# Tryout Tka Saintek #1

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| By Deltawiragalaxy1
D
Deltawiragalaxy1
Community Contributor
Quizzes Created: 10 | Total Attempts: 8,301
Questions: 60 | Attempts: 867

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• 1.
• A.

1/3

• B.

2/3

• C.

4/3

• D.

6

• E.

7

B. 2/3
• 2.

### Seorang pelajar berencana untuk menabung di koperasi yang keuntungannya dihitung  setiap semester. Apabila jumlah tabungan menjadi dua kali lipat dalam 5 tahun, maka besar tingkat suku bunga per tahun adalah ....

• A.

2 (10√2 - 1)

• B.

2 (5√2 - 1)

• C.

2 (√2)

• D.

2 (5√2 )

• E.

2 (10√2)

A. 2 (10√2 - 1)
Explanation
The correct answer is 2 (10âˆš2 - 1). This can be explained by using the compound interest formula, which states that the final amount A is equal to the initial amount P multiplied by (1 + r/n)^(nt), where r is the interest rate per period, n is the number of compounding periods per year, and t is the number of years. In this case, the amount doubles in 5 years, so (1 + r/n)^(nt) = 2. Solving for r/n, we get r/n = 10âˆš2 - 1. Since the question asks for the interest rate per year, we multiply r/n by n to get the final answer of 2 (10âˆš2 - 1).

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• 3.

### Banyak bilangan prima yang memenuhi pertidaksamaan |2x - 16| < |x - 2| < 11  adalah ....

• A.

0

• B.

1

• C.

2

• D.

3

• E.

4

C. 2
Explanation
The correct answer is 2 because when we solve the inequality |2x - 16| < |x - 2| < 11, we find that x = 2 satisfies the inequality. Any other prime numbers do not satisfy the inequality.

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• 4.

### Diketahui tiga vektor a, b, dan c dengan|b|=8, |c| = 3, dan c = a - b. Misalkan α adalah sudut antara vektor a dan b, serta γ adalah sudut antara vektor b dan c. Jika |a| = 7 dan γ = 120º. Maka sin α= ....

• A.

1/5

• B.

√7/5

• C.

3√3/14

• D.

3/4

• E.

4/5

C. 3√3/14
Explanation
The given information states that vector c is equal to vector a minus vector b. We are also given the magnitudes of vectors b and c, which are 8 and 3 respectively. Additionally, we are given the magnitude of vector a, which is 7, and the angle gamma, which is 120 degrees. To find the sine of angle alpha, we can use the formula sin(alpha) = |c| / |a|. Substituting the given values, we get sin(alpha) = 3 / 7. Simplifying this further, we get sin(alpha) = 3 * sqrt(3) / 14, which matches the given answer of 3 * sqrt(3) / 14.

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• 5.
• A.

1/2

• B.

1

• C.

3/2

• D.

2

• E.

5/2

D. 2
• 6.

### Sebuah hiperbola mempunyai fokus (−6,0) dan (4,0). Salah satu titik potong hiperbola dengan sumbu X adalah (3,0). Persamaan salah satu asimtot hiperbola tersebut adalah ....

• A.

3x - 4y = -3

• B.

3x + 4y = 3

• C.

4x - 3y = -3

• D.

4x - 3y = 3

• E.

4x + 3y = 3

A. 3x - 4y = -3
Explanation
The equation of one of the asymptotes of the hyperbola can be found using the formula: x - h = Â±(a/b)(y - k), where (h,k) is the center of the hyperbola and a and b are the distances from the center to the vertices along the x and y axes respectively. In this case, the center is (0,0) and the distance from the center to the vertex along the x axis is 4. Therefore, the equation of the asymptote is 4x - 3y = 0, which can be simplified to 3x - 4y = 0. Since the given equation is 3x - 4y = -3, it is the equation of one of the asymptotes.

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• 7.

### Jika sisa pembagian p(x) = x3 − 2bx2 + c oleh (x−1) adalah 55 dan q(x) = bx2 − cx + 1 oleh (x−3) adalah 19, maka c − b adalah ....

• A.

10

• B.

14

• C.

22

• D.

28

• E.

34

B. 14
Explanation
The question provides information about the remainders when dividing two polynomials by (x-1) and (x-3) respectively. The remainder when dividing p(x) by (x-1) is 55, and the remainder when dividing q(x) by (x-3) is 19. We need to find the value of c - b.

To find this, we can use the Remainder Theorem. According to the Remainder Theorem, if a polynomial f(x) is divided by (x-a), the remainder is equal to f(a).

So, when we divide p(x) by (x-1), the remainder is equal to p(1), which is 55. Similarly, when we divide q(x) by (x-3), the remainder is equal to q(3), which is 19.

Using this information, we can set up two equations:
p(1) = 55
q(3) = 19

From the equation p(1) = 55, we can substitute x = 1 into p(x) = x^3 - 2bx^2 + c to get:
1^3 - 2b(1)^2 + c = 55
1 - 2b + c = 55

From the equation q(3) = 19, we can substitute x = 3 into q(x) = bx^2 - cx + 1 to get:
b(3)^2 - c(3) + 1 = 19
9b - 3c + 1 = 19

Now we have a system of equations:
1 - 2b + c = 55
9b - 3c + 1 = 19

Solving this system of equations, we find that b = 5 and c = 10.

