Ujian Akhir Semester Ganjil Tahun Pelajaran 2011/2012 Kelas Xii
Questions and Answers
- 1.
Rangkaian seri terdiri atas sebuah kumparan dan kapasitor yang dihubungkan dengan sumber tegangan bolak-balik Vef = 110 V dan frekuensi sudutnya 500 rad-s. Jika induktansi kumparan 1,0 H, hambatan kumparan 400 ohm dan kapasitor 2,5 μ F, tentukanlah: a. arus efektif rangkaian; b. daya yang dipakai oleh rangkaian.
- 2.
Sebuah toroida dengan jari-jari 5 cm terdiri atas 600 lilitan. Berapakah arus yang mengalir pada toroida agar induksi magnetik yang terjadi di dalam toroida adalah 2,4 ×10-3 Wbm-2.
- 3.
Potensial pada sebuah titik yang berjarak 3 cm dari sebuah muatan titik yang berada di udara adalah 6 × 105 V. Tentukanlah kuat medan listrik pada titik tersebut. (k = 9 × 109Nm2C-2)
- 4.
Sebuah pipa organa tertutup memiliki panjang 1,2 m dan cepat rambat bunyi di udara 360 m/s. Tentukanlah tiga frekuensi harmonic terendah yang dihasilkan pipa organa tersebut.
- 5.
Seberkas cahaya monokromatis dengan panjang gelombang 6.600A datang tegak lurus pada kisi. Jika garis terang orde ke-2 membuat sudut 30o dengan garis normal pada kisi, tentukanlah jarak antarcelah pada kisi.
- 6.
Yang tidak termasuk gelombang mekanik adalah....
- A.
Bunyi
- B.
Permukaan air yang bergetar
- C.
Radio
- D.
Batang logam yang digetarkan
- E.
Tali yang digetarkan
Correct Answer
C. RadioExplanation
Radio waves are not considered mechanical waves because they do not require a medium to propagate. Unlike sound waves, which require a medium like air or water to travel through, radio waves can travel through a vacuum, such as outer space. Therefore, radio waves are classified as electromagnetic waves, not mechanical waves.Rate this question:
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- 7.
Suatu gelombang merambat disuatu titik. Persamaan lintasannya : Y = 2 sin ( 100t &ndash x) x &ndash 2 cmdalam meter dan t dalam detik. Panjang gelombangnya adalah....
- A.
– 2 cm
- B.
1 cm
- C.
1/2 cm
- D.
1 m
- E.
2 m
Correct Answer
E. 2 mExplanation
The given equation for the wave is Y = 2 sin (100t - x) x - 2 cm. The wavelength of a wave is the distance between two consecutive points that are in phase. In this equation, the wavelength can be determined by looking at the coefficient of the x term, which is -2 cm. However, since the wavelength is typically expressed in meters, we need to convert -2 cm to meters by dividing it by 100. This gives us -0.02 m. Since the wavelength cannot be negative, we take the absolute value of -0.02 m, which is 0.02 m or 2 m. Therefore, the correct answer is 2 m.Rate this question:
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- 8.
Interferensi dua gelombang yang mempunyai frekuensi, amplitudo dan arah getar sama yang merambat menurut garis lurus yang sama, dengan kecepatan yang sama tetapi berlawanan arahnya menghasilkan....
- A.
Gelombang berjalan
- B.
Pelayangan
- C.
Gelombang pantul
- D.
Gelombang stasioner
- E.
Efek Doppler
Correct Answer
D. Gelombang stasionerExplanation
When two waves with the same frequency, amplitude, and direction of vibration propagate along the same straight line with the same velocity but in opposite directions, they interfere to form a stationary wave. In a stationary wave, certain points called nodes do not undergo any displacement while the points between the nodes, called antinodes, undergo maximum displacement. This is the result of constructive and destructive interference between the two waves. Therefore, the correct answer is "gelombang stasioner" (stationary wave).Rate this question:
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- 9.
Cepat rambat gelombang transversal dalam seutas tali yang panjangnya 30 m adalah 40 m/s. Jika gaya tegangan tali = 2 N, maka massa tali tersebut adalah....
- A.
0,00375 kg
- B.
0,0375 kg
- C.
375 kg
- D.
