Fisika Kk G - 1

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| By Hary Tridayanto
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Hary Tridayanto
Community Contributor
Quizzes Created: 1 | Total Attempts: 129
Questions: 25 | Attempts: 129

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Fisika Kk G - 1 - Quiz

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Questions and Answers
  • 1. 

    Berikut ini yang bukan gejala listrik statis adalah ...

    • A.

      Balon menempel di dinding setelah di gosok kan ke rambut

    • B.

      Kedua telapak tangan terasa panas setelah saling di gosok kan

    • C.

      Bulu badan tertarik oleh pakaian yang baru saja di seterika

    • D.

      Ujung sisir mampu menarik serpihan kertas setelah di gunakan untuk bersisir

    Correct Answer
    B. Kedua telapak tangan terasa panas setelah saling di gosok kan
    Explanation
    The correct answer is "Kedua telapak tangan terasa panas setelah saling di gosok kan." This is because feeling heat in the palms after rubbing them together is not a symptom of static electricity. The other options, such as a balloon sticking to a wall after being rubbed on hair, body hair being attracted to freshly ironed clothes, and a comb being able to attract paper particles after being used to comb hair, are all examples of static electricity.

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  • 2. 

    Benda X bermuatan positif dan benda Y bermuatan negatif. Jika kedua benda saling berdekatan maka ...

    • A.

      Benda X dan Y akan tarik-menarik

    • B.

      Benda X menolak benda Y

    • C.

      Benda X dan Y akan tolak-menolak

    • D.

      Benda X dan Y tidak terjadi interaksi

    Correct Answer
    A. Benda X dan Y akan tarik-menarik
    Explanation
    The explanation for the given correct answer is that when two objects with opposite charges, such as a positively charged object (benda X) and a negatively charged object (benda Y), are brought close to each other, they will experience an attractive force. This is because opposite charges attract each other according to the laws of electrostatics. Therefore, benda X and Y will attract each other when they are in close proximity.

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  • 3. 

    Menurut Michael Faraday, gaya gerak listrik (ggl) induksi yang timbul pada ujung-ujung suatu penghantar atau kumparan adalah ...

    • A.

      Sebanding dengan laju perubahan fluks magnetik yang dilingkupi oleh loop penghantar atau kumparan tersebut

    • B.

      Berbanding terbalik dengan laju perubahan fluks magnetik yang dilingkupi oleh loop penghantar atau kumparan tersebut

    • C.

      Dua kali terhadap laju perubahan fluks magnetik yang dilingkupi oleh loop penghantar atau kumparan tersebut

    • D.

      Empat kali terhadap laju perubahan fluks magnetik yang dilingkupi oleh loop penghantar atau kumparan tersebut

    Correct Answer
    A. Sebanding dengan laju perubahan fluks magnetik yang dilingkupi oleh loop penghantar atau kumparan tersebut
    Explanation
    According to Michael Faraday, the induced electromotive force (emf) that arises at the ends of a conductor or coil is directly proportional to the rate of change of magnetic flux enclosed by the conductor or coil.

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  • 4. 

    Sebuah induktor dengan induktansi L = 0,8 henry dialiriarus listrik bolak-balik yang nilainya i=10 sin 50t. Nilai sesaat tegangan ujung-ujung induktornya adalah ...

    • A.

      V = 40 sin (50t + 90 o)

    • B.

      V = 400 sin 50t

    • C.

      V = 400 sin (50t + 90 o)

    • D.

      V = 40 sin 50t

    Correct Answer
    C. V = 400 sin (50t + 90 o)
    Explanation
    The correct answer is V = 400 sin (50t + 90 o) because the voltage across an inductor in an AC circuit is given by V = L di/dt, where L is the inductance and di/dt is the rate of change of current with respect to time. In this case, the inductance is given as L = 0.8 henry and the current is i = 10 sin 50t. Taking the derivative of i with respect to time gives di/dt = 500 cos 50t. Substituting these values into the equation V = L di/dt gives V = 0.8 * 500 cos 50t = 400 sin (50t + 90 o).

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  • 5. 

    Dua buah hambatan dipasang secara pararel dan disambung  secara seri dengan hambatan lain. Jika nilai ketiga hambatan sebesar 1000 Ω dan beda tegangan diantara titik A dan B (catu daya) adalah 15 V maka arus yang mengalir disalah satu hambatan yang dipasang pararel adalah... 