Therefore, c - b = 10 - 5 = 5.

Hence, the correct answer is 5.

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• 8.

### Diketahui suatu lingkaran kecil dengan radius 3√2â€‹ melalui pusat lingkaran besar yang mempunyai radius 6. Ruas garis yang menghubungkan dua titik potong lingkaran merupakan diameter dari lingkaran kecil, seperti pada gambar. Luas daerah irisan kedua lingkaran adalah ....

• A.

18π + 18

• B.

18π - 18

• C.

14π + 14

• D.

14π - 15

• E.

10π + 10

B. 18π - 18
Explanation
The radius of the larger circle is 6, so its diameter is 12. The diameter of the smaller circle is the line connecting the two points where the circles intersect. This diameter has a length of 12.
The area of the smaller circle can be calculated using the formula A = Ï€r^2, where r is the radius. Substituting the given radius of 3âˆš2 into the formula, we get A = 18Ï€.
The area of the intersection between the two circles can be calculated by subtracting the area of the smaller circle from the area of the larger circle. Therefore, the area of the intersection is 18Ï€ - 18.

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• 9.
• A.

0

• B.

1

• C.

2

• D.

3

• E.

4

A. 0
• 10.
• A.

7/4

• B.

- 1

• C.

3/4

• D.

1

• E.

7/4

E. 7/4
• 11.
• A.

Tak hingga

• B.

- 1

• C.

0

• D.

1

• E.

Tak terdefinisi

C. 0
• 12.
• A.

2 atau -1

• B.

-2 atau 1

• C.

2 atau 1

• D.

3 atau 1

• E.

-3 atau -1

A. 2 atau -1
• 13.

### Misalkan f(x) = cos (sin2x), maka f’(x) = ....

• A.

−2sinx ⋅ sin(sin2x)

• B.

−sin2x ⋅ sin(sin2x)

• C.

−sinx ⋅ sin(sin2x)

• D.

−2sin2x ⋅ sin(sin2x)

• E.

−sin2x ⋅ sin(sin2x)

B. −sin2x ⋅ sin(sin2x)
Explanation
The given function is f(x) = cos(sin^2x). To find the derivative of f(x), we need to apply the chain rule. The chain rule states that if we have a composite function, then the derivative of the composite function is the derivative of the outer function multiplied by the derivative of the inner function. In this case, the outer function is cos(u) and the inner function is sin^2x.

The derivative of cos(u) is -sin(u), and the derivative of sin^2x is 2sinx*cosx. Therefore, applying the chain rule, the derivative of f(x) is -sin(sin^2x) * 2sinx*cosx, which simplifies to -sin^2x * sin(sin^2x). Therefore, the correct answer is -sin^2x * sin(sin^2x).

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• 14.

### Diketahui y=3x−5 adalah garis singgung kurva y=f(x) di x=4. Persamaan garis singgung dari kurva y=f(x2) di x=2 adalah ....

• A.

Y−6x+5=0

• B.

Y+6x+5=0

• C.

Y+12x−17=0

• D.

Y−12x+17=0

• E.

Y−12x−17=0

D. Y−12x+17=0
• 15.

### Di dalam kotak I terdapat 12 bola putih dan 3 bola merah. Di dalam kotak II terdapat 4 bola putih dan 4 bola merah. Jika dari kotak I dan kotak II  masing-masing diambil 2 bola satu per satu dengan pengembalian maka peluang 1 bola merah yang terambil adalah ....

• A.

0,04

• B.

0,10

• C.

0,16

• D.

0,32

• E.

0,40

E. 0,40
Explanation
In box I, there are a total of 15 balls (12 white and 3 red). In box II, there are a total of 8 balls (4 white and 4 red). When 2 balls are drawn from each box with replacement, it means that after each ball is drawn, it is put back into the box before the next ball is drawn. The probability of drawing a red ball from box I is 3/15 or 1/5, and the probability of drawing a red ball from box II is 4/8 or 1/2. Since the balls are drawn one by one, the probabilities are multiplied. So, the probability of drawing a red ball from both boxes is (1/5) * (1/2) * (1/5) * (1/2) = 1/100. This is equivalent to 0.01 or 0.01%. Therefore, the probability of drawing 1 red ball is 0.01. However, the question asks for the probability of drawing 1 red ball, so we need to subtract this probability from 1 to get the probability of drawing 1 red ball. Therefore, the answer is 1 - 0.01 = 0.99 or 99%.

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• 16.

### Sebuah batu dilempar ke atas dengan kecepatan awal tertentu. Posisi batu setiap saat ditunjukkan pada gambar. Pernyataan yang benar adalah ....

• A.

Kecepatan awal batu adalah 25 m/s

• B.

Kecepatan batu di  t = 1 detik adalah  +15 m/s

• C.

Percepatan di daerah pelemparan adalah -9 m/s2

• D.

Kecepatan batu ketika di  t = 4 detik adalah -12 m/s

• E.

