CAPÍTULO III: DESCRIPCIÓN DEL INSTITUTO TECNOLÓGICO DE PUEBLA
3.1 ANTECEDENTES DEL INSTITUTO TECNOLÓGICO DE PUEBLA
Maximum Marks: 80 Time : 3 Hours
Q. No. Expected Value point Marks.
1. Reverse 1
2. No. 1
3. Because moon has no atmosphere ½
Green house gases like CO2 maintain moderate global temperature on earth. ½
4. 1
75 min or 60
75 s or 0.8s 1
5. Correct diagram 1
AB – isothermal expansion.
BC – adiabalie expansion. 1
CD – isotheronal compression.
DA – adiabalic compression
6. diagram 1
Labelling 1
7. Excess Pressure in a soap bubble = 4T
r ½
∴ A pressure in A is less than the pressure in B ½
fluids flow flow from higher to lower pressures. ½
∴ size of A will increase ½
A
B
D C
P(Pa)
V(m )3
Hot Coolant
Cold Coolant
Fuel rods Graphite
Block Shielded Chamber
Control rods
d
(i) When seperation between the source shts, ve; d decreases fringe width w increases ½ (ii) Sodium lamp is a mono chromatic source where as mercury lamp is a polychromatic source.
½
Due to inter ference only a few coloured fringes will be seen around a central white bright fringe.
9. NAND Gate is refresented as y′
10. The loaded truck has more mass ∴ it has lesser kinetic energy ½
Q Kinetic energy =
2
2 p
m and P is same for both the vehicles ½
According to Work – Energy theorem .
kε = f v ½
as F is same x ∝ κε a ∴ Truck will come to stop in smaller distance ½ 11. The law of parallelogram of vectors states; “If two vectors may be represented in magnitude and direction
by the two adjacent sides of a parallelogram then the magnitude and direction of the resultant vector will be given by the diogonal of the parallelogram passing through the common origin.”(figure) 1 In right ∆ACD
2
13. Let the speed of the projectile at the highest point
(efore splitting into two halves be v.
As a single piece it might have travelled a horizontal distance
2
R in reaching the highest point. ½
On splitting , a piece of mass 2
m returning bacle to the starting point (A) must also have
speed v (as it also has to travel a distance 2
R horizontally). 1
According to conservation of momentum, the other piece must have a speed = V1 (say) mv =
2 m v1 –
2
mv ⇒ v1 = 3v 1½
Hence this piece coill fall at a horizontal distance of 3
2R from or a distance of 2R from A. 1
14. (a) Hooke’s Law of elasticity statis, “Withins elastic limits stress is directly proportional to strain.”
i.e. stress = strain E
where E is a constant for the given material and is called as coefficient of elasticity of the
material. The coefficient of elasticity is a scolar quantity and has limit nm-2. 1 (b) Pascal’s Law of hydrostatics states; “Pressure applied on any part of a confined liquid
is transmitted equally in all direction.” 1
Thus a force F applied on an area A results in a pressure =F
p A which is transmitted through the liquid equally in all direction and acts normally on an exposed surface. If the area of the surface is 10 A them the force due to the pressure on the surface = 10A.P = 10F.
This way effect of force may be multiplied by increasing surface.
Pressure also is a scolar Quantity and its SI unit is Nm-2 (pascal). 1
15. (i) Stationary wave 1
16. (i) The two main defects of images formed by spherical lenses are :
(a) Chromatic aberration (b) Spherical aberration 1
(ii) (a) Chromatic aberration is caused by the dispersion in various parts of the lens due to which white light from a point source do not get focussed at one point, rather dispersed rays of different colours are focussed at different points.
(b) Spherical aberration is caused by the spherical shape of lens surfaces due to which the paraxial rays are focussed at a distant position and marginal rays at a near by position. 2 (iii) Suitable lens combinations may be formed to minimize these defects. 1 or any other suitable answer.
17. Construction
NS – strong permanent magnet with cylimelical pole pieces. 1
I – Cylindirical soft iron core fixed cooxially with the cylindricl air gap between the pole pieces, to
provide radial field. 1
C – a rectangular coil of large number of turns of insulated copper wire mounted on an insulated for me.
S1 and S2⇒ hair spring with the help of which the coil is kept in position at pirots P1 and
P2. The springs also help as lead for current terminal T1 to T2 through the coil.
P ⇒ Pointer attached to the centre of cal to move on horizontal scale.
Working : When current is passed through the coil it experiences a torgue Τ1 = BINA and gets deflected. As the coil turns the springs are twisted and apply a restoring torgue
T1
BINA = CO.
= C θ = θ
I K
BNA
Thus more current is passed more deflection is shown by the pointer. On reversing current the direction of deflection of the pointer on the scale is also reversed.
To convert the galvanometer of resistance Rg, which shows full scale deflection for current Ig, into a voltmeter of range O – V volts we will have to connect a high resistance
⎛ – ⎞
= ⎜ ⎟
⎝ ⎠
R V Rg
Ig , in series with the galvanometer coil, make it end zero and colibrate for
measuring voltages. 1
18. Correct symbol of pnp transistor ½
Correct orientation of transistor ½ Correct biasing of eb and cb junction. ½+½ Working of transistor as amplifier.
Small change in base current results in large
change in collector current. ½
Reb < Rbc resistance ½
Current gain and gain both are greater
than 1 hence substantial voltage gain. ½ Input and output graphsor idea of phase
reversal. ½
= Boltzmann constant = 3
2RT 1
In deriving relation (1) a monoatomic molecule having three degrees of freedom was cousidered
∴ Energy associated with each degree of freedom = 1
2RT . ½
ET = 5 × 1
At dc it is the renstance of the wire only which ultimately hinders the current.
∴ 100 100
= 1 = Ω
R 1
In ac in addition to resistance reactance of the coil also becaome effective and hence
impedence of the coil 1
(b) One the capacitor is charged and battery removed the charge on the plates will stay
unaltered. 1
(c) =Q
V C as Q remains same and C is halved V will be doubled. 1
(d) E= v
d as v and d both are doubled E stays unaltered 1
2
energy started per umit volume remains cost. 1
22. Correct indication A quantities and their units on axes . 1
Peaks at internals of 4v. 1
Rise from 1.1 mer/u to 7.7 Mev/u for A = 1 to A = 4 1
In fusion reaction 4 protons fuse together to form a helium nucleus in accordance with
the following reaction 1
1 4
1H⇒2He+2+1ot+25.6Mev 1
Total B. ε of four nucleous in 42H 4 × 7.7 = 30.8 Mev Total B. ε of four protons 4 × 1.1 4.4 Mev
∴ in the fusion reaction energy rebased = 26.4 Mev
Nearly 1 Mev energy is corresponding to two positions. 1