Have a query? Ask a Tutor!!

Access to best educators and exclusive paper discussions

Offsite Campus Turito Championship

Can’t find what you’re looking for?

Our tutors are from top schools around the world, and they are waiting for you 24/7, anytime, anywhere.

Homework- One to One Solutions
Understand concepts to solve better
Get your doubts solved instantly

Maths-

General

Easy

Question

This section contains 3 multiple choice questions numbered 14 to 16 .Each question has 4 choices (A), (B),(c) , and (D), out of which ONE or MORE may be correct
If $fraction numerator 3 plus i 2 s i n invisible function application theta over denominator 1 minus i 2 s i n invisible function application theta end fraction$ is real , then

$θ = k π, k \in i$
$θ = (2 k + 1) \frac{π}{2}, k \in I$
$θ = 0$
$θ = 2 k π, k \in I$

The correct answer is: $theta equals k pi comma k element of i$

$I m invisible function application left parenthesis 2 right parenthesis equals 0 rightwards double arrow s i n invisible function application theta equals 0$
$theta equals k pi comma k element of I$

Related Questions to study

A charge q is placed at the center of the open end of cylindrical vessel. The flux of the electric field through surface of the vessel is

A charge q is placed at the center of the open end of cylindrical vessel. The flux of the electric field through surface of the vessel is

Physics-General

The force between two charges 0.06 m apart is 5 N. If each charge is moved towards the other by 0.01 m, then the force between them will become

The force between two charges 0.06 m apart is 5 N. If each charge is moved towards the other by 0.01 m, then the force between them will become

Physics-General

Figure shows four charges q₁, q₂, q₃ and q₄ fixed in space. Then the total flux of electric field through a closed surface S, due to all charges q₁, q₂, q₃ and q₄ is

Figure shows four charges q₁, q₂, q₃ and q₄ fixed in space. Then the total flux of electric field through a closed surface S, due to all charges q₁, q₂, q₃ and q₄ is

Physics-General

parallel

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

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

Physics-General

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

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

Physics-General

The ratio of $phi subscript E end subscript$ passing through the surface S₁ and S₂ is

The ratio of $phi subscript E end subscript$ passing through the surface S₁ and S₂ is

Physics-General

parallel

A point charge q is placed at the origin. The net flux of electric field (E) passing through the curved surface S is

A point charge q is placed at the origin. The net flux of electric field (E) passing through the curved surface S is

Physics-General

Three charges Q, +q and +q are placed at the vertices of a right-angled isosceles triangle as shown in figure. The net electrostatic energy of the configuration is zero, if Q is equal to

Three charges Q, +q and +q are placed at the vertices of a right-angled isosceles triangle as shown in figure. The net electrostatic energy of the configuration is zero, if Q is equal to

Physics-General

A charge Q is shifted from the origin to a point P(a, a). the work done by the electrical force is

A charge Q is shifted from the origin to a point P(a, a). the work done by the electrical force is

Physics-General

parallel

The charge -Q is shifted slowly from C to D, vertices of square of side l. The work done by the external agent is

The charge -Q is shifted slowly from C to D, vertices of square of side l. The work done by the external agent is

Physics-General

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

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

Physics-General

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 $s i n to the power of negative 1 end exponent invisible function application x plus c o s to the power of negative 1 end exponent invisible function application x equals fraction numerator pi over denominator 2 end fraction comma text for all end text x element of R$

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 $s i n to the power of negative 1 end exponent invisible function application x plus c o s to the power of negative 1 end exponent invisible function application x equals fraction numerator pi over denominator 2 end fraction comma text for all end text x element of R$

parallel

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.

Physics-General

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

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

Physics-General

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

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

Physics-General

parallel

card img

With Turito Academy.

card img

With Turito Foundation.

card img

Get an Expert Advice From Turito.

Turito Academy

card img

With Turito Academy.

Test Prep

card img

With Turito Foundation.