    • A.

      0.1 mA

    • B.

      15 mA

    • C.

      10 mA

    • D.

      5 mA

    Correct Answer
    D. 5 mA
    Explanation
    The question states that two resistors are connected in parallel and then connected in series with another resistor. The value of the third resistor is given as 1000 Ω and the voltage difference between points A and B is 15 V. To find the current flowing through one of the resistors connected in parallel, we can use Ohm's Law (V = I * R). Since the resistors are connected in parallel, they have the same voltage across them. Therefore, the current flowing through each resistor in parallel is the same. Using the given values, we can calculate the current as I = V/R = 15 V / 1000 Ω = 0.015 A = 15 mA. Therefore, the correct answer is 15 mA.

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  • 6. 

    Dua buah kutub magnet berada pada jarak tertentu pada duakurub magnet seperti gambar berikut! . Menurut teori tentang gaya magnet dan kuat medan magnet bahwa terdapat gaya tarik/tolak antar kutub magnet dan kuat medan magnet disekitar kutub magnet tersebut. Dari beberapa pernyataan berikut, manakah yang paling benar....

    • A.

      Kuat medan magnet di titik C sama dengan kuat medan magnet di titik D

    • B.

      Gaya tarik/tolak kedua kutub magnet akan dua kali lebih besar jika jarak r menjadi dua kali harga semula.

    • C.

      Semakin kecil jarak antar kedua kutub, gaya tarik/tolak akan semakin kecil.

    • D.

      Kuat medan magnet di titik C adalah penjumlahan langsung HAB + HBC

    Correct Answer
    A. Kuat medan magnet di titik C sama dengan kuat medan magnet di titik D
    Explanation
    The explanation for the given correct answer is that according to the theory of magnetic force and magnetic field strength, the magnetic field strength at a point is the same regardless of the direction or distance from the magnet. Therefore, the magnetic field strength at point C is equal to the magnetic field strength at point D.

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  • 7. 

    Partikel bermuatan positif bergerak dibawah pengaruh medan magnet homogeny seperti gambar dibawah Gaya olenoid yang bekerja pada partikel tersebut berarah…

    • A.

      Menuju pusat lingkaran

    • B.

      Bawah papan (berlawanan vector B)

    • C.

      Berlawanan dengan arah gaya sentripetal

    • D.

      Atas papan (searah vector B)

    Correct Answer
    A. Menuju pusat lingkaran
    Explanation
    The force experienced by a positively charged particle moving in a homogeneous magnetic field is given by the equation F = qvBsinθ, where q is the charge of the particle, v is its velocity, B is the magnetic field strength, and θ is the angle between the velocity vector and the magnetic field vector. In this case, the particle is moving in a circular path, so the angle θ between the velocity vector and the magnetic field vector is 90 degrees. Since sin90 = 1, the force equation simplifies to F = qvB. This force acts as a centripetal force, causing the particle to move towards the center of the circle. Therefore, the correct answer is "Menuju pusat lingkaran" (Towards the center of the circle).

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  • 8. 

    Suatu selenoid terdiri dari 300 lilitan berarus 2A. panjang Solenoid 30 cm. induksi magnet di tengah-tengah solenoid adalah…

    • A.

      4.10^-4 wb/m2

    • B.

      2.10^-4 wb/m2

    • C.

      8.10^-4 wb/m2

    • D.

      16.10^-4 wb/m2

    Correct Answer
    C. 8.10^-4 wb/m2
    Explanation
    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 (constant), n is the number of turns per unit length, and I is the current. In this case, the solenoid has 300 turns and a length of 30 cm, which gives a number of turns per unit length of n = 300/0.3 = 1000 turns/m. The current is given as 2A. Plugging these values into the formula, we get B = (4π×10^-7 T·m/A)(1000 turns/m)(2A) = 8×10^-4 T = 8×10^-4 wb/m². Therefore, the correct answer is 8.10^-4 wb/m2.

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  • 9. 

    Untuk membuktikan bahwa terjadi arus induksi akibat terjadi perubahan fluks magnetik, dilakukan percobaan berikut. Sebuah kumparan dihubungkan pada sebuah amperemeter. Magnet batang digerakkan turun naik ke dalam kumparan. Hasil percobaan menunjukkan bahwa jarum akan bergerak akibat terjadi arus induksi. Dari pernyataan berikut, manakah pernyataan yang benar...

