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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 7801. |
Which of the following waves can be polarized (i) Heat waves (ii) Sound waves? |
| Answer» SOLUTION :HEALT waves can be polarized because HEAT waves are transverse in NATURE. | |
| 7802. |
Small mercury drops of the same size are charged to the same potential V. If n such drops coalesce to form a single large drop, then calculate its potential. |
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Answer» Solution :Let r be the radius of a small DROP and R that of the large drop. Then, since the volume remains conserved, `(4)/(3) pi R^(2)= (4)/(3) pi r^(2) n "" rArrR^(3) = r^(3) n ` R = `r^(3) (n)^(1//3)` Further, since the total CHARGE remains conserved, we have using Q = CV `C_("large") V = n C_("small"^(V))` Where V is the potential of the large drop. `4 pi epsilon_(0) RV = n (4 pi epsilon_(0) r)_(v)` V= `("nrv")/(R) = ("nrv")/(r(n)^(1//3))` `V = vn^(2//3)` |
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| 7803. |
Calculate the de-Broglie wavelength of the electron orbiting in the n= 2 state of hydrogen atom. |
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Answer» Solution :In n - 2 STATEOF hydrogen atom, as PER Bohr.s quantum condition, we have `mvr_(2) = (nh)/(2pi) = (4h)/(2pi)= (h)/(pi)rArrmv = (h)/(pi r_(2))` `therefore ` de- Broglie wavelenght` lambda = (h)/(MV) = (h)/((h//pi r_(2))) = pi r_(2)` , where `r_(2)`is the radiusof orbitcorrespoding to n = 2 state . |
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| 7804. |
As the beam enters the medium, it will: |
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Answer» travel as a cylindrical beam 
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| 7805. |
A loop is kept so that its center lies at the origin of the pointing along Z -axis as shown in the figure |
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Answer» No emf and current will be INDUCED in the loop if it ROTATES about Z-axis When it rotates about y axis its flux linkage CHANGES . However in insulators there can not be motional emf. If the loop is made of copper , it is conductive therefore induced current is set up. If the loop moves along the Z axis variation of flux linkage is zero . Therefore the emf and current will be equal to zero. |
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| 7806. |
Gauss's law helps in |
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Answer» DETERMINATION of ELECTRIC field due to SYMMETRIC charge distribution |
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| 7807. |
In the given circuit Fig the switch is closed at t = 0. Choose the correct answers. |
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Answer» CURRENT in the inductor when the circuit reaches the steady state is `4 A`. At `t gt gt 0, I_(L) = (12)/(3) = 4 A` `:. | PHI| = L [i_(f) - i_(i)] = 500 xx 10^(-3) xx 3= 1.5 Wb `:.` (a) and (b) are the corrent choices. |
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| 7808. |
What were contributions of JC Bose and marconi to em waves ? |
| Answer» SOLUTION :In 1895, Sir JC BOSE SUCCEEDED in producing and observing em waves of MUCH shorter wavelengths (2 mm to 5 mm). | |
| 7809. |
Two pipes closed at one end, 53 cm and 54 cm long, produce 3 beats per second when they are sounded together in their fundamental modes. Ignoring end correction, calculate the speed of sound in air. |
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Answer» |
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| 7810. |
The extension in a string obeying Hooke's law is x. The speed of transverse wave in the stretched string is v. If the extension in the string is increased to 1.5x, the speed of transverse wave will be |
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Answer» 1.22v |
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| 7811. |
(A) : The chief characteristic of series resonant circuit is current magnification (R): At resonance the voltage drop across inductance is equal to that on capacitor. |
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Answer» Both 'A' and 'R' are TRUE and 'R' is the CORRECT explanation of 'A'. |
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| 7812. |
What do you mean by resolving power of an optical instrument ? How does resolving power of a telescope depend on wavelength of light and diameter of objective lens ? |
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Answer» Solution :The ability of an instrument to separate or distinguish close ADJACENT images is CALLED the resolving power of that OPTICAL instrument . Resolving power and resolving limit are reciprocal. If wavelength of LIGHT is `lambda` and diameter of objective is d, then resolving limit of atelescope , `theta =(1.22lambda)/(d)` `therefore` Resolving Power `=1/(theta)=d/(1.22 lambda)` Hence , if wavelength of light decreases then resolving power oftelescope INCREASES and if diameter of objective lens is increased then the resolving power of telescope will also increase. |
