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For Gauss’s law mark the correct statements . (1) If we displace the enclosed charges (within a Gaussian surface) without crossing the boundary, then both E→ and ϕ remain same. (2) If we displace the enclosed charges without crossing the boundary, then E→ changes but ϕ remains the same. (3) If the charge crosses the boundary, then both E→ and ϕ would change. (4) If we charge the boundary, then ϕ changes but E remians the same.
01
Nov
For Gauss’s law mark the correct statements . (1) If we displace the enclosed charges (within a Gaussian surface) without crossing the boundary, then both E→ and ϕ remain same. (2) If we displace the enclosed charges without crossing the boundary, then E→ changes but ϕ remains the same. (3) If the charge crosses the [...]
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For Gauss's law mark the correct statements . (1) If we displace the enclosed charges (within a Gaussian surface) without crossing the boundary ,
then both E→ and ϕ remain same. (2) If we displace the enclosed charges without crossing the boundary ,
then both E→ and ϕ would change. (4) If we charge the boundary ,
then E→ changes but ϕ remains the same. (3) If the charge crosses the boundary ,
then ϕ changes but E remians the same. ,
A 10 C charge is given to a conducting spherical shell, and a – 3 C point charge is placed inside the shell. For this arrangement , find the correct statements.
01
Nov
A 10 C charge is given to a conducting spherical shell, and a – 3 C point charge is placed inside the shell. For this arrangement , find the correct statements. A 10 C charge is given to a conducting spherical shell and a - 3 C point charge is placed inside the shell. For [...]
Consider a Gaussian spherical surface covering a dipole of charge q and -q , then
01
Nov
Consider a Gaussian spherical surface covering a dipole of charge q and -q , then Consider a Gaussian spherical surface covering a dipole of charge q and -q then November 1, 2020 Category: Uncategorised (JEE Advanced Physics by BM Sharma + GMP Solutions) ,
Consider Gauss’s law: phie E. ds = q/Eo Then, for the situation shown in figure at the Gaussian surface
01
Nov
Consider Gauss’s law: phie E. ds = q/Eo Then, for the situation shown in figure at the Gaussian surface Consider Gauss's law: phie E. ds = q/Eo Then for the situation shown in figure at the Gaussian surface November 1, 2020 Category: Uncategorised (JEE Advanced Physics by BM Sharma + GMP Solutions) ,
A charge Q1 is placed at O inside a hollow conducting sphere having inner and outer radii as 10 m and 11 m as shown. The force experienced by Q2 at P is F1 and Force experienced by Q2 when Q1 is placed at O1 is F2. Then F1/F2 is equal to
01
Nov
A charge Q1 is placed at O inside a hollow conducting sphere having inner and outer radii as 10 m and 11 m as shown. The force experienced by Q2 at P is F1 and Force experienced by Q2 when Q1 is placed at O1 is F2. Then F1/F2 is equal to are separated by [...]
A particle executes linear simple harmonic motion with an amplitude of 2 cm . When the particle is at 1 cm from the mean position the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is
31
Oct
A particle executes linear simple harmonic motion with an amplitude of 2 cm . When the particle is at 1 cm from the mean position the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is A particle starts Simple harmonic motion from the mean position. Its [...]
A mass m is suspended from a spring. Its frequency of oscillation is f. The spring is cut into equal halves and the same mass is suspended from one of the two pieces of the spring. The frequency of oscillation is the mass will be
31
Oct
A mass m is suspended from a spring. Its frequency of oscillation is f. The spring is cut into equal halves and the same mass is suspended from one of the two pieces of the spring. The frequency of oscillation is the mass will be A particle starts Simple harmonic motion from the mean position. [...]
A particle starts Simple harmonic motion from the mean position. Its amplitude is a and total energy E. At on instant its kinetic energy is 4/3E . Its displacement at that instant is
31
Oct
A particle starts Simple harmonic motion from the mean position. Its amplitude is a and total energy E. At on instant its kinetic energy is 4/3E . Its displacement at that instant is A particle starts Simple harmonic motion from the mean position. Its amplitude is a and total energy E. At on instant its [...]
A particle starts Simple harmonic motion from the mean position. Its amplitude is a and total energy E. At on instant its kinetic energy is 4/3E . Its displacement at that instant is
31
Oct
A particle starts Simple harmonic motion from the mean position. Its amplitude is a and total energy E. At on instant its kinetic energy is 4/3E . Its displacement at that instant is
Two simple harmonic motions y1 = Asinωt and y2 = Acosωt are superimposed on a particle of mass m. The total mechanical energy of the particle is
31
Oct
Two simple harmonic motions y1 = Asinωt and y2 = Acosωt are superimposed on a particle of mass m. The total mechanical energy of the particle is Two simple harmonic motions y1 = Asinωt and y2 = Acosωt are superimposed on a particle of mass m. The total mechanical energy of the particle is October [...]