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IMD-2011 PHYSICS

SOLUTION

1. An object of mass m = 4.0 Kg, starting from rest slides down an incline plane of length l = 3.0 m. The plane is

inclined by an angle θ = 300 to the ground. The coefficient of kinetic friction 𝜇𝑘 = 0.3 until it comes to

rest. The goal of this problem is to find out how far the object slides along the rough surface. a. What is the work done by the friction force while the mass is sliding down the inclined plane? Is this

positive of negative?

b. What is the work done by the gravitational force while the mass is sliding down the inclined plane? Is

this positive or negative?

c. What is the kinetic energy of the mass just at the bottom of the inclined plane?

d. What is the work done by the friction force while the mass is sliding along the ground? Is this

positive or negative?

e. How far does the object slide along the rough surface?

(1) Free Body Diagram (FBD) of block in inclined plane

(a)

Let block slip with acceleration.

Then perpendicular component of FBD is 𝑚𝑔 (cos 𝜗) = 𝑁

Horizontal component of FBD is 𝑚𝑔(𝑠𝑖𝑛𝜗) − 𝑓𝑘 = 𝑚𝑎

Friction force on block during slipping 𝑓𝑘 = 𝜇𝑘𝑁 = −𝜇𝑘𝑚𝑔(𝑐𝑜𝑠𝜗)

So work done by friction force is 𝑾𝒇 = −𝒇𝒌. 𝒍 = −𝒇𝒌𝒍(𝒄𝒐𝒔𝝑)

So work done by friction force is negative.

(b)

The Gravitational potential energy of block at the highest point is P.E. = 𝑚𝑔ℎ = 𝑚𝑔𝑙(𝑠𝑖𝑛𝜗)

Work done by gravitational force = Total Gravitational potential energy of block

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IMD-2011 PHYSICS

SOLUTION

So, work done by gravitational force is 𝑾𝒈 = 𝒎𝒈𝒍(𝒔𝒊𝒏𝝑)

So work done by gravitational force is positive.

(c)

Change in the kinetic energy = Total work done

𝐾𝐵 − 𝐾𝑇 = 𝑊𝑓 + 𝑊𝑔 (Here 𝐾𝐵 Kinetic energy at bottom, and 𝐾𝑇 Kinetic energy at the top.)

𝐾𝐵 − 0 = −𝜇𝑘𝑚𝑔𝑙(𝑐𝑜𝑠𝜗) + 𝑚𝑔𝑙(𝑠𝑖𝑛𝜗)

So Kinetic energy at bottom 𝑲𝑩 = 𝒎𝒈𝒍(𝒔𝒊𝒏𝝑 − 𝝁𝒌𝒄𝒐𝒔𝝑)

(d)

Let at ground block sliding at the acceleration 𝑎′ .Then friction force during moving at the ground is

𝑓𝑘′ = 𝜇𝑘

′ 𝑁 = 𝜇𝑘′ 𝑚𝑔.

Suppose block slid at distance 𝑟 than

So work done by friction force is 𝑊𝑟 = 𝑓𝑘′. 𝑟 = 𝑓𝑘

′𝑟𝑐𝑜𝑠(180𝑜) = −𝑓𝑘′𝑟

Work done by friction force is 𝑾𝒓 = −𝝁𝒌′ 𝒎𝒈𝒓

So work done by friction force is negative.

(e)

At the ground Change in the kinetic energy = Total work done

𝐾𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑙 − 𝐾𝐵 = 𝑊𝑟 (Here 𝐾𝐵 Kinetic energy at bottom, and 𝐾𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑙 Kinetic energy at terminal point.)



IMD-2011 PHYSICS

SOLUTION

0 − 𝑚𝑔𝑙(𝑠𝑖𝑛𝜗 − 𝜇𝑘𝑐𝑜𝑠𝜗) = −𝜇𝑘′ 𝑚𝑔𝑟

Or 𝑟 =𝑚𝑔𝑙(𝑠𝑖𝑛𝜗−𝜇𝑘𝑐𝑜𝑠𝜗)

𝜇𝑘′ 𝑚𝑔

= 𝑙(𝑠𝑖𝑛𝜗−𝜇𝑘𝑐𝑜𝑠𝜗)

𝜇𝑘′ =

3.0×(sin 30𝑜−0.2×cos 30𝑜)

0.3≅ 3.3 𝑚

So block will side along the rough surface at 3.3 m distance.