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| 7813. |
Light of wavelength 424 nm is incident perpendicularly on a soap film (n = 1.33) suspended in air. What are the (a) least and (b) second least thicknesses of the film for which the reflections from the film undergo fully construc- tive interference? |
| Answer» SOLUTION :(a) 79.7 NM, (B) 239 nm | |
| 7814. |
A Fresnel's biprism is placed at a distance of 30 cm in front of a narrow slit illuminated by a monochromatic light . The virtual images formed by the prism are 0.30 cm apart . When the screen is placed at 120 cm from the biprism , the fringe width found to be 0.235 mm . find the wavelength of light used . |
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Answer» |
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| 7815. |
In Fraunhoffer diffraction by a single slit, the width of the slit is 0.01 cm. If the wavelength of the light incident normally on the slit is 5000 Å the angular distance of second maxima from the mid line of central maxima is ...... rad. |
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Answer» `0.125` `d sin theta=(2n+1)(lambda)/(2)` `=(5lambda)/(2d)"" [ :. n=2` `=(5xx5xx10^(-5))/(2xx0.01)` `=(25)/(2)xx10^(-3)=0.0125` RAD |
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| 7816. |
A constant current I flows thorugh a thick wire of circular cross-section having a radius 'r'. Magnetic field at a point on its surfae is given as B = …………………. . |
| Answer» SOLUTION :`(mu_0 I)/(2 PI R)` | |
| 7817. |
A photon has a wavelength 6400 Å. Calculate, i. its frequency, ii. Its energy, iiii. Its momentum. |
| Answer» Solution :`i. 4.7xx10^(14)Hz, ii. 31.06xx10^(-20)J, iii. 1.036xx10^(-27)kg ms^(-1)` | |
| 7818. |
Three waves A, B and C of frequencies 1600 kHz, 5 MHz, and 60 MHz respectively are to be transmitted from one place to another. Which of the following is the most appropriate mode of communication ? |
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Answer» A is transmitted VIA space wave while B and C are transmitted via SKY wave |
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| 7819. |
The centrifugal force is maximum at |
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Answer» Poles |
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| 7820. |
The concepts of communication are a) mode of communication b) need for modulation c) types of modulation d) detection of modulated wave |
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Answer» a, b, C are true |
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| 7821. |
In an LCR series circuit, thevoltage across each of the components L,C,R is 20V. The voltage across the L-C combination will be: |
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Answer» 20V |
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| 7822. |
A projectile is fired with velocity u making an angle θ with the horizontal. What is the angular momentum of the projectile at the highest point about the starting point ? (Given the mass of the projectile is m ) |
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Answer» `(m cos theta)/(2g)` `"Linear MOMENTUM "=m u cos theta " at the highest"` `"point ANGULAR momentum of the projectile at "` `"the highest point"="linear momentum"XX"perpendicular distance"` `=m u cos theta xx "h mu cos "theta=(u^(2)sin^(2)theta)/(2g)` `=(m u^(2)sin^(2)theta cos theta)/(2g)` |
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| 7823. |
The current in a conductor varies with time .t. as I= 3t+4t^(2). Where I in amp and t in sec. The electric charge flows through the section of the conductor between t = 1s and t = 3s |
| Answer» Answer :C | |
| 7824. |
Suppose that the electric field amplitude of an electromagnetic wave is E_(0)=120 N/C and that its frequency is upsilon = 50.0 MHz.a.Determine, B_(0), omega, k, and lambda.b.Find the expressions for E and B. |
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Answer» SOLUTION :`E=120 NC^(-1), upsilon =50 MHZ = 50xx10^(6)HZ` a.`B_(0)=(E_(0))/(c )=(120)/(3XX10^(8))=4xx10^(-7)T` `omega = 2pi upsilon = 2xx3.14xx50xx10^(6)=3.14xx10^(8)RAD s^(-1)` `c=(omega)/(k)therefore k=(omega)/(c )=(3.14xx10^(8))/(3xx10^(8))=1.05 rad m^(-1)` `lambda = (2pi)/(k)=(2xx3.14)/(1.05)=5.98 m` b.`E=E_(0)sin (k x - omega t)=120 sin (1.05 x-3.14xx10^(8)t)Vm^(-1)` `B=B_(0)sin (k x-omega t)=4xx10^(-7)sin(1.05x -3.14xx10^(8)t)T`. |
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| 7825. |
A graph between the square of the velocity of a particle and the distance(s) moved is shown in figure. The acceleration of the particle in kilometers per hour square is:- |
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Answer» 2250 `a=(v^(2)-u^(2))/(2S)=((900)-(4600))/(2xx0.6)=-(3700)/(1.2)` `=-3084 km//hr^(2)` |
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| 7826. |