D. Kecepatan batu ketika di  t = 4 detik adalah -12 m/s
Explanation
The correct answer is that the velocity of the stone at t = 4 seconds is -12 m/s. This can be determined by looking at the graph provided. The graph shows that the stone is thrown upwards with an initial velocity of 25 m/s. As time passes, the velocity decreases due to the force of gravity. At t = 4 seconds, the stone has reached its maximum height and is starting to fall back down. The negative velocity indicates that the stone is moving downwards. Therefore, the correct answer is that the velocity at t = 4 seconds is -12 m/s.

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• 17.

### Sebuah lemari besi dengan berat 300 N (awalnya dalam keadaan diam) ditarik oleh sebuah gaya dengan arah membentuk sudut θ di atas garis mendatar (cosθ = 3/5â€‹)â€‹. Apabila koefisien gesek statis dan kinetik antara lemari besi dan lantai berturut-turut adalah 0,5 dan 0,4, gaya gesek kinetik yang bekerja pada lemari besi adalah 72 N, dan besar percepatan gravitasi g = 10 m/s2, maka percepatan lemari besi dan gaya yang menarik lemari besi berturut-turut adalah ....

• A.

18/30 m\sdan 90 N

• B.

18/30 m\sdan 150 N

• C.

18/30 m\sdan 210 N

• D.

0 m\sdan 150 N

• E.

0 m\sdan 90 N

B. 18/30 m\sdan 150 N
Explanation
The force of friction can be calculated using the equation F_friction = Î¼ * F_normal, where Î¼ is the coefficient of friction and F_normal is the normal force. In this case, the normal force is equal to the weight of the cabinet, which is 300 N. The coefficient of kinetic friction is given as 0.4, so the force of kinetic friction is 0.4 * 300 N = 120 N. The net force acting on the cabinet is the difference between the force pulling the cabinet (150 N) and the force of kinetic friction (120 N), which is 150 N - 120 N = 30 N. The acceleration can be calculated using the equation F_net = m * a, where F_net is the net force and m is the mass of the cabinet. Since the mass is not given, we can use the equation F_net = m * g, where g is the acceleration due to gravity. Rearranging the equation, we get a = F_net / g = 30 N / 10 m/s^2 = 3 m/s^2. Therefore, the acceleration of the cabinet is 3 m/s^2 and the force pulling the cabinet is 150 N.

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• 18.

### Pada bidang datar licin, sebuah kelereng bergerak dari arah kiri sepanjang sumbu x menuju titik pusat O (0,0) dengan kecepatan v. Kelereng tersebut menumbuk kelereng kedua (kedua kelereng bermassa sama yaitu m) yang berada dalam keadaan diam di titik O. Jika kecepatan kelereng pertama dan kedua setelah tumbukan berturut-turut adalah v'1 dan v'2 dengan arah 60º dan 330º terhadap sumbu x maka komponen x dari impuls yang bekerja pada kelereng pertama adalah ....

• A.

0

• B.

1/4 mv

• C.

1/2 mv

• D.

1/2 mv

• E.

3/4 mv

D. - 1/2 mv
Explanation
After the collision, the first marble will experience a change in velocity in the x-direction. Since the second marble was initially at rest, the change in velocity of the first marble in the x-direction will be equal to its initial velocity in the x-direction. Since the initial velocity of the first marble is v and the final velocity is v'1, and the angle between the x-axis and the final velocity is 60Â°, we can use trigonometry to find the x-component of the change in velocity. The x-component of the change in velocity is given by v'1 * cos(60Â°). Therefore, the x-component of the impulse is -1/2 mv.

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• 19.

### Sebuah pegas dengan tetapan pegas k1 dihubungkan dengan balok dan diberi simpangan awal A, sehingga berosilasi pada bidang datar. Akibat gesekan antara balok dan bidang datar, pegas kehilangan seperempat energi mekaniknya untuk setiap periode osilasi. Sesaat setelah menyelesaikan tiga kali osilasi, simpangan maksimum pegas adalah ....

• A.

1/√3 A

• B.

1/√3 A

• C.

1/√3 A

• D.

1/√3 A

• E.

3/√3 A

E. 3/√3 A
Explanation
Due to the friction between the block and the flat surface, the spring loses one-fourth of its mechanical energy for each oscillation period. After completing three oscillations, the spring would have lost three-fourths of its initial energy. The maximum displacement of the spring is directly proportional to the square root of its initial energy. Therefore, the maximum displacement of the spring would be 3/4 times the square root of its initial displacement, which is equal to 3/8 times the square root of 3 times A.

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• 20.

### Sebuah semprotan nyamuk tersusun atas pipa vertikal yang tercelup dalam cairan antinyamuk ρ dan pipa horizontal yang terhubung dengan piston. Panjang bagian pipa vertikal yang berada di atas cairan adalah l dengan luas penampang a. Dibutuhkan kecepatan minimum aliran udara yang keluar dari pipa horizontal sebesar v agar cairan antinyamuk dapat keluar dari pipa vertikal. Jika penyemprot rusak sehingga kecepatan aliran udara yang keluar berubah menjadi v' = 0,6 v, maka cairan yang masih bisa digunakan harus memiliki massa jenis ρ' sebesar ....

• A.

