Physics Tuition Classes for College Level in Durgapur

We know that S.I unit of power lens 1dioptre,D= reciprocal of focal length(f) in metre So, D=1/f or, f=1/D=1/2= 0.5 metre= 50 c.m. So, focal length of this lens is = 50 c.m.

1Horse Power= 746 Watts

Cell is the active source or elements in electric circuit. It's main function is to give power supply to the closed circuit or for which current will flow through the circuit.

Distance of the object from optical centre of the convex lens (u)= 15 c.m. Focal length of the convex lens (v)= 30 c.m. Let's, distance of image formed from optical centre= v c.m. Now, applying lens formula, 1/v - 1/u = 1/f; or, 1/v -1/15= 1/30 or, 1/v= 1/15 - 1/30 or, 1/v= - 1/30 or,v= -30 So, from sign convention formula of lens image will be formed on same side of the object and 30 cm from optical centre. Image nature will be virtual and erect.

Here uniform electric field (E) is parallel to the axis of the cone . So, at any point on the cone surface uniform electric field(E) and area vector (dA) will be perpendicular to each other. So, total flux for the surface of the cone= ( vector E)•(vector dA)= E (dA) cos(theta)= E (dA) cos 90 = 0 Here correct answer is option (d).

Motion is important when force is applied.

Coulomb's law states that: The magnitude of the electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them. In order to obtain both the magnitude and direction of the force on a charge, q 1 at position, experiencing a field due to the presence of another charge, q 2 at position , the full vector form of Coulomb 's law is required.

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A cell functions as the main power supply in an electric circuit. It is a device, which converts chemical energy into electrical energy.and when switched on provides the same to all the circuit components.