In an experiment, a boy plots a graph between (v_(max))^(2)and (a_(max))^(2) for a simple pendulum for different values of (small) amplitudes, where v_(max)and a_(max) is the maximum velocity and the maximum acceleration respectively. He found the graph to be a straight line with a negative slope, making an angle of 30^(@)when the experiment was conducted on the earth surface. When the same experiment was conducted at a height h above the surface, the line was at an angle of 60^(@). The value of h is [radius of the earth = R] |
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Answer» 0.5R |
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| 7827. |
Which of the following represents a Z-polarised wave? |
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Answer» `y(x,t)= a SIN(kx- OMEGA t)` |
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| 7828. |
A proton and an electron have same de Broglie wavelength, which possesses more KE ? |
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Answer» SOLUTION :de Broglie wavelength of a particle of mass m and kinetic energy E is `LAMBDA = h/sqrt(2mE), mE = h^2/(2lambda^2), lambda` is then same for both PROTON ELECTRON . So, ,mE = a constant. The mass of proton `m_p gt` the mass of electron `m_e`. So, that energy of proton `E_P lt` the energy of electron `E_e`. ELECTRONS have more KE. |
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| 7829. |
A charge q is distributed uniformly on a ring of radius R. A sphere of equal radius R is constructed with its centre at the periphery of the ring, Find the flux of the electric field through the surface of the sphere |
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Answer» `(Q)/(in_(0))` |
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| 7830. |
The moment of inertia of a thin square plate ABCD, as shown A in the figure, of uniform thickness about an axis passing through the centre O and perpendicular to the plane of |
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Answer» `l_(1)+l_(2)` Let `I_(0)` = Moment of inertia about an axis perpendicular to square plate and passing through the CENTRE O. `thereforeI_(0)=I_(1)+I_(2)`, by theorem of perpendicular axes. Also `I_(0)=I_(3)+I_(4)`, by theorem of perpendicular axes. or `I_(0)+I_(0)=(I_(1)+I_(2))+(I_(3)+I_(4))` or `2I_(0)=I_(1)+I_(1)+I_(3)+I_(3)=2(I_(1)+I_(3))` or `I_(0)=I_(1)+I_(3)`. Hence options (a), (b) and (c) are CORRECT. |
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| 7831. |
The diagonal AC of a right angled prism ABC is silvered. A ray of light falls on face AB perpendicular to AC and finally comes out of the-face BC. What is the angle made by the outgoing ray with the normal to the face BC? Refractive index of the prism is 1.5 and ZBAC is 60^(@). |
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Answer» |
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| 7832. |
If the rate of change of current per second in one coil induces an e.m.f. of 1 volt in the neighboring coil, the mutual inductance of two coil is: |
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Answer» 1H |
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| 7833. |
The pressure of a medium is changed from 1.01 xx 10^(5) Pa to 1.165 xx 10^(5) Pa and change in volume is 10% keeping temperature constant. The Bulle modulus of the medium is: |
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Answer» `204.8xx10^(5)Pa` `K=(DeltaP)/((DELTAV)/V)=((1.165xx10^(5)-1.01xx10^(5)))/(1//10)` `=1.55xx10^(5)` pascal `therefore` CORRECT choice is (d). |
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| 7834. |
The slits in Young's double slit experiment are 0.5 mm apart and interference pattern is observed on a screen distant 100 cm from the slits. It is found that the 9th bright fringe is at a distance of 8.835 mm. From the second dark fringe. The wavelength of light will be : |
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Answer» `7529 Å` SOLVING, `lambda = 5890 Å` |
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| 7835. |
When the current through a solenoid increases at a constant rate, the induced current |
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Answer» Is CONSTANT and is in the direction of the INDUCING current |
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| 7836. |
Figure shows a plot of the transverse displacement of the particle of a string at t = 0 through which a travelling wave is passing in the positive x-direction. The wave speed is 20 cm/s. Find (a) the amplitude, (b) the wavelength, (c) the wave number and (d) the frequency of the wave. |
| Answer» SOLUTION :(a) 1.0 MIN (B) 4 CM (c ) `0.71 cm^(-1)`(d) 5Hz | |
| 7837. |
Which of the following is the correct diagram of a half-wave rectifier? |
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Answer»
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| 7838. |
fpressure P, velocity v and time Tare taken as the fundamental (base) quantities, Jind the dimensional formula of Jorce. |
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Answer» Solution :Let FORCE = F= `P^(a)V^(b)T^(c )` `(F)= M^(1)L^(1)T^(-2), (P) = M^(1)L^(-1)T^(-2)` `(V) = M^(0)L^(1)T^(-1), (T)= M^(0)L^(0)T^(-1)` `:. MLT^(-2)= (ML^(-1)T^(-2))^(a)(M^(0)L^(0)T^(-1))^(b)(M^(0)L^(0)T^(1))^(c)` `MLT^(-2) = M^(a)L^(-a+b)T^(-2a+b+c)` `:.` Comparing powers of M,L,T on both sides a=1 , -a+b=1 `:.` b=2 and `-2a-b +c =-2 , -2 -2+c=-2,c=2` `:.` Force (F) = `PV^(2)T^(2)`. |
