Recent content by Ananya0107

Hey Is my method right?

I think this is the maximum emf produced because when the pendulum rotates through an angle θ the area swept by it first decreases then increases again for the other half of the motion. Therefore the maximum emf would be produced when it sweeps through an angle θ.

Mgl(1-cosθ) = 1/2 ml2 ω2 Which gives ω= (2g(1-cosθ)/l)½ Which gives E = Bωl2 /2 E = Bl2sin(θ/2) (g/l)½

Yes.

Homework Statement A ring of radius R is kept in the xy plane and a constant uniform magnetic field exists of magnitude B in the -k direction (negative z direction ) . It is heated through a temperature T . If the resistance of the ring is R1find the final radius of the ring. Coefficient of...

The question contained the vectors a and b so it is easier to express the required vector x in terms of a , b and a×b , it is just easier for computation .

a, b, and a×b are a set of non coplanar vectors and any vector in three dimensional space can be expressed as a linear combination of three non-coplanar vectors,( just like i, j, k)

Yes, thanks

We have to find the sum...

Any hints for this : 1- (11C1/2.3 ).2^2 + (11C2/3.4 ). 2^3 ...so on up to 12 terms .

Tell me if this argument is right... I constructed a 1by 1 square on the coordinate plane vertices (0,0) ,(0,1),(1,0),(1,1). Then I said that to graph any continuous function , the graph must touch or cut the square's diagonal , that is y=x.

Let f:[0,1]→[0,1] be a continuous function . Show that there exists a point such that f(x) = x

True..

Actua actually I was thinking too much ...:sorry: Eccentricity of the hyperbola = 5/3 from the question , and it passes through (±3, 0) , its equation therefore is x^2/9 - y^2/b^2 =1 where 1+ b^2/9 = 25/9 therefore equation of hyperbola is x^2/9 - y^2/9 = 1 and its focii are ±5,0

How about dividing both sides by x^2 then put x-1/x = t, 1+ 1/x^2 dx = dt