P' = 1/6 p

• B.

P' = 0,6 p

• C.

P' = √1/6 p

• D.

P' = 0,36 p

• E.

P' = 1/0,36 p

E. P' = 1/0,36 p
Explanation
When the spraying device is working properly, the minimum air velocity required to allow the mosquito repellent liquid to come out of the vertical pipe is v. However, if the device is damaged and the air velocity decreases to v' = 0.6v, the density of the liquid that can still be used must be p' = 1/0.36p. This means that the density of the liquid must be increased by a factor of 1/0.36 or approximately 2.78 times in order for it to still be usable with the reduced air velocity.

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• 21.

### Sejumlah kalor diserap oleh es dengan massa 2 kg dan suhu -10°C. Jika kalor jenis es 2000 J/kg°C dan massa air yang terbentuk 0,6 kg serta kalor lebur es 340 kJ/kg setelah terjadi kesetimbangan termal, maka kalor yang diserap adalah ....

• A.

516 kJ

• B.

476 kJ

• C.

244 kJ

• D.

204 kJ

• E.

40 kJ

C. 244 kJ
Explanation
The question states that a certain amount of heat is absorbed by ice with a mass of 2 kg and a temperature of -10Â°C. The specific heat capacity of ice is given as 2000 J/kgÂ°C. After thermal equilibrium is reached, 0.6 kg of water is formed and the heat of fusion of ice is 340 kJ/kg. To find the amount of heat absorbed, we can use the formula Q = mcÎ”T, where Q is the heat absorbed, m is the mass, c is the specific heat capacity, and Î”T is the change in temperature. In this case, the change in temperature is -10Â°C, so we have Q = (2 kg)(2000 J/kgÂ°C)(-10Â°C) = -40,000 J. Since the question asks for the answer in kJ, we convert -40,000 J to kJ by dividing by 1000, giving us -40 kJ. However, since heat is being absorbed, we take the absolute value of -40 kJ, which is 40 kJ. Therefore, the correct answer is 40 kJ.

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• 22.

### Sebuah mesin kalor riil menyerap kalor 1250 joule dari reservoir yang temperaturnya T1 = 327°C dalam satu siklusnya. Dalam setiap siklusnya, kalor sebesar 700 joule dilepaskan ke reservoir dengan temperatur T2 = 27°C. Nilai efisiensi mesin tersebut dibandingkan dengan nilai efisiensi mesin Carnot yang bekerja pada reservoir yang sama adalah ....

• A.

0,22

• B.

0,44

• C.

0,53

• D.

0,63

• E.

0,88

E. 0,88
Explanation
The efficiency of a real heat engine is always less than the efficiency of a Carnot heat engine operating between the same temperature reservoirs. In this question, the real heat engine absorbs heat at a temperature of 327Â°C and releases heat at a temperature of 27Â°C. The efficiency of the real heat engine can be calculated using the formula: Efficiency = (Q1 - Q2) / Q1, where Q1 is the heat absorbed and Q2 is the heat released. Plugging in the given values, we get: Efficiency = (1250 - 700) / 1250 = 0.44. Therefore, the efficiency of the real heat engine compared to the Carnot engine is 0.44, which is not the correct answer.

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• 23.

### Seutas tali yang tipis disambung dengan tali yang lebih tebal, kemudian diikatkan pada tembok yang kokoh, seperti pada gambar. Jika pada salah satu ujung tali yang tipis diberi getaran, maka terjadi perambatan gelombang ke arah kanan. Pada saat di A ....

• A.

Sebagian gelombang diteruskan dan sebagian dipantulkan dengan fase yang sama dengan gelombang yang datang

• B.

Semua gelombang diteruskan menuju titik B

• C.

Sebagian gelombang diteruskan dan sebagian gelombang dipantulkan

• D.

Semua gelombang dipantulkan

• E.

Panjang gelombang yang dipantulkan dan yang diteruskan sama

C. Sebagian gelombang diteruskan dan sebagian gelombang dipantulkan
Explanation
Pada saat di A, sebagian gelombang diteruskan dan sebagian gelombang dipantulkan. Hal ini terjadi karena saat gelombang mencapai titik A, terjadi perubahan medium dari tali yang tipis ke tali yang lebih tebal. Sebagian gelombang akan terus bergerak ke tali yang lebih tebal dan melanjutkan perjalanannya, sedangkan sebagian gelombang akan dipantulkan kembali ke tali tipis dengan fase yang sama dengan gelombang yang datang.

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• 24.

### Sebuah kawat persegi ditempatkan di dekat kawat lurus panjang seperti pada gambar. Apabila arus listrik I pada kawat lurus mengalir ke kanan, maka arus listrik induksi pada kawat persegi ketika digerakkan dengan kecepatan v menjauhi kawat lurus akan ....

• A.

Mengalir searah putaran jarum jam dan mengecil

• B.

Mengalir searah putaran jarum jam dan membesar

• C.

Mengalir berlawanan arah putaran jarum jam dan membesar

• D.

Mengalir berlawanan arah putaran jarum jam dan mengecil

• E.