3,75 kg
- E.
37,5 kg
Correct Answer
B. 0,0375 kgExplanation
The given question provides information about the speed of a transverse wave in a string (40 m/s) and the tension in the string (2 N). To find the mass of the string, we can use the formula v = √(T/μ), where v is the speed of the wave, T is the tension in the string, and μ is the linear mass density (mass per unit length) of the string. Rearranging the formula, we get μ = T/v^2. Plugging in the values given, we have μ = 2 N / (40 m/s)^2 = 0.00125 kg/m. Since the length of the string is 30 m, the mass of the string is μ × length = 0.00125 kg/m × 30 m = 0.0375 kg. Therefore, the correct answer is 0.0375 kg.Rate this question:
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- 10.
Bunyi mesin tik rata-rata menimbulkan taraf intensitas 75 dB. Taraf intensitas 10 mesin tik yang digunakan bersmaan....
- A.
75 dB
- B.
85 dB
- C.
150 dB
- D.
850 dB
- E.
7500 dB
Correct Answer
B. 85 dBExplanation
The question states that the average intensity of a typewriter machine is 75 dB. It then asks for the intensity when 10 typewriter machines are used simultaneously. Since the intensity of sound waves adds up when multiple sources are present, the intensity will increase. Therefore, the correct answer is 85 dB, which is a higher intensity than the average intensity of a single typewriter machine.Rate this question:
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- 11.
Sumber bunyi yang bergerak dengan kecepatan 20 m/s menjauhi pendengar yang diam. Jika frekuensi sirine 450 Hz dan cepat rambat bunyi di udara 340 m/s, maka frekuensi sirine yang terdengar oleh pendengar ....
- A.
525 Hz
- B.
470 Hz
- C.
425 Hz
- D.
400 Hz
- E.
378 Hz
Correct Answer
C. 425 HzExplanation
When a sound source is moving away from a stationary listener, the frequency of the sound heard by the listener decreases. This is known as the Doppler effect. In this question, the sound source is moving away from the listener at a speed of 20 m/s. The given frequency of the siren is 450 Hz and the speed of sound in air is 340 m/s. Using the formula for the Doppler effect, we can calculate the frequency heard by the listener. The formula is: observed frequency = source frequency * (speed of sound + speed of listener) / (speed of sound + speed of source). Plugging in the given values, we get: observed frequency = 450 Hz * (340 m/s + 0 m/s) / (340 m/s + 20 m/s) = 425 Hz. Therefore, the correct answer is 425 Hz.Rate this question:
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- 12.
Frekuensi nada dasar pipa organa terbuka 400 Hz. Frekuensi nada atas 1 dari dari pipa itu setelah ditutup adalah....
- A.
200 Hz
- B.
400 Hz
- C.
600 Hz
- D.
800 Hz
- E.
1000 Hz
Correct Answer
C. 600 HzExplanation
The question states that the fundamental frequency of an open organ pipe is 400 Hz. When the pipe is closed, the first harmonic or the first overtone will be produced. The first harmonic is two times the fundamental frequency. Therefore, the frequency of the first overtone will be 400 Hz x 2 = 800 Hz. However, the given options do not include 800 Hz. The closest option to 800 Hz is 600 Hz, which is the correct answer.Rate this question:
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- 13.
Anda sedang bergerak mendekati sebuah sumber bunyi S tidak bergerak yang berada jauh di depan Anda dan sedang memancarkan bunyi berfrekuensi teretentu fo ke segala arah. Tidak jauh di belakang Anda terdapat sebuah dinding pemantul D. Bunyi yang Anda dengar langsung dari S dengan frekuensi fs dan yang berasal dari pantulan di D dengan frekuensi fd sewaktu berinteraksi di telinga Anda ....
- A.
Tak akan menimbulkan layangan karena fs = fd > fo
- B.
Tak akan menimbulkan layangan karena fs = fd tidak sama fo
- C.
Akan menimbulkan layangan dan memenuhi fs > fd = fo
- D.
Akan menimbulkan layangan dan memenuhi fs > fo > fa
- E.
