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101.

Draw equipotential surfaces for a single point charge O>0.

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SOLUTION :
Discussion
102.

Which among the following is a type of resources classified on the basis of exhaustibility?:

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BIOTIC and abiotic
Renewable and non-renewable
National and individual
Potential and reserves

Answer :B
Discussion
103.

A TV tower has a height of 100 m. The population density around the TV if the population covered is 60.288 lac, is

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`5 xx 10^(3)km^(-2)`
`1.5 xx 10^(3) km^(-2)`
`7.5 xx 10^(3) km^(-2)`
`10^(4) km^(-2)`

Answer :B
Discussion
104.

If Y = 2 xx 10^11 N/m^2 for a wire, the force required to double the length of a wire of the same material of cross section 1 mm ^2 is

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a)`2 XX 10^11 N`
B)`2 xx 10^5 N`
C)`10^11 N`
d)2N

Answer :B
Discussion
105.

A ray of light is incident normally on one face of a triangular prism of refracting angle 30^(@) and refractive index 1.5 Find the deviaton of the ray produced.

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`18.6^(@)`
`12.2^(@)`
`8.4^(@)`
`6.3^(@)`

Solution :(a) For normal incidence,I = 0 and r = 0
Now r + r. = A
`0 + r. = 30^(@), r. = 30^(@)`
`THEREFORE mu = (sin I.)/(sinr.), therefore I. = 1.5 sin 30^(@)`
`therefore I. = 48.6^(@)`
Now I + I. = A + D
`D = (i+I.) - A = 18.6^(@)`
Discussion
106.

A cubical frame is made of 12 rods each of mass m and length l. The moment of inertia of the cube about its face diagonal is

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`(20)/(3)ml^(2)`
`(14)/(3)ml^(2)`
`(43)/(6)ml^(2)`
`(23)/(3)ml^(2)`

SOLUTION :`I=((4)/(3)ml^(2)SIN^(2)45^(@))+((4)/(3)ml^(2)sin45^(@)+4ML^(2))+((2)/(3)ml^(2))+(2xx(5)/(6)ml^(2))`
Discussion
107.

A ball of material of specific gravity d_(1) falls from a height 'h' on the surface of a liquid of relative density d_(2) such that d_(2) > d_(1). The time for which the body will be falling into the liquid is :

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`(d_(1))/(d_(2))sqrt((2H)/G)`
`(d_(2))/(d_(1))sqrt((2h)/g)`
`(d_(1))/(d_(2)-d_(1))sqrt((2h)/g)`
`(d_(2)-d_(1))/(d_(2))sqrt((2h)/g)`

Solution :Velocity of FALL of the body just before striking
`v=sqrt(2gh)`
Retarding force F = V `(d_(2)- d_(1)) g`
`therefore` Retardation a =`F/m=(V(d_(2)-d_(1))/(Vd_(1))g`
`=(d_(2)-d_(1))/(d_(1))g`
Time of fall `t=v/a=(sqrt(2gh))/((d_(2)-d_(1))/(d_(1))g)`
`=(d_(1))/(d_(2)-d_(1))xxsqrt((2h)/g)`
`therefore` Correct CHOICE is (c).
Discussion
108.

In an experiment of determine acceleration due to gravity by simple pendulum, a student commits 1% positive error in the measurement of length and 3% negative error in the time then % error in g will be

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`3%`
`4%`
`7%`
`10%`

ANSWER :C
Discussion
109.

What type of lens is formed by a bubble inside water?

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Solution :AIR bubble has spherical surface and is SURROUNDED by medium (WATER) of higher refractive index. When light passes from water to air it gets diverged. So air bubble behaves as a concave LENS.
Discussion
110.

In a standing wave on a string.

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In ONE TIME period all the particles are simultaneously at rest TWICE.
All the particles MUST be at their positive extremes simultaneously once in one time period.
All the particles may be at their positive extremes simultaneously once in a time period.
All the particles are never at rest simultaneously.

Answer :A::C
Discussion
111.