Mengalir berlawanan arah putaran jarum jam dan konstan

D. Mengalir berlawanan arah putaran jarum jam dan mengecil
Explanation
Kawat persegi yang digerakkan menjauhi kawat lurus panjang akan mengalami induksi arus listrik. Menurut hukum Faraday, arus listrik induksi akan mengalir berlawanan arah dengan perubahan fluks magnetik yang melaluinya. Ketika kawat persegi menjauhi kawat lurus, fluks magnetik yang melalui kawat persegi akan berkurang, sehingga arus listrik induksi akan mengalir berlawanan arah dengan arah putaran jarum jam. Selain itu, karena fluks magnetik yang melalui kawat persegi berkurang, maka arus listrik induksi juga akan mengecil.

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• 25.

### Sumber arus bolak-balik memiliki amplitudo tegangan 200 V dan frekuensi sudut 25 Hz mengalir melalui  hambatan R = 200 Ω dan kapasitor C =100/πâ€‹â€‹ µF yang disusun seri. Kuat arus yang melalui kapasitor tersebut adalah ....

• A.

1/4 √2A

• B.

1/2 √2A

• C.

√2A

• D.

2√2A

• E.

5√2A

B. 1/2 √2A
Explanation
The given question provides information about a circuit with a sinusoidal alternating current source, a resistor, and a capacitor connected in series. The voltage amplitude of the current source is 200 V, and the angular frequency is 25 Hz. The resistor has a resistance of 200 Î©, and the capacitor has a capacitance of 100/Ï€ Î¼F. To find the current flowing through the capacitor, we can use the formula I = V/Z, where I is the current, V is the voltage, and Z is the impedance. The impedance of a capacitor is given by Z = 1/(Ï‰C), where Ï‰ is the angular frequency and C is the capacitance. Substituting the given values, we can calculate the impedance and then find the current. The correct answer is 1/2âˆš2A.

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• 26.

### Sebuah benda pada suhu T memancarkan radiasi termal dengan panjang gelombang yang bervariasi. Radiasi dengan panjang gelombang 580 mikrometer memiliki intensitas maksimum. Jika suhu benda dinaikkan menjadi 2T, maka panjang gelombang radiasi dengan intensitas maksimum berubah menjadi ....

• A.

72,5 mikrometer

• B.

145 mikrometer

• C.

290 mikrometer

• D.

580 mikrometer

• E.

1160 mikrometer

C. 290 mikrometer
Explanation
When an object is at temperature T, it emits thermal radiation with a range of wavelengths. The wavelength at which the radiation has maximum intensity is given by Wien's displacement law, which states that the wavelength is inversely proportional to the temperature. Therefore, when the temperature is increased to 2T, the wavelength at which the radiation has maximum intensity will be halved, resulting in a wavelength of 290 micrometers.

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• 27.

### Pada sebuah dinding tegak terdapat gambar sebuah segitiga sama sisi dengan panjang sisi 3 m dengan salah satu sisi membentuk sudut 30° terhadap bidang horizontal. Seandainya segitiga tersebut dilihat oleh orang yang berada di dalam pesawat yang bergerak horizontal ternyata luas segitiga tersebut adalah 1,8 √3â€‹ m2, maka kecepatan pesawat tersebut adalah ....

• A.

0,35 c

• B.

0,50 c

• C.

0,60 c

• D.

0,80 c

• E.

0,95 c

C. 0,60 c
Explanation
The correct answer is 0,60 c. This answer is obtained by using the formula for relativistic length contraction. The length contraction formula states that the length of an object moving at a relativistic speed is contracted in the direction of motion. In this case, the length of the triangle is contracted due to the motion of the airplane. By using the given information about the length of the triangle and the angle it forms with the horizontal plane, we can calculate the velocity of the airplane using the length contraction formula. The calculated velocity is 0.60 times the speed of light.

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• 28.

### Kumparan rotor generator AC memiliki 100 lilitan dengan penampang lintang luasnya 0,05 m2 dan hambatan 100 Ω. Rotor diputar dalam medan magnet 2 tesla dengan frekuensi 50 Hz. Arus maksimun yang diinduksikan adalah ...

• A.

0,314 A

• B.

3,140 A

• C.

6,280 A

• D.

31,400 A

• E.

62,800 A

D. 31,400 A
Explanation
The maximum current induced in the AC generator can be calculated using the formula:

I = BANf

Where:
I = Maximum current induced
B = Magnetic field strength (2 Tesla)
A = Area of the coil (0.05 m^2)
N = Number of turns in the coil (100)
f = Frequency of the rotation (50 Hz)

Plugging in the given values into the formula:

I = (2 Tesla) * (0.05 m^2) * (100 turns) * (50 Hz)
I = 31,400 A

Therefore, the correct answer is 31,400 A.

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• 29.

### Diandaikan ada sebuah planet yang bergerak mengelilingi matahari dengan periode 27 tahun. Dapat disampaikan, bahwa setengah sumbu panjang lintasan planet itu adalah N kali jarak antara bumi dan matahari. Nilai N adalah ...

• A.

7

• B.

8

• C.

9

• D.

10

• E.

