The differential equation of all circles which pass through the origin and whose centre lies on y-axis is

For such questions, we should know how to write the equation of a parabola. The axis of parabola can be parallel to x or y axis. So, we have to write the equation accordingly.

Solution of the differential equation $open parentheses fraction numerator d y over denominator d x end fraction close parentheses to the power of 2 end exponent minus fraction numerator d y over denominator d x end fraction open parentheses e to the power of x end exponent plus e to the power of negative x end exponent close parentheses plus 1 equals 0$ is given by

When we get multipliction of two brackets equal to zero, any of the bracket can be zero or both of the brackets can be zero. It is decided based on the conditions in the equation.

Solution of the differential equation $open parentheses fraction numerator d y over denominator d x end fraction close parentheses to the power of 2 end exponent minus fraction numerator d y over denominator d x end fraction open parentheses e to the power of x end exponent plus e to the power of negative x end exponent close parentheses plus 1 equals 0$ is given by

When we get multipliction of two brackets equal to zero, any of the bracket can be zero or both of the brackets can be zero. It is decided based on the conditions in the equation.

The differential equation of all ellipse centred at the origin and major and minor axes along coordinate axes is

Area of the region bounded by $y equals tan invisible function application x comma$ tangent drawn to the curve at $x equals fraction numerator pi over denominator 4 end fraction$ and the $x minus$ axis is

Maximum value of $x left parenthesis 1 minus x right parenthesis to the power of 2 end exponent$ , when $0 less or equal than x less or equal than 2$ , is

If x = $c o s to the power of negative 1 end exponent$ t, y = $square root of 1 minus t to the power of 2 end exponent end root$ , then $y subscript 2 end subscript equals$

Two capacitors $C subscript 1 end subscript equals 2 mu F$ and $C subscript 2 end subscript equals 6 mu F$ in series, are connected in parallel to a third capacitor $C subscript 3 end subscript equals 4 mu F$ . This arrangement is then connected to a battery of e.m.f. = 2V, as shown in the figure. How much energy is lost by the battery in charging the capacitors

The combination of capacitors with $C subscript 1 end subscript equals 3 mu F comma C subscript 2 end subscript equals 4 mu F$ and $C subscript 3 end subscript equals 2 mu F$ is charged by connecting AB to a battery. Consider the following statements
I. Energy stored in $C subscript 1 end subscript$ = Energy stored in $C subscript 2 end subscript$ + Energy stored in $C subscript 3 end subscript$
II. Charge on C₁ = Charge on C₂ + Charge on C₃
III. Potential drop across C₁ = Potential drop across C₂ = Potential drop across C₃
Which of these is/are correct

Consider a parallel plate capacitor of $10 mu rightwards arrow over short leftwards arrow F$ (micro-farad) with air filled in the gap between the plates. Now one half of the space between the plates is filled with a dielectric of dielectric constant 4, as shown in the figure. The capacity of the capacitor changes to

In the figure a capacitor is filled with dielectrics. The resultant capacitance is

Equivalent capacitance between A and B is

In the figure, three capacitors each of capacitance 6 $p F$ are connected in series. The total capacitance of the combination will be

Two capacitors A and B are connected in series with a battery as shown in the figure. When the switch S is closed and the two capacitors get charged fully, then

A parallel plate capacitor of area A, plate separation d and capacitance C is filled with three different dielectric materials having dielectric constants $k subscript 1 end subscript comma k subscript 2 end subscript$ and $k subscript 3 end subscript$ as shown. If a single dielectric material is to be used to have the same capacitance C in this capacitor, then its dielectric constant k is given by

In the circuit here, the steady state voltage across capacitor C is a fraction of the battery e.m.f. The fraction is decided by