2. Consider taling a monatomic ideal gas around the closed cycle depicted below. It consist of one isotherm at temperature T0, One change from constant pressure (from 2 V0 to V0), and one change at constant volume. The heat capacity of gas is Cv = 3/2 Nk .

a. Between which pair(s) of point heat added to the system? (We will denote the heat added as QH) Between which pair(s) of points is heat removed from the system? (We will denote the heat added as QC)

b. Calculate Wout, The work done by the gas after one cycle. Your answer should be expressed in terms of given quantities.

c. Calculate QH and QC in terms of given quantities. d. Calculate the efficiency η of this engine. Compare this to the efficiency of a Carnot Engine operating

between the highest and lowest temperature of cycle.

(2) Thermodynamics process

(a)

In Process 1 → 2 (Isothermal Process)



IMD-2011 PHYSICS

SOLUTION

In this Process gas expands, so the system perform work on the environment. This change occurs along an isothermal process, so the energy of the ideal gas does not change during this process. Therefore heat must be absorbed (added to) by the system.

In Process 2 → 3 (Isochoric Process)

In this Process volume decrease at constant pressure. The only way that can happen for an ideal gas is if the temperature decreases. So, the environment performs work on the gas while its energy goes down, Therefore heat is rejected (removed from) by the system.

In Process 3 → 1 (Isobaric Process)

In this Process the pressure increases at constant volume. No work is being done So, heat is absorbed (added to ) by the system.

(b)

In Process 1 → 2 (Isothermal Process )

Work done by the system is 𝑊𝑜𝑢𝑡 = 𝑅𝑇 ln (𝑉2

𝑉1)

𝑊12 = 𝑅𝑇𝑜 ln (2𝑉𝑜

𝑉𝑜) = 𝑅𝑇𝑜 ln 2

In Process 2 → 3 ( Isobaric Process )

Work done by the system is 𝑊𝑜𝑢𝑡 = ∫ 𝑃𝑑𝑉𝑉2

𝑉1

𝑊23 = ∫𝑅𝑇𝑜

𝑉𝑑𝑉

𝑉𝑜

2𝑉𝑜= −

𝑅𝑇𝑜

2

In Process 3 → 1 ( Isochoric Process )

Work done by the system is 𝑊𝑜𝑢𝑡 = 𝑃 ∫ 𝑑𝑉𝑉2

𝑉1

𝑊31 = 𝑃 ∫ 𝑑𝑉𝑉2

𝑉1= 0

Total 𝑊𝑜𝑢𝑡 = 𝑊12 + 𝑊23 + 𝑊31 = 𝑅𝑇𝑜 ln 2 −𝑅𝑇𝑜

2+ 0 = 𝑅𝑇𝑜 ln 2 −

𝑅𝑇𝑜

2

(c)

In Process 1 → 2 (Isothermal Process )

Heat absorbed by the system is 𝑄1→2 = 𝑈12 + 𝑊12

𝑄1→2 = 0 + 𝑅𝑇𝑜 ln 2 = 𝑅𝑇𝑜 ln 2

In Process 3 → 1 (Isochoric Process )

Heat absorbed by the system is 𝑄3→1 = 𝑈31 + 𝑊31



IMD-2011 PHYSICS

SOLUTION

𝑄3→1 = 𝐶𝑉∆𝑇 + 0

𝑄3→1 =3

2𝑁𝑘∆𝑇 + 0 =

3

2𝑅∆𝑇

At point 1 Temperature is 𝑇𝑜 and at point 2 the pressure is same as the at point 3 but the volume is doubled.

Hence the temperature at point 3 must be 𝑇𝑜/2 .