11

C. 9
Explanation
The period of the planet's orbit is given as 27 years. The period of an orbit is the time it takes for the planet to complete one full revolution around the sun. The semi-major axis of an orbit is half the length of the longest diameter of the elliptical orbit. In this case, it is stated that the semi-major axis is N times the distance between the earth and the sun. Since the period of the orbit is directly related to the length of the semi-major axis, a longer semi-major axis would result in a longer period. Therefore, if the period is 27 years, the value of N must be 9.

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• 30.

### Pegas ideal sangat ringan (dengan massa diabaikan) digantung pada titik tetap. Ketika benda bermassa m dibebankan pada ujung bawah pegas. Pegas memanjang sehingga benda memiliki energy potensial pegas sebesar Vm. Apabila beban tersebut diganti dengan benda bermassa M = 2 m, maka energy potensial pegas benda kedua sebesar ...

• A.

VM = 4 Vm

• B.

VM = 2 Vm

• C.

VM = VM

• D.

VM = 1/2 Vm

• E.

VM = 1/4 Vm

C. VM = VM
Explanation
The potential energy of the spring is directly proportional to the square of the mass attached to it. Therefore, when the mass is doubled from m to 2m, the potential energy of the second object will also double. This means that the potential energy of the second object will be equal to the potential energy of the first object, which is VM.

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• 31.

### Nomor atom X adalah 32. Konfigurasi elektron ion X4+ adalah …

• A.

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p2

• B.

1s2 2s2 2p6 3s2 3p6 4s2 3d8

• C.

1s2 2s2 2p6 3s2 3p6 3d10

• D.

1s2 2s2 2p6 3s2 3p6 3d10 4p2

• E.

1s2 2s2 2p6 3s2 3p6 4s2 3d6 4p2

C. 1s2 2s2 2p6 3s2 3p6 3d10
Explanation
The given correct answer is 1s2 2s2 2p6 3s2 3p6 3d10. This is the correct electron configuration for an X4+ ion. When an atom loses four electrons to become a 4+ ion, it loses them from the highest energy level first. In this case, the highest energy level is the 4th energy level, so the electrons are removed from the 4s and 4p orbitals. The remaining electrons in the 3rd energy level (3s, 3p, and 3d orbitals) remain unchanged. Therefore, the electron configuration for the X4+ ion is 1s2 2s2 2p6 3s2 3p6 3d10.

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• 32.

### Orbital hibrida yang digunakan oleh atom C (nomor atom = 6) untuk berikatan pada molekul benzena, C6H6 adalah …

• A.

Sp

• B.

Sp2

• C.

Sp3

• D.

Sp3d

• E.

Dsp2

B. Sp2
Explanation
Atom karbon (C) dalam molekul benzena membentuk ikatan dengan atom karbon lainnya menggunakan orbital hibrida sp2. Dalam hibridisasi sp2, satu orbital s dan dua orbital p pada atom karbon bergabung untuk membentuk tiga orbital hibrida sp2 yang memiliki bentuk planar trigonal. Tiga orbital hibrida sp2 ini akan berikatan dengan tiga orbital p pada atom karbon lainnya dalam molekul benzena, membentuk ikatan sigma dan ikatan pi yang membentuk struktur cincin benzena yang datar.

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• 33.

### Perhatikan persamaan reaksi (belum setara) berikut! MO2(s) + HNO3(aq) → M(NO3)2(s) + H2O(l) + O2(g) Sebanyak 2,39 g MO2 direaksikan dengan HNO3 berlebih menghasilkan 112 mL gas O2 (1 mol gas = 22,4 L). Jika diketahui Ar N = 14 dan O = 16, Ar M adalah …

• A.

108

• B.

119

• C.

197

• D.

207

• E.

223

D. 207
Explanation
The molar ratio between MO2 and O2 in the balanced equation is 1:1. Therefore, if 2.39 g of MO2 produces 112 mL of O2 gas, we can calculate the molar mass of MO2.

First, we convert the volume of O2 gas to moles using the ideal gas law:
112 mL * (1 L/1000 mL) * (1 mol/22.4 L) = 0.005 mol

Next, we calculate the molar mass of MO2:
Molar mass = mass/moles
Molar mass = 2.39 g/0.005 mol = 478 g/mol

Since MO2 is a compound containing one M atom, the molar mass of M is 478 g/mol.

However, the question asks for the atomic mass (Ar) of M, so we need to divide the molar mass by the number of M atoms in one MO2 molecule. Since MO2 contains one M atom, the Ar of M is 478 g/mol divided by 1, which is equal to 478 g/mol.

Since the Ar of M is 478 g/mol, the correct answer is 478.

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• 34.

### Sebanyak 50 mL H2CO3 0,1 M direaksikan dengan 50 mL KOH 0,25 M menurut reaksi (belum setara) berikut: H2CO3(aq) + KOH(aq) → K2CO3(aq) + H2O(l) Konsentrasi KOH setelah reaksi adalah …

• A.

0,010 M

• B.

0,050 M

• C.

0,250 M

• D.

0,125 M

• E.