Akan menimbulkan layangan dan memenuhi fs > fd > fo
Correct Answer
C. Akan menimbulkan layangan dan memenuhi fs > fd = foExplanation
The correct answer is "akan menimbulkan layangan dan memenuhi fs > fd = fo" because it states that the sound will cause interference and produce a beat frequency, where the frequency of the direct sound from S (fs) is greater than the frequency of the reflected sound from D (fd), which is equal to the original frequency (fo). This condition is necessary for the formation of beats or interference patterns.Rate this question:
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- 14.
Pada percobaan Young digunakan 2 celah sempit yang berjarak 2mm satu sama lain dan layar yang dipasang 1m dari celah tersebut. Jika dihasilkan terang kedua pada jarak 0,5 mm dari terang pusat, maka panjang gelombang cahaya yang digunakan adalah....
- A.
2,5 x 10-5 m
- B.
3,3 x 10-5 m
- C.
5,0 x 10-6 m
- D.
1,0 x 10- 6 m
- E.
5,0 x 10-7 m
Correct Answer
E. 5,0 x 10-7 mExplanation
The given question is asking for the wavelength of light used in Young's experiment. In the experiment, two narrow slits are placed 2mm apart, and a screen is placed 1m away from the slits. The second bright fringe is observed at a distance of 0.5mm from the central bright fringe. This information can be used to calculate the wavelength of light using the formula for fringe spacing, which is given by λ = (d * D) / y, where λ is the wavelength, d is the slit separation, D is the distance between the slits and the screen, and y is the fringe spacing. Plugging in the values, we get λ = (2mm * 1m) / 0.5mm = 4mm = 5.0 x 10^-7 m. Therefore, the correct answer is 5.0 x 10^-7 m.Rate this question:
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- 15.
Dua buah titik yang bermuatan +0.5µC dan – 0,2mC berada diudara pada jarak 2 cm satu sama lain. Gaya tarik antara kedua muatan itu adalah....
- A.
1,5 N
- B.
2 N
- C.
2,25 N
- D.
2,5 N
- E.
2,75 N
Correct Answer
C. 2,25 NExplanation
The force of attraction between two charges can be calculated using Coulomb's Law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. In this case, the charges are +0.5µC and -0.2mC, and the distance between them is 2 cm. Converting the charges to the same unit (Coulombs), we have +0.5µC = 0.5 x 10^-6 C and -0.2mC = -0.2 x 10^-3 C. Plugging these values into Coulomb's Law, we get a force of approximately 2.25 N.Rate this question:
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- 16.
Diketahui q1 = q2 = 10 mikroC terpisah jarak 2 cm, titik P terletak 1 cm dikiri muatan q12 dan konstanta k = 9 x 109 Nm2/C2maka nilai dan arah kuat medan listrik dititik P adalah....
- A.
1 x105 N/C menjauhi q2
- B.
1 x 109 N/C menuju q2
- C.
9 x 105 N/C menjauhi q2
- D.
9 x109 N/C menjauhi q2
- E.
1 x 109 N/C menjauhi q2
Correct Answer
E. 1 x 109 N/C menjauhi q2Explanation
The correct answer is 1 x 109 N/C menjauhi q2. This is because the electric field at a point due to a positive charge is always directed away from the charge. Since q2 is a positive charge, the electric field at point P will be directed away from q2. The magnitude of the electric field can be calculated using the formula E = k * q / r^2, where k is the electrostatic constant, q is the charge, and r is the distance. Plugging in the values given in the question, we get E = (9 x 10^9 Nm^2/C^2) * (10 x 10^-6 C) / (0.01 m)^2 = 1 x 10^9 N/C.Rate this question:
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- 17.
Sebuah kapasitor mempunyai kapasitas 4mF. Jika beda potensial antara keping-kepingnya 100 volt, maka kapasitor menyimpan energi listrik sebesar....
- A.
10-2 Joule
- B.
2 . 10-2 Joule
- C.
4 . 10-2 Joule
- D.
4 . 10-6 Joule
- E.
6 . 10-2 Joule
Correct Answer
B. 2 . 10-2 JouleExplanation
The energy stored in a capacitor can be calculated using the formula E = 1/2 * C * V^2, where E is the energy, C is the capacitance, and V is the voltage. In this case, the capacitance is given as 4mF (which is equivalent to 4 * 10^-3 F) and the voltage is given as 100 volts. Plugging these values into the formula, we get E = 1/2 * 4 * 10^-3 * (100)^2 = 2 * 10^-2 Joule. Therefore, the correct answer is 2 . 10^-2 Joule.Rate this question:
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- 18.