So, 𝑄3→1 =3

2𝑅(𝑇𝑜 − 𝑇𝑜/2) =

3

4𝑅𝑇𝑜

Total heat Added is 𝑄𝐻 = 𝑄1→2 + 𝑄3→1 = 𝑅𝑇𝑜 ln 2 +3

4𝑅𝑇𝑜

Heat Removed from the system = Total heat added – Total work done by the system

So, 𝑄𝐶 = 𝑄𝐻 − 𝑊𝑜𝑢𝑡 = 𝑅𝑇𝑜 ln 2 +3

4𝑅𝑇𝑜 − 𝑅𝑇𝑜 ln 2 +

𝑅𝑇𝑜

2=

5

4𝑅𝑇𝑜

(d)

Efficiency of the engine η = 𝑊𝑜𝑢𝑡

𝑄𝐻

So, η = 𝑅𝑇𝑜 ln 2−

𝑅𝑇𝑜2

𝑅𝑇𝑜 ln2+3

4𝑅𝑇𝑜

=ln 2−

1

2

ln 2+3

4

≅ 0.14

Or Efficiency of this engine η is 14% .

Maximum efficiency of Carnot engine 𝜂𝑐𝑎𝑟𝑛𝑜𝑡 when it operates between 0o and 100o C is only 27% .

So, 𝜼𝒄𝒂𝒓𝒏𝒐𝒕 > 𝛈

3. (i) The hub of a wheel is attached to a spring with spring constant k and negligible mass. The wheel has radius R and mass M. The mass of the spoke is negligibly small. The wheel rolls without slipping i.e, the wheel translates by the same distance that its circumference rotates. The Centre of Mass of the wheel oscillates (simple harmonic motion) in the horizontal direction about its equilibrium point x = 0



IMD-2011 PHYSICS

SOLUTION

a) Find an expression for the total energy in terms of k, M, R and x(t). Since the spoke has negligible mass

you may assume that moment of inertia for rotation about the axle is MR2

b) Using conservation of energy derive the differential equation of motion.

c) What is the angular frequency of small oscillations about equilibrium?

(ii) A uniform rod of mass m is bent in a circular arc with radius R. It is suspended at the middle and it freely

swing about point P (see Figure). The length of the arc is2

3𝜋𝑅. What is the period of small angle oscillation

about P?

(3) (i)

Let the center of the wheel moves 𝑥 distance, then wheel rotated an angle 𝜗.

Then the angle is =𝐴𝑟𝑐

𝑅𝑎𝑑𝑖𝑢𝑠=

𝑥

𝑅 . Let 𝑣 is translation velocity of the wheel and 𝜔 be the angular velocity.

(a)

The total energy of the wheel at 𝑥 point is

E = Translation motion energy + Rotational motion energy + Elastic Potential Energy of Spring

Total Energy 𝐸 =1

2𝑀𝑣2 +

1

2𝐼𝜔2 +

1

2𝑘𝑥2 ……(1)

Substitute 𝐼 = 𝑀𝑅2 and 𝜔 =𝑣

𝑅 in equation (1)

𝐸 =1

2𝑀𝑣2 +

1

2(𝑀𝑅2) (

𝑣

𝑅)2

+1

2𝑘𝑥2

𝐸 =1

2𝑀𝑣2 +

1

2(𝑀𝑣2) +

1

2𝑘𝑥2 = 𝑀𝑣2 +

1

2𝑘𝑥2 ……(2)

So, Total Energy is 𝑬 = 𝑴𝒗𝟐 +𝟏

𝟐𝒌𝒙𝟐

(b)

According to Conservation law of energy total energy of a system is constant E = Constant



IMD-2011 PHYSICS

SOLUTION

So, time derivative of Energy is zero 𝑑𝐸

𝑑𝑡= 0

Taking derivative of equation (2)

𝑑𝐸

𝑑𝑡= 2𝑀𝑣

𝑑𝑣

𝑑𝑡+

1

2𝑘(2𝑥)

