Maths-1999-Set II

Q1) If y = cot x, show that d2y/dx2 + 2y dy/dx = 0 (Marks 2)
Ans1) Here y = cot x
differentiating both sides w.r.t. x:-
dy/dx = -cosec2x
differentiating again w.r.t. x:-
d2y/dx2 = -d/dx (cosec2x)
d2y/dx2 = -2cosecx(-cosec x . cot x ) = +2cosec2x . cot x
or d2y/dx2 = -2 dy/dx . y
or d2y/dx2 + 2y dy/dx = 0.

Q3) If
A = 2 2
-3 1
4 0 B = 6 2
1 3
0 4

find the matrix C such that A + B + C is a zero matrix. (Marks 2)
Ans3) Here
A = 2 2
-3 1
4 0 B = 6 2
1 3
0 4

we want to find C such that A + B + C = 0
< => C = 0 – A – B = -A – B
= – 2 2
-3 1
4 0 – 6 2
1 3
0 4
= -2 -2
3 -1
-4 0 + -6 -2
-1 -3
0 -4
= -8 -4
2 -4
-4 -4

Q5) Construct a 3 x 2 matrix whose elements in the ith row and the jth coloumn are given by:-
aij = (3i + j)/2 (Marks 2)

Ans5) We want a 3 x 2 matrix
=> 1 i 3 and 1 j 2
where aij = (3i + j)/2
… a11 = (3 + 1)/2 = 2; a12 = (3 + 2)/2 = 5/2;
a21 = (6 + 1)/2 = 7/2; a22 = (6 + 2)/2 = 4;
a31 = (9 + 1)/2 = 5; a32 = (9 + 2)/2 = 11/2
… The reqd matrix is :-

a11 a12
a21 a22
a31 a32
=

2 5/2
7/2 4
5 11/2

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Maths-2000-Set I

Q1) Evaluate
{(1 – Cos x)/x2}

Ans1)
{(1 – Cos x)/x2} = (2Sin2x/2)/x2
= 2/4 (Sin(x/2)/(x/2))2 =2/4 . 12 = 1/2 (as Sinx/x=1)

Q2) A particle moves along a straight line so that S = t . Show that the acceleration is negative and is proportional to the cube of velocity.
Ans2) Here the equation of motion is
S = t
differentiating both sides w.r.t. ‘t’
v = dS/dt = 1/2t = 1/2S ( S = t )
or v = 1/2S
differentiating again w.r.t. ‘t’
accn = dv/dt = -1/2s2 ds/dt
= -2/4s2 ds/dt
= -(2/(1/v2)). v = -2v3
accn is negative and is proportional to cube of velocity.

Q3) Evaluate
Cos4xdx

Ans3) let
I = Cos4xdx = (Cos2x)2dx
= ((1 + Cos 2x)/2)2dx = 1/4 (1 + Cos22x + 2Cos 2x)dx
=1/4 [1 + (1 + Cos 4x)/2 + 2Cos 2x]dx
= 1/4[x + x/2 + (Sin 4x)/8 + (2Sin 2x)/2] + C
= 3x/8 + (Sin 4x)/32 + (1/4)Sin 2x + C

Q4) Evaluate
(1 + tan x)/(x + log sec x) dx

Ans4) Let
I = (1 + tan x)/(x + log sec x) dx
= log | x + log sec x | + C
[ as f'(x)/f(x) dx = log | f(x) | + C ]

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Maths-2000-Set II

Q2) Find a matrix X such that A + 2B + X = 0, where
A= 2 -1
3 5
B=-1 1
0 2

Ans2) Here
A= 2 -1
3 5
B= -1 1
0 2

A + 2B + X = 0
X = -A – 2B
= – 2 -1 – -1 1
3 5 2 0 2

= 0 -1
-3 -9

Q4) Calculate Spearman’s rank correlation from the following Data:
x 1 2 3 4 5 6
y 2 3 1 6 5 4

Ans4) Here :-
X Y RX RY di=RX – RY di2
1 2 6 5 1 1
2 3 5 4 1 1
3 1 4 6 -2 4
4 6 3 1 2 4
5 5 2 2 0 0
6 4 1 3 -2 4
di2 = 14