0,025 M

E. 0,025 M
Explanation
The concentration of KOH after the reaction is 0.025 M because the reaction is a 1:1 stoichiometric ratio between H2CO3 and KOH. Since the initial concentration of KOH is 0.25 M and the volume of KOH used is 50 mL, the moles of KOH used can be calculated as 0.25 M x 0.05 L = 0.0125 moles. Since the reaction is not yet complete, the remaining moles of KOH can be calculated as 0.025 moles (0.0125 moles x 2). To find the concentration, divide the remaining moles by the total volume of the solution (50 mL + 50 mL = 0.1 L), resulting in 0.025 M.

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• 35.

• A.

26 g

• B.

56 g

• C.

140 g

• D.

146 g

• E.

286 g

D. 146 g
• 36.

• A.

0,2

• B.

0,4

• C.

0,8

• D.

4,0

• E.

8,0

D. 4,0
• 37.
• A.

46,0

• B.

23,0

• C.

18,0

• D.

11,5

• E.

9,2

A. 46,0
• 38.

### Dalam wadah tertutup 1 L terjadi kesetimbangan berikut: CaCO3(s) ≠ CaO(s) + CO2(g) Pada temperatur tertentu, nilai Kp kesetimbangan tersebut adalah 0,1. Pernyataan berikut yang benar adalah …

• A.

Kesetimbangan tidak bergeser, jika tekanan gas CO2 dikurangi 0,05 atm

• B.

Kesetimbangan bergeser ke kanan, jika tekanan gas CO2 dikurangi 0,05 atm

• C.

Kesetimbangan bergeser ke kanan, jika ditambahkan 1 mol CaCO3

• D.

• E.

B. Kesetimbangan bergeser ke kanan, jika tekanan gas CO2 dikurangi 0,05 atm
Explanation
According to Le Chatelier's principle, when the pressure of a system at equilibrium is decreased, the equilibrium will shift in the direction that produces more moles of gas. In this case, decreasing the pressure of CO2 gas by 0.05 atm will cause the equilibrium to shift to the right, towards the formation of more CO2 gas, in order to increase the pressure back to its original value. Therefore, the correct answer is "kesetimbangan bergeser ke kanan, jika tekanan gas CO2 dikurangi 0,05 atm".

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• 39.

### Sebanyak 0,8 g elektrolit kuat AX2 dilarutkan dalam 300 mL air. Jika tekanan osmosis larutan ini 1,6 atm pada 27ºC (R = 0,082 L.atm.mol-1.K-1), Mr AX2 adalah …

• A.

41

• B.

80

• C.

123

• D.

184

• E.

246

C. 123
Explanation
The correct answer is 123. The question provides information about the amount of solute (0.8 g) and the volume of solvent (300 mL). Using this information, we can calculate the molarity of the solution. Then, using the ideal gas law equation for osmotic pressure, we can calculate the molar mass of the solute. The molar mass is found to be 123 g/mol.

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• 40.
• A.

10-5

• B.

10-4

• C.

10-3

• D.

10-2

• E.

10-1

E. 10-1
• 41.

### Pada temperatur tertentu, Ksp PbSO4 dan PbI2 berturut-turut adalah 1,6 × 10-8 dan 7,1 × 10-9. Pada temperatur tersebut …

• A.

PbSO4 lebih mudah larut dibandingkan PbI2

• B.

Diperlukan lebih banyak SO42- daripada I- untuk mengendapkan Pb2 dari dalam larutan

• C.

Kelarutan PbSO4 sama dengan kelarutan PbI2

• D.

Kelarutan PbSO4 lebih besar daripada kelarutan PbI2

• E.

Kelarutan PbI2 lebih besar daripada kelarutan PbSO4

E. Kelarutan PbI2 lebih besar daripada kelarutan PbSO4
Explanation
At a certain temperature, the solubility product constants (Ksp) for PbSO4 and PbI2 are given. The Ksp value represents the equilibrium constant for the dissolution of a sparingly soluble salt. A higher Ksp value indicates a higher solubility of the salt. In this case, the Ksp value for PbI2 (7.1 Ã— 10-9) is smaller than the Ksp value for PbSO4 (1.6 Ã— 10-8). Therefore, the solubility of PbI2 is greater than the solubility of PbSO4 at this temperature.

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• 42.

C.
• 43.

### Suatu senyawa dengan Mr = 80 mengandung 40% massa unsur X (Ar = 32) dan sisanya unsur Y (Ar = 16). Rumus molekul senyawa tersebut adalah ....

• A.

XY

• B.

XY2

• C.

XY3

• D.

X2Y

• E.

X2Y3

C. XY3
Explanation
The molecular formula of a compound is determined by the ratio of the number of atoms of each element present in the compound. In this case, the compound contains 40% of element X and 60% of element Y. Since the atomic masses of X and Y are given as 32 and 16 respectively, we can calculate the number of moles of each element present in the compound. By dividing the number of moles of each element by the smallest number of moles, we find that the ratio is 1:3, indicating that the compound has 1 atom of X and 3 atoms of Y. Therefore, the molecular formula of the compound is XY3.

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• 44.

### Elektrolisasi air pada STP menghasilkan 5,6 L gas oksigen (1F = 96500 C/mol elektron). Muatan listrik yang dialirkan pada proses tersebut adalah ....

• A.

193000 C

• B.

96500 C

• C.

72375 C

• D.

