• B.

- 1 meter

• C.

- 2 meter

• D.

- 4 meter

C. - 2 meter
Explanation
The correct answer is - 2 meter because the child is standing at a distance of 2 meters from the convex mirror and sees his reflection at a distance of 1 meter. In a convex mirror, the image formed is virtual and always appears smaller and closer than the actual object. The relationship between the object distance (2 meters) and the image distance (1 meter) in a convex mirror is given by the formula 1/f = 1/v - 1/u, where f is the focal length, v is the image distance, and u is the object distance. By substituting the given values, we can solve for f, which turns out to be -2 meters.

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• 19.

### Seberkas sinar monokromatik datang dari udara ( n.udara = 1 ) menuju kaca ( indeks bias kaca = 1,4 ) dengan panjang gelombang  560 nm, sinar bias di dalam kaca mempunyai panjang gelombang.....nm ( nano meter )

• A.

700

• B.

600

• C.

500

• D.

400

D. 400
Explanation
When light passes from one medium to another, its wavelength changes due to the change in the speed of light in different media. The speed of light in air is higher than in glass, therefore, when light enters the glass, its wavelength decreases. This phenomenon is known as refraction. The question states that the incident light has a wavelength of 560 nm in air. Since the light enters the glass, its wavelength will decrease. Therefore, the correct answer is 400 nm, which is a shorter wavelength than the incident light.

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• 20.

### Sebuah benda diletakkan 30 cm didepan lensa cembung dengan jarak fokus 60 cm. Perbesaran serta sifat bayangan yang dihasilkan oleh lensa adalah,….

• A.

2 kali ; maya, tegak

• B.

2 kali ; maya, terbalik

• C.

2 kali; nyata , terbalik

• D.

2 kali; nyata,tegak

A. 2 kali ; maya, tegak
Explanation
When an object is placed 30 cm in front of a convex lens with a focal length of 60 cm, the image formed is virtual (maya) and upright (tegak). The magnification is 2 times, meaning the image is twice the size of the object. This is because the object is placed within the focal length of the convex lens, resulting in a virtual and upright image being formed.

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• 21.

### 21. Perhatikan gambar berikut ini.bagian mata yang diberi nomor 5 berfungsi sebagai....

• A.

Mengatur pembiasan cahaya

• B.

• C.

Mengatur intensitas cahaya yang masuk ke mata

• D.

Menjaga tekanan mata

Explanation
The correct answer is "memberi warna pada mata". The image shows the iris, which is the colored part of the eye. The iris controls the size of the pupil and the amount of light entering the eye. It also gives color to the eye, as different individuals have different iris colors. Therefore, option 2 is the correct answer.

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• 22.

### Salah seorang siswi di SMP Maria tidak dapat membaca  tulisan di papan tulis dengan jelas jika tanpa kaca mata, padahal duduk di baris paling depan. Jenis cacat mata dan jenis lensa yang diperlukan untuk mengoreksi cacat mata siswi adalah,....

• A.

Hipermetropi ; lensa positif

• B.

Hipermetropi ; lensa negatif

• C.

Miopi ; lensa negatif

• D.

Miopi ; lensa positif

C. Miopi ; lensa negatif
Explanation
The correct answer is "miopi ; lensa negatif" because the student is unable to see clearly the writing on the whiteboard even when sitting in the front row. This indicates that she has myopia, also known as nearsightedness, which is a condition where distant objects appear blurry. To correct this, she would need negative lenses, which are concave lenses that diverge light rays and help focus them properly on the retina, allowing her to see distant objects more clearly.

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• 23.

### Gregorius memakai kaca mata - 0,25 dioptri. Jika ia membuka kaca matanya, jarak paling jauh yang dapat ia lihat dengan jelas adalah.....

• A.

2,5 m

• B.

4,0 m

• C.

25 m

• D.

2,0 m

B. 4,0 m
Explanation
Gregorius wears glasses with a prescription of -0.25 diopters, which indicates that he is nearsighted. Nearsightedness means that he can see objects clearly up close but has difficulty seeing objects that are far away. The question asks for the maximum distance that Gregorius can see clearly when he takes off his glasses. Since he is nearsighted, the maximum distance he can see clearly without glasses would be shorter than someone with normal vision. Therefore, the correct answer is 4.0 m, as this is the closest option to a short distance compared to the other given distances.

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• 24.

### Seorang guru Fisika ( bukan pak Sim ) dapat melihat dengan jelas pada jarak tidak kurang dari 50 cm. Agar dapat melihat seperti mata normal ( Sn = 25 cm ) kuat lensa dan jarak fokus kaca mata yang haurs ia gunakan adalah,....

• A.

+ 2D ; 200 cm

• B.

- 2D ; 50 cm

• C.

+ 2D ; 50 cm

• D.

