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2601.

The half-life of a radioactive substance is 100 years. Calculate in how many years the activity will decay to 1/10th of its initial value. .

Answer»

332.3 YEARS
232.3 years
432.3 years
532.3 years

ANSWER :A
Discussion
2602.

Two wheels which are rotated by some external source with constant angular velocity in opposite directions as shown in figure. A uniform plank of mass M is placed on it symmetrically. The friction co-efficient between each wheel and the plank is mu. find the frequency of oscillations, when plank is slightly displaced along its length and released.

Answer»


Solution :If plank is displaced by `X` toward RIGHT then

Let `N_(1), N_(2)` and `f_(1), f_(2)` are Normal and friction force at Point `A` and `B` by force balance
`N_(1) + N_(2) = Mg_(1)`
amd Torque balance
`Mg (l + X) = N_(2) xx 2l__(2)`
by equation `(1)` and `(2)`
`N_(1) = (Mg)/(2) - (MGX)/(2l), N_(2) = (Mg)/(2) + (Mgx)/(2l)`
So `f_(1) = MU((Mg)/(2) - (Mgx)/(2l))`
and `f_(2) = mu((Mg)/(2) + (Mgx)/(2l))`

`F.B.D` of `M`
`f_(2) - f_(1) = Ma`
`mu. (2Mgx)/(2l) = Ma = -(d^(2)x)/(dt^(2)).M`
`rArr (d^(2)x)/(dt^(2)) = (-mugx)/(l)`
TIME period `= 2pisqrt((l)/(mug))`
Discussion
2603.

In Young'sdouble slit experiment the intensity of light at a point on the screen where the path difference is lambdais K . The intensity of light at a point where the path difference is (lambda)/(3)[ lambdais the wavelength of light used ] is

Answer»

K / 4
K / 3
K / 2
K

SOLUTION :`K = I_(0) cos^(2) (pi) to (1)`
`I = I_(0) cos^(2) ((pi X)/(3)) to (2) theta = (2 pi x)/(3)`
Discussion
2604.

A plane electromagnetic wave of frequency 20 MHz travels through a space alongx direction. If the electric field vector at a certain point in space is 6 V m^(-1), what is the magnetic field vector at that point ?

Answer»

`2 xx 10^(-8) T`
`(1)/(2) xx 10^(-8)T`
2T
`(1)/(2)T`

SOLUTION :`E//B = C`
`B = E//C = (6)/(3 xx 10^(8)) = 2 xx 10^(-8) T`
Discussion
2605.

A glass cube of edge 1 cm and mu=1.5 has a spot at the centre. The area of the cube face that must be covered to prevent the spot from being seen is (in cm^(2))

Answer»

`SQRT(5)PI`
`5PI`
`(pi)/(sqrt(5))`
`(pi)/5`

ANSWER :D
Discussion
2606.

Electron, initially almost at rest, are accelerated by a potenetial difference of 200 volt. They then pass through an electric field and a magnetic field at right angle to each other and also at right angles to the electron beam. The magnitudes of two fields are so adjusted that the electrons emerge out straight without any deviation. Determine the specific charge, if the electric field is 1117.6 volt cm^(-1) and magnetic field is 14 gauss.

Answer»


ANSWER :`1.59 xx 10^(11) C G^(-1)`
Discussion
2607.

Give reasons for the following: (i) Long distance radio broadcasts use short-wave bands. (ii) The small ozone layer on top of the stratosphere is crucial for human survival. (iii) Satellites are used for long distance TV transmission.

Answer»

SOLUTION : (i) Because they can be reflected by the ionosphere of the Earth.s atmosphere and can be sent to longer distances.
(ii) The ozone LAYER absorbs the ultraviolet radiations coming from the Sun and prevents them from reaching the Earth.s surface.
(iii) As television signals are having HIGH frequency, THEREFORE, they cannot be reflected by ionosphere. To REFLECT these high frequency signals, satellites are required.
Discussion
2608.

Suppose that water drops are released from a point at the edge of a roof with a constant time interval Deltat between one water drop and the next. The drops fall a distance h to the ground. If Deltat is very short ie the number of drops falling through the air at any given instant is very large then the CM of the drops is very nearly at a height (above the ground ) of

Answer»

`h//2`
`h//3`
`2h//3`
`3h//4`

Solution :The droplets fall a DISTANCE `(1)/(2)g t^(2)` in time t, and the NUMBER of the droplets and hence their mass is proportional to `dt`. COMPUTING the CM of the droplets using the definition, we get the RESULT.
Discussion
2609.

Threebodiesof the samematerialand havingmassesm,mand3mare attemperature40^@ C , 50^@Cand60^@ Crespectively. Ifthebodiesare broughtin thermalcontactthe finaltemperaturewill be

Answer»

`45^@ `C
`54^@ C `
`52^@ C `
`48^@ C`

Solution :here`m_1=m_2-=m,m_2 = m m_3= 3m`
`s_1 =s_2= S_3= s,T_1= 40^@C,T_2= 50 ^@c, T_3=60^@ C `
LetT bethe finaltemperatureat the thermalequilibriumusingprnicipleof caloriemetry
Heathainedby `m_1and m_2`= heatlost byby `m_3 `
`m_1 s_1(T -T_1)+m_2 s_2(T-T_2)=m_3s_3(T_3 -T )`
`ms(t-40 )+ ms(T- 50 )= 3 ms( 60 - T)`
`t- 40+T- 50 = 3 (60 -T ) or2T- 90= 180- 3T `
5T = 270` thereforeT= 54 ^@C `
Discussion
2610.

The anglewhich thetotal magnetic field of earth makeswith thesurface of the called

Answer»


ANSWER :D
Discussion
2611.

In the accompanying diagram , if C_(1)=3muF,C_(2)=6muF,C_(3)=9muF,C_(4)=12muF,C_(5)=15muF and C_(6)=18muF , then the equivalent capacitor between the ends A and B is

Answer»

`1.22 MUF`
`5.16 muF`
`2.25 muF`
`2.51 muF`

ANSWER :D
Discussion
2612.

A particle of mass 1 gm and charge 1muC is held at rest on a frictionaless horizontal surface at distance 1 m from the fixed charge 2mc. If the particleis released, it will be repelled . The spedd of the particle when it is at a distance of 10m from the fixed charge is

Answer»

`60ms^(-1)`
`100ms^(-1)`
`90ms^(-1)`
`180ms^(-1)`

Solution :
Change in kinetic ENERGY = work done by ELECTRIC field
`(1)/(2)mv^(2)-0=(V_(A)-V_(B))Q=(Q)/(4piepsilon_(0))[(1)/(r_(1))-(1)/(r_(2))]xxq`
`(1)/(2)xx10^(-3)xxv^(2)=9xx10^(9)[1-(1)/(10)]xx10^(-6)xx2xx10^(-3)`
`v=180ms^(-1)`
Discussion
2613.

On a smooth inclined plane, a body of mass M is attached between two springs. The other ends of the springs are fixed to firm supports. If each spring has force constant K, the period of oscillation of the body (assuming the springs as massless) is :

Answer»

`2pi(M//2K)^(1//2)`
`2pi(2M//K)^(1//2)`
`2pi ("MG "SIN theta//2K)`
`2pi (2"Mg"//K)^(1//2)`

Answer :A
Discussion
2614.

Let R and S be two non-void relations on a set A. Which of the following statement is false?

Answer»

A rheostat can be used as a POTENTIAL divider.
Kirchhoff's second LAW represents ENERGY conservation
Wheatstone bridge is the most sensitive when all the four RESISTANCE are of the same order of magnitude.
In a balanced Wheatstone bridge if the CELL and the galvanometer are exchanged the null point is disturbed.

Solution :In a balanced Wheatstone bridge if the cell and the galvanometer are exchanged the null point is disturbed.
Neutral point will not be changed.
Discussion
2615.

Arrange the following communication frequency bands in the increasing order of frequencies. 1. A.M Boradcast2. Cellular mobile radio 3. F.M. Broadcast4. Television UHF 5. Satelite communication

Answer»

1,3,4,2,5
1,2,3,4,5
5,2,4,3,1
1,3,2,4,5

Solution :REFER to table 10(a).1, PAGE `10//6`
Discussion
2616.

Where was Mini sitting?

Answer»

In her room
In her balcony
In ANJU's house
None of the above

Answer :B
Discussion
2617.

Two seconds after projection, a projectile is moving at 30^(@) above the horizontal. After one more second it is moving horizontally. If g=10ms^(2), the velocity of projection is

Answer»

`10sqrt3ms^(-1)`
20 m/s
10 m/s
`20sqrt3m//s`

ANSWER :D
Discussion
2618.

5J of work is done in moving a positive charge of 0.5C between two points. What is the potential difference between the points?

Answer»

2.5V
10V
0.1 V
5.5V

Answer :B
Discussion
2619.

The spectral lines of Pfund series lies in the ...... region.

Answer»

ULTRAVIOLET
INFRARED
visible LIGHT
NONE of these

SOLUTION :infrared
Discussion
2620.

Assume three concentric, conducting spheres where charge 4, and q, have been placed on inner and outer sphere where as middle sphere has been earthed. Find the charge on the outer surface of middle spherical conductor

Answer»

`(B)/(C ) q_2`
`-q_1`
`-q_2`
`(b)/(a)q_1`

ANSWER :A
Discussion
2621.

A ray of light passes through an equilateral prism such that the angle of incidence is equal to 3/4 th angle of emergence and each one is equal to the angle of prism. The angle of deviation is

Answer»

`45^(@)`
`39^(@)`
`60^(@)`
`30^(@)`

ANSWER :D
Discussion
2622.

A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30° with the direction of a uniform horizontal magnetic field of magnitude 0.8 T. What is the magnitude of the torque experienced by the coil?

Answer»

SOLUTION :N=20 , `A=100 XX 10^(-4) m^2. 1=12 A`
R=0.8 T `theta = 30^@` Torque `TAU = NI( bar A xx bar B)`
`= 20 xx 12 xx 100 xx 10^(-4) xx 0.8 xx sin 30^2@`
`= 20 xx 12 xx 100 xx 10^(-4) xx 0.8 xx 1/2 = 0.96 Nm`
Discussion
2623.

Can we take one slab of p-type semiconductor and physically join it to another n-type semiconductorto get p-n junction ?

Answer»

Solution :No ! Any slab,howsoever FLAT, will haveroughness much large than the inter-ATOMIC crystal spacing`( ~2` to `3A^(@))` and hence continuous contact at the atomic LEVEL will notbe possible . Thejunction will behave as a discontinuity for the flowing CHARGE CARRIERS.
Discussion
2624.

Calculate the velocity of sound in a mixture of oxygen, nitrogen and argon at 0^(@) when their masses are in the ratio 2:7: 1. The molecular weights of gases are 32, 2S and 40 respectively.

Answer»


ANSWER :B::C
Discussion
2625.

In an oscillating LC circuit, when 75.0% of the total energy is stored in the inductor's magnetic field, (a) what multiple of the maximum charge is on the capacitor and (b) what multiple of the maximum current is in the inductor?

Answer»


ANSWER :(a)0.500 ; (B) 0.886
Discussion
2626.

Draw a ray diagram to obtain the virtual image formation in (i) a concave mirror and (ii) a convex mirror.

Answer»

SOLUTION :
Discussion
2627.

An electric dipole is placed in an uniform electric field of a point charge, then ........

Answer»

the RESULTANT force ACTING on the dipole is ALWAYS ZERO.
the resultant force acting on the dipole MAY be zero.
torque acting on it may be zero.
torque acting on it is always zero.

Answer :A
Discussion
2628.

In Fraunhoffer diffraction draw the graph of intensity of light on screen versus diffraction angle theta and write the characteristics of it.

Answer»

Solution :In Fraunhoffer diffraction the graph of the intensity of light on screen versus diffraction angle `THETA` is shown in graph.
WIDTH="80%">
From the graph it can be said that the intensi of central maxima is highest.
(i) All minima have zero light intensity.
(ii) The intensity of maxima decreases wit INCREASE the angle of diffraction, thus i: intensity decreases as the order increase and the width of maxima ALSO decreases.
(iii) As ratio `(lamda)/(a)` small, the diffraction will be and it is large, the diffraction will be more
(iv) The angular width of central maxima inversely proportional to the width of SLIT.
Discussion
2629.

KMnO_4 oxidises S_2O_3^(2-)to SO_4^(2-) in medium x and NO_2^(-) to NO_3^(-) in medium y,x and y are respectively.

Answer»

acidic,basic
acidic,acidic
acidic,neutral
neutral,acidic

SOLUTION :In neutral or faintly alkaline solution
`8MnO_4^(-) +3S_2O_3^(-2) +H_2O to8MnO_2+6SO_4^(-2) +2OH^(-)`
In acidic medium,
`2MnO_4^(-) +5NO_2^(-)+6H^(+) to2Mn^(+2) +5NO_3^(-) +3H_2O`
Hence OPTION (d) is correct.
Discussion
2630.

When a beam a of high energy electrons is bombarded on a metal target __________are produced.

Answer»

SOLUTION :X - RAYS
Discussion
2631.

If a satellite is travelling in the same direction as the rotation of earth i.e., west to east, what is the interval between two successive times at which it will appear vertically overhead to an observer at a fixed point on the equator? (R = 6400 km, h = 1400 km)

Answer»


ANSWER :A
Discussion
2632.

Acetic acid on heating with NH_3 forms A. When A reacts with LiAH_4 followed by hydrolysis gives B. When B is heated with chloroform in KOH medium gives C. What are B and C respectively?

Answer»

`CH_3CONH_2,CH_3 CH_2 NC`
`CH_3 CH_2 NH_2, CH_3 CH_2 NC`
`CH_3 CH_2 NH_2, CH_3 COOH`
`CH_3 CH_2 CH_2 NH_2, CH_3 CH_2 NC`

Solution :
Thus, OPTION (B) is CORRECT.
Discussion
2633.

When two spheres having 4Q and - 2Q charge are placed at a certain distance, the force acting between them is F. Now they are connected by a conducing wire and again separated from each other. Now they are kept at a distance half of the previous one. The force acting between them is.......

Answer»

F
`F/4`
`F/2`
`F/8`

SOLUTION :`F =K((4Q)(-2Q))/r^(2) =-(8kQ^(2))/r^(2)`
`F. = k((Q)(Q))/(r.)^(2) =(4kQ^(2))/r [THEREFORE r.=r/2]`
`therefore F^(.)/F =(4kQ^(2))/r^(2) xx (-r^(2))/(8kQ^(2)) =-1/2`
`therefore F. = F/2, therefore |F.| = |F/2|`
Discussion
2634.

What is the work done in taking a north pole of strength m around a long and straight conductor in a circular path at a perpendicular distance of r from the straight conductor?

Answer»

Solution :`mu_0` im, since the FORCE acting on the north pole is `F = mB = (m mu_0 i)/(2pir)` along the tangent and the work DONE `= vec (F),vec(S)` = FS COS 0 as both force and displacmentare long the same direction.
Hence , work done `= (mu_0 i)/(2PI r) (2pi r) = mu_0 `im.
Discussion
2635.

A cube is soldered together írom identical pieces of wire each of a resistance r. It is connected to the circuit at the corners lying on opposite ends of a body diagonal Find the equivalent resistance.

Answer»


Solution :It is clear from considerations of symmetry that the potentials at points 2, 4 and 5 coincide, i.e. `Psi_(2)= Psi_(4)= Psi_(6)=Psi` Hence the potentials at points `delta, delta and 3` coincide as WELL: `Psi_(5)=Psi_(5)=Psi_(3)=Psi..`. If we SHORT circuit the points at equal potential, i.e. connect the points by means of conductors of NEGLIGIBLE resistance, the resistance of the circuit remains unchanged. The circuit obtained in this WAY is shown in Fig. 26.51. Its resistance is equal to tho sum of the resistances of three series connected sections of three, six, and three parallel connected branches each.
Discussion
2636.

1/(sqrt9-sqrt8)is equal to -

Answer»

`1/2(3-2sqrt2)`
`1/(3+2sqrt2)`
`(3-2sqrt2)`
`(3+2sqrt2)`

ANSWER :D
Discussion
2637.

Which of the following represents pressure ?

Answer»

`ML^-1T^-2`
`MLT^-2`
`M L^-2T^-2`
`M LT^-3`

ANSWER :A
Discussion
2638.

In the relation P=(alpha)/(beta)e^(-alpha z//ktheta) where {:(P="pressure"),(z= "distance"), (k = "Boltzman's constant"), (theta = "Temperature"):} The dimensional formula of beta will be :

Answer»

`M^(0)L^(2)T^(0)`
`ML^(2)T`
`ML^(0)T^(-1)`
`M^(0)L^(2)T^(-1)`

SOLUTION :As exponential expression is dimensionless
`(alphaz)/(ktheta)=M^(0)L^(0)T^(0)`
`alpha=(ktheta)/(z)=(ML^(2)T^(-2)theta^(-1)xxtheta^(1))/(L^(1))`
`:.alpha=ML^(1)T^(-2).`
Now `(alpha)/(BETA)` will have dimension of `P`.
`:.beta=(alpha)/(P)=(ML^(1)T^(-2))/(ML^(-1)T^(-2))=L^(2).`
HENCE correct CHOICE is `(a)`.
Discussion
2639.

Find the ratio of magnetic field magnitudes at a distance 10 m along the axis and at 60^@ from the axis, from the centre of a coil of radius 1 cm, carrying a current 1 amp.

Answer»

SOLUTION :`4 //SQRT(7)`
Discussion
2640.

In a arrangement shown in the fig.the distance D is latge compared to the separation d between this slits ,find: The minimum value of d for which there is a dark fringe at the point O.

Answer»

`SQRT(lambdaD//2)`
`sqrt(Dlambda//3)`
`sqrt(lambdaD)`
cannot be calcualted

Answer :B
Discussion
2641.

A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom make a transition from the upper level to the lower level ?

Answer»

SOLUTION :It is given here `E_(n_(1)) - E_(n_(2)) = 2.3 eV = 2.3 xx 1.6 xx 10^(-19)` J
`therefore ` radiation frequecy ` V= (E_(n_(1)) - E_(n_(2)))/(h) = (2.3 xx 1.6 xx 10^(-19))/(6.63 xx 10^(-34)) = 5.6 xx 10^(14)Hz`
Discussion
2642.

In a arrangement shown in the fig.the distance D is latge compared to the separation d between this slits ,find: The position of first bright fringe for the minimum value of dis

Answer»

`d//2`below
`d` above
`3d//2`below
`3d//2`above

Answer :C
Discussion
2643.

What is the nature of force between two parallel conductors carrying current in the same direction?

Answer»

SOLUTION :ATTRACTIVE
Discussion
2644.

In a arrangement shown in the fig.the distance D is latge compared to the separation d between this slits ,find: The fringe width is

Answer»

`(3Dlambda)/(4D)`
`(3Dlambda)/(2D)`
`(Dlambda)/(d)`
`(2Dlambda)/(d)`

ANSWER :B
Discussion
2645.

Statement I : The pitch of the sound wave depends upon its frequency. Statement II: The loudness of the sound wave depends upon its amplitude.

Answer»

Both the statements are INDIVIDUALLY TRUE and Statement II is the correct explanation of Statement I
Both the statements are individually true but Statement II is not the-correct explanation of Statement I
Statement I is the true but Statement II is false
Statement I is false but Statement II is true

ANSWER :B
Discussion
2646.

The electric field in a region is given by vec(E ) = E_0 x/L hati . Find the charge contained inside a cubical volume bounded by the surface x = 0, x = L, y = 0 , y = L, z = 0 and z = L.

Answer»

Solution :At x = 0 , E = 0 and at `x = l , vec(E ) = E_0 HATI`
The direction of the field is along the x-axis , so it will cross the YZ- face of the cube. The FLUX of this field.
`phi = phi_("left face") + phi_("right face") = 0 + E_0 L^2 = E_0 L^2`
By Gauss.s law, `phi = q/(in_0) :. q = in_0 phi = in_0 E_0L^2`.
Discussion
2647.

A metallic rod of 1 m length is rotated with a frequency of 50 rev/s, with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius 1 m, about an axis passing through the centre and perpendicular to the plane of the ring (Fig.). A constant and uniform magnetic field of 1 T parallel to the axis is present everywhere. What is the emf between the centre and the metallic ring?

Answer»

Solution :Method I
As the rod is rotated, free electrons in the rod move towards the outer end due to Lorentz force and get distributed over the ring. Thus, the resulting separation of charges produces an emf across the ends of the rod. At a certain value of emf, there is no more flow of electrons and a steady state is reached. Using Eq. (6.5), the magnitude of the emf generated across a length dr of the rod as it moves at RIGHT angles to the magnetic field is given by ` d epsi = Bv d r` . Hence,
`epsi = int d epis = int_(0)^(R) Bv d r = int_(0)^(R) B OMEGA r dr = (B omega R^2)/(2)`
Note that we have used `v = omega r `. This gives.
`epsi = 1/2 xx 1.0 xx 2PI xx 50 xx (1^2)`
`= 157 V`
Method II
To calculate the emf, we can imagine a closed loop OPQ in which point O and P are connected with a resistor R and OQ is the rotating rod. The potential difference across the resistor is then equal to the induced emf and EQUALS B `x×` (rate of change of area of loop). If `theta` is the angle between the rod and the radius of the circle at P at time t, the area of the sector OPQ is given by
`pi R^2 xx (theta)/(2pi) = 1/2 R^2 theta`
where R is the radius of the circle . Hence , the induced emf is
`epsi = B xx d/(dt) [1/2 R^2 theta] = 1/2 BR^2 (d theta)/(dt) = (B omegaR^2)/(2)`
[Note : `(d theta)/(dt) = omega = 2pi v`]
This expression is identical to the expression obtained by Method I and we get the same value of `epsi`.
Discussion
2648.

The dispersive power of crown and flint glasses are 0.03 and 0.05 respectively. If the difference in the refractive indices of blue and red colours are 0.015 for crown glass and 0.022 for flint glass, calculate the angles of the two prisms for a deviation 2° without dispersion.

Answer»

SOLUTION :`10^@ ,6.8^@`
Discussion
2649.

Animals that rely on heat from the environment rather than of metabolism, to raise their body temperature are in the strict sense called

Answer»

Ectothermic
Poikilothermic
Homoeothermic
Endothermic

Answer :A
Discussion
2650.

What speed should be galaxy move with respect to us to the sodium line at 589.0 nm is observed at 589.6nm?

Answer»

Solution :Since `v LAMDA=c, (DELTA v)/v=-(Delta lamda)/(lamda) `(for small changes is v and `lamda`) for `Delta lamda=589.6-589.0=0.6nm`
`:.(Deltav)/v=-(Delta lamda)/(lamda)=(v_("RADIAL))/c`
or `v_("radial")~=+c(0.6/58.0)=+3.06xx10^(5)ms^(-1)=306km//s`
Therefore, the galaxy is moving away from US.
Discussion