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Physics NCERT Exemplar Solutions Class 12th Chapter Five Physics Class 12th physics ncert exemplar solutions class 12th chapter two Physics Electrostatic Potential and Capacitance

A positive charge particle of 100mg is thrown in opposite direction to a uniform electric field of strength 1 × 10⁵ NC^-1, If the charge on the particle is $40 μ C$ and the initially velocity is 200 ms^-1, how much distance it will travel before coming to the rest momentarily:

Option 1 -

1 m

Option 2 -

5 m

Option 3 -

10 m

Option 4 -

0.5 m

4 Views | Posted 2 months ago

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2 Answers

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 4

Detailed Solution:

a = $\frac{q E}{m} = \frac{40 * 1 0^{? 6} * 1 0^{5}}{100 * 1 0^{? 3} * 1 0^{? 3}}$

= 40 * 10³ m/sec²

v² = u² + 2a s

$? 0 = {(200)}^{2} ? 2 * 40 * 1 0^{3} s$

$s = \frac{200 * 200}{2 * 40 * 1 0^{3}} = 0.5 m$

a = <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1025" DrawAspect="Content" ObjectID="_1814269418"> </o:OLEObject></xml><! [endif]->  <math> <mrow> <mfrac> <mrow> <mi>q</mi> <mi>E</mi> </mrow> <mrow> <mi>m</mi> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mn>4</mn> <mn>0</mn> <mo>*</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mo>? </mo> <mn>6</mn> </mrow> </msup> <mo>*</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mn>5</mn> </mrow> </msup> </mrow> <mrow> <mn>1</mn> <mn>0</mn> <mn>0</mn> <mo>*</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mo>? </mo> <mn>3</mn> </mrow> </msup> <mo>*</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mo>? </mo> <mn>3</mn> </mrow> </msup> </mrow> </mfrac> </mrow> </math>                       = 40 * 103 m/sec2              v2 = u2 + 2a s <math> <mrow> <mo>? </mo> <mn>0</mn> <mo>=</mo> <msup> <mrow> <mrow> <mo> (</mo> <mrow> <mn>2</mn> <mn>0</mn> <mn>0</mn> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>? </mo> <mn>2</mn> <mo>*</mo> <mn>4</mn> <mn>0</mn> <mo>*</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mn>3</mn> </mrow> </msup> <mi>s</mi> </mrow> </math> <math> <mrow> <mi>s</mi> <mo>=</mo> <mfrac> <mrow> <mn>2</mn> <mn>0</mn> <mn>0</mn> <mo>*</mo> <mn>2</mn> <mn>0</mn> <mn>0</mn> </mrow> <mrow> <mn>2</mn> <mo>*</mo> <mn>4</mn> <mn>0</mn> <mo>*</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mn>3</mn> </mrow> </msup> </mrow> </mfrac> <mo>=</mo> <mn>0</mn> <mo>.</mo> <mn>5</mn> <mi>m</mi> </mrow> </math>              <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1026" DrawAspect="Content" ObjectID="_1814269419"> </o:OLEObject></xml><! [endif]->               <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1027" DrawAspect="Content" ObjectID="_1814269420"> </o:OLEObject></xml><! [endif]->

0 Upvotes 0 Downvotes

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 4

Detailed Solution:

a = $\frac{q E}{m} = \frac{40 * 1 0^{? 6} * 1 0^{5}}{100 * 1 0^{? 3} * 1 0^{? 3}}$

= 40 * 10³ m/sec²

v² = u² + 2a s

$? 0 = {(200)}^{2} ? 2 * 40 * 1 0^{3} s$

$s = \frac{200 * 200}{2 * 40 * 1 0^{3}} = 0.5 m$

a = <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1025" DrawAspect="Content" ObjectID="_1814269418"> </o:OLEObject></xml><! [endif]->  <math> <mrow> <mfrac> <mrow> <mi>q</mi> <mi>E</mi> </mrow> <mrow> <mi>m</mi> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mn>4</mn> <mn>0</mn> <mo>*</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mo>? </mo> <mn>6</mn> </mrow> </msup> <mo>*</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mn>5</mn> </mrow> </msup> </mrow> <mrow> <mn>1</mn> <mn>0</mn> <mn>0</mn> <mo>*</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mo>? </mo> <mn>3</mn> </mrow> </msup> <mo>*</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mo>? </mo> <mn>3</mn> </mrow> </msup> </mrow> </mfrac> </mrow> </math>                       = 40 * 103 m/sec2              v2 = u2 + 2a s <math> <mrow> <mo>? </mo> <mn>0</mn> <mo>=</mo> <msup> <mrow> <mrow> <mo> (</mo> <mrow> <mn>2</mn> <mn>0</mn> <mn>0</mn> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>? </mo> <mn>2</mn> <mo>*</mo> <mn>4</mn> <mn>0</mn> <mo>*</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mn>3</mn> </mrow> </msup> <mi>s</mi> </mrow> </math> <math> <mrow> <mi>s</mi> <mo>=</mo> <mfrac> <mrow> <mn>2</mn> <mn>0</mn> <mn>0</mn> <mo>*</mo> <mn>2</mn> <mn>0</mn> <mn>0</mn> </mrow> <mrow> <mn>2</mn> <mo>*</mo> <mn>4</mn> <mn>0</mn> <mo>*</mo> <mn>1</mn> <msup> <mrow> <mn>0</mn> </mrow> <mrow> <mn>3</mn> </mrow> </msup> </mrow> </mfrac> <mo>=</mo> <mn>0</mn> <mo>.</mo> <mn>5</mn> <mi>m</mi> </mrow> </math>              <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1026" DrawAspect="Content" ObjectID="_1814269419"> </o:OLEObject></xml><! [endif]->               <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1027" DrawAspect="Content" ObjectID="_1814269420"> </o:OLEObject></xml><! [endif]->

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A positive charge particle of 100mg is thrown in opposite direction to a uniform electric field of strength 1 × 10⁵ NC^-1, If the charge on the particle is $40 μ C$ and the initially velocity is 200 ms^-1, how much distance it will travel before coming to the rest momentarily:

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A positive charge particle of 100mg is thrown in opposite direction to a uniform electric field of strength 1 × 105 NC-1, If the charge on the particle is 4 0 μ C and the initially velocity is 200 ms-1, how much distance it will travel before coming to the rest momentarily:

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A positive charge particle of 100mg is thrown in opposite direction to a uniform electric field of strength 1 × 10⁵ NC^-1, If the charge on the particle is $40 μ C$ and the initially velocity is 200 ms^-1, how much distance it will travel before coming to the rest momentarily: