. The equivalent capacitance between and in the circuit given below is : 6F F 5F 5F 4F ns. () () 3.6F ().4F (3) 4.9F (4) 5.4F 6F 5F Simplified circuit 6F F D D E E D E F 5F E F E E 4F F 4F 5F F eq 5F 5 eq.4 F c 6 6 5 Tel:+9-0-466500 Page No.
. charge Q is placed at a distance. The electric flux through the square the surface is a/ above the centre of the square surface of edge a as shown in the figure. P a/ a () ns. () Q 6 0 Q () Q 0 (3) Q 3 0 (4) Q 0 a/ a charged particle can be onsidered at centre of a cube of side a, and given surface represents its one side. Q So flux = 6 0 3. uniform rod is suspended from a point X, at a variable distance x from, as shown. To make the rod horizontal, a mass m is suspended from its end. set of (m,x) values is recorded. The appropriate variables that given a straight line, when plotted, are : X ns. (3) () m,x. () x m m, (3) x m, (4) m,x x Tel:+9-0-466500 Page No.
x /-X Mg m alancing torque w.r.t. point of suspension mg x = Mg x mx = M Mx m = M M x y = equation of a straight line x 4. The energy required to remove the electron from a singly ionized Helium atom is. times the energy required to remove an electron from Helium atom. The total energy required to ionize the Helium atom completely is : () 34eV () 0eV (3) 79eV (4) 09eV ns. (3) Energy required to remove e from singly ionized helium atom = 54.4 ev Energy required to remove e form helium atom = x ev given 54.4 ev =.x x = 4.73 ev Energy required to ionize helium atom = 79. ev 5. solution containing active cobalt 60 7 o having activity of 0.8 i and decay constant is injected in an animal's body. If cm 3 of blood is drawn from the animal's body after 0 hrs of injection, the activity found was 300 decays per minute. What is the volume of blood that is flowing in the body? (i = 3.7 0 0 decays per second and at t = 0 hrs e t = 0.84) () 4 liters () 6 liters (3) 5 liters (4) 7 liters ns. (3) Let total volume of blood is v, initial activity 0 = O.Q ci its activity at time t = = 0e t activity of x 0 volume = x x e V V 0 V x e t 3 8 0 3.7 0 300 60 = 4.97 0 3 cm 3 = 4.97 liter 7 0 t V = (cm ) 0.84 Tel:+9-0-466500 Page No.3
6. In a common emitter configuration with suitable bias, it is given that L is the load resistance and E is small signal dynamic resistance (input side). Then, voltage gain, current gain and power gain are given, respectively, by : is current gain,, and E are respectively base, collector and emitter currents. L L L E L L L L L (),, (),, (3),, (4),, E ns. () From NET urrent gain = I I E VE L Voltage gain v = I Power gain p = v = E E L E E E 7. body of mass m is moving in a circular orbit of radius about a planet of mass M. t some instant, it splits into two equal masses. The first mass moves in a circular orbit of radius. and the other mass, in a circular orbit of radius () Gm () 6 3. The difference between the final and initial total energies is : GMm (3) E E GMm (4) 6 ns. (3) GMm Ei GMm / GMm / GMm GMm 4GMm Mm Ef = 3 6 6 3 GMm GMm E f E i = 3 6 E GMm 8. Helmholtz coil has a pair of loops, each with N turns and radius. They are placed coaxially at distance and the same current I flows through the loops in the same direction. The magnitude of magnetic field at P, midway between the centres and, is given by [efer to figure given below]: E E P () ns. (4) 8N 0 () / 5 4N 0 (3) 3 / 5 4N 0 (4) / 5 8N 0 5 3 / Tel:+9-0-466500 Page No.4
NI = 4 NI 5 8 0 0 3 / 3 / 8 NI 5 0 = 3 / 9. thin uniform tube is bent into a circle of radius r in the vertical plane. Equal volumes of two immiscible liquids, whose densities are and ( > ), fill half the circle. The angle between the radius vector passing through the common interface and the vertical is : () = tan (3) = tan ns. () () = tan (4) = tan P P equating pressure at point, g (cos sin) = g(sin + cos) sin cos tan cos sin tan tan = + tan ( + ) tan= = tan Tel:+9-0-466500 Page No.5