A squareframe carryinga current I = 0.90 A is located in the same plane as a long straight wire carrrying a current I_0 = 5.0 A. The frome side has a length a = 8.0 cm. The axis of the frame passing through the mid-points of the opposite sides is parallel to the wire and is separated from it by the distance which is eta = 15 times greater than the side of the frame. Find :

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Solution :(a) Force of attraction between parallel currents
`F_(1) = (mu_0)/(4pi) ( 2I I_0)/((2eta -1 )(a)/(2)`
Similar force of repulsion between antiparallel currents
`F_(2) = (mu_0)/(4pi) (4I I_0)/((2 eta + 1)a)`
Net force of repulsion between the square frame and the long straight wire
`F = F_1 - F_2 = (2 mu_0)/(pi a) (I I_0)/(2 eta^2 -1)`
Putting values,`F = 0.05 mu N.`
b) Work perfomed in turning the frame through `180^@`
`W = int_0^(pi) tau d theta,` But `d tau = bar(dM) XX bar (B)(r)`
`dM = IdA - = 2a dr n B(r) = (mu_0)/(4 pi) (2I_0)/(r) n`.
n = unit normal to the frame into the paper
`therefore d tau = dM B sin theta`
`theta` = angle between B(r) and dA during the rotation process.
`thereforetau = int d tau (mu_0)/(4 pi). 2I I_(0) sin theta int_( eta a - a//2)^(eta a+a//2) (dr//r)`
`= (mu_0)/(4pi) 2 I I_0a sin theta log_e (2eta + 1)/(2 eta-1) therefore` Workdone
` W = int_(0)^(pi) taud theta = (mu_0)/(4pi)2 I I_0 a LOG ""(2eta +1)/(2 eta - 1) sin theta d theta`
` = (mu_0)/(pi) I I_0 a log ""(2 eta + 1)/(2 eta -1) = 0.1 44 mu J`
Discussion
112.

List the uses of polaroids.

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Solution :Uses of polaroids
1. Polaroids are used in goggles and camerasto avoid glare of light.
2. Plaroids are useful in three dimensionalmotion pictures i.e. in HOLOGRAPHY.
3. Polaroids are used to IMPROVE constrast in OLE oil paintings.
4. Polaroidsare used in optical STRESS analysis.
5. Polaroids are used as window glasses to control the intensity of incoming light.
Discussion
113.

Explain AC circuit with only capacitor.

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SOLUTION :A pure capacitor connected in AC circuit is shown in figure. Capacitor is connected with ac voltage `V =V_(m) sin OMEGA t`.

When a capacitor is connected to a voltage source in a dc circuit, current will flow for the short time required to charge the capacitor.
As charge accumulates on the capacitor PLATES, the voltage across them increases opposing the current.
When the capacitor is fully charged, the current in the circuit FALLS to zero.
When the capacitor is connected to an across source, it limits or regulates the current but does not completely prevent the flow of charge.
The capacitor is alternately charged and discharged as the current reverses each half cycle.
Let q be the charge on the capacitor at any time t. The instantaneous voltage V across thecapacitor is `V = ( q)/( C )` where C is the capacitance of capacitor.
From the Kirchhoff.s loop rule,
`V _(m)sin omega t- ( q )/( c ) = 0`
`:. V_(m) sin omegat= ( q)/( C )`
`:. CV_(m) sin omega tq `
NOw `I = ( dq)/( dt )`
`:. I = ( d )/( dt) [ V_(m) C sin omega t ]`
`:. I = V_(m) C omega cos omega t `
`:. I = ( V _(m))/((1)/( omega C )) . cos omega t `
Now putting `cos omega t = sin ( omega t + ( pi )/( 2))`
`:. I = I_(m) sin ( omega t + (pi )/( 2))` .......(2)
where, `I_(m) = ( V_(m))/((1)/( omega C ))` is the amplitude of the oxcillating current.
Comparing this equation to `I_(m) = ( V_(m))/( R )`
`(1)/( omega C )` plays the role of resistance. It is called capacitance reactance and is denoted by `X_(C )`.
`:. X_(C ) = ( 1)/( omega C )`
So that the amplitude of the current is `I_(m) = ( V_(m))/( X_(C ))`
SI unit of capacitance reactanceis Ohm `( Omega )` .
The capacitive reactance limits the amlitude of the current in the purely capacitive circuit in the same way as the resistance limits the current in a purely resistive circuit.
In a pure capacitive circuit amplitude of current is inversely proportional to the frequency and the capacitance.
A comparison of `V = V(m) sin omega t ` and `I = I_(m) sin ( omega t + ( pi )/(2))` shows the current is `( pi )/(2)` ahead of voltage.
Below figure shows the phasor diagram at an instant `t_(1)`.