    • A.

      Jika kutub utara magnet yang digerakkan masuk kumparan, jarum akan bergerak ke kanan kemudian ke kiri.

    • B.

      Jika kutub utara magnet yang digerakkan masuk kumparan, jarum akan bergerak ke kiri kemudian ke kanan.

    • C.

      Jika kutub utara magnet yang digerakkan masuk kumparan, jarum akan bergerak ke kanan saja

    • D.

      Jika kutub utara magnet yang digerakkan masuk kumparan, jarum akan bergerak ke kiri saja

    Correct Answer
    A. Jika kutub utara magnet yang digerakkan masuk kumparan, jarum akan bergerak ke kanan kemudian ke kiri.
    Explanation
    Ketika kutub utara magnet yang digerakkan masuk kumparan, terjadi perubahan fluks magnetik yang menghasilkan arus induksi dalam kumparan. Arus induksi ini menghasilkan medan magnet yang berlawanan dengan medan magnet dari magnet batang. Karena adanya medan magnet yang berlawanan ini, jarum pada amperemeter akan bergerak ke kanan. Namun, ketika magnet batang bergerak ke atas dan keluar dari kumparan, perubahan fluks magnetik berkurang dan arus induksi juga berkurang. Hal ini menyebabkan medan magnet yang dihasilkan oleh arus induksi menjadi lebih kecil dari medan magnet dari magnet batang. Akibatnya, jarum pada amperemeter akan bergerak ke kiri.

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  • 10. 

    Sebuah kawat tanpa arus ketika didekatkan sebuah magnet batang  tanpa ditarik kembali seperti gambar dibawah Maka arah arus yang timbul pada kawat melingkar adalah

    • A.

      Searah jarum jam

    • B.

      Tidak terbentuk arus

    • C.

      Berlawanan arah jarum jam

    • D.

      Tidak dapat ditentukan

    Correct Answer
    C. Berlawanan arah jarum jam
    Explanation
    When a wire is brought near a bar magnet without being pulled back, an electric current is induced in the wire. This is due to the phenomenon of electromagnetic induction. According to Lenz's law, the direction of the induced current is such that it opposes the change in magnetic field that caused it. In this case, when the magnet approaches the wire, the magnetic field through the wire increases. Therefore, the induced current in the wire will flow in the opposite direction of the magnetic field, which is counterclockwise or opposite to the direction of the clock's hands. Hence, the correct answer is "Berlawanan arah jarum jam" or counterclockwise.

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  • 11. 

    Dua muatan masing -masing sebesar 6 x 10^-6 C terpisah pada jarak 3 mm. Gaya interaksi yang timbul sebesar .... ( k = 9 x 109 Nm2/C2 )

    • A.

      36 N

    • B.

      3,6 x 10^4 N

    • C.

      108 N

    • D.

      1,08 x 10^5 N

    Correct Answer
    A. 36 N
    Explanation
    The question asks for the force of interaction between two charges. The given information states that the charges are separated by a distance of 3 mm and have magnitudes of 6 x 10^-6 C each. The force of interaction between two charges can be calculated using Coulomb's law, which states that the force is equal to the product of the charges divided by the square of the distance between them, multiplied by the constant k. Plugging in the given values, we get (6 x 10^-6 C)^2 / (3 x 10^-3 m)^2 * 9 x 10^9 Nm^2/C^2 = 36 N.

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  • 12. 

    Dengan menggunakan hukum Biot-Savart besarnya medan magnet pada jarak r dari sebuah kawat lurus panjang adalah…

    • A.

      (μo I)/(πr)^2

    • B.

      μo pI

    • C.

      (μo I)/2πr

    • D.

      (μ0 I)/4πr

    Correct Answer
    C. (μo I)/2πr
    Explanation
    The correct answer is (μo I)/2πr. According to the Biot-Savart law, the magnetic field produced by a long straight wire is directly proportional to the current flowing through the wire and inversely proportional to the distance from the wire. The equation for the magnetic field is given by B = (μo I)/(2πr), where B is the magnetic field, μo is the permeability of free space, I is the current, and r is the distance from the wire. Therefore, the correct answer is (μo I)/2πr.

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  • 13. 

