Question 1: A student measures the time period of 100 oscillations of a simple pendulum four times. The data set is 90s, 91s, 95 s and 92 s. If the minimum dividion in the measuring clock is 1s, then the reported mean time should be:

(1) 92+- 2 s

(2) 92+- 5.0 s

(3) 92+-1.8 s

(4) 92+-3 s

Solotion:

Mean value of time period T_m= (90+91+95+92)/4 = 92 s

Taking T_m as the true value, the absolute errors in different observations are,

DeltaT_1 =92-90=+2s

DeltaT_2 =92-91=+1s

Deltat_3 =92-95=-3

Deltat_4 =92-92= 0s

Mean absolute error =(|DeltaT_1|+|DeltaT_2|+|DeltaT_3|+|DeltaT_4|)/4

=> (2+1+3+0)/4= 1.5

Value = 92+-1.5

Since least count = 1s, therefore:

Value = 92+-2s

Option (1) is correct.

Question 2: A particle of mass m is moving along the side of a square of side 'a', with a uniform speed v in the x-y plane as shown in the figure

Which of the following statements is false for the angular momentum vec L about the origin ?

(1) vec L =-(mv)/sqrt(2) R hat k when the particle is moving from A to B.

(2) vec L = mv[R/sqrt(2) -a] hat k when the particle is moving from C to D

(3) vec L = mv[R/sqrt(2) +a] hat k when the particle is moving from B to C

(4) vec L = (mv)/sqrt(2) R hat k when the particle is moving from D to A.

Solution:

vecL = vecrxxvecp

D-> A, Angular momentum,vecL = mv (R/sqrt2) (-hatk)

 A-> B , Angular momentum,vecL = mv (R/sqrt2) (-hatk)

C-> D, Angular momentum,vecL = mv (R/sqrt2+a) hatk

B-> C, Angular momentum,vecL = mv (R/sqrt2+a) hatk

Option (2) and (4) are correct.

Question 3: A point particle of mass m, moves along the uniformly rough track PQR as shown in the figure. The coefficient of friction, between the particle and the rough track equal mu. The particle is released, from rest, from the point P and it comes to rest at a point R. The energies, lost by the ball, over the parts, PQ and QR, of the track, are equal to each other, and no energy is lost when particle changes direction from PQ and QR.

The values of the coefficcient of frcition mu and the distnace x(=QR), are, respectively close to:

(1) 0.2 and 6.5 m

(2) 0.2 and 3.5 m

(3) 0.29 and 3.5 m

(4) 0.29 and 6.5 m

Solution:

Length of PQ, l = h/(cos30^0) = 2/(1/2)= 4 m

Energy loss is due to workdone by friction

Energy lost in PQ = (mu m gcos30^0)xxl = mu mgcos30^0 xx4 ....(1)

Energy lost in QR = mu mg xx X....(2)

Given that both energies lost by ball are equal,

So from equation (1) and (2), we get:

 mu mgcos30^0 xx4 = mu mg xx X

=> sqrt3/2 xx4 = X

=> X = 2sqrt3 m = 3.46 = 3.5 m

Loss in GPE = W_(PQ)+ W_(QR)= 2W_(QR)

=> mgh = 2(mu mg X)

=> mu = h/(2X)= 1/X = 1/(2sqrt3) = 0.288 = 0.29

Option (3) is correct.

Question 4: A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up? Fat supplies 3.8 xx 10^7J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g = 9.8 ms^(-2)

(1) 2.45xx 10^(-3) kg

(2) 6.45xx 10^(-3) kg

(3) 9.89xx 10^(-3) kg

(4) 12.89xx 10^(-3) kg

Solution :

m =10 kg and h = 1m

Work done against gravity = (mgh)1000 [in lifting 1000 times]

W => 10xx9.8xx1xx1000= 9.8xx10^4 J

20% efficiency is to converts fat into energy.

