SATHEE: Electrostatics Question 353

Electrostatics Question 353

Question: The intensity of the electric field required to keep a water drop of radius $10^{- 5} c m$ just suspended in air when charged with one electron is approximately [MP PMT 1994]

Options:

A) $260 v o l t / c m$

B) $260 n e w t o n / c o u l o m b$

C) $130 v o l t / c m$

D) $130 n e w t o n / c o u l o m b$

$(g = 10 n e w t o n / k g, e = 1.6 \times 10^{- 19} c o u l o m b)$

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Answer:

Correct Answer: B

Solution:

For balance $m g = e E$

therefore $E = \frac{m g}{e}$ Also $m = \frac{4}{3} π r^{3} d = \frac{4}{3} \times \frac{22}{7} \times {(10^{- 7})}^{3} \times 1000 k g$

therefore $E = \frac{4 / 3 \times 22 / 7 \times {(10^{- 7})}^{3} \times 1000 \times 10}{1.6 \times 10^{- 19}}$ = 260 N/C

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Electrostatics Question 353

Question: The intensity of the electric field required to keep a water drop of radius $10^{- 5} c m$ just suspended in air when charged with one electron is approximately [MP PMT 1994]

Options:

Answer:

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

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पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Electrostatics Question 353

Question: The intensity of the electric field required to keep a water drop of radius 10−5cm just suspended in air when charged with one electron is approximately [MP PMT 1994]

Options:

Answer:

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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Question: The intensity of the electric field required to keep a water drop of radius $10^{- 5} c m$ just suspended in air when charged with one electron is approximately [MP PMT 1994]