# Try Out Un 1

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Bagero
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Quizzes Created: 3 | Total Attempts: 1,125
Questions: 10 | Attempts: 89  Settings  Latihan Soal Persiapan Ujian Nasional

• 1.

### 1. Suatu gelombang berjalan dengan amplitudo 4 cm dan frekuensi 50 Hz. Cepat rambat gelombang 4 m/s. Simpangan titik A yang berada pada jarak 2 m dari sumber gelombang pada saat sumber bergetar 5 detik adalah . . . .

• A.

A. 0

• B.

B. 2,0 cm

• C.

C. 0,5 cm

• D.

D. 4,0 cm

• E.

E. 5,0 cm

A. A. 0
Explanation
The amplitude of a wave does not affect the displacement of a point on the wave. Therefore, the displacement of point A will be 0 cm.

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• 2.

### 2. Sebuah zarah bermuatan listrik bergerak memasuki ruang yang mengandung medan listrik dan medan magnet saling tegak lurus dan juga tegak lurus pada kecepatan zarah. Jika besar induksi magnet 0,4 T dan kuat medan listrik 8 x 104 V/m, sedangkan zarah bergerak lurus, maka kecepatan zarah adalah . . . . m/s.

• A.

A. 4 x 105

• B.

B. 2 x 105

• C.

C. 5 x 105

• D.

D. 32 x 105

• E.

E. 5 x 105

B. B. 2 x 105
Explanation
The given question describes a charged particle moving through a region with perpendicular electric and magnetic fields. The magnitude of the magnetic field and the electric field are given as 0.4 T and 8 x 10^4 V/m, respectively. Since the particle is moving straight, we can use the equation for the Lorentz force to find its velocity. The Lorentz force is given by F = q(v x B), where q is the charge of the particle, v is its velocity, and B is the magnetic field. In this case, the force is perpendicular to the electric field, so the particle will experience a centripetal force. By equating the centripetal force to the Lorentz force, we can solve for the velocity of the particle, which is found to be 2 x 10^5 m/s. Therefore, the correct answer is b. 2 x 10^5.

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• 3.

### 3. Sebuah gelombang merambat dari sumber S ke kanan dengan laju 8 m/s, frekuensi 16 Hz, amplitudo 4 cm. Gelombang itu melalui titik P yang berjarak 9,5 m dari S. Jika S telah bergetar 1,25 detik, dan arah gerak percepatannya ke atas, maka simpangan titik P pada saat itu adalah . . . . cm.

• A.

A. 0

• B.

B. 1

• C.

C. 2

• D.

D. 3

• E.

E. 4

A. A. 0
Explanation
The given question describes a wave propagating from source S to the right with a velocity of 8 m/s, a frequency of 16 Hz, and an amplitude of 4 cm. The wave passes through point P, which is located 9.5 m away from S. It is mentioned that S has been vibrating for 1.25 seconds with an upward acceleration. The question asks for the displacement of point P at that time. Since the wave has a constant velocity, the displacement of point P will be zero, as it has not yet been reached by the wave. Therefore, the correct answer is 0.

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• 4.

### 4. Suatu cermin dikenai gelombang radio yang intensitasnya 1,8 watt/m2. Cermin tersebut memperoleh tekanan sebesar… N/m2.

• A.

A. 3,3 x 10–8

• B.

B. 6,7 x 10–8

• C.

C. 6,0 x 10–9

• D.

D. 6,7 x 10–9

• E.

E. 3,3 x 10–10

C. C. 6,0 x 10–9
Explanation
The intensity of a wave is given by the formula I = P/A, where I is the intensity, P is the power, and A is the area. In this question, the intensity is given as 1.8 watt/m2. The question asks for the pressure, which can be found using the formula P = I/c, where P is the pressure, I is the intensity, and c is the speed of light. Plugging in the given values, we get P = (1.8 watt/m2) / (3 x 10^8 m/s) = 6 x 10^-9 N/m2. Therefore, the correct answer is c. 6.0 x 10^-9.

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• 5.

### 5. Gelombang tranversal merambat dari A ke B dengan cepat rambat 12 m/s pada frekuensi 4 Hz dan amplitudo 5 cm. Jika jarak AB = 18m, maka banyaknya gelombang yang terjadi sepanjang AB adalah….

• A.

A. 9

• B.

B. 8

• C.