𝑑𝑥

𝑑𝑡= 0

2𝑀𝑣𝑑𝑣

𝑑𝑡+ 𝑘𝑥𝑣 = 0 Or 𝑀𝑣

𝑑2𝑥

𝑑𝑡2+ 𝑘𝑥 = 0

Differential Equation of Motion is 𝒅𝟐𝒙

𝒅𝒕𝟐= −

𝒌𝒙

𝟐𝑴

(c)

Differential Equation of motion can be written as 𝑑2𝑥

𝑑𝑡2= −𝜔2𝑥 Here (𝜔2 =

𝑘

2𝑀)

Angular frequency for small oscillation is 𝜔 = √𝑘

2𝑀

Time period for small oscillation is T = 2𝜋

𝜔= 2𝜋√

2𝑀

𝑘

(ii)

Let consider the simple case of two mass rigid Pendulum both of whose masses are equidistance from pivot point

at P . All three points are line in a circle of diameter D and subtend at angle 𝛼 at the pivot.



IMD-2011 PHYSICS

SOLUTION

Let the distance of each mass from the pivot point be .

Then the moment of inertia of two masses together 𝐼 =𝑀𝑙2

2+

𝑀𝑙2

2= 𝑀𝑙2 .

At equilibrium position of each mass is 𝑙𝑐𝑜𝑠 (α

2) =

𝑙2

𝐷 below Point P.

The gravitational potential energy of the system, after being displaced over a small angle 𝜗 is 𝑈 =𝑀𝑔𝑙2𝜃2

2𝐷

Total energy of the system 𝐸 =𝑀𝑔𝑙2𝜃2

2𝐷+

1

2𝑀𝑙2 (

𝑑𝜃

𝑑𝑡)2

So, 𝑑𝐸

𝑑𝑡=

1

2𝐷𝑀𝑔𝑙22𝜃

𝑑𝜃

𝑑𝑡+

1

2𝑀𝑙22 (

𝑑𝜃

𝑑𝑡) (

𝑑2𝜃

𝑑𝑡2) = 0

Or 𝑔

𝐷𝜃 +

𝑑2𝜃

𝑑𝑡2 = 0 Or 𝑑2𝜃

𝑑𝑡2 = −𝑔

𝐷𝜃 = −𝜔2𝜃 Here (𝜔2 =

𝑔

𝐷)

Angular frequency for small oscillation is 𝜔 = √𝑔

𝐷

Time period for small oscillation is T = 2𝜋

𝜔= 2𝜋√

𝐷

𝑔

Hence the period is independent of mass 𝑀 and angle . It only depends on the diameter 𝐷 of the circle.

So, Now Considering the circular arc system. We can see that is a collection of many such two mass

pendulums. Since the period of those pendulums is the same 𝑇 = 2𝜋√𝐷

𝑔

So the period of arc is also 𝑇 = 2𝜋√𝐷

𝑔= 2𝜋√

2𝑅

𝑔

Time period of small angle oscillation is 𝐓 = 𝟐𝛑√𝟐𝐑

𝐠

4. (i) A beam of unpolarized light of 500nm I air is incident on a plate of glass. The angle of incident is 40o (this is the angle between the direction of the incoming light and the normal to the glass). The index of refraction of the glass is 1.5. a) Which fraction of incoming 10kW is reflected off the front face of the glass? b) What is the degree of linear polarization of the reflected light?

(ii) Circularly polarized light with intensity IO is incident on a glass prism. The prism is completely surrounded on all sides by air. The surface indicated by AB, BC and AC are perpendicular to the plane of the paper. Angle ABC = 90o, the other the angle are both 45o. The index of refraction of the glass is 1.5. The light strike at the surface (AB) at right angle (normal incidence).



IMD-2011 PHYSICS

SOLUTION

a) What percentage of light intensity reflected off by the surface AB? b) Is this light still circularly polarized? Give your reasons. c) Show that the sum of the intensities of the reflected light and that of the light that enters in the

prism IO. The light that penetrates the glass will now be incident on the surface of AC. d) What percentage of this light is reflected and what emerges into the air? Make a clear sketch

and calculate the relevant angles of reflection and refraction. The reflected light (at surface AC) will now be incident on the surface BC.

e) What percentage of this light is reflected and what percentage emerges into the air at the surface BC? Make a clear sketch and calculate the relevant angle of reflection and refraction.

