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To estimate the charge induced on each of the identical styofoam (or pith) balls suspended in a vertical plane by making use of coulom's law

Posted by Khushi Ekka 1 year, 4 months ago

CBSE > Class 12 > Physics

2 answers

Ayansh Pandey 1 year, 4 months ago

To estimate the charge induced on each of the identical Styrofoam (or pith) balls suspended in a vertical plane, you can use Coulomb's Law, which states that the electrostatic force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between their centers. The formula for Coulomb's Law is

$F = \frac{{k \cdot q_1 \cdot q_2}}{{r^2}}$ , where: -

$F$ is the electrostatic force, -

$k$ is Coulomb's constant (

$8.99 \times 10^9 \ \text{N m}^2/\text{C}^2$ ), -

$q_1$ and

$q_2$ are the charges on the two objects, and -

$r$ is the separation between their centers. To estimate the charge, you would rearrange the formula as

$q = \sqrt{\frac{{F \cdot r^2}}{{k}}}$ . Measure the force between the balls, the separation distance, and then use these values in the formula to estimate the charge on each ball. Keep in mind that this is an estimation and real-world factors such as air resistance may affect the accuracy of the results.

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Shivam Singh 1 year, 4 months ago

On the face of it MCQs

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Which mirror has the wide field of view convex or cylindrical

Posted by Ghahsjshdhdjdkgx Sbkyotsu8E 1 year, 4 months ago

CBSE > Class 12 > Physics

3 answers

D Chahal 1 year, 3 months ago

Convex mirror

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Kamal Kanwasi 1 year, 4 months ago

convex bro

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Lahar . 1 year, 4 months ago

Since the image is diminished convex mirror has wider field of view .

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Two small identical electric dipoles AB and CD, each of dipole moment p are kept at an angle of 120° to each other in an external electric field E pointing along the x-axis as shown in the figure. Find the dipole moment of the arrangement and magnitude and direction of the net torque acting on it.

Posted by Sakshi Singh 1 year, 4 months ago

CBSE > Class 12 > Physics

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why does the light bend when it passes from one medium to another medium.

Posted by Ajeet Chauhan 1 year, 4 months ago

CBSE > Class 12 > Physics

5 answers

D Chahal 1 year, 3 months ago

When light passes from rarer medium to denser medium it bend towards normal..... Or. ...... It can also be termed that when light passes from denser to rarer medium it bends away from its normal , this phenomenon is called refraction.

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Ankit Singh 1 year, 4 months ago

Because due to medium change , speed of light decreases and it goes to denser medium and hence light bend

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Abhishek Babu 1 year, 4 months ago

Because of refraction

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Prashant Singh 1 year, 4 months ago

Light pass from one medium to another is bouncing of light

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Account Deleted 1 year, 4 months ago

Refraction.

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Explain the refraction of light by hygens theory of secondary wavlets

Posted by Anmol Sharma 1 year, 4 months ago

CBSE > Class 12 > Physics

0 answers

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Physical quantity

Posted by Shobhit Dixit 1 year, 4 months ago

CBSE > Class 12 > Physics

1 answers

Shobhit Dixit 1 year, 4 months ago

A quantity which can be measured and expressed in the form of law is called a physical quantity.

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Light of wavelength 2000A°. Falls on an aluminium surface. In aluminium 4.2ev are required to remove an electron. What is the kinetic energy of(a) fastest (b) the slowest photoelectron?

Posted by Anjali Singh 1 year, 4 months ago

CBSE > Class 12 > Physics

2 answers

Shobhit Dixit 1 year, 4 months ago

Thanks

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Rahul Kumar 1 year, 4 months ago

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Three cells of EMF andE,2E and 5E having internal resistance are 2 hour and 3 hour variable resistance r is shown in the figure find the expression for the current

Posted by Sadika Parveen 1 year, 4 months ago

CBSE > Class 12 > Physics

0 answers

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Water components

Posted by Mayank Sharma 1 year, 5 months ago

CBSE > Class 12 > Physics

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Derive an expression for capacitance of an isolated sphere of radius 'r' ?