Here the current phasor I is `( pi )/(2)` ahead of the voltage phasor V as they ROTATE counter clockwise. ( Lead)
Figure (b) shows that variation of voltage and current with time. The current reaches its maximum value earlier than the voltage by one -fourth of a period.
One-fourth of a period.
`(T)/(4) = ( T )/( 2pi ) xx ( 2pi )/( 4) = ( (pi )/( 2)) /( omega ) = ( pi )/( 2 omega )`
Discussion
114.

Assertion:An induced emf is generated when magnet is withdrawn from the solenoid. Reason: The relative motion between magnet and solenoid induces emf.

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If both assertion and REASON are t rue and the reason is the CORRECT explanation of the assertion.
If both assertion and reason are true but reason is not the correct explanation of the assertion
If assertion is true but reason is FALSE.
If the assertion and reason both are false

Answer :A
Discussion
115.

Whichofthe four the graphsmay bestrepresentthecurrent-deflectionrealationin a tangent galvanormetre?

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ANSWER :C
Discussion
116.

The momentum of a body is numerically equal to its kinetic energy. The velocity of the body is :

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`1 MS^(-1)`
`2 ms^(-1)`
`4 ms^(-1)`
`8 ms^(-1)`

Solution :K.E. = momentum.
`:. 1/2mv^(2)=MV`
`implies v=2 ms^(-1)`
Discussion
117.

The time period of a simple pendulum of infinte length is

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INFINITE
`2pisqrt(R//g)`
`2pi sqrt(g//R)`
`(1)/(2pi) sqrt(R//g)`

ANSWER :B
Discussion
118.

Radio waves are electromagnetic waves used for communication purpose. In standard radio broadcasting which modulation is commonly used.

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SOLUTION :AMPLITUDE MODULATION
Discussion
119.

An object is kept at a distance 40 cm from the pole of a mirror and an inverted image 5 times the size of object is formed. Find the nature and the focal length of the mirror.

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SOLUTION :CONCAVE `(100)/(3)` CM
Discussion
120.

In Figassuming the diodes to be ideal :

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`D_(1)` is forward biased and `D_(2)` is reverse biased and HENCE current FLOW form A to B
`D_(2)` is forward biased and `D_(1)` is reverse biased and hence no current flows from B to A and vice versa
`D_(1) and D_(2)` are both forward biased and hence current flows from A to B
`D_(1) and D_(2)` areboth reverse biased and hence no current flows from A to B and vice-versa

Answer :B
Discussion
121.

Electric fuse is a protective device used in series with an electric circuit or an electric appliance to save it from damage due to overheating porduced by strong current in the circuit or application. Fuse wire is generally made from an alloy of lead and tin, which has hifh resistance and low meltingpoint. It is connected in series in an electric installation. If a circuit gets accidentally short-circuited, a large current flows, then fuse wire melts away, which causes a break in the circuit. The power through fuse (F') is equal to heat energy lost per unit area per unit time (h) (neglecting heat loses from ends of the wire). P = I^2R = h xx 2pirl [R = (rhol)/(pir^2)] where r and l are the length and radius of fuse wire, respectively. A Battery is described by its emf (E) and internal resistance (r). Efficiency of battery (eta) is defined as the ratio of the output to the input power. eta = (Output power)/(Input power) xx 100% But I= E//(R+r),input power = EI. Output power = EI - I^2r, then. eta = ((EI - I^2r)/(EI)) xx 10 (1-(Ir)/E) xx 100 =1 - (E/(R+r)) (r/E) xx 100 = (R/(R+r)) xx 100 We know that output power of a source is maximum when the external resistance is equal to internal resistance, i.e., R = r . Two fuse wires of same potential material have length ratio 1:2 and ratio 4:1. Then respective ratio of their current rating will be

