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4651.

A metal of work function 4eV is exposed to a radiation of wavelength 140xx10^(-9) m. Find the stopping potential developed by it.

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Solution :`E=(HC)/(lamda)`
`E=(6.62xx10^(-34)xx3xx10^(8))/(140xx10^(-9)xx1.6xx10^(-19))eV=8.86eV`
work function `W_(0)=4eV`
`eV_(0)=E-W_(0)=8.86-4=4.86V`
`:.` Stopping POTENTIAL `V_(0)=4.86V`
4652.

A thin prism P_(1) with angle 4^(@) and made for glas of refractive index 1.54 is combined with another thin prism P_(2) made from glass refractive index 1.72 to produce dispersion without deviation. The angle of prism P_(2) is

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ANSWER :`3`
4653.

A thin plano-convex lens.of focal length f is split into two halves : One of the halves is shifted along the optical axis (figure)The separation between object and image planes is 1.8m.The magnification of the image formed by one of the half lensis 2. Find the focal length of the lens and separation between the two halves. Draw the ray diagram for image formation.

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ANSWER :40 CM, 60 cm
4654.

When a dielectric is introduced between two charges the electric field between them

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Increases
Remains the same
Decreases
Becomes infinte

Answer :C
4655.

A boy sitting in a car moving at constant velocity throws a ball straight up into the air. Where will the ball fall ?

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Behind him
Into his hands
In FRONT of him
Towards the left

Solution :Due to inertia of MOTION the ball covers the same horizontal distance as is covered by the car.
Hence CORRECT choice is (b).
4656.

67^(@)C is equal to ………………… K

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`298 K`
`287 K`
`308 K`
`340 K`

ANSWER :D
4657.

An almost inertia-less rod of length l3.5 m can rotate freely around a horizontal axis passing through its top end. At the bottom end of the roa a small ball of mass m andat the mid-point another small ball of mass 3m is attached. Find the angular frequency (in SI units) of small oscillations of the system about the equilibrium position. Gravitational acceleration is g=9.8 m//s^(2).

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`2.0 rad//s`
`2.5 rad//s`
`3.0 rad//s`
`3.5 rad//s`

ANSWER :C::D
4658.

When two capacitors C_1 and C_2 are connected in parallel, the ratio of charges and potential differences across two capacitors are ____________ and ___________ respectively.

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SOLUTION :`C_1 : C_2, 1 : 1`.In PARALLEL grouping of CAPACITORS, charges on the capacitors are directly proportional to capacitances and POTENTIAL difference across all the capacitors is same.
4659.

ABC is an equilateral triangle of side 10 m and D is the midpoint of BC. Charges of +100 C, -100 C " and +75 C are placed at B, C and D respectively. What is the force experienced by + cC charge placed at point A ?

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ANSWER :`1.27 XX 10^(10)N,`"at aangle of" `45^(@)`" with the HORIONTAL"
4660.

Name the parts of the electromagnetic spectrum which is used to treat muscular strain. Write in brief, how these waves can be produced.

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Solution :INFRAREDinfrared are PRODUCED by the vibrating MOLECULES and atoms in HOT BODIES.
4661.

If two straight current carrying wires are kept perpendicular to each otheralmost touching then the wires

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ATTRACT each other
Repel each other
Remain stationary
Become PARALLEL to each other

Answer :D
4662.

In the figure shown two loops ABCD & EFGH are in the same plane.The smaller loop carries time varying current I=b t, where b is a positive constant and t is time.The resistance of the smalller loop is r and that of the larger loop is R.:(Neglect the self inductance of large loop) The magnetic force on the loop EFGH due to loop ABCD is (mu_(0)^(2) Iab)/(x pi^(2)R)ln 4/3.Findout value of x.

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Solution :`i'=(EMF)/R=-(dphi)/(Rdt)=-(dintBds)/(Rdt)=(mu_(0)AB)/(2piR) "ln" 4/3rArrF=intBIdl=(mu_(0)^(2)Iab)/(12pi^(2)R)"ln"4/3`
4663.

Flint glass lens is made of refractive index 1.5. When it is placed in a liquid of refractive index 1.25, then its focal length will be ....