R, the rank Correlation
= 1 – (6di2/n(n2 – 1))
= 1 – (6 x 14)/6(35)
= 1 – 14/35 = 21/35 = 0.6

Q7) Evaluate
1+ Cot x dx
x + log Sin x

Ans7) Let
I= 1 + Cot x dx
x + log Sin x
= log | x + log Sin x | + C [ f'(x)/f(x) dx = log|f(x)| + c and d/dx(x + logSin x) = 1+Cot x ]

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Physics-1996-Set I

Q 1. What is the work done in moving a charge of 10 nC between two points on an equipotential surface ? (1 mark)
Ans 1. dw = qdv
on equipotential surface dv = 0

dw = 0 i.e,no work is done in moving a change on an equipotential surface.

Q 2. Name the device used for measuring the internal resistance of a secondary cell ? (1 mark)
Ans 2. Internal resistance of a secondary cell can be measured by a potentiometer.

Q 3. What is the nature of magnetic field in a moving coil galvanometer ? (1 mark)
Ans 3. A radial magnetic fields exists in a moving coil galvanometer.

Q 4.If a rate of change of current of 2 A/s induces an e.m.f.of 10mV in a solenoid, what is the self -inductance of the solenoid ? (1 mark)

Ans 4. Given dI = 2 A/S , 10mv = 10-3V
dt
| | = L dI
dt

or 10 x 10

-3 = L x 2

or L = 10-2 = 5 x 10-3 H = 5mH
2

Q 5. What type of magnetic material is used in making permanent magnets ? (1 mark)
Ans 5. A substance like steel which as a large retentivity and a large coercivity is used in making permanent magnets.

Q 6. Two metals A, B have work-functions 2eV, 4eV respectively. Which metal has lower threshold wavelength for photoelectric effect ? (1 mark)
Ans 6. Work function W = h v

0 = hc where is threshold wavelength.

x 1
W
Metal B has higher value of W & hence lower threshold wavelength.

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Physics-1996-Set II

[Only those questions and answers which are different from set 1 have been given here.]

Q.1.If a rate of change of current of 4 A/s induces an e.m.f. of 20 mV in a solenoid, what is the self-inductance of the solenoid ? (1 mark)

Ans 1.

…………….. dI = 4A/S
…………… …….dt
…………… ……. = 20mv = 20 x 10-3 V
…………… ……. | | = LdI
…………… ……………….dt
20 x 10-3 = L x 4 or L = 20 x 10-3 = 5 x10-3 H
…………… …………………. …………… 4

Q 4. What is the work done in moving a charge of 20 nC between two points on an equipotential surface ? (1 mark)
Ans 4. The work done in moving a change zone between two points on equipotential service is zero work done = q dv
…………… …dv = 0
…….work done = 0
Q 5. Two stars A and B have magnitudes +1 and -1 respectively. Which of these two appears brighter? (1 mark)
Ans 5. More negative the magnitude,brighter is the star. Hence B is brighter than A.

Q 6. Define transconductance of a transistor. (1mark)
Ans 6. Transconductance of a transistor is the ratio of change in collector current to the corresponding change in base-emitter voltage.

Q 7. Which of the following has the shortest wavelength : microwaves,ultra-violet rays X-rays ? (1 mark)
Ans 7. X-rays have the shortest wave-length among the three given wavelengths.

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Physics-1996-Set III

[Only those questions and answers which are different from Set I have been given here.]

Q 2. If a rate of change of current of 3 A/s induces an e.m.f. of 30 mV in a solenoid,what is self-inductance of the solenoid ?(1 mark)
Ans 2. dI = 3A/S
dt
= 30 mv = 30 x 10-3
| | = L dI
dt
30 x 10-3 = L x 3 or L = 10-2 H.

Q 4. Name the device used for measuring the e.m.f. of a cell.
(1 mark)

Ans 4 Potentiometer is used for measuring the emf of a cell.

