Suggested solution:
If R is the resistance of the microphone and r is the resistance of the ammeter, the..
If R is the resistance of the microphone and r is the resistance of the ammeter, the..Electrostatic Potential and Capacitance Problems - 01
01. - Calculate the absolute potential at the point P in the figur..
01. - Calculate the absolute potential at the point P in the figur..Suggested solution:
(a) The terminal voltage V (the voltage drop across the terminals of the battery) is related to the EMF of the battery (open circuit voltage drop) by the relation, V = E - Ir Now, E = 12 V, I = 90 A or V = 12 - 4.5 = 7.5 V (b) The maximum current from a battery can be drawn..
(a) The terminal voltage V (the voltage drop across the terminals of the battery) is related to the EMF of the battery (open circuit voltage drop) by the relation, V = E - Ir Now, E = 12 V, I = 90 A or V = 12 - 4.5 = 7.5 V (b) The maximum current from a battery can be drawn..Magnetic Field due to a Straight Conductor Carrying Current
Consider a straight conductor XY carrying a current I as shown. To find the magnetic field at P, consider a small current element of length dl. If be the position of point P from current element q the angle between Let CO = I According to Biot and Savart's law, field at P due to current f..
Evaluation of Magnetic Field
Magnetic Field due to a Straight Conductor Carrying Current - Consider a straight conductor XY carrying a current I as shown. To find the magnetic field at P, consider a small current element of length dl. If be the position of point P from current element q the angle between Let CO = I According t..
Suggested answer:
02. An electric dipole consists of +5C and -5C separated by a distance 3cm. Find the flux passing through a sphere of radius 6cm enclosing the dipol..
02. An electric dipole consists of +5C and -5C separated by a distance 3cm. Find the flux passing through a sphere of radius 6cm enclosing the dipol..Suggested answer:
Total potential at the centre O of the circle is given by o..
Total potential at the centre O of the circle is given by o..Magnetic Field Intensity due to a Bar Magnet Along the Axis
Consider a point P located on the axial line of a short bar magnet of magnetic length '2l' and pole strength m. Let us find at a point p which is at a distance d from the centre of the magnet. Magnetic flux density at p due to N-pole ..
Consider a point P located on the axial line of a short bar magnet of magnetic length '2l' and pole strength m. Let us find at a point p which is at a distance d from the centre of the magnet. Magnetic flux density at p due to N-pole ..Alternative Method
Double the current for double the time will deposit 0.5 x 2 x 2 mole i.e., 2 mole of silver. ..
Double the current for double the time will deposit 0.5 x 2 x 2 mole i.e., 2 mole of silver. ..Suggested solution:
The valency of silver is one. So, silver is monoatomic. Thus, the chemical equivalent of silver is the same as the molar mass of silver. Since charge required to liberate one chemical equivalent of silver is 96500 C, therefore, the charge q required to liberate 0.5 chemical equivalent of silver is..
The valency of silver is one. So, silver is monoatomic. Thus, the chemical equivalent of silver is the same as the molar mass of silver. Since charge required to liberate one chemical equivalent of silver is 96500 C, therefore, the charge q required to liberate 0.5 chemical equivalent of silver is.. Result
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