[20% "of" 3.8xx10^7 J] = 9.8 xx 10^4

=> m = (4.9 xx 10^4)/(3.8xx10^7)

=> x = 12.89xx10^(-3) kg

Option (4) is correct.

Question 5: A roller is made by joining together two cones at their vertices O. It is kept on two ralis AB and CD which are placed asymmetrically (see figure), with its axis perpendicular to CD and its center O at the centre of line joining AB and CD (see figure). It is given a light push so that it starts rolling with its center O moving parallel to CD in the direction shown. As it moves, the roller will tend to:

(1)Turn left

(2)Turn right

(3) Go straight

(4) Turn left and right alternatively

Solution: Say the distance of central line from instantaneous axis of rotation is r.Then r from the point on left becomes lesser than that for right.

So v_"left"  point = omega r'< omega r = v_"right" point
So the roller will turn to left.
Option (1) is correct

Question 6: A satellite is revolving in a circular orbit at a height 'h' from the earth's surface (radius of earth R; h<

(1) sqrt(2gR)

(2) sqrt(gR)

(3)sqrt(gR//2)

(4) sqrt(gR) (sqrt2 -1)

Solution:

The orbital velocity in a circular orbit cloase to the earth is v= sqrt(gR)

The velocity require to escape is v_e = sqrt(2gR)

=> v_e -v = sqrt(2gR)= sqrt(gR)= sqrt(gR)(sqrt2-1)

Option (4) is correct.

Question 7: A pendulum clock loses 12s a day if the temperature is 40^0 C and gains 4s a day if the temperature is 20^0 C. The temperature at which the clock will show correct time, and the co-efficient of linear expansion (alpha) of the metal of the pendulum shaft are respectively:

(1) 25^0 C; alpha = 1.85 xx10^(-5) //^0C

(2) 60^0 C; alpha = 1.85 xx10^(-4) //^0 C

(3)30^0 C; alpha = 1.85 xx10^(-3) //^0 C

(4) 55^0 C; alpha = 1.85 xx10^(-2) //^0 C

Solution:

Time lost in one day = (1/2 alpha theta)xx 86400

At temperature, T_1 = 40^0

Time lost in one day = (1/2 alpha (40-T))xx 86400= 12.....(1)

AT temperature, T_2 = 20^0

Time lost in one day = (1/2 alpha (T-20))xx 86400= 4......(2)

From equation (1) and (2), we get,

=> (40-T)/(T-20)= 12/4 = 3

=> 40-T = 3T-60

=> T = 25^0

From the equation (2), we get

1/2 alpha (T-20)xx 86400= 4

=> 1/2 alpha (25-20)xx 86400 =4

=> alpha = 1.85xx 10^-5 /^0 C

Option (1) is correct.

Question 8: An ideal gas undergoes a quasi static reversible process in which its molar heat capacity C remains constant. If during this process the relation of pressure P and volume V is given by PV^n = constant, then n is given by (Here C_p and C_v are molar specitfic heat at constant pressure and constant volume, respectively)

(1) n = C_p/C_v

(2) n = (C- C_p)/(C- C_v)

(3) n = (C_p- C)/(C- C_v)

(4) n = (C- C_v)/(C- C_p)

Solution:

deltaQ = mCdT, dU = mC_v dT and deltaW = PdV.....(1)

Given that: PV^n=constant

=> P (nV^(n-1)dV)+ VdP = 0

=> nPdV= -VdP...(2)

Also,  PV= mRT

=> PdV+ VdP = mRdT

From equation (2), we get:

=>PdV -nPdV = mRdT

=> PdV= deltaW = (mRdT)/(1-n)....(3)

We know that, detlaQ = dU + deltaW

=> mCdT = mC_vdT + (mRdT)/(1-n)

=> C= C_v+ R/(1-n) [R= C_p -C_v]

=> C -C_v = (C_p -C_v)/(1-n)

=> 1-n = (C_p -C_v)/(C- C_v)

=> n= (C- C_p)/(C- C_v)

Option (2) is correct.