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1.25 f
2.5 f
1.2 f
1.3 f

Solution :Let, FOCAL LENGTH is f,
`therefore 1/f=(""_amu_g-1)((1)/(R_1)-(1)/(R_2))`
`=(1.5-1)((1)/(R_1)-(1)/(R_2))`
`1/f=0.5((1)/(R_1)-(1)/(R_2))`
`therefore (1)/(0.5 f)=(1)/(R_1)-(1)/(R_2)`…...(1)
Let, after INSERTING it in LENS, it BECOMES f.,
`therefore (1)/(f)=(""_lmu_g-1)((1)/(R_1)-(1)/(R_2))`
`((1.5)/(1.25)-1)((1)/(R_1)-(1)/(R_2))`
`(1)/(f.)=((0.25)/(1.25))xx(1)/(0.5f)`
`therefore(1)/(f.)=(1)/(2.5f)`
`therefore f. =2.5` f
4664.

Find the angular momentum of an oxygen molecule whose rational energy is E=2.16 meV and the distance between the nuclei is d=121p m.

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Solution :The angular momentum is `SQRT(2E)=M`
Now `I=(md^(2))/(4) (m = "mass of" O_(2) "molecule")=1.9584xx10^(-39) GMC^(2)`
So `M= 3.68xx10^(-27) ERG sec.=3.49 ħ`
(This corresponding to `J=3)`
4665.

Find the Rydberg correction for the 3P term of a Na atom whose first exitation potential is 2.10V and whose valance electron in the normal 3S state has the binding energy 5.14 eV.

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Solution :The ENRGY of the `3p` state must be `-(E_(0)-E varphi)` where `-E_(0)` is the ENERGY of the `3S` state. Then
`E_(0)-evarphi_(1)=( ħR)/((3+alpha_(1))^(2))`
so `alpha_(1)=sqrt(( ħR)/(E_(0)-e varphi_(1)))-3 =-0.885`
4666.

A box is projected up a long ramp (incline angle with the horizontal=30^(@)) with an initial speed of 8m/s. if the surface of the ramp is very smooth (essentially frictionless), how high up the ramp will the box go? What distance along the ramp will it slide?

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Solution :Because friction is negligible, we can apply conservation of mechanical ENERGY. Calling the BOTTOM of the ramp our H=0 reference level, we write
`K_(i)+U_(i)=K_(f)+U_(f)`
`(1)/(2)mv_(0^(2)+0=0+mgh`
`h=((1)/(2)v_(0)^(2))/(g)` ltBrgt `=((1)/(2)(8m//s)^(2))/(10m//s^(2))`
=3.2m
Since the INCLINE angle is `theta=30^(@)`, the distance, d, it slides up the ramp is found in the following way.

`h=dsintheta`
`d=(h)/(sintheta)=(3.2m)/(sin30^(@))=6.4m`.
4667.

The energy of the fast neutron released in the nuclear fission process is almost?

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2MeV
2 KeV
10 MeV
20 MeV

Solution : KNOWLEDGE BASED QUESTION
4668.

For the circuit shown in the figure. The equivalent resistance between point A and Bfor the two cases (i) V_A > V_B, (ii) V_B > V_A respectively is……….Omega and ……….Omega respectively. (D_1 and D_2 are ideal diodes)

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`25, OO`
`50 , oo`
`oo, 25`
`25,25`

ANSWER :A
4669.

A fish in a water tank sees the outside world as if it (the fish) is at the vertex of cone such that the circular base of the cone coincides with the surface of water. Given the depth of water where fish is located being h and the critical angle for water - air interface being i_(C), find out by drawing a suitable ray diagram the relationship between the radius of the cone and the height h.

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ANSWER :`R=(h)/(sqrt(N^(2)-1))`
4670.

According to Bohr's atomic model the radii of stationary orbits are directly proportional to cube of the quantum number n.

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SOLUTION :RADII of variousorbits are directly proportional to SQUARE of the QUANTUM number.n..
4671.

Proceeding from the fundamental equation of relativistic dynamics, find: (a) under what circumstances the acceleration of a particle coincides in direction with the force F acting on ti, (b) the proportionality factors relating the force F and the acceleration w in the cases when F_|_v and F||v, where v is the velocity of the particle.

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SOLUTION :`vecF=(d)/(DT)((m_0vecv)/(SQRT(1-v^2/c^2)))=m_0(vecv)/(sqrt(1-v^2/c^2))+m_0vecv/c^2vecv*vecv(1)/((1-v^2/c^2)^(3//2))`
Thus `vecF_(_|_)=m_0(vecw)/(sqrt(1-beta^2)), vecw=vecv, vecw_|_vecv`
`vecF_(||)=m_0(vecw)/((1-beta^2)^(3//2)), vecw=vecv, vecw_(||)vecv`
4672.