0. The velocity-time graphs of a car and a scooter are shown in the figure. (i) The difference between the distance travelled by the car and the scooter in 5 s and (ii) the time at which the car will catch up with the scooter are, respectively. 45 car Velocity (ms ) 30 5 E F G 0 O D 0 5 0 5 0 5 Time in (s) (3) 5.5 m and 0 s (4).5 m and 5 s ns. () Distance travelled by car in 5 sec = (45) (5) = 675 m, Distance traveled by scooter in 5 seconds = 30 5 = 450 Let car catches scooter in time t; 675 45(t 5) 30t 337.5 + 45t 675 = 30t 5t = 337.5 t =.5 sec. 3 monochromatic beam of light has a frequency v 0 Hz and is propagating along the direction î ĵ. It is polarized along the kˆ direction. The acceptable form the magnetic field is : () cos î ĵ E 0 4 î ĵ 0.r 3 0 t () kˆ cos E 0 4 î ĵ 0.r 3 0 t E î ĵ (3) 0 4 î ĵ cos0.r 3 0 t E î ĵ kˆ (4) 0 cos0 3 î ĵ.r 3 0 t 4 Tel:+9-0-466500 Page No.6
ns. (3) Eˆ ˆ should give direction of wave propagation ˆi ˆj K ˆ II ˆi ˆj ˆj ( ˆi) ˆi ˆj ˆi ˆj option ˆK option - and 4 does not satisfy this. wave propagation vector ˆK should be along ˆ i ˆ j So correct option is 3.. Take the mean distance of the moon and the sun from the earth to be 0.4 0 6 km and 50 0 6 km respectively. Their masses are 8 0 kg and 0 30 kg respectively. The radius of the earth is 6400km. Let F be the difference in the forces exerted by the moon at the nearest and farthest point on the earth and F be the difference in the force exerted by the sun at the nearest and farthest points F on the earth. Then, the number closest to is : F () 6 () 0 (3) (4) 0.6 ns. (3) GM m F e r GM Ms F e r GMem GMeMs F = r 3 F = 3 r r r 3 3 F mr r m r r 3 3 F r Msr Ms r r using r = r = earth m = 8 0 kg M s = 0 30 kg r = 0.4 0 6 km r = 50 0 6 km F we get F 3. planoconvex lens becomes an optical system of 8cm focal length when its plane surface is silvered and illuminated from left to right as shown in fig- If the same lens is instead silvered on the curved surface and illuminated from other side as in fig., it acts like an optical system of focal length 0 cm. The refractive index of the material of lens is: Fig. Fig. ().55 ().50 (3).75 (4).5 ns. () Tel:+9-0-466500 Page No.7
ase- + f f = 8 P = P + P 8 ase- + f f f= 0cm P = P + P 0 0 8 8 0 8 80 = 80 9 cm 9 8 80 = 5 9 5 4 =.55 9 9 4. One mole of an ideal monatomic gas is compressed isothermally in a rigid vessel to double its pressure at room temperature, 7. The done on the gas will be : () 300 () 300 In (3) 300 In 6 (4) 300 In 7 ns. () p f Work done on gas = nt n = (300) n() = 300n pi 5. n automobile, travelling at 40 km/h, can be stopped at a distance of 40 m by applying brakes. If the same automobile is travelling at 80 km/h, the minimum stopping distance, in metres, is (assume no skidding) : () 50m () 00m (3) 75m (4) 60m ns. (4) S = S S u a u u u S = S = () (40) = 60 m u Tel:+9-0-466500 Page No.8
6. carnot's engine works as a refrigerator between 50K and 300K. It receives 500 cal heat from the reservoir at the lower temperature. The amount of work done in each cycle to operate the refrigerator is () 77 J () 40 J (3) 00 J (4) 50 J ns. () T W Efficiency = T Q W 50 W 300 Q W Q 500 4. W J 40J 5 5 7. In a screw gauge, 5 complete rotations of the screw cause it to move a linear distance of 0.5 cm. There are 00 circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of 4 main scale divisions and 30 circular scale divisions. ssuming negligible zero error, the thickness of the wire is : () 0.4300 cm () 0.350 cm (3) 0.0430 cm (4) 0.50 cm ns. (4) 0.5 Least ount = cm = 5 0 4 cm 5 00 eading = 4 0.05cm + 30 5 0 4 cm = (0. + 0.050) cm = 0.50 cm 8. The number of amplitude modulated broadcast stations that can be accommodated in a 300 khz band width for the highest modulating frequency 5 khz will be : () 5 () 0 (3) 8 (4) 0 ns. (4) If modulating frequency is 5 KHz then band width of one channel = 30 khz No of channels accommodate = 300kHz 0 30kHz 9. n ideal capacitor of capacitance 0. F is charged to a potential difference of 0 V. The charging battery is then disconnected. The capacitor is then connected to an ideal inductor of self inductance 0.5 mh. The current at a time when the potential difference across the capacitor is 5V is : () 0.5 () 0.7 (3) 0.34 (4) 0.5 ns. () Using energy conservation 0. 0 6 0 0 = 0. 0 5 0.5 0 3 I I 3 0 = 0.7 0. Light of wavelength 550 nm falls normally on a slit of width.0 0 5 cm. The angular position of the second minima from the central maximum will be (in radians) : () () (3) (4) 4 8 6 ns. () Tel:+9-0-466500 Page No.9