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`8:1`
`2:1`
`1:8`
`4:1`

Solution :a. `I prop r^(3//2)`
`I_1/I_2 = (4R)^(3//2)/(r^(3//2) = 8/1`
`P = v^2/ R or 2 XX 10^3 = V^2/20`
or `V = 200V`
`ETA = R/(R+r) xx 100 `, where R = r, EFFICIENCY will be MAXIMUM.
Discussion
122.

Electric fuse is a protective device used in series with an electric circuit or an electric appliance to save it from damage due to overheating porduced by strong current in the circuit or application. Fuse wire is generally made from an alloy of lead and tin, which has hifh resistance and low meltingpoint. It is connected in series in an electric installation. If a circuit gets accidentally short-circuited, a large current flows, then fuse wire melts away, which causes a break in the circuit. The power through fuse (F') is equal to heat energy lost per unit area per unit time (h) (neglecting heat loses from ends of the wire). P = I^2R = h xx 2pirl [R = (rhol)/(pir^2)] where r and l are the length and radius of fuse wire, respectively. A Battery is described by its emf (E) and internal resistance (r). Efficiency of battery (eta) is defined as the ratio of the output to the input power. eta = (Output power)/(Input power) xx 100% But I= E//(R+r),input power = EI. Output power = EI - I^2r, then. eta = ((EI - I^2r)/(EI)) xx 10 (1-(Ir)/E) xx 100 =1 - (E/(R+r)) (r/E) xx 100 = (R/(R+r)) xx 100 We know that output power of a source is maximum when the external resistance is equal to internal resistance, i.e., R = r . The maximum power rating of a 20.0 Omega fuse wire is 2.0 kW.This fuse wire can be connected safely to a DC source (negligible internal resistance) of

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300 V
190V
250 V
220 V

Solution :B. `I prop r^(3//2)`
`I_1/I_2 = (4r)^(3//2)/(r^(3//2) = 8/1`
`P = v^2/ R or 2 xx 10^3 = V^2/20`
or `V = 200V`
`eta = R/(R+r) xx 100 `, where R = r, EFFICIENCY will be MAXIMUM.
Discussion
123.

Show that the avearge energy density of the electric field vecE equals the average energy density of the magnetic field vecB, in electromagnetic waves.

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SOLUTION :`mu_e=1/2e_0E^2 & u_B=1/2 B^2/mu_0`
`mu_E=1/2E_0E^2=1/2 e_0(CB)^2 As c=E/B`
`=1/2 e_0 B^2/(mu_0e_0) c=1/sqrt(mu_0E_0)`
`=B^2/(2mu_0)`
`mu_B`
Discussion
124.

How are each of the following quantities of a SHM affected by doubling the amplitude ? Time period

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SOLUTION :TIME PERIOD is INDEPENDENT of the AMPLITUDE
Discussion
125.

An electron of mass m_(e) and a proton of mass m_(p) are moving with thesame speed. The ratio of their deBroylie's wavelength lamda_(e)//lamda_(p) is :

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1
1836
`(1)/(1836)`
918

Solution :de-Broglie.s WAVELENGTH,
`lambda=(h)/(mv)`
For electron,
`lambda_(E)=(h)/(m_(e)v)`
for photon
`lambda_(p)=(h)/(m_(p)v)`
`:.(lambda_(e))/(lambda_(p))=(h//m_(e)v)/(h//n_(p)v)`
`(lambda_(e))/(lambdal_(p))=(m_(p))/(m_(e))`
`(lambda_(e))/(lambda_(p))=(1.67xx10^(-27))/(9.1xx10^(-31))`
Discussion
126.