Posted by Puspendu Patel 1 year, 5 months ago

CBSE > Class 12 > Physics

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What is valence band

Posted by Angel Maria 1 year, 5 months ago

CBSE > Class 12 > Physics

3 answers

D Chahal 1 year, 3 months ago

Valence band is the place where all the electrons are exist

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Xxxxxxxx Xxxxxx 1 year, 4 months ago

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Diya Bhaneriya 1 year, 5 months ago

Bond between conductor and insulators

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Posted by Kriti Singh 1 year, 5 months ago

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Pollaride

Posted by Tarannum Ahmad 1 year, 5 months ago

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Posted by Shivam Chaudhary 1 year, 5 months ago

CBSE > Class 12 > Physics

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Relation between Xm and Xr

Posted by Suraj Singh Meena Meena 1 year, 5 months ago

CBSE > Class 12 > Physics

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How many interference fringes will be seen if the size of each slit is 1/5 the separation of between two slits

Posted by Rekha Sharma 1 year, 5 months ago

CBSE > Class 12 > Physics

2 answers

Xxxxxxxx Xxxxxx 1 year, 4 months ago

Don't post personal information, mobile numbers and other details.
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Xxxxxxxx Xxxxxx 1 year, 4 months ago

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IC=ID

Posted by Mohit Tanwar 1 year, 5 months ago

CBSE > Class 12 > Physics

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Biot's savarts law formula

Posted by Goutam Singh Chouhan 1 year, 5 months ago

CBSE > Class 12 > Physics

3 answers

Priyanshu Singh 1 year, 5 months ago

Munot upon 4pis idl sintetaupon r square

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Manish Yadav 1 year, 5 months ago

Idlsintheta by r square

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Krish Agrawal 1 year, 5 months ago

dB = (μ₀ / 4π) * (Idl × r) / r³

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(a) If one of two identical slits producing interference in Young's experiment is covered with glass, so that the light intensity passing through it is reduced to 50 %, find the ratio of the maximum and minimum intensity of the fringe in the interference pattern. (b) What king of fringes do you expect to observe if white light is used instead of monochromatic light?

Posted by Ekta Singla 1 year, 5 months ago

CBSE > Class 12 > Physics

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What are the 2 postulates of Rutherford atomic module

Posted by Suraj Negi 1 year, 5 months ago

CBSE > Class 12 > Physics

1 answers

Krish Agrawal 1 year, 5 months ago

Ernest Rutherford's atomic model, also known as the Rutherford model, was proposed in the early 20th century. It is primarily based on two key postulates: 1. **The Nucleus**: Rutherford postulated that an atom consists of a tiny, dense, and positively charged core at its center, known as the nucleus. This nucleus contains most of the atom's mass and is where the positively charged protons are located. Electrons orbit around the nucleus. 2. **Electron Orbits**: Rutherford proposed that electrons orbit the nucleus in well-defined, circular or elliptical paths. These orbits are sometimes referred to as electron shells or energy levels. Electrons in different shells have different energy levels, and they can jump from one energy level to another by absorbing or emitting energy in the form of electromagnetic radiation. Rutherford's model was groundbreaking because it replaced the earlier Thomson model, which depicted atoms as a "plum pudding" with electrons distributed throughout a positively charged sphere. Rutherford's experiments, most notably the famous gold foil experiment, provided evidence for the existence of the atomic nucleus and laid the foundation for our modern understanding of atomic structure. However, Rutherford's model had limitations and was eventually refined by Niels Bohr and others to account for the quantization of energy levels in electrons and the behavior of atoms in more detail.