Two waves having intensities in the ratio 9:1 produce interference. What will be the ratio of intensity of maximumto minima ?

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Solution :LET `I_1` and `I_2` be the INTENSITIES of two BEAMS having amplitudes `a_1` and `a_2.`
`I_1/I_2=9/1`
But,`I_1/I_2=a_1^2/a_2^2 `
`therefore a^2_1/a_2^2=9/1` or, `a_1/a_2=3/1`
If `a_max` and `a_min` are the amplitudes are MAXIMA and minima.
`a_max/a_min =a_1+a_2/a_1-a_2=3+1/3-1=4/2=2/1`
`I_max/I_min=a^2_max/a^2_min=4/1`
4673.

A jet plane passes over you at a height of 4800 m and a speed of Mach 1.5. (a) Find the Mach cone angle (the sound speed is 331 m/s). (b) How long after the jet passes directly overhead does the shock wave reach you?

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SOLUTION :`(a) 42^@ , (B) 11 s`
4674.

Time taken by the light to travel through x cm of air. R.I. of glass is 1.5, then x is

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7.5 cm
1.33 cm
9 cm
6 cm

Solution :`d_(G) = 5CM, d_(a) x cm, t_(g) = t_(a), mu_(g) = 1.5, x = ?`
When TIME is same then
`mu_(g) = (d_(a))/(d_(g))`
`:. d_(a) = mu_(g) xx d_(g) = 1.5 xx 5 = 7.5 cm`
4675.

Due to a vertical temperature gradient in the atmosphere, the index of refraction varies. Suppose index of refraction varies as n=n_(0)sqrt(1+ay), where n_(0) is the index of refraction at the surface and a=2.0xx10^(-6)m^(-1) . A person of height h=2.0 m stands on a level surface. Beyond what distance will he not see the runway?

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Solution :As refractive index is changing along y-direction, we can assum a number of thing layers of AIR PLACED parallel to x-axis. Let O be the distant object just visible to the man. Consider a layer of air at a distance y from the ground. Let P be a point on the trajectory of the ray. Form Fig. , `theta =90^(@)-i` .
The slope of trangent at point P is `tan theta=dy//dx =cot i` .
From SNELL's law, n sin i= constant
At the surface, `n =n_(0)` and `i= 90^(@)` .
`n_(0)sin90^(@)=nsini=(n_(0)sqrt(1+ay))sini`
`sin i=(1)/(sqrt(1+ay))rArr cot i=(dy)/(dx)=sqrt(ay)`
`int_(0)^(y)(dy)/(sqrt(ay))=int_(0)^(x)dxrArr x=2sqrt((y)/(a))`
On substituting `y=2.0 m` and `a=2xx10^(-1)` , we have
`x_(max)=2sqrt((2)/(2xx10^(-6)))=2000m`
4676.

A road runs between two parallel rows of buildings. A motorist moving with speed of 36 km/h sounds the horn. He hears the echo one second after he sounded the horn. If speed of sound is 330 m/s then the distance between two rows of buildings is :

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330m
`2sqrt((165)^(2)-(5)^(2))m`
165M
`SQRT((165)^(2)-(5)^(2)m)`

ANSWER :B
4677.

Figure 8.6 shows a capacitor made of two circular plates each of radius 12 cm, and separated by 5.0 cm. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 0.15A. (a) Calculate the capacitance and the rate of change of potential difference between the plates. (b) Obtain the displacement current across the plates. (c) Is Kirchhoff’s first rule (junction rule) valid at each plate of the capacitor? Explain.

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Solution :(a) `C=epsi_(0)A//d =80.1 PF`
`(dQ)/(dt) =C ""(dV)/(dt)`
`(dV)/(dt) =(0.15)/(80.1xx10^(-12))=1.87xx10^(9) V s^(-1)`
(b) `i_(d) =epsi_(0) ""(d)/(dt) Phi _(E)`. Now across the capacitor `Phi_(E)=EA`, ignoring and corrections.
THEREFORE, `i_(d)=epsi_(0)A""(d Phi_(E))/(dt)`
Now, `E=(Q)/(epsi_(0)A)`. Therefore, `(dE)/(dt)=(i)/(epsi_(0) A)`, which implies `i_(d)=i=0.15 A.`
(c) Yes, provided by ‘current’ we mean the sum of conduction and DISPLACEMENT currents.
4678.