A ball is dropped from the top of a tower 100m high. Simultaneously another ball is thrown upward with a speed of 50m/s . After what time do they cross each other ?

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1 s
2 s
3 s
4 s

Answer :B
Discussion
127.

The deceleration experienced by a moving motorboat, after it's engine is cut of is given by dv/dt = -kv^3 , where k is constant.If v_0 is the magnitude of the velocity at cut off the magnitude of velocity at time t after cut off is

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`v_0/2`
`v_0`
`v-0 E(-KT)`
`v_o/sqrt ((v_o)^2 kt + L)`

ANSWER :D
Discussion
128.

Moseley.s law states that

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`sqrtv=a(z-b)`
`v=a^(2)(z-b)`
`v=sqrt(a(z-b))`
`sqrtv=a(z-b)^(2)`

ANSWER :A
Discussion
129.

Two inductors L_(1) (inductance 1 mH, internal resistance 3Omega)and L_(2) (inductance 2 mH, internal resistance 4Omega), and a resistance R (resistance 12Omega)are all connected in parallel across a 5V battery . The circuit is switched on at time t=0. The ratio of the maximum to the minimum current (I_(max)//I_(min)) drawn from the battery is :

Answer»


ANSWER :8
Discussion
130.

A sinusoidal voltage of rms value 220V is applied to a diode and a resistor R in the circuit shown in fig. Show that half wave rectifaction occurs. If the diode is ideal, what is the rms voltage across?

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ANSWER :`110sqrt(2)V`
Discussion
131.

Assume (unrealistically ) that a TV station acts as a point source broadcasting isotropically at 1.0 MW. What is the intensity of the transmitted signal reaching proxima Centauri the star nearest our solar system 4.3 ly away ? A light -year (ly) is the distance light travels in one year .

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SOLUTION :`4.8xx10^(-29) W//m^(2)`
Discussion
132.

The equation of wave travelling in a medium is y = 5 cos pi (100t -x). The wavelength of the wave is

Answer»

1 UNIT
3 unit
5 unit
2 unit

Answer :D
Discussion
133.

A 2 H coil, a 10 μF resistor, and a 5 V battery are all connected in parallel. After a current is well established through the resistor and inductor, the battery is disconnected from the circuit by opening a switch as shown in the following figure. Immediately after the battery is disconnected, Lenz's law would predict

Answer»

The voltage across R is zero
The current through R reverses direction
The current through the INDUCTOR reverses direction, but its magnitude TENDS to remain the same
The magnitude and direction of the voltage across the inductor tend to remain the same

Answer :B
Discussion
134.

Figure shows a student, again sitting on a stool that can rotate freely about a vertical axis. The student, initially at rest, is holding a bicycle wheel whose rim is loaded with lead and whose rotational inertia I_(wh) about its central axis is 1.2kg*m^(2). The wheel is rotating at an angular speed omega_(wh) of 3.9 rev/s, as seen from overhead, the rotation is counterclockwise. The axis of the wheel is vertical, and the angular Momentum vecL_(wh) of the wheel points vertically upward. The student now inverts the wheel so that, as seen from overhead, it is rotating clockwise. Its angular momentum is now -vecL_(wh). The inversion results in the student, the stool, and the wheel's center rotating together as a composite rigid body about the stool's rotation axis, with rotational inertia I_(b)=6.8kg*m^(2). With what angular speed omega_(b) and in what direction does the composite body rotate after the inversion of the wheel?

Answer»