If angular position of nd maxima from central maxima is then 3 3 550 0 sin a 0 9 7 ~ 8 rad. force of 40 N acts on a point at the end of an L-shaped object as shown in the figure. The angle that will produce maximum moment of the force about point is given by : 4m m F () tan = 4 () tan = 4 (3) tan = (4) tan = ns. (3) Moment of force will be maximum when line of action of force is perpendicular to line. 4m tan 4 m. tuning fork vibrates with frequency 56 Hz and gives one beat per second with the third normal mode of vibration of an open pipe. What is the length of the pipe? (Speed of sound in air is 340 ms ) () 0 cm () 00 cm (3) 90 cm (4) 80 cm ns. () Organ pipe will have frequency either 55 or 57 Hz Using 55Hz 3V 3 340 55 m 55 = 00 cm. Tel:+9-0-466500 Page No.0
3. body of mass M and charge q is connected to a spring of spring constant k. It is oscillating along x-direction about its equilibrium position, taken to be at x = 0, with an amplitude. n electric field E is applied along the x-direction. Which of the following statements is correct? () The total energy of the system is m q E +. k () The new equilibrium position is at a distance qe from x = 0. k (3) The new equilibrium position is at a distance qe from x = 0. k (4) The total energy of the system is m q E. k ns. () Equilibrium position will shift to point where resultant force = 0 kx eq = qe x eq = qe k Energy qe m k = q E m k 4. given object takes n times more time to slide down a 45 rough inclined plane as it takes to slide down a perfectly smooth 45 incline. The coefficients of kinetic friction between the object and the incline is : () () n n ns. () (3) n (4) n Time taken to slide along smooth surface s gsin45 t t s g Time taken to slide along rough surface S (gsin45 gcos 45 )t t = s g( ) t = nt s s n g( ) g = n = n 5. The relative error in the determination of the surface area of a sphere is. Then the relative error in the determination of its volume is : ns. () () 3 () 3 (3) (4) 5 Tel:+9-0-466500 Page No.
s r s r v 3 v v r 3 v r 6. In a meter bridge as shown in the figure it is given that resistance Y =.5 and that the balance is obtained at a distance 39.5 cm from end (by Jockey J). fter interchanging the resistances X and Y a new balance point is found at a distance l from end. What are the values of X and l? X Y 39.5 G 00 (.) attery Key () 9.5 and 39.5 cm () 8.6 and 60.5 cm (3) 8.6 and 39.5 cm (4) 9.5 and 60.5 cm ns. () For a balanced meter bridge y 39.5 = x (00 39.5) x =.5 39.5 8.6 60.5 when x & y are interchanged and (00 ) will also interchange = 60.5 cm. 7. The -H curve for a ferromagnet is shown in the figure. The ferromagnet is placed inside a long solenoid with 000 turns/cm. The current that should be passed in the solenoid to demagnetise the ferromagnet completely is : (T) -00.0.0-00 00 00.0.0 H (/m) () m () 0 (3) m (4) 40 ns. (3) oercivity of Ferro magnet H = 00 /m ni = 00 00 I = 5 m 0 Tel:+9-0-466500 Page No.
8. In the given circuit all resistances are of value of ohm each. The equivalent resistance between and is : ns. (4) () 5 () 3 (3) 5 3 (4) eq = 9. particle is oscillating on the x-axis with an amplitude cm about the point x 0 = 0cm with a frequency. concave mirror of focal length 5cm is placed at the origin (see figure). x = 0 x 0 = 0cm Identify the correct statements? () The image executes periodic motion () The image executes non-periodic motion () The turning points of the image are asymmetric w.r.t. the image of the point at x = 0 cm. (D) The distance between the turning points of the oscillation of the image is 00 cm. () (, ) () (,, D) (3) (, D) (4) (, D) ns. () Tel:+9-0-466500 Page No.3
When object is at 8 cm f u 40 V cm u f 3 When object is at cm 60 V cm 7 00 Separation = V V cm So, and D are correct 30. Two electrons are moving with non-relativistic speeds perpendicular to each other. If corresponding de roglie wavelengths are and their de roglie wavelength in the frame of reference attached to their centre of mass is : () ns. () M () M = (3) M = = (4) M = Momentum of each electron Velocity of centre of mass h h V ˆ ˆ cm i j m m h h î & ĵ Velocity of st particle about centre of mass h h V ˆ ˆ cm i j m m h cm h h 4 4 Tel:+9-0-466500 Page No.4