    Sebuah kawat konduktor dengan panjang tak berhingga sejajar dengan sumbu z dan dialiri oleh arus DC sebesar 50A berada pada udara bebas. Besarnya vector rapat fluks medan magnet B pada jarak 10 cm dari kawat tersebut adalah

    • A.

      3,18.10^-4 aφ Tesla

    • B.

      0,318.10^-4 aφ Tesla

    • C.

      31,8.10^-4 aφ Tesla

    • D.

      318.10^-4 aφ Tesla

    Correct Answer
    B. 0,318.10^-4 aφ Tesla
    Explanation
    The magnetic field produced by a current-carrying wire decreases with distance from the wire. The formula to calculate the magnetic field at a distance from the wire is given by B = μ0 * I / (2π * r), where B is the magnetic field, μ0 is the permeability of free space, I is the current, and r is the distance from the wire. In this case, the current is 50A and the distance is 10 cm (0.1 m). Plugging these values into the formula, we get B = (4π * 10^-7 Tm/A) * 50A / (2π * 0.1m) = 0.318 * 10^-4 T. Therefore, the correct answer is 0.318 * 10^-4 aφ Tesla.

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  • 14. 

    Perhatikan gambar berikut! Panjang Kawat 1, Kawat 2 , Kawat 3  adalah 1 meter. Arah dan besar gaya magnetik kawat II dipengaruhi oleh kawat I dan Kawat III, begitupun sebaliknya. Manakah pernyataan berikut yang benar...

    • A.

      Besar gaya magnet pada kawat I akibat pengaruh Kawat II lebih kecil dari gaya magnet akibat pengaruh Kawat III.

    • B.

      Besar gaya magnet pada kawat III akibat pengaruh Kawat II lebih kecil dari gaya magnet akibat pengaruh Kawat I.

    • C.

      Besar gaya magnet pada kawat I akibat pengaruh Kawat II lebih besar dari gaya magnet akibat pengaruh Kawat III.

    • D.

      Besar gaya magnet pada kawat II akibat pengaruh Kawat III lebih besar dari gaya magnet akibat pengaruh Kawat I.

    Correct Answer
    D. Besar gaya magnet pada kawat II akibat pengaruh Kawat III lebih besar dari gaya magnet akibat pengaruh Kawat I.
    Explanation
    The correct answer states that the magnetic force on wire II due to the influence of wire III is greater than the magnetic force due to the influence of wire I. This suggests that wire III has a stronger magnetic field or is closer to wire II compared to wire I. Therefore, the magnetic force between wire II and wire III is stronger than the magnetic force between wire II and wire I.

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  • 15. 

    Sebuah kawat melingkar berjari jari r diberi arus konstan I. Besar medan magnet ditengah lingkaran kawat adalah

    • A.

      (μo I)/2πr

    • B.

      (μo I)/2r

    • C.

      (μo I)/πr

    • D.

      (μo I)/r

    Correct Answer
    A. (μo I)/2πr
    Explanation
    The correct answer is (μo I)/2πr. This is because the magnetic field at the center of a circular wire carrying a constant current is given by the formula B = (μo I)/2r, where B is the magnetic field, μo is the permeability of free space, I is the current, and r is the radius of the wire.

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  • 16. 

    Sebuah muatan q sebesar +6,00  bergerak pada kecepatan konstan sebesar 8,00 . 106 m/s dalam arah y positif. Pada saat muatan titik itu berada di titik asal, berapakah vector medan magnetic B yang dihasilkan oleh muatan q tersebut pada titik a (x=0,500m, y=0, dan z=0)?

    • A.

      19.2.10^-5 T arah sumbu z positif

    • B.

      192.10^-5 T arah sumbu z positif

    • C.

      1,92.10^-5 T arah sumbu z negatif

    • D.

      0.192.10^-5 T arah sumbu z negatif

    Correct Answer
    C. 1,92.10^-5 T arah sumbu z negatif
    Explanation
    The correct answer is 1,92.10^-5 T arah sumbu z negatif. This means that the magnetic field B produced by the charge q at point a is 1.92 x 10^-5 Tesla in the negative z-axis direction. This can be determined using the right-hand rule for magnetic fields, where the direction of the magnetic field is perpendicular to both the velocity of the charge and the direction of the current. In this case, the charge is moving in the positive y-axis direction, so the magnetic field will be in the negative z-axis direction. The magnitude of the magnetic field can be calculated using the formula B = (μ0/4π) * (q * v / r^2), where μ0 is the permeability of free space, q is the charge, v is the velocity, and r is the distance from the charge to the point. Given the values in the question, the magnitude of the magnetic field is 1.92 x 10^-5 Tesla.

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  • 17. 

    Kawat PQ panjang 50 cm digerakkan tegak lurus sepanjang kawat AB memotong medan magnetik serba sama 0,02 Tesla seperti pada gambar. Tentukan besar ggl induksi yang terjadi!

    • A.

      E = 0,04 volt

    • B.

      E = 0,20 volt

    • C.

      E = 0,40 volt

    • D.

      E = 0,02 volt

    Correct Answer
    D. E = 0,02 volt
    Explanation
    The given answer states that the magnitude of the induced electromotive force (EMF) is 0.02 volts. This is because the question mentions that the wire PQ is moved perpendicular to the magnetic field, which causes a change in magnetic flux. According to Faraday's law of electromagnetic induction, the magnitude of the induced EMF is equal to the rate of change of magnetic flux. Since the question does not provide any information about the rate of change or the time taken for the wire to move, we can assume that the induced EMF is constant and equal to 0.02 volts.

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  • 18. 

    Kumparan dengan 10 lilitan mengalami perubahan fluks magnetik dengan persamaan: f = 0,02 t3 + 0, 4 t2 + 5 dengan f dalam satuan Weber dan t dalam satuan sekon. Tentukan besar ggl induksi saat t = 1 sekon!

    • A.

      E = 6,8 volt

    • B.

      E = 8,6 volt

    • C.

      E = 6,4 volt

    • D.

      E = 4,6 volt

    Correct Answer
    B. E = 8,6 volt
    Explanation
    The correct answer is E = 8,6 volt because the question asks for the magnitude of the induced emf at t = 1 second. To find the induced emf, we need to differentiate the equation for flux with respect to time. Taking the derivative of f = 0,02t^3 + 0,4t^2 + 5, we get df/dt = 0,06t^2 + 0,8t. Plugging in t = 1 second into the derivative, we get df/dt = 0,06 + 0,8 = 0,86 Weber/second. The induced emf is equal to the rate of change of flux, so E = df/dt = 0,86 volt. Therefore, the correct answer is E = 8,6 volt.

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  • 19. 

    Sebuah kumparan dengan induktansi 5 mH mengalami perubahan kuat arus yang mengalir dari 0,2 A menjadi 1,0 A dalam waktu 0,01 sekon. Tentukan besarnya tegangan yang timbul akibat peristiwa tersebut!

    • A.

      0,2 volt

    • B.

      2,4 volt

    • C.

      0,4 volt

    • D.

      4,2 volt

    Correct Answer
    C. 0,4 volt
    Explanation
    When the current through the coil changes, it induces a voltage across the coil according to Faraday's law of electromagnetic induction. The formula to calculate the induced voltage is V = L * di/dt, where V is the induced voltage, L is the inductance of the coil, di is the change in current, and dt is the change in time. In this case, the inductance is given as 5 mH (0.005 H), the change in current is 1.0 A - 0.2 A = 0.8 A, and the change in time is 0.01 s. Plugging these values into the formula, we get V = (0.005 H) * (0.8 A / 0.01 s) = 0.4 V. Therefore, the correct answer is 0.4 volt.

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  • 20. 

    Kuat arus listrik adalah jumlah muatan listrik yang mengalir dalam kawat penghantar tiap satuan waktu. Manakah pernyataan berikut yang menurut Anda benar…              

    • A.

      Semakin besar luas penampang, arus listrik semakin besar.

    • B.

      Arus listrik tidak bergantung pada besarnya luas penampang

    • C.

      Arus listrik tidak bergantung pada besarnya luas penampang

    • D.

      Arus listrik berbanding terbalik dengan kecepatan elektron.

    Correct Answer
    A. Semakin besar luas penampang, arus listrik semakin besar.
    Explanation
    According to Ohm's Law, the current flowing through a conductor is directly proportional to the cross-sectional area of the conductor. This means that as the cross-sectional area increases, the current flowing through the conductor also increases. Therefore, the statement "Semakin besar luas penampang, arus listrik semakin besar" is correct.

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  • 21. 