Question: 9 'n' moles of an ideal gas undergoes a process A-> B as shown in the figure. The maximum temperature of the gas during the process will be:

(1) (9P_0 V_0)/(4nR)

(2) (3P_0 V_0)/(2nR)

(3) (9P_0 V_0)/(2nR)

(4) (9P_0 V_0)/(nR)

Solution :

Equation of line AB: (P-P_1) = (P_2 -P_1)/(V_2-V_1) (V-V_1)
P_1= P_0, P_2 = 2P_0, V_2 = V_0 and V_1 = 2V_0

=> (P-P_0) = (2P_0- P_0)/(V_0-2V_0)(V-V_0)

=> (P-P_0) = -(P_0/V_0)(V-V_0)

=> P = 3P_0 -(P_0/V_0)V [PV= nRT]

=> (nRT)/V = 3P_0 -(P_0/V_0)V

=> T = 1/(nR)[3P_0V -P_0/V_0 V^2]

For maximum T, (dT)/(dV)= 0

(dT)/(dV)= 1/(nR) [3P_0 - P_0/V_0 (2V) ]= 0

=> V = (3V_0)/2

T_("max") = 1/(nR)[3P_0((3V_0)/2) -P_0/V_0 ((3V_0)/2)^2]

T_("max") =1/(nR) 9/4 P_0V_0

Option (1) is correct.

Question: 10 A particle performs simple harmonic motion with amplitude A. Its speed is trebled at the instant that it is at a distance (2A)/3 from equilibruim position. The new amplitude of the motion is:

(1) A/3 sqrt(41)

(2) 3A

(3) A sqrt3

(4) (7A)/3

Solution : We know that: V = omega sqrt(A^2- x^2)

Speed at x= (2A)/9

=> V = omega sqrt(A^2- (4A^2)/9) = (omega Asqrt5)/3...(1)

Given that: V^' = 3V = omega Asqrt5 (From equation (1))

V^' = omega sqrt(A'^2- (4A^2)/9) = omega Asqrt5

=> A'^2- (4A^2)/9 = 5A^2

=> A'^2 = (49A^2)/9

=> A' = (7A)/3

Option (d) is correct.

Question 11: A unifrom string of length 20m is suspended from a rigid support. A short wave pulse is inroduced at its lowest end. It starts moving up the string. The time taken to reach the support is: (take g = 10 ms^(-2) )

(1)  2pisqrt2 s

(2) 2 s

(3) 2sqrt2 s

(4) sqrt2 s

Solution :

Velocity at point P, v = sqrt ((((m//L) gx))/(m//L))= sqrt (gx)

(dv)/(dx)= sqrtg (1/(2sqrtx))

=> a = V (dv)/(dx)= (sqrt (gx))xx(sqrtg (1/(2sqrtx)))= g/2

S= 1/2at^2

20 = 1/2 (10/2)t^2

=>t^2 = 8

t = 2sqrt2s

Option (3) is correct.

Question 12: The region between two concentric spheres of radii 'a' and 'b', respectively (see figure), has volume charge density,  rho = A/r, where A is a constant and r is the distance from the center. At centre of the spheres is a point Q. The value of A such that the electric field in the region between the spheres will be constant is:

(1)  Q/(2pia^2)

(2) Q/ (2pi(b^2-a^2))

(3)  (2Q)/ (pi(b^2-a^2))

(4) (2Q)/ (pia^2)

Solution: Gaussian surface at distance r from center

E (4piepsilon_0r^2) = Q + int_a^r (A/r) 4pir^2 dr

=> E (4piepsilon_0r^2) = Q + A 4pi (r^2-a^2)/2

=> E = 1/(4piepsilon_0)[Q/r^2 + A2pi(r^2 -a^2)/r^2].....(1)

=> E = 1/(4piepsilon_0) (A2pi).....(2)

From equation (1) and (2), we get:

1/(4piepsilon_0) (A2pi) = 1/(4piepsilon_0)[Q/r^2 + A2pi(r^2 -a^2)/r^2]

=> Q = 2piA a^2

A = Q/(2pia^2)

Option (1) is correct.