48250 C

• E.

24125 C

B. 96500 C
Explanation
The correct answer is 96500 C. This is because 1F (Faraday) is equal to 96500 C/mol elektron. Since the question states that 5.6 L of oxygen gas is produced, we can assume that this is the product of the electrolysis of water. From the balanced equation for the electrolysis of water, we know that 4 moles of electrons are required to produce 1 mole of oxygen gas. Therefore, the total charge required is 4 times the charge of 1 mole of electrons, which is 4 * 96500 C = 386000 C. However, since the question asks for the charge required for 5.6 L of oxygen gas, we need to adjust the value accordingly. The molar volume of a gas at STP is 22.4 L/mol, so the charge required for 5.6 L of oxygen gas is (5.6/22.4) * 386000 C = 96500 C.

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• 45.

### Gas asetilen (Mr = 26) sebagai gas penera-ngan dapat diperoleh dari reaksi: CaC2(s) + 2H2O(l) → Ca(OH)(aq) + C2H2(g) jika penerangan 1 jam membutuhkan 112 L gas asetilen pada STP, mka CaC2 (Mr = 64) yang diperlukan untuk penerangan selama 4 jam adalah ....

• A.

0,32 kg

• B.

0,64 kg

• C.

0,96 kg

• D.

1,28 kg

• E.

2,56 kg

D. 1,28 kg
Explanation
The balanced equation shows that 1 mole of CaC2 reacts with 2 moles of H2O to produce 1 mole of C2H2. Since the molar mass of CaC2 is 64 g/mol, 1 mole of CaC2 weighs 64 grams.

To find the amount of CaC2 needed for 4 hours of illumination, we can use the stoichiometry of the reaction.

1 mole of C2H2 produces 112 L of gas at STP, so 1 mole of CaC2 is needed to produce 112 L of C2H2.

Therefore, for 4 hours of illumination, 4 moles of CaC2 is needed.

4 moles of CaC2 is equal to 4 x 64 g = 256 g = 0.256 kg.

So, the correct answer is 0.256 kg, which is closest to 0.32 kg.

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• 46.

### Pernyataan yang tepat tentang pengelompokkan mikroorganisme adalah sebagai berikut ....

• A.

Virus hepatitis B tidak termasuk mahluk hidup

• B.

Eschericia coli termasuk ekuariota

• C.

Volvox globator termaksuk prokariota

• D.

Saccharomyces cerevisiae termasuk bryophyta

• E.

Plasmodium vivax termaksuk ascomycota

A. Virus hepatitis B tidak termasuk mahluk hidup
Explanation
The correct statement about the classification of microorganisms is that the hepatitis B virus is not included as a living organism. This is because viruses are considered acellular entities that cannot carry out metabolic processes or reproduce on their own. They are instead classified as infectious agents that require a host cell to replicate and cause disease.

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• 47.

### Beberapa tumbuhan memiliki nilai ekonomi yang tinggi karena memiliki kayu yang harum. Salah satu tumbuhan tersebut adalah cendana. cendana berbau harum, karena ....

• A.

Batangnya ditumbuhi lumut kerak, sehingga menghasilkan senyawa berbau harum

• B.

Batangnya dihuni oleh serangga yang menghasilkan feromon berbau harum

• C.

Memiliki simbion berupa bakteri yang menghasilkan resin yang berbau harum

• D.

Batangnya mengandung minyak atsiri yang berbau harum

• E.

Mempunyai jamur yang menghasilkan senyawa berbau harum

D. Batangnya mengandung minyak atsiri yang berbau harum
Explanation
Cendana memiliki kayu yang harum karena batangnya mengandung minyak atsiri yang berbau harum. Minyak atsiri merupakan senyawa yang terdapat dalam berbagai tumbuhan dan memiliki aroma yang khas. Pada cendana, minyak atsiri tersebut terkandung dalam batangnya, sehingga memberikan aroma harum pada kayu cendana.

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• 48.

### Protista yang dapat dimasukkan ke dalam kelompok autotrof dan juga heterotrof adalah ....

• A.

Trypanosome

• B.

Euglena

• C.

Amoeba

• D.

Giardia

• E.

Leishmania

B. Euglena
Explanation
Euglena is a protist that can be classified as both autotrophic and heterotrophic. It has the ability to perform photosynthesis and produce its own food using sunlight, similar to plants. However, when sunlight is not available, Euglena can switch to a heterotrophic mode and obtain nutrients by ingesting organic matter from its environment. This unique ability to switch between autotrophy and heterotrophy allows Euglena to survive in various environmental conditions.

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• 49.

### Perhatikan gambar sel neuron di atas! Sel neuron yang berfungsi menghantarkan rangsangan dari alat indra ke otak adalah ....

• A.

I

• B.

II

• C.

III

• D.

III dan IV

• E.

I dan IV

B. II
Explanation
The correct answer is II because the neuron labeled II is responsible for transmitting stimuli from the sensory organs to the brain. Neurons I, III, and IV may have other functions, but they are not specifically involved in transmitting sensory stimuli to the brain.

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• 50.

• A.

A dan B

• B.

A dan G

• C.

C dan F

• D.

B dan D

• E.

A dan E