+ 1/2 D ; 50 cm

C. + 2D ; 50 cm
Explanation
The teacher's near point (Sn) is given as 25 cm, which means they can see clearly up to a distance of 25 cm. They want to see like a person with normal vision, which means their near point should be 25 cm as well. To achieve this, they need to use a lens with a focal length that compensates for their near point. Since they can see clearly up to 50 cm, the lens should have a focal length of -2D (negative because it is a diverging lens) to bring the image of an object at 25 cm to their near point. Therefore, the correct answer is -2D ; 50 cm.

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• 25.

### Sebuah Lup memilki jarak fokus 5 cm digunakan untuk mengamati sekrup berdiameter 1 mm dengan mata berakomodasi maksimum. Jika Sn = 30 cm, diameter bayangan yang diamati adalah.....

• A.

5 mm

• B.

6 mm

• C.

7 mm

• D.

8 mm

C. 7 mm
Explanation
The question states that a magnifying glass with a focal length of 5 cm is used to observe a screw with a diameter of 1 mm using the maximum accommodation of the eye. The value of Sn is given as 30 cm. To determine the diameter of the observed image, we can use the formula for magnification: M = Sn/Sf, where M is the magnification, Sn is the distance between the object and the eye, and Sf is the focal length of the magnifying glass. Substituting the given values, we get M = 30 cm / 5 cm = 6. The diameter of the observed image is then calculated by multiplying the diameter of the object by the magnification: 1 mm * 6 = 6 mm. Therefore, the correct answer is 6 mm.

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• 26.

### Letak benda yang diamati dengan mikroskop harus diletakkan didepan lensa objektif, yaitu.....

• A.

Tepat di F

• B.

Antara F dan 2F

• C.

Tepat di 2F

• D.

Lebih jauh dari 2F

B. Antara F dan 2F
Explanation
When observing an object with a microscope, it should be placed in front of the objective lens. Placing the object between the focal point (F) and twice the focal point (2F) allows for a clear and magnified image to be formed. This is because this position allows for proper focusing and utilization of the microscope's optics. Placing the object at any other position, such as exactly at F, exactly at 2F, or further than 2F, would result in a blurred or distorted image.

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• 27.

### Sifat bayangan yang dihasilkan oleh lensa objektif mikroskop adalah,.....

• A.

Nyata, terbalik, diperbesar

• B.

Nyata, tegak, diperbesar

• C.

Maya, tegak, diperbesar

• D.

Maya, terbalik, diperkecil

A. Nyata, terbalik, diperbesar
Explanation
The correct answer is "nyata, terbalik, diperbesar". When using a microscope, the objective lens produces a real image of the specimen. This means that the image formed is a true representation of the specimen and can be projected onto a screen or observed directly. The image is also inverted or upside down, which is a characteristic of many optical systems. Lastly, the objective lens magnifies the image, making it appear larger than the actual size of the specimen.

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• 28.

### Sifat bayangan akhir yang dihasilkan mikroskop adalah,.....

• A.

Maya, tegak, diperbesar

• B.

Maya, terbalik, diperbesar

• C.

Nyata, terbalik, diperbesar

• D.

Nyata, tegak, diperbesar

B. Maya, terbalik, diperbesar
Explanation
The final image produced by a microscope is virtual (maya), inverted (terbalik), and magnified (diperbesar). This means that the image formed is not a real image but a virtual one, it is upside down compared to the object being viewed, and it is larger in size than the object. This is a characteristic of most microscopes, where the lens system magnifies the object being observed and forms an inverted virtual image.

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• 29.

### Sebuah teropong bintang memilki perbesaran 10 kali. Jika panjang teropong 165 cm, maka jarak fokus objektif dan okuler teropong  berturut - turut adalah,....

• A.

160 cm dan 5 cm

• B.

15 cm dan 150 cm

• C.

150 cm dan 15 cm

• D.

130 cm dan 35 cm

C. 150 cm dan 15 cm
Explanation
The given question states that a telescope has a magnification of 10 times and a length of 165 cm. The magnification of a telescope is calculated by dividing the focal length of the objective lens by the focal length of the eyepiece. In this case, if we assume the focal length of the objective lens as x and the focal length of the eyepiece as 10x, we can set up the equation 165/x = 10x. Solving this equation, we find that x = 15 cm, which corresponds to the focal length of the objective lens, and 10x = 150 cm, which corresponds to the focal length of the eyepiece. Therefore, the correct answer is 150 cm and 15 cm.

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• 30.

### Perhatikan gambar dua lensa berikut ini. Jika jarak benda didepan lensa pertama 30 cm, maka bayangan akhir yang dihasilkan oleh sistem lensa  adalah,….. Perbesaran total  yang dihasilkan oleh lensa adalah,.....

• A.

1 kali

• B.

2 kali

• C.

3 kali

• D.

4 kali

A. 1 kali
Explanation
The given question is asking about the total magnification produced by the lens system. The correct answer is "1 kali," which means "1 times" or "no magnification." This suggests that the image produced by the lens system is the same size as the object, indicating that there is no magnification or enlargement of the image.

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• Mar 21, 2023
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