(4) (i)

(a) A beam of Unpolarized light incident on plate of glass at angle 40o



IMD-2011 PHYSICS

SOLUTION

According to Snell’s law sin𝜗1

sin𝜗2=

𝑛2

𝑛1

So, Refracted angle is 𝜗2 = sin−1 (𝑛1

𝑛2sin 𝜗1) = sin−1 (

1

1.5sin 40𝑜) ≅ 24.5𝑜

Since the incident light is unpolarized we can assume that the intensity of parallel and perpendicular components each 50 % of the incident light intensity.

The reflectivity for parallel components are 𝑟∥ =−tan(𝜗1−𝜗2)

tan(𝜗1+𝜗2)=

−tan(15.5𝑜)

tan(64.5𝑜)≅ −0.12 Or 𝑟∥

2 ≅ 0.014

The reflectivity for perpendicular components are 𝑟⊥ =−sin(𝜗1−𝜗2)

sin(𝜗1+𝜗2)=

−sin(15.5𝑜)

sin(64.5𝑜)≅ −0.28 Or 𝑟⊥

2 ≅ 0.077

The fraction of incoming 10 kW that is reflected is 0.5(0.014 + 0.077) = 4.55%

So the 𝟒. 𝟓𝟓% fraction of incoming 10 kW light is reflected off.

(b)

Degree of linear polarization is=𝑟⊥2−𝑟∥

2

𝑟⊥2+𝑟∥

2 =0.077−0.014

0.077+0.014≅ 0.69 ≅ 70%

So, 𝟕𝟎% of linear polarization of the reflected light.

(ii)

A circularly polarized light with intensity Io incident from air to a glass prism. The prism having refractive index 1.5.

(a)

Reflection of light at normal incidence is

𝐸𝑟

𝐸𝑖=

𝑛1−𝑛2

𝑛1+𝑛2=

1−1.5

1+1.5= −0.2



IMD-2011 PHYSICS

SOLUTION

Or Reflected Light is (𝑬𝒓

𝑬𝒊)𝟐

= 𝟎. 𝟎𝟒 = 𝟒%

(b)

The reflected and transmitted light is still circularly polarized as r and t are the same for parallel and perpendicular component of light.

(c)

The Intensity of light is the product of poynting vector and the cross sectional area of light beam.

⟨𝑆⟩ =⟨ × ⟩

𝜇𝑜 And |𝐵| =

|𝐸|

𝑣=

|𝐸|𝑛

𝑐

As the light enters in the prism, the cross sectional area is the same as that of the incident beam.

Poynting vector for Incident component ⟨𝑆𝑖⟩ =⟨𝐸𝑖 ×𝐵𝑖 ⟩

𝜇𝑜=

⟨𝐸𝑖2⟩𝑛1

𝜇𝑜𝑐

Poynting vector for transmitted component ⟨𝑆𝑡⟩ =⟨𝐸𝑡 ×𝐵𝑡 ⟩

𝜇𝑜=

⟨𝐸𝑡2⟩𝑛2

𝜇𝑜𝑐

So ⟨𝑆𝑡⟩

⟨𝑆𝑖⟩=

⟨𝐸𝑡2⟩𝑛2

⟨𝐸𝑖2⟩𝑛1

= ⟨𝐸𝑡

𝐸𝑖⟩2

𝑛2

𝑛1= (

2𝑛1

𝑛1+𝑛2)2 𝑛2

𝑛1= (

2×1.0

1.0+1.5)2 1.5

1.0= 0.96

So 96% light enters.

So only 4% light is reflected and 𝟗𝟔% light enter in the prism and incident on AC.

(d)

At surface AC 100% of the light will be reflected.

