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# Tutor profile: Rakesh Varma M.

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Rakesh Varma M.
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## Questions

### Subject:Physics (Newtonian Mechanics)

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Question:

A body of mass 50 kgs is resting on a floor. Friction exists between the bottom layer of the body and the floor, the coefficient of static friction ($$\mu$$) is 0.1. A time dependent horizontal force is applied on the body as, F = 5t newtons. Find the time at which the body starts to move. Assume the acceleration due to gravity to be 10$$\frac{m}{s^2}$$

Inactive
Rakesh Varma M.
Answer:

The Normal reaction (N) faced by the body due to the force applied by the floor on it is equal to the weight of the body. $$\therefore\ N\ = \ mg$$ => $$N$$ = $$50 kgs*10\frac{m}{s^2}$$ = 500 Newtons. The maximum static friction that the body faces is equal to $$\mu*N$$ $$\mu*N\ =\ 0.1*500$$ = 50 Newtons. so, the body moves only when any force greater than 50 Newtons is applied horizontally on it. so, this horizontal force acting on it, should be greater than the maximum static friction, for it to move. $$\therefore\ F\ =\ 5t$$ > 50 Newtons => 5t > 50 => t > 10 seconds $$\therefore$$ The body starts to move just after 10 seconds.

### Subject:Physics (Electricity and Magnetism)

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Question:

A body of mass M has a charge Q, it is projected into a uniform magnetic field of intensity $$\vec{B}$$ with a velocity of $$\vec{V}$$. The velocity of the body just before it entered the field is perpendicular to the magnetic field. Describe the behavior of the body after its entrance into the magnetic field.

Inactive
Rakesh Varma M.
Answer:

As the body is projected into the magnetic field, the Lorentz force experienced by it is $$\vec{F} = Q( \vec{V} *\vec{ B} )$$ => $$F = Q*V*B*\sin\Theta$$ Where $$\Theta$$ is the angle between $$\vec{V} and \vec{B}$$. As $$\vec{V} and \vec{B}$$ are perpendicular to each other $$\Theta$$ = 90$$^0$$ $$Sin\ \Theta\ =\ \sin\ 90^0\ =\ 1$$ $$\therefore\ F\ = Q*V*B$$ --- eq(1) The force acting on the body is always perpendicular to the velocity vector and the magnitude of velocity is not changed, but its direction is changed continuously, this is due to the influence of this force. so, it can be inferred that the body moves in a circular path. This lorentz force acts as the centripetal force and the centrifugal force experienced by the body is $$\frac{MV^2} {R}$$, where R is the radius of the circular path. Both of the forces are equal in magnitude and opposite in direction. $$\therefore\ F\ =\ Q*V*B\ = \frac{MV^2} {R}$$ --- from eq(1) => $$R\ = \frac{MV}{QB}$$ ---eq(2) So, the body moves in a circular path of radius R given by eq(2), with the magnitude of velocity remaining unchanged from the initial.

### Subject:Geometry

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Question:

Prove that, if the midpoint of hypotenuse and midpoint of one of the legs(sides other than the hypotenuse) of a right angled triangle are joined, the line segment thus obtained is parallel to the other leg of the triangle.

Inactive
Rakesh Varma M.
Answer:

Proof: Let us consider a right angled triangle AOB, right angled at O. let us mark the midpoints of hypotenuse AB and leg AO as D and P respectively and join them using a line segment. As, D is the midpoint of AB $$\frac{AD}{AB} = \frac{1}{2}$$ => AB = 2AD --- eq(1) As, P is the midpoint of AO $$\frac{AP}{AO} = \frac{1}{2}$$ => AO = 2AP --- eq(2) Let $$\angle ADP = \Theta_1$$ $$\sin\Theta_1 = \frac{AP}{AD}$$ ---eq(3) Let $$\angle ABO = \Theta_2$$ $$\sin\Theta_2 = \frac{AO}{AB}$$ --- eq(4) As AO = 2AP and AB = 2AD from eq(1) and eq(2) $$\frac{AO}{AB} = \frac{2AP}{2AD} = \frac{AP}{AD}$$ $$\therefore \sin\Theta_2 = \sin\Theta_1$$ from eq(3) and eq(4) => $$\Theta_1 = \Theta_2$$ (angles in a right triangle are acute) If we consider AB as the transversal cutting the line segments DP and BO, $$\angle ADP$$ which is $$\Theta_1$$ and $$\angle ABO$$ which is $$\Theta_2$$ act as the corresponding angles between the two lines and the transversal. As the corresponding angles are equal, it implies that the lines are parallel $$\therefore DP \Vert BO$$.(DP is the line joining the midpoint of hypotenuse and a leg, BO is the other leg) Hence, proved

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