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Physics-

General

Easy

Question

A Gaussian surface in the figure is shown by dotted line. The electric field on the surface will be

due to q₁ and q₂ only
due to q₂ only
zero
due to all

The correct answer is: due to all

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A Gaussian surface S enclosed two charges q₁ = q, q₂ = -q and q₃ = q. The net flux passing through S is

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The field pattern is shown due to two point charges 1 and 2. Then q₁/q₂ =

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Statement : 1 The value of k for which the equation $x to the power of 2 end exponent plus k x minus fraction numerator pi over denominator 2 end fraction plus s i n to the power of negative 1 end exponent invisible function application open parentheses x to the power of 2 end exponent minus 6 x plus 10 close parentheses plus c o s to the power of negative 1 end exponent invisible function application open parentheses x to the power of 2 end exponent minus 6 x plus 10 close parentheses equals 0$ Has a real solution is -3.
Statement : 2 $s i n to the power of negative 1 end exponent invisible function application x plus c o s to the power of negative 1 end exponent invisible function application x equals fraction numerator pi over denominator 2 end fraction comma text for all end text x element of R$

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Six charges of equal magnitude, 3 positive and 3 negative are to be placed on PQRSTU corners of a regular hexagon, such that field at the centre is double that of what it would have been if only one +ve charge is placed at R.

Six charges of equal magnitude, 3 positive and 3 negative are to be placed on PQRSTU corners of a regular hexagon, such that field at the centre is double that of what it would have been if only one +ve charge is placed at R.

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The ratio of q and Q so as to make the system in equilibrium is

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A positive charge Q is placed at the centre of a square of side l. Then the net force acting on the central charge Q due to the other four charges Q, 2Q ,3Q and 4Q placed at the vertices of the square is

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Let $alpha comma beta comma gamma greater than 0$ and $alpha plus beta plus gamma equals fraction numerator n over denominator 2 end fraction$
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Reason (R) : tan $alpha t a n invisible function application beta$ +tan $beta$ tan $gamma$ + tan $y$ tan $C$ =1

Let $alpha comma beta comma gamma greater than 0$ and $alpha plus beta plus gamma equals fraction numerator n over denominator 2 end fraction$
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Consider the function f(x) = cosx + 2cos kx, k Q, k 0
STATEMENT1 : The equation f(x) = 3 has infinite number of solutions. because
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parallel

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