Solution :1. The angular speed `omega_(b)` we seek is related to the final angular MOMENTUM `vecL_(b)` of the composite body about the stool.s ROTATION AXIS by `(L=Iomega)`.
2. The initial angular speed `omega_(wh)` of the wheel is related to the angular momentum `vecL_(wh)` of the wheel.s rotation about its center by the same equation.
3. The vector ADDITION of `vecL_(b)andvecL_(wh)` gives the total angular momentum `L_("tot")` of the system of the student, stool, and wheel.
4. As the wheel is inverted, no net external torque acts on that system to change `vecL_("tot")` about any vertical axis. So, the system.s total angular momentum is conserved about any vertical axis, including the rotation axis through the stool.
Calculations: The conservation of `vecL_("tot")` is represented with vectors in Fig. We can also write this conservation in terms of components along a vertical axis as
`L_(bf)+L_(wh.f)=L_(b,i)+L_(wh,i)`.
Where i and f refer to the initial state and the final state. Because inversion of the wheel inverted the angular momentum vector of the wheel.s rotation, we substitute `-L_(wh,i)" for "L_(wh,f)`. Then, if we set `L_(b,i)=0`.
`L_(b,f)=2L_(wh,i)`.
Using Equation, we next substitute `I_(b)omega_(b)" for "L_(b.f)andI_(wh)omega_(wh)" for "L_(wh,i)` and solve for `omega_(b)`, finding
`omega_(b)=(2I_(wh))/I_(b)omega_(wh)`
= `((2)(1.2kg*m^(2))(3.9rev//s))/(6.8kg*m^(2))=1.4rev//s`.
Discussion
135.

Self energy of conducting sphere of radius .r. carrying charge .Q. is ______

Answer»

`(Q^(2))/(8pi epsilon_(0)r)`
`(Q^(2))/(4PI epsilon_(0)r)`
`(Q^(2))/(6PI epsilon_(0)r)`
`(Q^(2))/(2PI epsilon_(0)r)`

Answer :A
Discussion
136.

Capacitance of a parallel plate capacitor can be increased by

Answer»

a) increasing the distance between the plates
b) increasing the THICKNESS of the plates
C) increasing the thickness of the plates
d) decreasing the distance between the plates

ANSWER :A::B::C::D
Discussion
137.

Answer the following: (which help you understand the difference between Thomson's model and Rhutherford's model better). (a) Is the average angle of deflection of alpha particles by a thin gold ofil predicted by Thomson's model much less, about the same or much greater than that predicted by Rutherford's atom model? (b) Is the probability of backward scattering (i.e., scattering of alpha particles at angles greater than 90^(@)) predicted by Thomson's model, much less, about the same, or much greater than that predicted by Rutherford's model? (c) Keeping other factors fixed, it is found experimentally,that for a small thickness t, the number of alpha particles scattered at moderate angles is propotional to t. What clues does this linear dependence on t provide? (d) In which atom model, is it completely wrong to ignore multiple scattering for the calculation of average angle of scattering of alpha particles by a thin ofil?

Answer»

Solution :(a) About the same. This is because we are talking of average ANGLE of deflection.
(b) Much less, because in Thomson's model, there is no such massive central core called the nucleusas in Rutherford's model.
(c) This suggests that scattering is predominantly due to a single collision, because chance of a single collision increase with the NUMBER of target atoms which increases linearly with the thickness of the ofil.
(d) In Thomson model, positive charge is uniformaly DISTRIBUTED in the spherical atom. Therefore, a single collision causes very little deflection. Therefore, average scattering angle can be explained only by considering multiple scattering. Thus it is WRONG to ignore multiple scattering in Thomson's model.
On the CONTRARY, in Rutherford's model, most of the scattering comes form a single collision. Therefore, multiple scattering may be ignored as a first approximation.
Discussion
138.

A thick hollow metallic sphere of Inner and outer radius 2a and 4a, respectively, encloses a sold concentric metallic sphere of radius a'. These spheres are connected to the earth through switches S_1 and S_2 as shown in the figure First, switch S_1 is closed and after some time S_1 is opened. Then, S_2 is closed and after some time S_2 is opened. This completes one cycle. Initially, charge Q is given to the outer sphere with both S_1 and S_2 open. The charge on the inner sphere after n cycles of earthing the inner and outer spheres is

Answer»

`-Q/3`
`-Q/3^n`
`-(2/3)^2Q`
`Q/3^n`

ANSWER :B
Discussion
139.

The work function for the following metals is given: Na:2.75eV,K:2.30eV,Mo:4.17eV,Ni:5.15eV. Which of these metals will not give photoelectric emission for a radiation of wavelength 3300 Å from a He-Cd laser placed 1 m away from the photocell ? What happens if the laser is brought nearer and placed 50 cm away ?

Answer»

Solution :Here `lamda=3300Å=3300xx10^(-10)m=3.3xx10^(-17)m`
`therefore`Energy of radiation photon in eV
`E=(hc)/(elamda) eV=(6.63xx10^(-34)x3xx10^(8))/(1.6xx10^(-19)xx3.3xx10^(-7))eV=3.767eV`
For photoelectric emission energy of radiation photon must be greater than or equal to the work function of given metal. therefore, we conclude that in present CASE photoelectric emission from NA and K is possible but from Mo and Ni no photoelectric emission will take place.
There is no effect on photoelectric emission from Mo and Ni even when laser source is brought nearer to the PHOTOCELL. HOWEVER, on bringing the source nearer, the intensity of radiation increases. As a result, in case of Na and K, the value of photoelectric current will rise in proportion to the rise in intensity of radiation.
Discussion
140.

A conducting shell having no charge has radius R. A point charge Q is placed in front of it at a distance r_0 from its centre. Find potential due to charge induced on the surface of the shell at a point P inside the shell. Distance of point P from point charge Q is r.

Answer»


ANSWER :`(KQ)/(r_(0))-(KQ)/(R)`
Discussion
141.

एक अवतल दर्पण में वस्तु (बिम्ब) की स्थिति फोकस और ध्व बे ठीच हो, तो प्राप्त प्रतिबिंब होगा

Answer»

वास्तविक और बड़ा
वास्तविक और छोटा
आभासी (काल्पनिक) और बड़ा
आभासी और छोटा

Answer :C
Discussion
142.

Show that the total energy is conserved during LC oscillations.

Answer»

Solution :i. In inductors, the energy is stored in the form of magnetic field while in capacitors , it is stored as the electric field.
ii. Whenever energy is given to a circuit containing a pure INDUCTOR of inductance L and a capacitor of capacitance C, the energy oscillates back and FORTH between the magnetic field of the capacitor. Thus the electrical osicillations of definite frequency are generated. These oscillations are CALLED LC oscillations.
Generation of LC oscillations
i. Let us assume that the capacitor is fully charged with maximum charge Qm at the initial stage. So that the energy stored in the capacitor is maximum and is given by `U_(E)=(Q_(m)^(2))/(2C)` As there is no current in the inductor, the energy stored in it is zero i.e.,`U_(B)=0.` Therefore, the total energy is wholly electrical. This is shown in Figure.
ii. The capacitor now begins to dischange through the inductor that establishes current i in clockwise direction. This current produces a magnetic field around the inductor and the energy stored in the inductor is given by `U_(B)=(Li^(2))/(2).` As the charge in the capacitor decreases, the energy stored in it also decreases and is given by `U_(E)=(Q^(2))/(2C).`
iii. Thus there is a transfer of some part of energy from the capacitor to the inductor. At that instant, the total energy is the sum of electrical and magnetic energies (figure).
iv. When the charges in the capacitor are exausted, its energy becomes zero i.e., `U_(E)=0.` The energy is fully transfered to the magnetic field of the inductor and its energy is maximum.
v. This maximum energy is given by `U_(B)=(LI_(m)^(2))/(2)` where Im is the maximum current flowing in the circuit. The total energy is wholly magnetic (Figure(c)).
vi. Even though the charge in the capacitor is zero, the current will continue to flow in the same direction because the inductor will not allow it to stop immediately.
vii. The current is made to flow with decreasing magnitude by the collapsing magnetic field of the inductor.
viii. As a result of this, the capacitor begins to charge in the opposite direction. A part of the energy is transfered from the inductor back to the capacitor. The total energy is the sum of the electrical and magnetic energies (Figure (d)).
ix. When the current in the circuit reduces to zero, the capacitor becomes fully charged in the oppsite direction.

x. The energy stored in the capacitor becomes maximum. Since the current is zero, the energy stored in the inductor is zero. The total energy is wholly electrical (Figure (e)).
xi. The state of the circuit is similar to the initial state but the difference is that the capacitor is charged in opposite direction. The capacitor then startto discharge through the inductor with anti-clockwise current. The total energy is the sum of the electrical and magnetic energies (Figure (f)).
xii. As already explained, the PROCESS are repeated in opposite direction (Figure (g) and (h)). Finally, the circuit returns to the intial state (Figure (a)). Thus, when the circuit goes through these stages, an alternating current flows in the circuit. As this process is repeated again and again, the electrical oscillations of definite frequency are generated. These are known as LC oscillations.
Discussion
143.

What is the polarizing angle of a medium of refractive index sqrt3?

Answer»

SOLUTION :`n=sqrt3`
`n=tantheta`
`theta=tan^(-1)n=tan^(-1)sqrt3=60^@`
Discussion
144.

A disc of mass m_(0) rotates freely about a fixed horizontal axis passing through its centre. A thin cotton pad is fixed to its rim, which can absorb water. The mass of water dripping onto the pad per unit time is mu. After what time will the angular velocity of the disc get reduced to half its initial value?

Answer»

`(2m_(0))/(MU)`
`(3m_(0))/(mu)`
`(m_(0))/(mu)`
`(m_(0))/(2MU)`

ANSWER :D
Discussion
145.

If the electron drift speed is so small(~ 10 ^(-3)m//s)and the electron’s charge is very small, how can we still obtain a large amount of current in a conductor.

Answer»

Solution :The electronnumberdensity is ofthe order of`10 ^(29) m^(-3) , RARR`thenet currentcan beveryhighevenifthe driftspreadis low.
Discussion
146.

Four photon of lambda gt lambda_(0) are incident on metal surface ……..photo -electron will be emitted.

Answer»

Zero
One
Two
Four

Solution :The condition for the EMISSION of photo-electrons
`vgtv_(0)`
`therefore LAMBDA LT lambda_(0)[because lambda=(c )/(v)]`
`therefore` photo-electron will not be EMITTED,
Discussion
147.

The specific heat c of a solid at low temperature shows temperature dependence according to the relation c = DT^(3)where D is a constant and T is the temperature in kelvin. A piece of this solid of mass m kg is taken and its terriperature is raised from 20 K to 30 K. The amount of heat required in the process in energy units

Answer»

`5 XX 10^(4) Dm`
`(33//4) xx 10^(4) Dm`
`(65//4) xx 10^(4) Dm`
`(5//4) xx 10^(4) Dm`

Solution :The amount of heat supplied or gained by a body is directly proportional to mass ofbody and amount of change in temperature , i.e., `Delta Q = mc Delta T`
Therefore , amount of heat supplied or gaind `Q = INT Delta Q = underset(T_(1)) OVERSET (T_(2)) int mc Delta T `
Here, `c = DT^(3) , T_(1) = 20 K " and " T_(2) = 30 K`
` :. Q = underset(20)overset(30) int mDT^(3) dT = mD underset(20)overset(30) int T^(3) dT = mD [T^(4)/4]_(20)^(30)`
` = (mD)/4 [(30)^(4) - (20)^(4)] = (mD)/4 xx 10^(4) [81 - 16]`
` = (65/4) xx 10^(4) Dm`
Discussion
148.

A string that is 10 cm long is fixed at both ends. At t = 0, a pulse travelling from left to right at 1 cm/s is 4.0 cm from the right end as shown in figure. Determine the next two times when the pulse will be at that point again. State in each case whether the pulse is upright or inverted.

Answer»

Solution :8 sec, INVERTED, 20 sec UPRIGHT
Discussion
149.

The angular magnification of a simple microscope can be increased by

Answer»

INCREASING the focal length of the LENS
increasing the size of the OBJECT
increasing the apertureof the lens
increasing thepower of the lens

Answer :D
Discussion
150.

What is the kinetic energy of a baseball (mass=0.15kg) moving with a speed of 20 m/s?

Answer»

SOLUTION :From the definition,
`K=(1)/(2)mv^(2)=(1)/(2)(0.15kg)(20m//s)^(2)=30J`
Discussion