The graph shown here, shows the variation of the total energy ( E) stored in a capacitor against the value of the capacitance (C ) itself. Which of the two the charge on the capacitance or the potential used to charge it is kept constant for this graph?

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SOLUTION :Energy stored in a capacitor,
`E=(Q^2)/(2C)= (1)/(2)CV^2`
The NATURE of the graph suggests that it is a plot of the equation,
`E= (Q^2)/(2C)`
Hence in this graph the CHARGE on the capacitor (Q) is kept constant.
4679.

The best ideal black body is

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lamp of charcoal HEATED to a HIGH temperature
metal COATED with a black dye
glass surface coated with colter
hollow ENCLOSURE blackened inside and having a SMALL hole

Answer :D
4680.

Obtain the resonant frequency omega_(r) of a series LCR circuit with L = 2.0 H, C = 32 muF and R = 10 Omega.What is the Q-value of this circuit?

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SOLUTION :Here, L = 2.0 H, C = 32 `muF= 32 xx 10^(-6)` F and `R = 10 OMEGA`
`therefore` Resonant frequency `omega_(r) =1/sqrt(LC) = 1/sqrt(2.0 xx 32 xx 10^(-6)) = 125 s^(-1)`
and Q-value `=(Lomega_(r))/R = (2.0 xx 125)/10 = 25`
4681.

P - V diagram of an ideal gas is shown. The gas undergoes from initial state A to final state B such that initial and final volumes are same. Select the correct alternative for given process AB.

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WORK DONE by GAS is positive
Work done by gas is negative
Temperature of gas INCREASES continuously
Process is isochoric

Answer :B
4682.

Three pure inductances each of 3H are connected as shown in figure. The equivalent inductance between points A and B is

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1H
2H
3H
9H

Answer :A
4683.

At temperature 300 K number density of electrons and holes in pure silicon is 1.5xx10^(16)m^(-3). The number density of hole increase by 4.5xx10^(22)m^(-3) dopping the indium impurity so calculate number denstiyof electron.

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SOLUTION :`n_(E )=5XX10^(9)m^(-3)`
4684.

Two charges 2 c and 6 c are separated by a finite distance. If a charge of - 4c is added to each of them, the initial force of 12 xx 10^3N will change to

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`4 xx10^3N `repulsion 
`4 xx10^2N` repulsion
`6 xx10^3N `ATTRACTION 
`4 XX 10^3N `attraction 

ANSWER :D
4685.

Aprojectile thrown at an angle 25^@ has a certain range. The other angle of projection having the same range for the velocity of projection will be

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`20^@`
`35^@`
`65^2`
`155^@`

ANSWER :C
4686.

In the figure shown, there is friction between the blocks P and Q but the constact between the block Q and lower surface is frictionless. Initially the block Q with block P over it lies at x = 0, with spring at its natural length. The block Q is pulled to right and then released. As the spring -blocks system undergoes SHM with amplitude A, the block P tends to slip over Q, P is more likely to slip at

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X=0
`x = +A`
`x = + (A)/(2)`
`x = + (A)/(SQRT2)`

ANSWER :B
4687.

A transformer has 2100 tums in primary and 4200 turns in secondary. An' ac source of 120V, 10A is connected to its primary. The secondary voltage and current are-

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240V, 5A
120V, 10A
240V, 10A
120V, 20A

Answer :A
4688.

What is the radius of gyration of a thin uniform rod ofb length l about an axis perpendicular to its length and passing through its centre?

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`l/sqrt12`
`l/sqrt3`
`l/sqrt2`
`l/sqrt5`

ANSWER :C
4689.

An object falls from the car and the observer in the train notices that the car has moved on for one minute, turned back, and moved with a speed of 10 km/h and picked up the object two minutes after turning. Find (a) the velocity of the train relative to the ground and (b) the velocity of the car during its forward and reverse journeys. Assume that the object comes to rest immediately after fall from the point of view of the observer on the ground.

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ANSWER :(a) 3.34 km/hr (B) 13.34 km/hr,6.67 km/hr
4690.

