Physics-
General
Easy

Question

A current IA is flowing in the sides of equilateral triangle of side 4.5 cross times 10 to the power of negative 2 end exponentm The magnetic induction at centroid of the triangle is

  1. 4 cross times 10 to the power of negative 5 end exponent T    
  2. 40 T    
  3. 0.4 cross times 10 to the power of negative 3 end exponent T    
  4. 4 cross times 10 to the power of negative 2 end exponent T    

The correct answer is: 4 cross times 10 to the power of negative 5 end exponent T

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The magnetic field at the centre of circular loop in the circuit shown below is

The magnetic field at the centre of circular loop in the circuit shown below is

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Is this quadrilateral a trapezoid?

Is this quadrilateral a trapezoid?

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A thin 50 cm long metal bar with mass 750 g rests on, but is not attached to two metallic supports in a uniform 0.45T magnetic field as shown in Fig .A battery and a 25capital omega blankresistor in series are connceted to the supports. The largest voltage the battery can have without breaking the circuit at the supports (units are in ”V”) is

A thin 50 cm long metal bar with mass 750 g rests on, but is not attached to two metallic supports in a uniform 0.45T magnetic field as shown in Fig .A battery and a 25capital omega blankresistor in series are connceted to the supports. The largest voltage the battery can have without breaking the circuit at the supports (units are in ”V”) is

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What is the area of the trapezoid?

What is the area of the trapezoid?

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The radius of a circle is 7 meters. What is area of the circle?

The radius of a circle is 7 meters. What is area of the circle?

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Multiply: 9 × 4 = 

Multiply: 9 × 4 = 

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Solve the quadratic equation below using the Quadratic Formula x2+5x14=0

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In the figure, C A B equals 90 to the power of ring operator end exponent text  and  end text A D perpendicular B C. If AC = 75 cm AB = 100 cm and BD = 1.25 cm then AD = 

we can use the property that ratio of sides remains same in similar triangles.

In the figure, C A B equals 90 to the power of ring operator end exponent text  and  end text A D perpendicular B C. If AC = 75 cm AB = 100 cm and BD = 1.25 cm then AD = 

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we can use the property that ratio of sides remains same in similar triangles.

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text In the figure  end text fraction numerator A O over denominator O C end fraction equals fraction numerator B O over denominator O D end fraction equals fraction numerator 1 over denominator 2 end fraction text  and  end text A B equals 5 c m. text  The value of  end text D C text  is end text

we can use the property that ratio of sides remains same in similar triangles.

text In the figure  end text fraction numerator A O over denominator O C end fraction equals fraction numerator B O over denominator O D end fraction equals fraction numerator 1 over denominator 2 end fraction text  and  end text A B equals 5 c m. text  The value of  end text D C text  is end text

Maths-General

we can use the property that ratio of sides remains same in similar triangles.

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In the figuretext ,  end text stack D E with bar on top divided by divided by stack B C with bar on top text end textand areatext end text left parenthesis triangle A D E right parenthesis equals text  area  end text left parenthesis triangle B C E D right parenthesis. text end textThe value oftext end text fraction numerator B D over denominator A B end fraction equals

In the figuretext ,  end text stack D E with bar on top divided by divided by stack B C with bar on top text end textand areatext end text left parenthesis triangle A D E right parenthesis equals text  area  end text left parenthesis triangle B C E D right parenthesis. text end textThe value oftext end text fraction numerator B D over denominator A B end fraction equals

maths-General
General
Maths-

In the figure, angle A equals angle C E D, AB = 9 cm, AD = 7 cm, CD = 8 cm and CE = 10 cm Then DE = ?

if 2 angles are same in a triangle, then the triangles are similar. we can use the property that ratio of sides remains same in similar triangles.

In the figure, angle A equals angle C E D, AB = 9 cm, AD = 7 cm, CD = 8 cm and CE = 10 cm Then DE = ?

Maths-General

if 2 angles are same in a triangle, then the triangles are similar. we can use the property that ratio of sides remains same in similar triangles.

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In the figuretext ,  end text stack A B with bar on top divided by divided by stack Q R with bar on top text  . If  end text A B equals 3 c m comma P B equals 2 c m text  and end text PR = 6 cm then QR = ? cm

we can use the property that ratio of sides remains same in similar triangles

In the figuretext ,  end text stack A B with bar on top divided by divided by stack Q R with bar on top text  . If  end text A B equals 3 c m comma P B equals 2 c m text  and end text PR = 6 cm then QR = ? cm

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we can use the property that ratio of sides remains same in similar triangles

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In the figuretext ,  end text stack A B with bar on top divided by divided by stack C D with bar on top text  and  end text stack A C with bar on top intersection stack B D with bar on top equals 0. text  If  end text O A equals 3 x minus 1 text  , end text OB = 2x + 1, OC = 5x – 3, OD = 6x – 5 then AC = ? units.

solving the quadratic equations by the factorization method is used. in this method, the linear term is broken down into 2 terms so that we can take out the common factors from the terms and convert the equation into product form.

In the figuretext ,  end text stack A B with bar on top divided by divided by stack C D with bar on top text  and  end text stack A C with bar on top intersection stack B D with bar on top equals 0. text  If  end text O A equals 3 x minus 1 text  , end text OB = 2x + 1, OC = 5x – 3, OD = 6x – 5 then AC = ? units.

Maths-General

solving the quadratic equations by the factorization method is used. in this method, the linear term is broken down into 2 terms so that we can take out the common factors from the terms and convert the equation into product form.

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Two line segments text end text stack A B with rightwards arrow on top text  and  end text stack C D with bar on top text end text intersect at E such that triangle A C E tilde operator triangle B D E. If AE = 4cm, BE = 3cm, CE=6cm and DE = x cm then x = ?

We can also use trigonometry to solve this question since all the angles are same in similar triangles. Angle AEC = angle BED.

Two line segments text end text stack A B with rightwards arrow on top text  and  end text stack C D with bar on top text end text intersect at E such that triangle A C E tilde operator triangle B D E. If AE = 4cm, BE = 3cm, CE=6cm and DE = x cm then x = ?

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We can also use trigonometry to solve this question since all the angles are same in similar triangles. Angle AEC = angle BED.

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From the adjacent figure ,the values of x and y are

From the adjacent figure ,the values of x and y are

maths-General
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