The angle of incidence 45o is larger than the critical angle which is sin 𝜃𝐶𝑅𝐼𝑇𝐼𝐶𝐴𝐿 =𝑛2

𝑛1=

1

1.5

So 𝜃𝐶𝑅𝐼𝑇𝐼𝐶𝐴𝐿 = 41.8o



IMD-2011 PHYSICS

SOLUTION

(e)

The reflected light r is = 𝑛1−𝑛2

𝑛1+𝑛2=

1.5−1

1.5+1= 0.2 or 4% light will be reflected.

Thus 96% light will emerge in air.

5. In the diagram below, four capacitors have the same capacitance; the battery provides 120V.

Consider two cases, starting in both cases with unchanged capacitors. Case (1) a) While switch B is kept open, switch A is closed and then opened after C1, C2, C3 are fully

charged. What is now the electric potential differences across each capacitor? b) Subsequently switch B is closed. What is now the electric potential difference across each

capacitor?

Case (2)

c) Switch A is open Switch B is first closed. What is now the electric potential difference across the capacitor?

d) Subsequently switch A is closed. What is now the potential differences across the capacitor?

(5)

Let assume each capacitor having capacitance is C. Which are connected with 120 V battery.



IMD-2011 PHYSICS

SOLUTION

CASE (I)

(a)

As switch A is close and B is open then capacitor C1, C2, and C3 will be in series and Capacitor C4 is out of picture. Then the circuit looks like as shown in figure at right side.

Then the total voltage must be V = V1 + V2 + V3

And all three Capacitor must have same charge ±𝑄 on their plates. Since all capacitor have same capacitance C. Then charge Q = CV1 = CV2 = CV3 Or V1 = V2 = V3

Then voltage on each capacitor is V/3 = 120/3 = 40 V

And Once the capacitor is charged in this manner, the potential Across each capacitor will remain same even after the opening Switch A.

(b)

Opening switch A is taken the circuit branch containing battery out of the game, but not before charging capacitor C1, C2, and C3 as described.

Now closing switch B brings C4 back into play and then the circuit looks like

As shown in figure at right side. Capacitor C1 and C3 will still each carry the same charge Q as before. However the charge Q that originally resided

On C2 alone will now distribute between C2 and C4.

As C2 = C4 by symmetry each have Q/2 charge and the voltage across C2 and

C4 will be half the original voltage on C2.

So Voltage across C1 and C3 is = 40 V

And Voltage across C2 and C4 is = 20 V

CASE (II)

(c)

As switch A is open then circuit is incomplete so if we close switch B there is zero potential on each capacitor till switch A is open.

(d)



IMD-2011 PHYSICS

SOLUTION

Now we close switch A, with switch B is previously closed . Then the circuit looks like as shown in right side.

Then the effective capactance of parallel capacitor C2 and C4 is

C24 = C2 + C4 = 2C

Then the equivalent capacitance of entire four capacitor will be

Cequivalent =(1

𝐶1+

1

𝐶24+

1

𝐶3)−1

= (1

𝐶+

1

2𝐶+

1

𝐶)−1

=2

5𝐶

Then total charge drawn from battery to each capacitor is

𝑄′ = 𝐶𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡𝑉 =2

5𝐶𝑉

So Voltage across capacitor C1 and C3 is = 𝑸′

𝑪=

𝟐

𝟓𝑽 = 𝟒𝟖 𝑽

And Voltage across capacitor C2 and C4 is = 𝑸′

𝟐𝑪=

𝟏

𝟓𝑽 = 𝟐𝟒 𝑽

6. (a) Find the value of therionic current density In apm/cm2 for a platinum filamentwhose work function is 4.1 eV and the temperature is 2000 K. The emission constant A=120amp/cm2K2 (k = 1.38 x 10-23 joule/K, 1 eV = 1.6 x 10-19 joule) (b) A germanium diode draws 40 mA with a forward bias 0.25 V. The junction is at room temperature 20o C. Calculate the reverse saturation of the diode. (c) When the emittor current of transistor is changed by 1 mA, its collector currrent changes by 0.995 mA. Calculate (i) its common base short circuit current gain (α) and (ii) Its common emmitor short circuit current gain (β) (d) A half wave rectifiew uses a diode whose forward resistance is 10 Ω and a transformer whose secondary winding resistance is 25 Ω. The rectifier is supplied with 120 V AC through a transformer having a turn of 2:1. If the value of the load resistor is 200 Ω. Estimate the optput DC voltage and average DC load current.