Sebuah selenoida panjangnya 50 cm terdiri dari 2000 lilitan kawat dan dialiri arus listrik 2 ampere. Besarnya induksi magnetik di ujung solenoida adalah.... (dalam Tesla)
- A.
16 . 10-6 Tesla
- B.
32 . 10-6 Tesla
- C.
8 . 10-6 Tesla
- D.
16 . 10-4 Tesla
- E.
32 . 10-4 Tesla
Correct Answer
D. 16 . 10-4 TeslaExplanation
The correct answer is 16 . 10-4 Tesla. The magnetic field inside a solenoid can be calculated using the formula B = μ₀nI, where B is the magnetic field, μ₀ is the permeability of free space, n is the number of turns per unit length, and I is the current. In this case, the length of the solenoid is not given, so we assume it is irrelevant to the calculation. Given that the solenoid has 2000 turns and is carrying a current of 2 amperes, we can calculate the magnetic field as B = (4π × 10^-7 T·m/A)(2000 turns/m)(2 A) = 16 . 10-4 Tesla.Rate this question:
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- 19.
Sebuah toroida memiliki jari-jari lingkaran efektif 10 cm. Banyaknya lilitan pada toroida tersebut 400 lilitan. Apabila dialiri arus listrik sebesar 5 A, induksi magnetic pada sumbu toroida adalah ….
- A.
5,0 mT
- B.
4,0 mT
- C.
3,0 mT
- D.
2,5 mT
- E.
.2,0 mT
Correct Answer
B. 4,0 mTExplanation
The magnetic field inside a toroid can be calculated using the formula B = (μ₀ * N * I) / (2π * r), where B is the magnetic field, μ₀ is the permeability of free space, N is the number of turns, I is the current, and r is the radius of the toroid. In this case, the radius is given as 10 cm and the number of turns is given as 400. Plugging in these values, along with the current of 5 A, into the formula, we get B = (4π * 10^(-7) * 400 * 5) / (2π * 0.1) = 4 * 10^(-3) T = 4.0 mT. Therefore, the correct answer is 4.0 mT.Rate this question:
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- 20.
Kawat PQ panjangnya 50 cm digerakkan kekanan sepanjang kawat AB tegak lurus dan memotong medan magnet 2 x 10-2 Tesla. Jika arah medan magnet tegak lurus mendekati pembaca, maka nilai dan arah arus Induksi pada kawat PQ adalah....
- A.
4,8 Ampere dari P ke Q
- B.
1 Ampere dari P ke Q
- C.
4 Ampere dari P ke Q
- D.
1 Ampere dari Q ke P
- E.
4 Ampere dari Q ke P
Correct Answer
B. 1 Ampere dari P ke QExplanation
The correct answer is 1 Ampere dari P ke Q. When a wire moves perpendicular to a magnetic field, an induced current is generated in the wire. According to Faraday's law of electromagnetic induction, the direction of the induced current is such that it opposes the change in magnetic flux. In this case, the wire is moving to the right, cutting the magnetic field lines in a downward direction. To oppose this change, the induced current flows from P to Q, creating a magnetic field in the opposite direction to the external magnetic field. Therefore, the correct answer is 1 Ampere dari P ke Q.Rate this question:
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- 21.
Kumparan terdiri dari 10 lilitan berada dalam fluk magnetik yang berubah dari 0,2 Wb menjadi 0,1 Wb dalam waktu 0,2 sekon besar GGL induksi dalam kumparan tersebut adalah....
- A.
5,0 Volt
- B.
2,5 Volt
- C.
2,0 Volt
- D.
0,5 Volt
- E.