A non - conducting piston of mass m and are S_(0) divides a non - conducting, closed cylinder as shown in the figure. A piston having mass m is connected with the top wall of the cylinder by a spring of force constant k. The top part is evacuated and the bottompart is evacuated and the bottom part contains an ideal gas at a pressure P_(0) in the equilibrium position.Adiabatic exponent gamma and in equilibrium length of each part is l. (neglect friction). Find the angular frequency for small oscillation.

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`(sqrt(KL+gammaP_(0)S_(0)))/(2ml)`
`(sqrt(3kl+gammaP_(0)S_(0)))/(ML)`
`sqrt((k)/(m))`
`(sqrt(kl +gammaP_(0)S_(0)))/(ml)`

ANSWER :D
4691.

A capacitor with an initial potential difference of 80.0 V is discharged through a resistor when a switch between them is closed at t=0. At t= 10.0s, the potential difference across the capacitor is 1.00V (a) What is time constant of the circuit? (b) What is the potential difference across the capacilor

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ANSWER :(a) 2.28 s; (B) 46.5mV
4692.

A pair of coherent sources may be

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ONE VIRTUAL and the other real
both real
both virtual
1 and 3

Answer :D
4693.

If the nuclear radius of ""^(27)Al is 3.6 fermi, the approximate nuclear radius of ""^(64)Cu is

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2.4
1.2
4.8
3.6

Solution :`R_(Al)/R_(CU)=(27)^(1//3)/(64)^(1//3)=3/4`
`R_(Cu)=4/3R_(Al)=4/3xx3.6` fermi
`R_(Cu)`=4.8 fermi
4694.

The capacitance of a parallel platecapacitor is C when the region between the plate has air. This region is now filled with a dielectric slab of dielectric constant k. The capacitor is connected to a cell of emf E, and the slab is taken out

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charge CE(k - 1) flows through the CELL
energy `E^(2)C(k - 1)`is absorbed by the cell.
the energy stored in the capacitor is reduced by `E^2C(k-1)`
the external AGENT has to do `1/(2)E^(2)C(k-1)`amount of WORK to take the slab out.

Answer :A::B::D
4695.

The rms value of current in a 50 Hz AC circuit is 6A . The average value of AC current over a cycle is

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`6sqrt(2)`
`(3)/(pisqrt(2))`
ZERO
`(6)/(pisqrt(2)`

Solution :AVERAGE value of AC over a cycle is zero.
4696.

The electric field part of an electromagnetic wave is a medium is represented by, E_x = 0, E_y=2.5 (N)/( c ) cos [(2pixx10^6rad//m]t-(pixx10^(-2)rad//s)^x], E_x=0. The wave is:

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Moving in X direction with FREQUENCY `10^6` Hz and wave length 100 m.
Moving along x direction with frequency `10^6` Hz and WAVELENGTH 200 m
Moving along -x direction with frequency `10^6` Hz and wavelength 300 m
Moving along y-direction with frequency `2pixx10^6` Hz and wavelength 200 m

Answer :B
4697.

Earth is revolving around the sun. If the distance of the earth from the sun is reduced to 1/4th of the present distance then the day length reduced to :

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`1//4`
`1//8`
`1//2`
`1//6`

Solution :By Kepler.s third LAW `T^(2) prop r^(3)`
`therefore (T_(2))/(T_(1))=((r_(2))/(r_(1)))^(3//2)=((r//4)/(r ))^(3//2)=(1)/(8)`
`therefore T_(2)=(T_(1))/(8)`
Thus correct choice is (b).
4698.

Ultraviolet light by wavelength 200 nm is incident on the polished surface of Fe(Iron). The work function of the surface is 4.71 eV. What will be its stopping potential ? (h = 6.626xx10^(-34) J s, 1 eV = 1.6 xx10^(-19)J,)

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1.5 V
2.5 V
0.5 V
none of these

ANSWER :A
4699.

What is the modulation index if an audio signal of amplitude one half of the career amplitude is used in AM

Answer»

1
0
1.5
gt1

Answer :C
4700.

A: When currents vary with time, Newton's third law is valid only if momentum carried by the electromagnetic field is taken into account. R: Magnetic field lines always form closed loops.

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If both Assertion & REASON are true and the reason is the correct EXPLANATION of the assertion then mark (1)
If both Assertion & Reason are true but the reason is not the correct explanation of the assertion, then mark 2.
If assertion is true statement but Reason is FALSE, then mark (3)
If both Assertion and Reason are false STATEMENTS, then mark (4)

Answer :B