Q.No.12.3: - How can you identify that which plate of a capacitor is positively charged?

Q.No.12.3: - How can you identify that which plate of a capacitor is positively charged?

Ans: - For this purpose we can use Gold Leaf Electroscope (GLE). We bring the disc of a positively charged electroscope close to the plate of the capacitor. If the divergence of the gold leaf increases, then the plate is positively charged and if the divergence in the leaf decreases, then the plate is negatively charged.

                                                  Or

Rub together a plastic rod and a piece of fur, both initially uncharged, the rod acquires a negative charge (since it takes electrons from the fur) and the fur acquires a positively charge of the same magnitude ( since it has lost as many electrons as the rod has gained). Now bring the plastic rod to one of the capacitors plate is there is a repulsion between the plastic rod and the capaticor’s plate then it means that the plate is negatively charged and the other has positive charge and vice versa.

Q.No.12.2: - Suppose that you follow an electric field line due to a positive point charge. Do electric field and the potential increase or decrease?

Q.No.12.2: - Suppose that you follow an electric field line due to a positive point charge. Do electric field and the potential increase or decrease?

Ans: - If we follow an electric field line due to a positive point charge then we will move away from the positive point charge because the electric field produced by a positive charge is away from charge. Electric field and electric potential both inversely dependent on the distance from the charge to the selected point (say P). So as we follow the positive charge then we will move away from the charge. The formula of electric field  and electric potential is

                    E    =    qq′/4πЄ₀r²      and       V     =     q/4πЄ₀r    

So as we follow then electric field and electric potential will decrease.

Q.No.12.1: - The potential is constant throughout a given region of space. Is the electrical field zero or nonzero in this region? Explain.

Q.No.12.1: - The potential is constant throughout a given region of space. Is the electrical field zero or nonzero in this region? Explain.

Ans: - The electric field has a relation with change in electrical potential which is

                                                E   =   -ΔV/Δr

In this case ΔV is the change in electrical potential and if the change in potential is zero then E will be zero and ΔV will be zero only when there is no change in electrical potential. This is given in question that electrical potential is constant throughout the given region of space hence, E = 0 when ΔV=0

Mathematically,

                                  E   =   -ΔV/Δr   =    -0/Δr    =   0

Test F.Sc Physics 1st year Chapter No.3 and 4. Long Questions

Q.No.3: - Attempt any two                               2   x  5   =  10

(i) Calculate the (a) time of flight (b) height of projectile (c) range of projectile

(ii) Derive the relation of absolute of P.E at the surface of Earth.

(iii) State and explain law of conservation of energy.

(iv) Find the final velocities of balls after an elastic collision.

Test F.Sc Physics Chapter no.3 and 4 short questions

Q.No.2: -Write the answer of the following questions.   2   x   5  =  10

(i) Prove that the units of impulse and momentum is same.

(ii) State isolated system and also describe that why is it important for law of conservation of momentum?

(iii) If two balls collide and after collision the magnitude of their velocities will remain same but their direction changes. Explain which type of collision both the balls will suffer.

(iv) Let two balls collide in such a way that their collision is perfectly elastic and both the balls are of same mass and the velocity of second ball is zero. Find the velocities of both the balls after collision?

(v) If you move with a bag in your hand and you move in the forward direction then what will be work done by your force and gravitational force? Explain it.

Test F.Sc Physics 1st year Chapter No.3 and 4. M.C.Qs

(i) The dimensions of impulse is that of

(a) Power            (b) velocity             (c) momentum            (d) acceleration

(ii) The time rate of change of momentum is called

(a) Velocity          (b) distance             (c) momentum           (d) force

(iii) The time rate of change speed is called

(a) Acceleration     (b) jerk                  (d) velocity                 (d) non of these

(iv) When we projected the ball with some velocity vi at an angle with the x-axis then the value of velocity at the highest point of its flight will be equal to its

(a) x-component    (b) y-component    (c) (d) non of these

(v) At which angle the range of projectile will be equal

(a) ϴ=45⁰ , 30⁰      (a) ϴ=30⁰ , 60⁰      (a) ϴ=60⁰, 90⁰            (d) non of these

(v) We can calculate the momentum by using

(vi) The work done will be zero if ϴ is

(a) 270⁰, 90⁰            (b) 0⁰, 180⁰            (b) 90⁰, 0⁰             (b) 0⁰, 360⁰

(vii) The work done by gravitational force is

(a) Independent of path (d) dependent of path(c) both (a) and (b).

(d) Non of these

(viii) The dimension of power is



(ix) is the dimension of

(a) Power                 (b) velocity                (c) work                  (d) K.E

(x) The absolute P.E at infinite distance will be

(a) ∞                         (b) 0                            (c) maximum            (d) non of these

Test Chapter No.13 current electricity 2nd year F.Sc Long questions

                                                                Long Questions

Q.No.3: -  Explain the following question.  Attempt all  ( Mark 3+3=6 each question)

Q.No.1:- (i) Find the power dissipation in a resistor also give its units.

(ii) How many electron pass through an electric bulb in one minute if the 300mA current is passing through it?

Q.No.2: -(i) State and explain electromotive force and internal resistance of a source of emf. Also calculate the emf for a source of emf. Why potential difference is less than the emf. 

(ii) A charge of 90 C passes through a wire in 1 hour and 15 minutes. What is the current in the wire?