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A transformer has 500 primary turns and 10 secondary turns. If the secondary has resistive load of 15 Ω, the currents in the primary and secondary

Posted by Gangana Pandey 1 year, 5 months ago

CBSE > Class 12 > Physics

1 answers

Krish Agrawal 1 year, 5 months ago

To calculate the currents in the primary and secondary of a transformer, you can use the transformer equation, which relates the primary and secondary voltages and currents. The equation is: (Vp / Vs) = (Np / Ns) Where: - Vp is the primary voltage. - Vs is the secondary voltage. - Np is the number of turns in the primary coil. - Ns is the number of turns in the secondary coil. In your case, you have: - Np = 500 - Ns = 10 Now, let's calculate the primary voltage (Vp) and the secondary voltage (Vs). The primary voltage (Vp) is typically the voltage applied to the primary coil. To find it, you would need more information, such as the source voltage. Assuming you have the primary voltage (Vp), you can use the equation to find the secondary voltage (Vs): (Vp / Vs) = (Np / Ns) Vs = Vp * (Ns / Np) Now, we need to calculate the currents in the primary and secondary. Use Ohm's Law: I = V / R Where: - I is the current. - V is the voltage. - R is the resistance. For the secondary: - Vs is the voltage (which you've found). - The resistive load is 15 Ω. Isolate the secondary current (Is) and calculate it: Is = Vs / Rs Is = Vs / 15 Ω For the primary current (Ip), you can use the transformer equation again: (Vp / Vs) = (Np / Ns) Ip = Is * (Ns / Np) Substitute the values you have to calculate Ip. Please note that to get precise values, you would need the primary voltage (Vp). Without that information, you won't be able to calculate the exact currents in the primary and secondary coils.

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R1 in a lens is positive or negative?

Posted by Tushar Dulloo 1 year, 5 months ago

CBSE > Class 12 > Physics

1 answers

Krish Agrawal 1 year, 5 months ago

The sign conventions for the radii of curvature (R1 and R2) for lenses are as follows: 1. **Convex Lens (Diverging Lens)**: - R1 (the radius of curvature of the first surface) is considered positive if the center of curvature is on the same side as the incident light (opposite to the direction of incoming light). - R2 (the radius of curvature of the second surface) is considered negative. 2. **Concave Lens (Converging Lens)**: - R1 (the radius of curvature of the first surface) is considered negative. - R2 (the radius of curvature of the second surface) is considered positive if the center of curvature is on the same side as the incident light (opposite to the direction of incoming light). These conventions are used in lens formula and mirror formula calculations to determine the sign of focal lengths and object distances.

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Question 1: A lens has a refractive index of 1.5 and a radius of curvature of 15 cm for one surface and 20 cm for the other surface. Calculate the difference in curvature between the two surfaces using the lens maker's formula Question 2: A concave lens with a refractive index of 1.4 is required to have a focal length of -12 cm. Determine the radil of curvature for the two lens surfaces using the lens maker's formula. Question 3: A biconvex lens with a refractive index of 1.6 has a known focal length of 25 cm. Calculate the difference in radii of curvature for its two surfaces using the lens maker's formula. Question 4 (Easy): A plano-concave lens has a refractive index of 1.2. The radius of curvature of the curved surface is 18 cm, and the flat side has a radius of curvature of infinity, Calculate the focal length of the lens using the lens maker's formula. Question 5 (Easy): A converging lens has a refractive index of 1.6. The radius of curvature of one surface is 12 om, and the radius of curvature of the other surface is 15 cm. Calculate the focal length of the lens using the lens maker's formula.

Posted by Tushar Dulloo 1 year, 6 months ago

CBSE > Class 12 > Physics

1 answers

Krish Agrawal 1 year, 5 months ago

To answer each of your questions, we can use the lens maker's formula, which is given by:

$\frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$ Where: -

$f$ is the focal length of the lens. -

$n$ is the refractive index of the material. -

$R_1$ is the radius of curvature of the first surface. -

$R_2$ is the radius of curvature of the second surface. Let's solve each of your questions: **Question 1:** Given

$n = 1.5$ ,

$R_1 = 15 \, \text{cm}$ , and

$R_2 = 20 \, \text{cm}$ , we need to calculate the difference in curvature between the two surfaces. Using the lens maker's formula, we can calculate the focal length, and then the difference in curvature. 1. Calculate

$\frac{1}{f}$ using the formula with the given values. 2. Once you find

$\frac{1}{f}$ , calculate the focal length

$f$ by taking the reciprocal. **Question 2:** Given

$n = 1.4$ and

$f = -12 \, \text{cm}$ , we need to determine the radii of curvature for the two lens surfaces. Using the lens maker's formula, you can calculate