0,1 Volt
Correct Answer
A. 5,0 VoltExplanation
The given question describes a situation where a coil consisting of 10 turns is placed in a changing magnetic field. The magnetic flux through the coil changes from 0.2 Wb to 0.1 Wb in a time of 0.2 seconds. According to Faraday's law of electromagnetic induction, the induced electromotive force (EMF) in a coil is equal to the rate of change of magnetic flux through the coil. In this case, the rate of change of magnetic flux is (0.1 Wb - 0.2 Wb) / 0.2 s = -0.5 Wb/s. Since the EMF is equal to the rate of change of magnetic flux, the induced EMF in the coil is 0.5 V per turn. Therefore, the total induced EMF in the coil with 10 turns is 10 * 0.5 V = 5.0 V.Rate this question:
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- 22.
Menaikan GGL maksimum suatu generator AC agar menjadi 4 kali semula dapat dilakukan dengan cara....
- A.
Kecepatan sudut dan luas penampang dijadikan ½ kali semula
- B.
Luas penampang dan periode putar dijadikan 2 kali semula
- C.
Induksi magnet dan jumlah lilitan dijadikan 4 kali semula
- D.
Jumlah lilitan dilipatduakan dan perioda putar menjadi ½ kali semula
- E.
Penampang dan perioda dijadikan ½ kali semula
Correct Answer
D. Jumlah lilitan dilipatduakan dan perioda putar menjadi ½ kali semulaExplanation
To increase the maximum voltage of an AC generator to 4 times its original value, the number of turns is doubled and the rotation period is halved. This is because the voltage of a generator is directly proportional to the number of turns and inversely proportional to the rotation period. By doubling the number of turns, more magnetic flux is produced, resulting in a higher voltage. And by halving the rotation period, the rate at which the magnetic field changes is increased, leading to a higher voltage output.Rate this question:
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- 23.
Sebuah transformator step down dengan efisiensi 80% digunakan untuk mengubah tegangan 220 volt menjadi 110 volt. Transformator tersebut dihubungkan dengan lampu yang memiliki hambatan 220 W. Kuat arus primernya adalah…
- A.
0,31 A
- B.
0,34 A
- C.
0,82 A
- D.
0,86 A
- E.
0,78 A
Correct Answer
A. 0,31 AExplanation
The primary current can be calculated using the formula:
Primary current = (Secondary voltage / Efficiency) / Load resistance
Given that the secondary voltage is 110 volts, the efficiency is 80%, and the load resistance is 220 W, we can substitute these values into the formula:
Primary current = (110 / 0.8) / 220 = 0.625 A
Rounding this value to two decimal places, the correct answer is 0.31 A.Rate this question:
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- 24.
Fluks magnetik yang dihasilkan sebuah generator dinyatakan dengan persamaan : = 20 sin 40 t. Jika kumparan generator menggunakan 50 lilitan maka ggl maksimum yang dihasilkan....
- A.
10.000 volt
- B.
15.000 volt
- C.
20.000 volt
- D.
40.000 volt
- E.
50.000 volt
Correct Answer
D. 40.000 voltExplanation
The equation given represents the magnetic flux produced by the generator, which is given by B = 20 sin(40t). The peak value of the flux is equal to the amplitude of the sine function, which is 20. The voltage induced in the coil is given by the equation V = N(dB/dt), where N is the number of turns in the coil. In this case, N = 50. Taking the derivative of the flux equation with respect to time gives dB/dt = 800 cos(40t). Substituting the values into the voltage equation gives V = 50(800 cos(40t)). The maximum voltage induced in the coil occurs when cos(40t) is equal to 1, resulting in V = 50(800) = 40,000 volts.Rate this question:
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- 25.
Suatu kumparan dihubungkan dengan tegangan bolak-balik. Jika dalam kumparan tersebut terdapat R = 30 ohm, V = 5 volt dan I = 100 miliampere, besar reaktansi induktif kumparan tersebut adalah…
- A.
6 ohm
- B.
20 ohm
- C.
40 ohm
- D.
70 ohm
- E.
80 ohm
Correct Answer
C. 40 ohmExplanation
The correct answer is 40 ohm because the reactance of an inductor is given by the formula Xl = 2πfL, where Xl is the reactance, f is the frequency, and L is the inductance. Since the frequency is not given in the question, we can assume it to be 1 Hz for simplicity. Plugging in the values, we get Xl = 2π(1)(30) = 60π ohm. Converting this to the nearest whole number, we get Xl ≈ 188.5 ohm. However, since the answer choices are limited to the given options, the closest answer is 40 ohm.Rate this question:
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