If the roots of the equation&nbsp;<math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mn>1</mn><mrow><mi mathvariant="normal">x</mi><mo>+</mo><mi mathvariant="normal">p</mi></mrow></mfrac><mo>+</mo><mfrac><mn>1</mn><mrow><mi mathvariant="normal">x</mi><mo>+</mo><mi mathvariant="normal">q</mi></mrow></mfrac><mo>=</mo><mfrac><mn>1</mn><mi mathvariant="normal">r</mi></mfrac></math>are equal in magnitude but opposite in sign, then the product of the roots will be

A particle of charge&nbsp;<math xmlns="http://www.w3.org/1998/Math/MathML"><mo>−</mo><mn>16</mn><mtext>  </mtext><mo>×</mo><mtext>  </mtext><msup><mn>10</mn><mrow><mo>−</mo><mn>18</mn></mrow></msup><mtext>  </mtext><mi mathvariant="normal">C</mi></math>&nbsp;moving with velocity&nbsp;<math xmlns="http://www.w3.org/1998/Math/MathML"><mn>10</mn><mtext>  </mtext><mi mathvariant="normal">m</mi><mtext>  </mtext><msup><mi mathvariant="normal">s</mi><mrow><mo>−</mo><mn>1</mn></mrow></msup></math>&nbsp;along x-axis &nbsp;enters a region where a magnetic field of induction B exists along the y-axis and an electric &nbsp;field of magnitude 104 V/m exists along the negative z-axis. If the charged particle continues moving along the x-axis, the magnitude of B is

A proton moves at a constant velocity of&nbsp;<math xmlns="http://www.w3.org/1998/Math/MathML"><mn>50</mn><mo>&nbsp;</mo><mi>m</mi><mo>/</mo><mi>s</mi></math>&nbsp;along the x-axis, in uniform electric and magnetic fields. The magnetic field is&nbsp;<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>B</mi><mo>=</mo><mfenced><mrow><mn>2</mn><mo>.</mo><mn>0</mn><mo>&nbsp;</mo><mi>m</mi><mi>T</mi></mrow></mfenced><mover><mi>j</mi><mo>^</mo></mover></math>. What is the electric field?

In a crossed field, the magnetic field induction is 2.0T and electric field intensity is&nbsp;<math xmlns="http://www.w3.org/1998/Math/MathML"><mn>20</mn><mo>×</mo><msup><mn>10</mn><mn>3</mn></msup><mi mathvariant="normal">v</mi><mo>/</mo><mi mathvariant="normal">m</mi></math>. At which velocity, the electron will travel in a straight line without the effect of electric and magnetic fields ?

A charged particle of mass m, carrying a charge Q is projected from the origin with a velocity&nbsp;<math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi mathvariant="normal">V</mi><mo>→</mo></mover><mo>=</mo><msub><mi mathvariant="normal">V</mi><mi mathvariant="normal">o</mi></msub><mfenced><mrow><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover><mo>+</mo><mover><mi mathvariant="normal">j</mi><mo>^</mo></mover></mrow></mfenced></math>. A uniform magnetic field&nbsp;<math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi mathvariant="normal">B</mi><mo>→</mo></mover><mo>=</mo><mi mathvariant="normal">B</mi><mover><mi mathvariant="normal">i</mi><mo>^</mo></mover></math>&nbsp;exists in that region. Then maximum distance of the particle from x-axis is

The electric field acts along positive x-axis. A charged particle of charge q and mass m is released from origin and moves with velocity&nbsp;<math xmlns="http://www.w3.org/1998/Math/MathML"><mover accent="true"><mi mathvariant="normal">v</mi><mo>→</mo></mover><mo>=</mo><msub><mi mathvariant="normal">v</mi><mn>0</mn></msub><mover accent="true"><mi mathvariant="normal">j</mi><mo>^</mo></mover></math>&nbsp;under the action of electric field and magnetic field,&nbsp;<math xmlns="http://www.w3.org/1998/Math/MathML"><mover accent="true"><mi mathvariant="normal">B</mi><mo>→</mo></mover><mo>=</mo><msub><mi mathvariant="normal">B</mi><mn>0</mn></msub><mover accent="true"><mi mathvariant="normal">i</mi><mo>^</mo></mover></math>&nbsp;The velocity of particle becomes&nbsp;<math xmlns="http://www.w3.org/1998/Math/MathML"><mn>2</mn><msub><mi>v</mi><mn>0</mn></msub></math>&nbsp;after time&nbsp;<math xmlns="http://www.w3.org/1998/Math/MathML"><mfrac><mrow><msqrt><mn>3</mn></msqrt><mtext> </mtext><msub><mi>mv</mi><mn>0</mn></msub></mrow><mrow><msqrt><mn>2</mn></msqrt><mtext> </mtext><msub><mi>qE</mi><mn>0</mn></msub></mrow></mfrac><mo>.</mo></math>&nbsp; Find the electric field.

If the roots of the equation $\frac{1}{x + p} + \frac{1}{x + q} = \frac{1}{r}$ are equal in magnitude but opposite in sign, then the product of the roots will be

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If the roots of the equation 1x+p+1x+q=1rare equal in magnitude but opposite in sign, then the product of the roots will be

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If the roots of the equation $\frac{1}{x + p} + \frac{1}{x + q} = \frac{1}{r}$ are equal in magnitude but opposite in sign, then the product of the roots will be