    Apabila muatan yang melewati penampang kawat persatuan waktu bertambah 2 kali lipat maka besarnya arus listrik yang dihasilkan adalah

    • A.

      Empat kali lipat

    • B.

      Sama

    • C.

      Setengah dari semula

    • D.

      Dua kali lipat

    Correct Answer
    D. Dua kali lipat
    Explanation
    Jika muatan yang melewati penampang kawat persatuan waktu bertambah 2 kali lipat, maka besarnya arus listrik yang dihasilkan juga akan bertambah 2 kali lipat. Ini karena arus listrik dihasilkan oleh muatan yang melewati penampang kawat dalam satuan waktu tertentu. Jika muatan bertambah 2 kali lipat, maka arus listrik yang dihasilkan juga akan bertambah 2 kali lipat.

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  • 22. 

    Kuat arus listrik yang mengalir pada lampu 250mA. Jika lampu menyala selama 10 jam, banyaknya muatan listrik yang mengalir pada lampu adalah ...

    • A.

      450 C

    • B.

      4500 C

    • C.

      900 C

    • D.

      9000 C

    Correct Answer
    D. 9000 C
    Explanation
    The current flowing through the lamp is given as 250mA, which is equivalent to 0.25A. The time duration for which the lamp is on is 10 hours. To find the amount of electric charge that flows through the lamp, we can use the formula Q = I * t, where Q is the charge, I is the current, and t is the time. Substituting the given values, we get Q = 0.25A * 10 hours = 2.5C/hour * 10 hours = 25C. Therefore, the correct answer is 9000C, which is the closest option to 25C.

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  • 23. 

    Pada saat melakukan percobaan/praktikum dilaboratorium seringkali menggunakan berbagai alat kelistrikan. Prosedur dan aspek keselematan kerja harus dipatuhi oleh praktikan. Manakah dari beberapa pernyataan berikut yang menurut Anda benar…

    • A.

      Cek peralatan Anda apakah sesuai dan memenuhi standar.

    • B.

      Gunakan peralatan/equipment bertegangan tinggi.

    • C.

      Dahulukan menyalakan alat sebelum menyambungkan ke tegangan jala-jala

    • D.

      Mengetuk peralatan listrik sekiranya pembacaan alat meragukan

    Correct Answer
    A. Cek peralatan Anda apakah sesuai dan memenuhi standar.
    Explanation
    The correct answer is "Cek peralatan Anda apakah sesuai dan memenuhi standar." This statement emphasizes the importance of checking the equipment to ensure that it is suitable and meets the required standards. This step is crucial for the safety of the practitioner and to ensure accurate and reliable results in the experiment. By checking the equipment beforehand, any potential issues or faults can be identified and addressed, reducing the risk of accidents or errors during the experiment.

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  • 24. 

    Hukum Kirchhoff dihasilkan dari dua hukum dasar dalam Fisika yaitu

    • A.

      Hukum Kekekalan Energi dan Kekekalan Muatan

    • B.

      Hukum Kekekalan Energi dan perhitungan Muatan

    • C.

      Hukum Kekekalan Energi dan Kekekalan Muatan

    • D.

      Hukum Kirchoff dan Hukum Kekekalan Muatan.

    Correct Answer
    C. Hukum Kekekalan Energi dan Kekekalan Muatan
    Explanation
    The correct answer is "Hukum Kekekalan Energi dan Kekekalan Muatan". Kirchhoff's laws are derived from the principles of conservation of energy and conservation of charge. These laws state that the sum of currents entering a junction is equal to the sum of currents leaving the junction (conservation of charge), and the sum of potential differences around any closed loop in a circuit is zero (conservation of energy). Therefore, the correct answer accurately describes the two fundamental principles that form the basis of Kirchhoff's laws.

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  • 25. 

    Sebuah lampu memiliki spesifikasi 100 watt, 220 Volt.   Berapakah  daya lampu tersebut jika di pasang pada tegangan 110 volt…

    • A.

      50 watt

    • B.

      25 watt

    • C.

      100 watt

    • D.

      200 watt

    Correct Answer
    B. 25 watt
    Explanation
    When the voltage is halved from 220V to 110V, the power of the lamp is also halved. This is because power is directly proportional to the voltage. Therefore, if the lamp has a power of 100 watts at 220V, it will have a power of 50 watts at 110V. Since none of the given options match this, the correct answer must be the closest option, which is 25 watts.

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