xii cbse i mid term solutions

7/31/2019 Xii Cbse i Mid Term Solutions

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I MIDTERM EXAMINATION - July 2012XII PHYSICS

Time Allowed: 1 hr 30 min MaximumMarks:35

1. What will be the minimum electrostatic force between two chargedparticles placed at a distance of 1m apart? (1 Mark)(Ans)

2. What is the ratio of resistances of two conductors of same materialhaving length l and 2l and radius r and r/2 respectively? (1 mark)Ans.

3. A current is flowing through a copper pipe. What is the magnitude ofmagnetic field inside the pipe? (1 Mark)Ans.

Since inside the pipe current will be zero, hence using Amperes law

that is B=0

4. An electric dipole consisting of two charges 0f 0.2 separated by a

distance of 2.0 cm is placed in an external field of Whatmaximum torque does the field exert on the dipole? (2 Mark)Ans.


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now,

5. The two graphs drawn below, show the variations of electrostaticpotential (V) with 1/r (r being the distance of field point from the pointcharge) for two point charges q1 and q2.

(i) What are the signs of the two charges.(ii) Which of the two charges has the larger magnitude and why? (2

Marks)

Ans.

(i) The potential due to positive charge is positive and due to negative charge,it is negative, so, q1 is negative and q2 is positive.

(ii)

As the magnitude of slope of the line due to charge q2 is greater than that dueto q1, q2 has larger magnitude


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6. In given circuit the cells E1 and E2 have emfs 4 V and 8 V and the internalresistance 0.5 ohm and 10 ohm respectively. Calculate the current ineach resistance. (2 Marks)

Ans.

Effective emf of the cell =

7. A circular coil of 400 turns has a radius 3.5 m and carries a current of4A.What is the magnetic field at a point on the axis of the coil from thecentre?Ans.

The magnetic field at a point on the axis of the coil is given by

If a = x


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Here I = 4A, n = 400, a = 3.5

= T

8. Use mirror formula to show that for an object lying between the pole andfocus of a concave mirror, the image formed is always virtual in nature.

9. A parallel plate condenser with plate area A is filled with two dielectricK1 and K2 each occupying equal space lengthwise. If the separationbetween two plates is 't' for each dielectric , then ,find the capacity ofthe condenser.


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(3 Marks)(Ans)Given A= total are of each plate .

One half of the area is with dielectric medium of constant K1 and other half is withdielectric medium of constant K2 .

10.In the following circuit, 5 amperes current enters node B throughResistor 1 and 2.5 amperes flows through resistor 3 from Node B toNode C.

a) How many amperes flow through Resistor 2?b) How much charge flows through resistor 4 in 4 seconds? Give youranswer in coulombs.Ans.a)According to the Kirchhoffs circuit law,

I1 =I3 + I2

5 = 2.5 + I2


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I2 = 5 - 2.5

I2 = 2.5 Amperes

b) Current flowing through I4 is 2.5 A

Therefore,

11.Derive an expression for maximum energy gained by the positive ion

using a cyclotron. (3 Marks)Ans.

If = maximum velocity

and = maximum radius of the circular path, followed by the positive ion incyclotron.

Then,

or

Where, B is the strength of magnetic field.

And, m and q are the mass and charge of positive ion respectively.

Then maximum energy acquired by the ion.

or

12.The image formed of an object, placed 20 cm from a concave mirror, isfour times the size of the image formed of the same object when placed

45 cm from the mirror. In both cases, the image formed was real. What isthe focal length of the mirro? (3 Marks)


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13.Using Gauss's law obtain the expression for the electric field due to auniformly charged thin spherical shell of radius R a point outside theshell. Draw a graph showing the variation of electric field with r, for r>Rand r


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between and is zero at each point. Hence, electric flux through

Gaussian surface =

Now, Gaussian surface is outside the given charged shell, so chargeenclosed by the Gaussian surface is Q.

Hence, by Gauss's theorem

Thus, electric field outside a charged thin spherical shell is same as if thewhole charge Q is concentrated at the centre.

Graphically,

For r R, there is no strength of electric field inside a charged spherical shell.

For r R, electric field outside a charged thin spherical shell is same as if the

whole charge Q is concentrated at the centre.14.Explain the principle behind the operation of a Van de Graff generator.


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Ans.

Consider a large spherical conducting shell of radius R1 with uniform chargedensity and total charge q1. Theelectric field inside the shell is zero. Since the electric field is zero, the inside

of the spherical shell has V=0, i.e., it is an equipotential surface andthe

Electric potential at any point inside the shell will be same everywhere and isequal to the potential at the surface

Now consider a small sphere of radius R2 carrying a uniform surface chargeq2.

The electric field at any point at a distance r outside due to the charge q2 is

radially outwards from the centre and its magnitude is , so theelectric field in the region between R1 and R2 is only due to charge q2 and ispointing radially outwards so the potential must be higher on smaller spherethan the larger sphere because the electric field lines always point from theregion of higher potential to lower potential.Also a positive charge always feels force from higher to lower potential. So ifwe connect the two spheres, the positive charges will move from smallersphere to the larger spherical shell.

The smaller sphere remains at a higher potential compared to the largersphere regardless of the charge on the two spheres.


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If somehow we put some charge on the smaller sphere and then connect itwith the larger sphere, the charge will continue to flow from inner to outersphere and large amount of charge q1 can build up on the outer sphere. Byaccumulating a large amount of charge, the potential on the outer sphere canbecome very large of the order of millions of volts. This is the principle of Van

de Graff generator.

15.Write the principle of working of a potentiometer. Describe briefly, withthe help of a circuit diagram, how a potentiometer is used to determinethe internal resistance of a given cell.Ans.

Principle of working: The voltage drop along the wire is directly proportional tothe length of the wire. The potentiometer works without drawing any currentfrom the voltage source.

We can also use a potentiometer to measure internal resistance of a cell. Forthis the cell (emf E) whose internal resistance (r) is to be determined isconnected across a resistance box through a

Key K2, as shown in the figure.

With key K2 open, balance is obtained at length L1 (AN1).

Then,

E = L1

When key K2 is closed, the cell sends a current (I ) through the resistance box(R). If V is theterminal potential difference of the cell and balance is obtainedat length L2 (AN2), then

V = L2


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So, we have E/V = L1/L2

But, E = I (r + R) and V = IR. This gives

E/V = (r + R)/R

On comparing the above equations, we get,

(R + r)/R = L1/L2

Using the above equation, we can find the internal resistance of agiven cell.

16.In a meter bridge, the null point is found at a distance of 40 cm from A. Ifa resistance of 12 is connected in parallel with S, the null point occursat 50.0 cm from A. Determine the values of R and S.

Ans.

When 12 resistance is not connected in parallel to S than we can write,


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Where l1 = 40 cm.

Therefore, R = S (40/60) = (2/3)S..(i)

When 12 _ is connected in parallel to S, then

The effective resistance can be written as

1/S1 = 1/S + 1/12 = (12+S)/12S

i.e. S1 = 12S/(12+S)

Now we can write,

Where l' = 50 cm.

i.e. R = 12S/(12+S) (50/50) = 12S/(12+S) ..(ii)

Now using (i) and (ii) we get,

(2/3)S = 12S/(12+S)

S = 6

Therefore from (i), R = (2/3) (6) = 4 .

xii cbse i mid term solutions

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