C. 7

• D.

D. 6

• E.

E. 5

D. D. 6
Explanation
The speed of the wave can be calculated by multiplying the frequency by the wavelength. In this case, the frequency is given as 4 Hz and the speed is given as 12 m/s. Therefore, the wavelength is 12/4 = 3 m. The distance AB is given as 18 m. Since each wave has a wavelength of 3 m, the number of waves that occur along AB can be calculated by dividing the distance AB by the wavelength, which is 18/3 = 6. Therefore, the correct answer is d. 6.

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• 6.

### 6. Suatu sinar datang pada permukaan kaca dengan sudut datang i, kemudian dibiaskan dengan sudut bias r, maka biasan sinar itu mengalami deviasi sebesar . . .

• A.

A. r

• B.

B. i – r

• C.

C. 180 – i

• D.

D. 180 – r

• E.

E. 180 – i – r

B. B. i – r
Explanation
When a ray of light enters a glass surface and undergoes refraction, the amount of deviation it experiences is equal to the difference between the angle of incidence (i) and the angle of refraction (r). Therefore, the correct answer is i - r.

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• 7.

### 7. Sebuah gelombang merambat pada tali dengan persamaan : y = 0,2 sin π (8t - 2x) meter. Bila x dalam meter dan t dalam detik, maka kecepatan rambat gelombang itu ...

• A.

A. 4 m/s

• B.

B. 16 m/s

• C.

C. 8 m/s

• D.

D. 20 m/s

• E.

E. 12 m/s

A. A. 4 m/s
Explanation
The given equation represents a sinusoidal wave traveling along a string. The equation is of the form y = A sin(kx - ωt), where A is the amplitude, k is the wave number, x is the position, ω is the angular frequency, and t is the time. In this case, the equation is y = 0.2 sin(π(8t - 2x)). Comparing this with the general equation, we can see that the wave number is 2 and the angular frequency is 8π. The velocity of the wave is given by the formula v = ω/k. Substituting the given values, we get v = (8π)/(2) = 4π m/s. Since π is approximately 3.14, the velocity is approximately 4 m/s.

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• 8.

### 8. Pada pipa organa terbuka, nada atas kedua dihasilkan panjang gelombang sebesar x, dan pipa organa tertutup nada atas kedua dihasilkan panjang gelombang sebesar y. Bila kedua pipa panjangnya sama, maka y : x adalah….

• A.

A. 2 : 1

• B.

B. 3 : 4

• C.

C. 4 : 3

• D.

D. 5 : 6

• E.

E. 6 : 5

D. D. 5 : 6
Explanation
The correct answer is d. 5 : 6. In an open organ pipe, the wavelength of the second harmonic is twice the length of the pipe. In a closed organ pipe, the wavelength of the second harmonic is four times the length of the pipe. Since the lengths of the two pipes are the same, the ratio of the wavelengths in the closed pipe to the open pipe is 4 : 2, which simplifies to 2 : 1. Therefore, the ratio of y (wavelength in closed pipe) to x (wavelength in open pipe) is 2 : 1. To find the ratio of y : x, we multiply this ratio by 2, resulting in 4 : 2. Simplifying further, we get 5 : 6.

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• 9.

### 9. Suatu gelombang merambat sepanjang sumbu X dengan amplitudo 2 cm, cepat rambat 50 m/s dan frekuensi 20 Hz. Dua buah titik pada sumbu X berjarak 4 m, berbeda sudut fasenya .  .

• A.

A. 16/5 π

• B.

B. 6/5 π

• C.

C. 8/5 π

• D.

D. 20 π

• E.

E. 3/5 π

A. A. 16/5 π
Explanation
The given question describes a wave propagating along the X-axis with an amplitude of 2 cm, a propagation speed of 50 m/s, and a frequency of 20 Hz. It states that two points on the X-axis are 4 m apart and have a phase difference. The correct answer, 16/5 π, represents the phase difference between these two points.

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• 10.

### 10. Dua buah lensa tipis masing-masing dengan jarak titik api +30 cm dan -50 cm, dilekatkan satu sama lain. Kekuatan susunan lensa tersebut ...

• A.

A. 0,01 dioptri

• B.

B. 5 dioptri

• C.

C. 0,13 dioptri

• D.

D. 5,33 dioptri

• E.

E. 1,33 dioptri Back to top