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Ask QuestionPosted by Aman Pandey 5 years, 1 month ago
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Posted by Mohit Kumar 5 years, 1 month ago
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Yogita Ingle 5 years, 1 month ago
Effect of rotation of earth on acceleration due to gravity, gλ=ge−Rew2cos2λ where w is the angular velocity of the rotation of earth and λ is the lattitude.
Now at poles, λ=90o
⟹ gpoles= ge = constant
Thus as the earth stops suddenly then gλ increases by an amount of Rew2cos2λ everywhere except at poles.
Posted by Aditya Kumar Verma 5 years, 1 month ago
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Yogita Ingle 5 years, 1 month ago
Atomis the fundamental building block of matterhaving a confined positively charged nucleus at the center, surrounded by negatively charged electrons.Every inorganic, organic, or even synthetic object is made up of atoms.
Posted by Prachi Lariya 5 years, 1 month ago
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Garima Garg 5 years, 1 month ago
Posted by Tarini Sahu 5 years, 1 month ago
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Gaurav Seth 5 years, 1 month ago
Given that,
Two plane mirrors are inclined to each other at a certain angle.
Incident angle = 10°
Let the angle between the mirror is θ.
On second mirror ray of light retraces its path
After five reflection it will be same.
So, ray of light falls normally on the second mirror
The normal angle is
We know that,
From snell's law,
Incident angle is equal to the reflection angle.
We need to calculate the angle b
Using formula for b
Put the value into the formula
We need to calculate the angle between the mirror
Using formula of angle
Put the value into the formula
Hence, The angle between the mirror is 10°.
Posted by Pradeep Solanki 5 years, 1 month ago
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Posted by Priyansh Goyal 5 years, 1 month ago
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Posted by Mani Prakash 5 years, 1 month ago
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Posted by Aradhana Singh 5 years, 1 month ago
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Chaitra Chavhan 5 years, 1 month ago
Arvind Kumar 5 years, 1 month ago
Posted by Ranu Patidar 5 years, 1 month ago
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Yogita Ingle 5 years, 1 month ago
Rydberg Constant Value
R∞ and electron spin g-factor are the most accurately measured physical constants.
| Rydberg Constant Value,R∞ | 10973731.568508(65) m-1 |
| Rydberg Constant Value,R∞ | 1.097 x 107m-1 |
Posted by Aradhana Singh 5 years, 1 month ago
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Vansh Prajapati 5 years, 1 month ago
Chaitra Chavhan 5 years, 1 month ago
Gaurav Seth 5 years, 1 month ago
1 coulomb = 1/1.602 × 10^-19 = 6.24 × 10^18 electrons. An electric current of 1 ampere is equal to 1 coulomb of charge passing a point in a circuit every second: Therefore a current of 1 ampere = 6.242 × 10^18 electrons moving past any point in a circuit every second.
Yogita Ingle 5 years, 1 month ago
1 microampere contains number of electrons.
The charge on one electron is coulombs. This can be used to deduce the number of electrons that 1 microampere contains.
Since, 1 coulomb = electrons
And we know that, “1 ampere” is equal to “1 coulomb per second”,
1 ampere = electrons/sec
Posted by Prince Tomar 5 years, 1 month ago
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Chaitra Chavhan 5 years, 1 month ago
Gaurav Seth 5 years, 1 month ago
- Paramagnetic substances are weakly attracted on application of magnetic field.
- It is due to presence of one or more unpaired electrons that gets attracted by the magnetic field.
- Application of a magnetic field magnetizes the paramagnetic substances in the same direction.
- They lose their magnetism in the absence of magnetic field.
- O2, Cu2+, Fe3+, Cr3+ are some examples of such substances.
Posted by Vijay Solanki 5 years, 1 month ago
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Posted by Aradhana Singh 5 years, 1 month ago
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Yogita Ingle 5 years, 1 month ago
Current density is a vector quantity having both a direction and a scalar magnitude. The electric current flowing through a solid having units of charge per unit time is calculated towards the direction perpendicular to the flow of direction.
Posted by Jagdish Kumawat 5 years, 1 month ago
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Arvind Kumar 5 years, 1 month ago
Posted by Varinder Singh 5 years, 1 month ago
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Gaurav Seth 5 years, 1 month ago
As conductor is connected with the cell , the potential difference across two paths (two possible paths i.e. one upside and one downside path) is same.
V1 = V2
I1R1=I2R2
I1dl1=I2dl2 (as resistivity and area of cross section same )
Now,
field, due to one path, is opposite due to another part. So magnetic field at the center is

[sin 90 = 1 , i1dl1 = i2dl2]
= 0 (proved)
As conductor is connected with the cell , the potential difference across two paths (two possible paths i.e. one upside and one downside path) is same.
V1 = V2
I1R1=I2R2
I1dl1=I2dl2 (as resistivity and area of cross section same )
Now,
field, due to one path, is opposite due to another part. So magnetic field at the center is

[sin 90 = 1 , i1dl1 = i2dl2]
= 0 (proved)
Posted by Priyansh Goyal 5 years, 1 month ago
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Posted by Riya Kodan 5 years, 1 month ago
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Gaurav Seth 5 years, 1 month ago
Long distance radio broadcasts use shortwave bands because only these bands can be refracted by the ionosphere. I.e they can be reflected by the ionosphere of the earth's atmosphere and thus can be send to longer distances.
Posted by Vani Singh 5 years, 1 month ago
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Arvind Kumar 5 years, 1 month ago
Posted by Vani Singh 5 years, 1 month ago
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Arvind Kumar 5 years, 1 month ago
Posted by Vani Singh 5 years, 1 month ago
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Arvind Kumar 5 years, 1 month ago
Posted by Bishnu Prajapati 5 years, 1 month ago
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Gaurav Seth 5 years, 1 month ago
Electron charge 1.6 × 10-19 C << 1 C
∴ Coulomb is bigger
= 6.25 × 1018 electrons are required
Or
Given that
Charge (Q) = 1 C
Electron (e) = 1.6× 10^-19
Number of electron (n) = ?
Q= ne
1 C = n × 1.6×10^-19 C
n = 1/(1.6×10^-19)
n = 6.25 × 10^18 electron
Posted by Shreepal Chouhan 5 years, 1 month ago
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Gaurav Seth 5 years, 1 month ago
According to Gauss’s law, the total of the electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity. The total electric flux through a closed surface is zero if no charge is enclosed by the surface.
- Gauss’s law is true for any closed surface, no matter what its shape or size.
- The term q on the right side of Gauss’s lawincludes the sum of all charges enclosed by the surface. The charges may be located anywhere inside the surface.
- In the situation when the surface is so chosen that there are some charges inside and some outside, the electric field [whose flux appears on the left side of Eq. (1.31)] is due to all the charges, both inside and outside S. The term q on the right side of Gauss’s law, however, represents only the total charge inside S.
- The surface that we choose for the application of Gauss’s law is called the Gaussian surface. The Gaussian surface can pass through a continuous charge distribution.
- Gauss’s law is useful for the calculation of the electrostatic field for a symmetric system.
- Gauss’s law is based on the inverse square dependence on distance contained in the Coulomb’s law. Any violation of Gauss’s law will indicate departure from the inverse square law.
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Manish Kushram 5 years, 1 month ago
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