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Tiny spherical oil drop carrying net charge q is balanced in still air with vertical uniform e field of strength 81/pie × 10 ^5 V/m .when the f uel I'd switch off .drop is observed to fall with terminal vel. 2× 10^-3 m/s .viscosity of air 1.8 ×10 -5 N sm-2 .and density of oil is 900 kg/m^3 .magnitude of q is ?.ans is 3.2 × 10 ^-19 C?
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Tiny spherical oil drop carrying net charge q is balanced in still air...
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Tiny spherical oil drop carrying net charge q is balanced in still air...
Given Data:
- Electric field strength (E) = 81/π × 10^5 V/m
- Terminal velocity (v) = 2 × 10^-3 m/s
- Viscosity of air (η) = 1.8 × 10^-5 N s/m^2
- Density of oil (ρ) = 900 kg/m^3
Calculating Charge on the Oil Drop:
- The force due to the electric field is given by Fe = qE, where q is the charge on the oil drop.
- The force due to gravity is given by Fg = mg, where m is the mass of the drop and g is the acceleration due to gravity.
- At terminal velocity, the net force on the oil drop is zero. Therefore, Fe = Fg.
- qE = mg
- q = mg/E
Calculating Mass of the Oil Drop:
- The force of gravity on the drop is balanced by the viscous drag force at terminal velocity.
- The viscous drag force is given by Fd = 6πηrv, where r is the radius of the drop.
- The weight of the drop is given by mg = Vg, where V is the volume of the drop.
- Since the drop is spherical, V = (4/3)πr^3 and m = ρV.
- Substituting the expressions for Fd and mg into the equation at terminal velocity gives 6πηrv = ρVg.
- Solving for the mass of the drop gives m = (4/3)πr^3ρ.
Substitute and Solve:
- Substituting m = (4/3)πr^3ρ and q = mg/E into the equation q = mg/E gives q = (4/3)πr^3ρg/E.
- Substituting the given values of ρ, g, and E gives q = 3.2 × 10^-19 C.
Therefore, the magnitude of the charge on the oil drop is 3.2 × 10^-19 C.
- Electric field strength (E) = 81/π × 10^5 V/m
- Terminal velocity (v) = 2 × 10^-3 m/s
- Viscosity of air (η) = 1.8 × 10^-5 N s/m^2
- Density of oil (ρ) = 900 kg/m^3
Calculating Charge on the Oil Drop:
- The force due to the electric field is given by Fe = qE, where q is the charge on the oil drop.
- The force due to gravity is given by Fg = mg, where m is the mass of the drop and g is the acceleration due to gravity.
- At terminal velocity, the net force on the oil drop is zero. Therefore, Fe = Fg.
- qE = mg
- q = mg/E
Calculating Mass of the Oil Drop:
- The force of gravity on the drop is balanced by the viscous drag force at terminal velocity.
- The viscous drag force is given by Fd = 6πηrv, where r is the radius of the drop.
- The weight of the drop is given by mg = Vg, where V is the volume of the drop.
- Since the drop is spherical, V = (4/3)πr^3 and m = ρV.
- Substituting the expressions for Fd and mg into the equation at terminal velocity gives 6πηrv = ρVg.
- Solving for the mass of the drop gives m = (4/3)πr^3ρ.
Substitute and Solve:
- Substituting m = (4/3)πr^3ρ and q = mg/E into the equation q = mg/E gives q = (4/3)πr^3ρg/E.
- Substituting the given values of ρ, g, and E gives q = 3.2 × 10^-19 C.
Therefore, the magnitude of the charge on the oil drop is 3.2 × 10^-19 C.
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Tiny spherical oil drop carrying net charge q is balanced in still air with vertical uniform e field of strength 81/pie × 10 ^5 V/m .when the f uel I'd switch off .drop is observed to fall with terminal vel. 2× 10^-3 m/s .viscosity of air 1.8 ×10 -5 N sm-2 .and density of oil is 900 kg/m^3 .magnitude of q is ?.ans is 3.2 × 10 ^-19 C?
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Tiny spherical oil drop carrying net charge q is balanced in still air with vertical uniform e field of strength 81/pie × 10 ^5 V/m .when the f uel I'd switch off .drop is observed to fall with terminal vel. 2× 10^-3 m/s .viscosity of air 1.8 ×10 -5 N sm-2 .and density of oil is 900 kg/m^3 .magnitude of q is ?.ans is 3.2 × 10 ^-19 C? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Tiny spherical oil drop carrying net charge q is balanced in still air with vertical uniform e field of strength 81/pie × 10 ^5 V/m .when the f uel I'd switch off .drop is observed to fall with terminal vel. 2× 10^-3 m/s .viscosity of air 1.8 ×10 -5 N sm-2 .and density of oil is 900 kg/m^3 .magnitude of q is ?.ans is 3.2 × 10 ^-19 C? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Tiny spherical oil drop carrying net charge q is balanced in still air with vertical uniform e field of strength 81/pie × 10 ^5 V/m .when the f uel I'd switch off .drop is observed to fall with terminal vel. 2× 10^-3 m/s .viscosity of air 1.8 ×10 -5 N sm-2 .and density of oil is 900 kg/m^3 .magnitude of q is ?.ans is 3.2 × 10 ^-19 C?.
Tiny spherical oil drop carrying net charge q is balanced in still air with vertical uniform e field of strength 81/pie × 10 ^5 V/m .when the f uel I'd switch off .drop is observed to fall with terminal vel. 2× 10^-3 m/s .viscosity of air 1.8 ×10 -5 N sm-2 .and density of oil is 900 kg/m^3 .magnitude of q is ?.ans is 3.2 × 10 ^-19 C? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Tiny spherical oil drop carrying net charge q is balanced in still air with vertical uniform e field of strength 81/pie × 10 ^5 V/m .when the f uel I'd switch off .drop is observed to fall with terminal vel. 2× 10^-3 m/s .viscosity of air 1.8 ×10 -5 N sm-2 .and density of oil is 900 kg/m^3 .magnitude of q is ?.ans is 3.2 × 10 ^-19 C? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Tiny spherical oil drop carrying net charge q is balanced in still air with vertical uniform e field of strength 81/pie × 10 ^5 V/m .when the f uel I'd switch off .drop is observed to fall with terminal vel. 2× 10^-3 m/s .viscosity of air 1.8 ×10 -5 N sm-2 .and density of oil is 900 kg/m^3 .magnitude of q is ?.ans is 3.2 × 10 ^-19 C?.
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