$R_1$ and

$R_2$ . 1. Rearrange the lens maker's formula to solve for either

$R_1$ or

$R_2$ . 2. Plug in the known values and solve for

$R_1$ and

$R_2$ . **Question 3:** Given

$n = 1.6$ and

$f = 25 \, \text{cm}$ , we need to calculate the difference in radii of curvature for its two surfaces. Again, use the lens maker's formula to calculate

$R_1$ and

$R_2$ . **Question 4:** Given

$n = 1.2$ ,

$R_1 = 18 \, \text{cm}$ , and

$R_2 = \infty$ for a flat side, calculate the focal length using the lens maker's formula. **Question 5:** Given

$n = 1.6$ ,

$R_1 = 12 \, \text{cm}$ , and

$R_2 = 15 \, \text{cm}$ , calculate the focal length using the lens maker's formula. For each question, perform the calculations as described above, and you will have the answers you need. If you have any specific values to calculate or encounter any issues, feel free to ask for further assistance.

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The equation of progressive wave is y=0.01sin(100t-x)

Posted by Patcharu Sandeep 1 year, 6 months ago

CBSE > Class 12 > Physics

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What do you mean by magnetic confinement?

Posted by Gaurav Choudhary 1 year, 6 months ago

CBSE > Class 12 > Physics

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कंसीडर थे फॉलोइंग रिएक्शन 1 h2 + 1h3 गिव्स तो he4 और + न + के कैलकुलेट द एनर्जी रिलीज इफ द मास्स ऑफ़ 1h2 इस 2.014102u और 1s 3 इस 3.0160949u मास ऑफ़ न्यूट्रॉन इस 1.008665 यू मास ऑफ़ 2 है 4 इस 4.00

Posted by Anshika Saxena 1 year, 6 months ago

CBSE > Class 12 > Physics

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Explain how OH group attached to a carbon in the Benzene ring activate the benzene towards electrophilic reaction

Posted by Haleema Khatoon 1 year, 6 months ago

CBSE > Class 12 > Physics

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Tan inverse(25/12) ki value

Posted by Sunil Prajapat 1 year, 6 months ago

CBSE > Class 12 > Physics

2 answers

Ankit Singh 1 year, 5 months ago

64° approx

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Gaurav Prabhakar 1 year, 6 months ago

Pta nahi

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1m in cm

Posted by Yuvraj Singh 1 year, 6 months ago

CBSE > Class 12 > Physics

4 answers

Krish Agrawal 1 year, 5 months ago

100cm

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Rahul Chauhan 1 year, 6 months ago

100 cm

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Faizan Raja 1 year, 6 months ago

1M=100cm

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Sunil Prajapat 1 year, 6 months ago

100 cm

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Power of combination of two concave lens derivation and formula

Posted by Soni Kushwaha 1 year, 6 months ago

CBSE > Class 12 > Physics

2 answers

Krish Agrawal 1 year, 5 months ago

The power of a combination of two concave lenses can be determined using the lens formula and the power formula for lenses. When lenses are in contact (i.e., their curved surfaces are facing each other), the formula to calculate their combined power is:

$P_{\text{total}} = P_1 + P_2$ Where: -

$P_{\text{total}}$ is the total power of the combination of lenses. -

$P_1$ is the power of the first lens. -

$P_2$ is the power of the second lens. The power

$P$ of a lens is given by:

$P = \frac{1}{f}$ Where: -

$P$ is the power of the lens. -

$f$ is the focal length of the lens. If you have two concave lenses with focal lengths

$f_1$ and

$f_2$ , you can calculate their individual powers as

$P_1 = \frac{1}{f_1}$ and

$P_2 = \frac{1}{f_2}$ , respectively. Then, you can find the total power by adding these individual powers. So, to calculate the total power of a combination of two concave lenses, simply add their individual powers using the formula

$P_{\text{total}} = P_1 + P_2$ . This is applicable when the lenses are in contact. If they are separated by a distance, you would need to consider the separation between the lenses when calculating their combined power.

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Rahul Chauhan 1 year, 6 months ago

1/f1+1/f2=1/F

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