Q47. A tiny spherical oil drop of mass 8 mg is kept suspended by applying a potential difference of

Q: What is the correct answer for CUET UG Physics Q47?

The correct answer is c. For the oil drop to be suspended in the electric field, the upward electric force (F_e) must balance the downward gravitational force (F_g).F_e = F_gThe electric force is F_e = qE, where q is the charge on the oil drop and E is the electric field strength. The electric field between two parallel plates is E = V/d, where V is the potential difference and d is the plate separation. So, F_e = qV/d.Th

Q: Q47. A tiny spherical oil drop of mass 8 mg is kept suspended by applying a potential difference of 100 V between two metallic plates kept separated by a distance 1 cm. The number of electrons on the…

The correct answer is c. For the oil drop to be suspended in the electric field, the upward electric force (F_e) must balance the downward gravitational force (F_g).F_e = F_gThe electric force is F_e = qE, where q is the charge on the oil drop and E is the electric field strength. The electric field between two parallel plates is E = V/d, where V is the potential difference and d is the plate separation. So, F_e = qV/d.Th

Q: What are the options for this CUET UG Physics question?

A. 5 x 1010 · B. 5 x 108 · C. 5 x 1012 · D. 5 x 1013

Question

Q47. A tiny spherical oil drop of mass 8 mg is kept suspended by applying a potential difference of 100 V between two metallic plates kept separated by a distance 1 cm. The number of electrons on the oil drop are: (Take g = 10ms-2)
(a) 5 x 1010
(b) 5 x 108
(c) 5 x 1012
(d) 5 x 1013

FirstInTest · Accepted Answer

For the oil drop to be suspended in the electric field, the upward electric force (F_e) must balance the downward gravitational force (F_g).F_e = F_gThe electric force is F_e = qE, where q is the charge on the oil drop and E is the electric field strength. The electric field between two parallel plates is E = V/d, where V is the potential difference and d is the plate separation. So, F_e = qV/d.The gravitational force is F_g = mg, where m is the mass of the oil drop and g is the acceleration due to gravity.Equating the forces: qV/d = mg => q = mgd/V.Given: m = 8 mg = 8 × 10⁻⁶ kg, g = 10 m/s², d = 1 cm = 1 × 10⁻² m, V = 100 V.q = (8 × 10⁻⁶ kg * 10 m/s² * 1 × 10⁻² m) / 100 V = (8 × 10⁻⁷) / 100 C = 8 × 10⁻⁹ C.The charge on the oil drop is quantized, q = ne, where n is the number of electrons and e is the elementary charge (1.6 × 10⁻¹⁹ C).n = q/e = (8 × 10⁻⁹ C) / (1.6 × 10⁻¹⁹ C) = 5 × 10¹⁰ electrons.Based on the given values, the calculated number of electrons is 5 × 10¹⁰ (option a). However, the provided correct answer is (c) 5 × 10¹² electrons. To obtain 5 × 10¹² electrons, the calculated charge q would need to be 8 × 10⁻⁷ C, which implies a mass of 800 mg for the oil drop, not 8 mg. There appears to be a mismatch between the provided problem statement and the intended answer.

Answer

5 x 1010

Answer

5 x 108

Answer

5 x 1012

Answer

5 x 1013

Related Questionsin Physics

More from CUET UG

Browse CUET Questions by Subject

Practice More CUET UG Questions