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| 7839. |
The following table gives the wavelength range of some constituents of the electromagnetic spectrum : Select the wavelength range and name the electromagnetic waves that are (i) widely used in the remote switches of household electronic devices, (ii) produced in nuclear reactions. |
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Answer» SOLUTION :(i) Infrared rays of wavelength range 1 mm to 700 NM. (ii) Gamma rays of wavelength range `lambda=10^(-3)nm`. |
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| 7840. |
A prism of refracting angle of 60^@has minimum angle of deviation of 30^@. What must be the angle of incidence for this case? |
| Answer» Answer :C | |
| 7841. |
Which of the following product of e,h,mu, G (where mu is the permeability ) be taken so that the dimensions of the product are same as that of speed of light ? |
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Answer» `he^(-2)mu^(-1)G^(0)` `a-2c=0` `a-b-2c-2d=-1` `b+c-d=0` `2b+c+3d=1` Solving for `.a., .b., .c.` and `.d.` we get `a=-2,b=1,` `c=-1,d=0.` Thus `v=e^(-2)hmu^(-1)G^(0).` Hence correct choice is `(a)`. |
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| 7842. |
The speed of an electron in first Bohr orbit is (c )/(137), where c is the velocity of light in vacuum, then the speed of electron in second orbit is ………….. |
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Answer» `(1)/(2)((C)/(137))` `mvr_(n)=(nh)/(2PI)` `:V=(nh)/(2pimr_(n))` but `r_(n)LTN^(2)` `:.r_(n)=Kn^(2)` (K constant) `:.v=(nh)/(2pimKn^(2))` `:.vprop(1)/(n)` (All others terms are constant) `:.(v_(2))/(v_(1))=(n_(1))/(n_(2))` `:.(v_(2))/(v_(1))=(1)/(2)` `:.v_(2)=(v_(1))/(2)` `:.v_(2)=(1)/(2)((c)/(137))` |
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| 7843. |
What we call the time interval after which the intanseous current in the circuit gets repeated ? |
| Answer» SOLUTION :TIME PERIOD | |
| 7844. |
A monochromatic source emitting light of wavelength 600 nm has a power output of 66 W. Calculate the number of photons emitted by this source in 2 minutes. |
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Answer» Solution :Energy of one photon `E=(hc)/(lambda)` `E=(6.6 xx 10^(-34) xx 3 xx 10^(8))/(6 xx 10^(-7))` `=3.3 xx 10^(-19)` J `E_(1)="energy emitted by the source in one second"` = 66 J `:.` Number of PHOTONS emitted by the source in 1 sec., `n=(66)/(3.3 xx 10^(-19))` `= 2 xx 10^(20)` `:.` TOTAL number of photons emitted by source in 2 minutes, `N=n xx 2 xx 60` `= 2 xx 10^(20) xx 120` `= 2.4 xx 10^(22)` photons |
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| 7845. |
If a current is passed through a spring, then the spring will |
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Answer» expand |
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| 7846. |
कोई वस्तु उदासीन होती है क्योंकि - |
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Answer» धनात्मक आवेश , ऋणात्मक आवेश से अधिक होता है। |
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| 7847. |
When an iron wire of diameter 1 cm is copper plated uniformly. Resistance of iron reduces to 1/3 of its original value. Calculate the thickness of copper plating, Resistivities of copper and iron are 1.8 times 10^-6 Omega.cm and 1.98 times 10^-5 Omega.cm respectively. |
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Answer» Solution :Let RESISTANCE of iron wire be r and that of copper plating x. As they are in PARALLEL, so equivalent resistance , `R_(eq)=(rx)/(r+x)` `therefore(rx)/(r+x)=r/3or,3rx=r^2+rxor,2rx=r^2` `therefore x=r/2` Now, `r=rhol/A=(1.98times10^-5timesl)/(pi(0.5)^2)` Similarly , `x=(1.8times10^-6timesl)/(2pi(0.5)d)` (l=length , d= thickness) SOLVING for r and x, it is observed that d=0.045 (approx) |
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| 7848. |
A mixture of light, consisting of wavelength 590 nm and unknown wavelength, illuminates Young's double slit and gives rise to two overlapping interference patterns on the screen.The central maximum of both lights coincide.Futther, it is observed that the third bright fringe of known light coincides with the 4th bright firnge of the unknown light.From this data, the wavelength of the unknown light is : |
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Answer» `442.5 nm` `(3 lambda_(1) - 4 LAMBDA)^(2)` `lambda_(2) - 3/4 lambda_(1) = 3/4 XX 590 = 442.5 nm`. |
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| 7849. |
A man weasring glasses of focal length +1 m cannot clearly see beyond 1 m |
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Answer» if he is farsighted |
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| 7850. |
A proton and alpha - particle are projected with the same velocity in a perpendicular magnetic field. The ratio of radii of their circular trajectory is |
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Answer» `1:2` |
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