Q 5. What is the work done in moving a charge 40nC between two points on an equipotential surface ?(1 mark)
Ans 5 dw = q dv
on an equipotential surface dv = 0
dw = 0

Q 6. Two stars A and B have magnitudes +1 and -1 respectively.Which of these two appear less bright? (1mark)
Ans 6. Star with positive magnitude i.e,+1 will appear less bright

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Physics-1997-Set I

Q1) Horizontal component of Earth’s magnetic field at a place is 3 times the vertical component. What is the value of angle of dip at this place? (Marks 1)
Ans1) tan = BV/BH = 1/3
or = 30o

Q2) Force between two point electric charges kept at a distance d apart in air is F. If these charges are kept at the same distance in water, how does the force between them change? (Marks 1)
Ans2) F = (1/4or) q1q2/r2
where r is the relative permittivity.
r = 1 for air & 81 for water.
Fa/Fw = 81/1
or Fw = 81 Fa

Q3) Give any two factors on which thermo-electric emf produced in a thermo-couple depends. (Marks 1)
Ans3) Thermoemf depends on
(i) Temp difference between the junctions
(ii) Nature of metals forming thermocouple

Q4) The electric current in a wire in the direction from B to A is decreasing. What is the direction of induced current in the metallic loop kept above the wire as shown in the figure? (Marks 1)
Ans4)

Induced current will be clockwise

Q5) Name the electromagnetic radiations used for viewing objects through haze and fog. (Marks 1)
Ans5) Infrared rays.

Q6) Give the ratio of the number of holes and the number of conduction electrons in an intrinsic semiconductor. (Marks 1)
Ans6) One

Q7) In the given diagram, is the diode D forward or reversed biased? (Marks 1)
Ans7) The diode is reverse biased since n-type is at a potential higher than p-type.

Q8) Name the planet which has maximum value of albedo.
(Marks 1)
Ans8) Venus

Q9) Two point electric charges of unknown magnitude and sign are placed a distance ‘d’ apart. The electric field intensity is zero at a point, not between the charges but on the line joining them. Write two essential conditions for this to happen. (Marks 2)
Ans9) (i) They should be of opposite signs
(ii) They must be of unequal magnitude.

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Physics-1997-Set-II

Q1) Is the ratio of number of holes and the number of conduction electrons in a n-type extrinsic semiconductors more than, less than or equal to 1? (Marks 1)
Ans1) nh/ne in N type < 1 as holes are minority charge carriers & electrons are majority charge carriers.

Q9) Use mirror formula to show that a convex mirror always produces a virtual image independent of the location of the object. (Marks 2)

Ans9) According to mirror formula
1 / f = 1 / v + 1/u
For convex mirror f > 0 (being +) & u < 0 (being - ve)
1 / v = 1 / f - 1 / u = 1 / |f| + 1 / |u|
v is always +ve i.e. , image will be behind the mirror. Thus convex mirror always gives virtual & erect image.

Q10) An astronomical telescope of magnifying power 10 consists of two thin lenses 55 cm apart. Calculate the focal lengths of the lenses. (Marks 2)
Ans10) m = 10, L = 55 cm
m = fo/fe or fo = mfe
L = fo + fe = mfe + fe = (m + 1) fe
55 = (10 + 1) fe or fe = 5 cm
fo = 10 fe = 50 cm.

Q11) An -particle and a proton are accelerated through same potential differences. Calculate ratio of linear momentum acquired by the two. (Marks 2)
Ans11) Now qV = 1/2 mv2
For particle
(2e)V = 1/2 mv2 – (i)
For proton
(e)V = 1/2 mpvp2 – (ii)
Dividing (i) by (ii)
2 = mv2/mpvp2
m/mp = 4
=> v2/vp2 = 2 x 1/4 =1/2
or v/vp = 1/2
The ratio of linear momentum is
P/Pp = mv/mpvp = 4 x 1/2 = 22

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Physics-1997-Set-III

Q1) Horizontal component and vertical component of Earth’s magnetic field at a place are equal. what is the value of angle of dip at this place? (Marks 1)
Ans1) BV = BH
tan = BV/BH = 1
or = 45o

Q5) Is the ratio of number of holes and the number of conduction electrons in a p-type extrinsic conductor more than, less than or equal to 1? (Marks 1)
Ans5) The ratio of number of holes to the number of conduction electrons in a p-type extrinsic conductor more than one as holes are the majority charge carriers.