(6)

(a)

According to Richardson-Dushman the current density of thermionic emission is 𝑖 = 𝑛𝑒 = 𝐴𝑇2𝑒−𝑊

𝑘𝑇

Here 𝑊 is work function, 𝑇 is temperature, and 𝑘 is Boltzmann’s constant

So current density of platinum filament is 𝑖 = 120 × (2000)2 × 𝑒−4.1𝑒𝑉

8.61×10−5𝑒𝑉×2000𝐾 ≅ 0.134

current density of platinum filament is 𝒊 ≅ 𝟎. 𝟏𝟑𝟒 Am/cm2



IMD-2011 PHYSICS

SOLUTION

(b)

Standard Current equation for diode is 𝐼𝐷 = 𝐼𝑆 (𝑒𝑞𝑉

𝑛𝑘𝑇 − 1)

Here 𝐼𝑆 is reverse saturation current, 𝐼𝐷 Current through diode, 𝑇 is temperature, 𝑞 is charge, 𝑉 is voltage,

and 𝑘 is Boltzmann’s constant.

For germanium n = 2 and if V < 1 then Current equation is 𝐼𝐷 = 𝐼𝑆(𝑒40𝑉 − 1)

So reverse saturation current 𝐼𝑆 is = 𝐼𝐷

(𝑒40 𝑉−1)=

40 𝑚𝐴

(𝑒40×0.25−1)≅ 1.81 𝜇𝐴

Reverse saturation current 𝑰𝑺 for germanium at 200 C is 1.81 µA.

(c)

In transistor changed Emitter current 𝐼𝐸 = 1 mA

changed Collector current 𝐼𝐶 = 0.995 mA

(i) Common base short circuit current gain 𝜶 =𝐼𝐶

𝐼𝐸=

0.995 𝑚𝐴

1.0 𝑚𝐴= 𝟎. 𝟗𝟗𝟓

(ii) Common emitter short circuit current gain 𝜷 =𝐼𝐶

𝐼𝐵 =

𝐼𝐶

𝐼𝐸−𝐼𝐶 =

0.995 𝑚𝐴

(1−0.995)𝑚𝐴= 𝟏𝟗𝟗

(d)

For a transformer 𝑁2

𝑁1=

𝑉2

𝑉1 Here N2 number of turns in secondary coil, N1 is Number of turns in Primary coil

So, 1

2=

𝑉2

120 𝑉 or V2 = 60 V

Maximum value of secondary voltage Vm = (√2)𝑉2 = √2 × 60 𝑉 ≅ 84.85 𝑉

For a half wave rectifier output DC voltage is VDC = IDC × RL =𝐼𝑚

𝜋𝑅𝐿 =

𝑉𝑚𝑅𝐿

𝜋(𝑅𝑓+𝑅𝐿)

Here RF is forward bias diode resistance, RL is Load resistance, Vm is maximum voltage.

Output DC voltage is VDC = 𝑉𝑚𝑅𝐿

𝜋(𝑅𝑓+𝑅𝐿)=

84.85 𝑉×200Ω

𝜋(10Ω+200Ω)≅ 25.72 𝑉

Output DC voltage is 25.72 V

For a half wave rectifier output DC current is IDC = 𝐼𝑚

𝜋=

𝑉𝑚

𝜋(𝑅𝑓+𝑅𝐿)

Or output DC current is 𝐼𝐷𝐶 =𝑉𝐷𝐶

𝑅𝐿=

25.72 𝑉

200Ω≅ 0.13 𝐴𝑚𝑝

Average value of output DC load current is 0.13 Amp


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