Question: 13 A combination of capacitors is set up as shown in the figure. The magnitude of electric field, due to a point charge Q (having a charge equal to the sum of the charges on the 4muF and 9muF capacitors), at a point distant 30 m from it, would equal:

(1)  240 N//C

(2)  360 N//C

(3)  420 N//C

(4)  480 N//C

Solution:

Q =24+18 =42 muC

E = (KQ)/r^2 = (9xx10^9xx42xx10^-6)/900 = 420 N//C

Option (3) is correct.

Question 14: The temperature dependence of resistances of Cu and undoped Si in the temperature range 300-400 K, is best described by:

(1) Linear increase for Cu, linear increase for Si.

(2) Linear increase for Cu, exponential increase for Si.

(3) Linear increase for Cu, exponential decrease for Si.

(4) Linear decrease for Cu, linear decrease for Si.

Solution: Cu->  Temperature dependence and resistance increases with temperature

Si-> Resistance decrease exponentially with temperature

Option (3) is correct.

Question 15: Two identical wires A and B, each of length 'L', carry the same current I. Wire A is bent into a circle of radius and wire B is bent to from a square of side 'a'. If B_A and B_B are the values of magnetic field at the centres of the circle and square respectively, then the ratio B_A/B_B is:

(1) pi^2/8

(2) pi^2/(16sqrt2)

(3) pi^2/16

(4) pi^2/(8sqrt2)

Solution:

For (A) 2pir = L => r = L/(2pi)

Magnetic field at the center of the circle, B_A = (mu_0I)/(2r)

=> B_A = (mu_0I)/(2(L/(2pi)))

=> B_A = (mu_0Ipi)/L

For (B) 4a = L=> a = L/4

=> B_B = (4mu_0I)/(pia) 1/sqrt2

=> B_B = 2sqrt2 (mu_0I)/(pia)= 8sqrt2 (mu_0I)/(pia)

=> B_B / B_A = pi^2/ (8sqrt2)

Option (4) is correct.

Question 16: Hysteresis loops two magnetic materials A and B are given below:

The materials are used to make magnets for electric generators, transformer core and electromagnet core. Then it is proper to use:

(1) A for electric generators and transformers.

(2) A for electromagnets and B for electric generators

(3) A for transformers and B for electric generators

(4) B for electromagnets and transformers.

Solution: For electromagent and transformers, we require the core that can be magnitised and demagnetised quickly when subjected to alternating current. From the given graphs, graph B is suitable.

Option (4) is correct.

Question 17: An arc lamp requires a direct current of 10 A at 80V to function. If it is connected to a 220 V(rms), 50 Hz AC supply, the series inductor needed for it to work is close to:

(1) 80H

(2) 0.08H

(3) 0.044 H

(4)0.065H

Solution: X_L = Lomega = L xx (2pif)= L xx(100pi)

Z = sqrt (8^2+(L xx(100pi))^2)

I_(rms = V_(rms)/Z

 10 = 220 / sqrt (8^2+(L xx(100pi))^2)

=>484 = 64 +10^4 (pi^2L^2)

=> L^2 = 0.0042

L = 0.0648 = 0.065 H

Option (4) is correct

Question 18: Arrange the following electromagnetic radiations per quantum in theorder of increasing energy:

A: Energy light; B: Yellow light; C: X-ray; D: Radiowave

(1) D,B,A,C

(2) A,B,D,C

(3) C,A,B,D

(4) B,A,D,C

Solution: E_"radio wave" < E_"yellow" < E_"blue"< E_"X-ray"

=>D< B< A< C

Option (1) is correct

Question 19: An observer looks at a distant tree of height 10m with a telescope of magnifying power of 20. To the observer the tree appears:

(1) 10 times taller

(2) 10 times nearer

(3) 20 times taller

(4) 20 times nearer

Solution: "Magnification = "Focal length in cm"/"Eyepiece focal length "

Telescope resolves and brings objects closer. Hence telescope with magnifying power of 20, the tree appears 20 times nearer.

