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Most Popular Test Series of SSC Scientific Assistant with ... · PDF fileSSC Scientific Assistant with ... Consider taling a monatomic ideal gas around the closed cycle depicted below.
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Most Popular Test Series of SSC Scientific Assistant with
Detailed Solution
Package Details EXAM BRANCH No of Questions No of tests Price Offer Price
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
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.
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.
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
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).
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
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.
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.
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.
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