Q9) What percentage of a given mass of a radioactive substance will be left undecayed after four half-life periods? Justify your answer. (Marks 2)
Ans9) N/No = (1/2)n, where n is no. of half-lives.
n = 4
… N = No(1/2)4
or N/No = (1/2)4 = 1/16
or N/No = 1/16 x 100% = 6.25%

Q12) Use mirror formula to show that for an object lying between F and 2F of a concave mirror, a real image is formed beyond 2F.
(Marks 2)

Ans12) According to mirror formula
1/f = 1/v + 1/u
For a concave mirror f is -ve , u is -ve
… -1/|f| = 1/v – 1/|u|
or 1/v = 1/|u| – 1/|f| – (i)
object lies between F and 2F
So |f| < |u| < |2f|
or 1/|f| > 1/|u| > 1/|2f| – (ii)
or 0 > 1/|u| – 1/|f| > -1/|2f|
using Equation (i)
0 > 1/v > -1/|2f|
1/v < 0 or v < 0 i.e. v is -ve, hence a real image is formed.
Also 1/v > -1/|2f|
or v < -|2f| or |v| > |2f| i.e. image is formed beyond 2F.

Q13) An astronomical telescope of magnifying power 7 consists of two thin lenses 40 cm apart. Calculate the focal length of the lenses. (Marks 2)
Ans13) 7 = fo/fe or fo = 7fe
Also L = fo + fe
40 = 7fe + fe = 8fe
or fe = 5 cm
& fo = 7fe = 35 cm.

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Physics-1998-Set-I

Q1) Name the physical quantity whose SI unit is Coulomb/Volt. (Marks 1)
Ans1) Capacitance

Q2) Write the frequency limit of visible range of electromagnetic spectrum in kHz. (Marks 1)
Ans2) 8 x 1011 HHz to 4 x 1011 KHz

Q3) How does the conductance of a semi-conducting material change with rise in temperature? (Marks 1)
Ans3) Increases.

Q4) The force experienced by a particle of charge e moving with velocity in a magnetic field is given by. = e( x ) Of these, name the pairs of vectors which are always at right angles to each other. (Marks 1)
Ans4) = e( x )
| and |

Q5) Two wires A and B are of same metal, have the same area of cross-section and have their lengths in the ratio 2 : 1. What will be the ratio of currents flowing through them respectively when the same potential difference is applied across length of each of them?
(Marks 1)

Ans5) Resistance length
RA/RB = 1/2
Current is inversely proportional to resistance
… IA/IB = 1/2

Q6) Calculate the rms value of the alternating current shown in the figure. (Marks 1)

Ans6) Irms = i2 = ((22 + (-2)2 +22)/3) = (12/3) = 2A

Q7) The image of an object formed by a lens on the screen is not in sharp focus. Suggest a method to get clear focusing of the image on the screen without disturbing the position of the object the lens or the screen. (Marks 1)
Ans7) The reason for such image is spherical aberration. Clearer focusing can be got by using the other central position of the lens & blocking the inter portion or using other proton & blocking central portion.

Q8) Two points A and B are placed between two parallel plates having a potential difference V as shown in the figure.

Will these protons experience equal or unequal force? (Marks 1)
Ans8) Both A and B will experience equal force as field between the plates is uniform.

Q9) Define the terms ‘threshold frequency’ and ’stopping potential’ for photo-electric effect. Show graphically how the stopping potential, for a given metal, varies with frequency of the incident radiations. Mark threshold frequency on this graph. (Marks 2)
Ans9) Threshold frequency: It is the maximum frequency of the incident radiation below which no emission of photoelectrons takes place.
Stopping potential : It is the maximum negative potential required to stop the fastest photoelectrons i e, photoelectric current becomes zero.

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