Option (4) is correct.

Question: 20 The box of a pin hole camera, of length L, has a hole of radius a, it is assumed that when the hole is illuminated by a parallelbeam of light of wavelength lambda the spread of the spot (obtained on the opposite wall of the camera) is the sum of its geometrical spread and the spread due to diffraction. The spot would then have its minimum size (say b_(min)) when:

(1)  a = (lambda^2)/L and b_(min) = ((2lambda^2)/L)

(2)  a = sqrt(lambdaL) and b_(min) = ((2lambda^2)/L)

(1)  a = sqrt(lambdaL) and b_(min) = sqrt(4lambda^2L)

(1)  a = (lambda^2)/L and b_(min) = sqrt(4lambdaL)

Solution: The diffreaction angle lambda/a causes a spreading of (L lambda)/a in the size of the spot. This becomes large when a is small. Adding the two kinds of spreading. We get a spot size a+(L lambda)/a Or we can write like this: sqrt ((a -(L lambda)/a)^2 + 4L lambda)

This is minimum when a -(L lambda)/a = 0 => a = sqrt(L lambda)

The minimum value is b_"min" = sqrt(4L lambda)

Option (3) is correct

Question: 21 Radiation of wavelength lambda, is incident on a photocell. the fastest emitted electron had speed v. If the wavelength in changed to (3lambda)/4, the speed of the fastest emitted electron will be:

(1) > v(4/3)^(1/2)

(2) < v(4/3)^(1/2)

(3) = v(4/3)^(1/2)

(4)  =v(3/4)^(1/2)

Solution: We know that: KE = h nu -phi

=> 1/2 m(V_("max"))^2 = (hc)/lambda -phi

=> (V_("max"))^2= v^2 = (2hc)/(lambda m)- (2phi)/m ...(1)

When lambda = (3lambda)/4

V'^2 =(2hc)/((3lambda)/4m)- (2phi)/m

=> V'^2= 4/3 (2hc)/(lambda m) - 4/3 (2phi)/m + (2phi)/m 1/3

=> V'^2 =4/3 ((2hc)/(lambda m)- (2phi)/m) + constant

From the equation (1), we get:

=> V'^2 = (sqrt(4/3)v)^2 + constant

=> V'^2 > (sqrt(4/3)v)^2

=> V' > sqrt(4/3)v

Option (1) is correct.

Question: 22 Half-lives of two radioactive elements A and B are 20 minutes and 40 minutes, respectively. Initially, the sample have equal number of nuclei. After 80 minutes, the ratio of decayed number of A and B nuclei will be:

(1) 1:16

(2) 4:1

(3) 1:4

(4) 5:4

Solution: We know that  N = N_0 e^(-lambdat)

half-life time T_(1/2)= ln2/ lambda

lambda_A = ln2 / 20 and lambda_B = ln2 / 40

At 80 mins,

N_A = N_0 e^((ln2 / 20 )xx80) = N_0 e^(-ln(2^4))= N_0/16

N_B = N_0 e^((ln2 / 40 )xx80) = N_0 e^(-ln(2^2)) = N_0/4

N_"decayed A" = N_0 - N_0/16= 15/16 N_0......(1)

N_"decayed B" = N_0 - N_0/4= 3/4 N_0....(2)

From equation (1) and (2), we get:

Ratio = (15/16)/(3/4)= 5/4

Option (4) is correct.

Question: 23 If a,b,c,d are inputs to a gateand x is the output, then, as per the following time graph, the gate is:

(1) NOT

(2) AND

(3) OR

(4) NAND

Solution:

Question: 24 Choose the correct statement:

(1) In amplitude modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.

(2) In amplitude modulation the frequency of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.

(3) In frequency modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the amplitude of the audio signal.

(4) In frequency modulation the amplitude of the high frequency carrier wave is made to vary in proportion to the ferquency of the audio signal.

