Home›MPPSC Forest›State Forest Service Mains Exam›Section 'B': Forestry and General Science›UNIT-07: Physics›Electricity and its effects: Electric intensity, potential, potential difference, electric current, Ohm's law, resistance, specific resistance, factors influencing resistance, combination of resistance (numerical examples), thermal effect of current and its use, calculation of power and electrical energy spent (numerical), precautions in electric experiments›Two positive point charges of 12 \mu C and 8 \mu C (micro coulomb)

Two positive point charges of 12 μ\muμ C and 8 μ\muμ C (micro coulomb) respectively are 10 cm apart. The work done in bringing them 4 cm closer so that they are 6 cm apart is
(1 / (4 π\piπ\varepsilon_0) = 9∗1099 * 10^99∗109 mF^-1)

20213M#mppsc-forest#pyq#mcq#2021
A.5.8 J
B.2.8 J
C.1.5 JCorrect
D.0 J
Explanation & Solution

Correct Answer: (C) 1.5 J

The work done in moving two point charges closer is equal to the change in potential energy: W = U_final - U_initial = (k * q_1 * q_2) * (1/r_final - 1/r_initial). Substituting the values gives 1.5 J.

हिंदी में प्रश्न एवं आदर्श उत्तर

12 μ\muμ C और 8 μ\muμ C के दो धनात्मक बिंदु आवेश 10 cm की दूरी पर हैं। उन्हें 4 cm करीब, (ताकि वे 6 cm दूर हों) लाने में किया गया कार्य है
(1 / (4 π\piπ\varepsilon_0) = 9∗1099 * 10^99∗109 mF^-1)

A.5·8 J
B.2·8 J
C.1·5 Jसही
D.0 J

सही उत्तर: (C) 1·5 J

दो आवेशों को एक-दूसरे के करीब लाने में किया गया कार्य उनकी स्थितिज ऊर्जा में परिवर्तन के बराबर होता है, जिसकी गणना करने पर कार्य का मान 1.5 J प्राप्त होता है।

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