Solution: Carrier wave y_k = A_c sin omega_c t

Message y_n = A_n sin omega_c t

y = (A_c+ A_n) sin omega_c t

Option (1) is correct.

Question: 25 A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the thickness of a thin sheetof Aluminium. before starting the measurement, it is found that when the twp jaws of the screw guage are bought in contact, 45^("th") division coincides with the main scale line and that the zero of the main scale is barely visisble. What is the thickness of the sheet if the main scale reading is 0.5mm and the 25^("th") division coincides with the main scale line?

(1) 0.75 mm

(2) 0.80 mm

(3) 0.70 mm

(4) 0.50 mm

Solution: Least count =0.5/50 = 0.01 mm

Zero error = - 5 xx0.01 = -0.05 mm

Reading = 0.5 + 25/100 = 0.75 mm

Thickness =0.75 -(-0.05)= 0.80 mm

Option (2) is correct.

Question: 26 A pipe open at both ends has a fundamental frequency f in air. The pipe is dipped vertically in water so that half of it is in water. The fundamental frequency of the air column is now:

(1) f/2

(2) (3f)/4

(3) 2f

(4) f

Solution:

For open end fundamental frequency f_0 = V/(2L)=f

Fundamental frequency of closed end,

f'_0 = V/4L'

Given that L'= L/2

Therefore, f'_0 = V/(4(L/2))= V/ (2L) =f

Option (4) is correct.

Question: 27 A galvanometer having a coil resistance of 100 Omega gives a full scale deflection, when a current of 1 mA is passed through it. The value of the resistance, which can convert this galavanometer into ammeter giving a full scale deflection for a current of 10A, is:

(1) 0.01 Omega

(1) 2 Omega

(1) 0.1 Omega

(1) 3 Omega

Solution:

G =100 Omega, i_g = 10^(-3) A, i =10A

i_g G = (i-i_g)s

 10^-3 (100) = (10 - 0.001)s

=> s = 0.1/(10 - 0.001) = 0.01 Omega

Option (1) is correct.

Question: 28 In an experiment for determination of refrative index of glass of a prism by i-delta, plot, it was found that a ray incident at angle 35^0, suffers a deviation of 40^0 and that it emerges at angle 79^0. In that case which of the following is closest to the maximum possible value of the refractive index?

(1) 1.5

(2) 1.6

(3) 1.7

(4) 1.8

Solution: i = 35^0, delta = 40^0, e =79^0

delta = i+e-A

=> 40 = 35+79-A

=> A = 74^0

and r_1 +r_2 = A = 74^0

delta_"min" < 40^0

=> mu < sin((A+delta)/2)/ sin (A/2) =sin((74+40)/2)/sin(74/2)

=> mu < sin114 / sin37 = 1.44

mu < 1.5

mu_"max"= 1.5

Option (1) is correct.

Question: 29 Identity the semiconductor devices whose characteristics are given below, in the order (a),(b),(c),(d):

(1) Simple diode, Zener diode, solar cell, Light dependent resistance

(2) Zener diode, Simple diode, Light dependent resistance, Solar cell

(3) Solar cell, Light dependent resistance, Zener diode, Simple diode

(4) Zener diode, Solar cell, Simple diode, Light dependent resistance

Solution: Zener diode works in breaks down region.

Option (1) is correct.

Question: 30 For a common emitter configuration, if  alpha and beta their usual meanings, the incorrect relationship between alpha and beta is:

(1) 1/alpha= 1/beta +1

(2) alpha = beta/(1-beta)

(3) alpha = beta/(1+beta)

(4) alpha = beta^2/(1+beta^2)

Solution: We know that I_E = I_C+ I_B....(1)

alpha = I_C/I_E and beta = I_C/I_B

From equation (1) we get,

I_E/I_C = I_C/I_C+ I_B/I_C

=> 1/alpha = 1+1/beta

1/alpha = (beta+1)/beta

alpha = beta/(beta+1)

Option (2) and (4) are correct.