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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.
| 39301. |
The thermal capacity of 10 gm of a substance is 8 cal //^@C . Then it's specific heat is |
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Answer» 0.8 |
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| 39302. |
Activity of radioactive element decreased to one third of original activity I_0in 9 years. After further 9 years, its activity will be |
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Answer» `I_0` |
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| 39303. |
A zener diode has |
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Answer» HEAVILY doped p-side and LIGHTLY doped n-side. |
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| 39304. |
Refer to V-I graph shown here, the emf of voltage supplyis ______and internal resistance is _____. |
| Answer» SOLUTION :3V, `1.5 OMEGA` | |
| 39305. |
A source of yellow light placed in air is observed by a person swimming under water. If the wavelength of yellow light in air is 6000Å , then determine its velocity, wavelength and colour as observed by the person. Take ve locity of light in air (c=3 xx 10^8 m//s^(-1)) |
| Answer» SOLUTION :`2.25 XX 10^8 m//s ,4500 Å ` YELLOW | |
| 39306. |
A simple pendulum oscillates with an amplitude of 2xx10^(-2)m. If the force acting on it at extreme position is 4 N. Then the time that another simple pendulum of length 81 cm takes to execute the same number of oscillations is |
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Answer» 4N `=((A)/(2))/(A)=(1)/(2)` `F_(m)=(F_(e))/(2)=(4)/(2)=2N` |
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| 39307. |
The maximum number of possible interference maxima for slit separation equal to twice the wavelength in Young.s double slit experiment . |
| Answer» Answer :C | |
| 39308. |
The refractive indices of crown glass prism for C, D and F lines are 1.527, 1.530 and 1.535 respectively. Find the dispersive power of the crown glass prism. |
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Answer» 0.01509 |
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| 39309. |
A coil of area A is held parallel to the magnetic field B. The magnetic flux link with it is: |
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Answer» Zero |
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| 39310. |
A 100 muF capacitor in series with a 40 omega resister connected to a 110 V, 60 Hz supply What is the time lag between the current maximum and the voltage maximum. |
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Answer» SOLUTION :`tan phi = X_c/R = 1/(RomegaC)` `=(1/(40 XX 2PI xx 60 xx100 xx 10 ^(-6)))`= 0.6628 |
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| 39311. |
A 100 muF capacitor in series with a 40 omega resister connected to a 110 V, 60 Hz supply What is the maximum current in the circuit ? |
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Answer» SOLUTION :`X_c = sqrt(R^2 + (1/(OMEGAC))^2` `= sqrt ((40)^2 + (1/(2PI xx 60 xx 100 xx 10^(-6)))^20mega` = 48 W |
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| 39312. |
When due to change of magnetic fiels currents flows in closed circuits, what we call them? |
| Answer» SOLUTION :EDDY CURRENTS. | |
| 39313. |
Read the following passageon the basis of your understanding of the following paragraph and the related studied concepts: A freely suspended bar magnet capable of rotation in a horizontal plane in its equilibrium state, rests in the direction of earth's magnetic field at the place. If the magnet is slightly rotated from its equilibrium direction and then released, it executes angular oscillations of period T given as per relation T = 2pi sqrt((I)/(mB_H)) Here I moment of inertia of bar magnet about its oscillation axis, magnetic moment of bar magnet and B_H horizontal component of earth's magnetic field at the place. An oscillation magnetometer is designed on this principle and is employed to compare magnetic moments of magnets and for determining earth's magnetic field at a place. Two bar magnets of same dimensions and same mass but different magnetic moments are placed together in an oscillation magnetometer with their like poles together and the combination completes 20 oscillations per minute. Subsequently, the two magnets are placed with their unlike poles together in the oscillation magnetometer. Now the combination completes 10 oscillations per minute only. Compare the ratio of magnetic moments of two given magnets. |
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Answer» `5/3` |
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| 39314. |
Read the following passageon the basis of your understanding of the following paragraph and the related studied concepts: A freely suspended bar magnet capable of rotation in a horizontal plane in its equilibrium state, rests in the direction of earth's magnetic field at the place. If the magnet is slightly rotated from its equilibrium direction and then released, it executes angular oscillations of period T given as per relation T = 2pi sqrt((I)/(mB_H)) Here I moment of inertia of bar magnet about its oscillation axis, magnetic moment of bar magnet and B_H horizontal component of earth's magnetic field at the place. An oscillation magnetometer is designed on this principle and is employed to compare magnetic moments of magnets and for determining earth's magnetic field at a place. Two bar magnets A and B of same configuration and mass have oscillation periods of 2s and 2.5 s respectively at a given place. The ratio of magnetic moments of A and B is |
| Answer» SOLUTION :`(m_A)/(m_B) = ((T_B)/(T_A))^2 = ((2.5)/(2))^2 = (25)/(16)` | |
| 39315. |
Read the following passageon the basis of your understanding of the following paragraph and the related studied concepts: A freely suspended bar magnet capable of rotation in a horizontal plane in its equilibrium state, rests in the direction of earth's magnetic field at the place. If the magnet is slightly rotated from its equilibrium direction and then released, it executes angular oscillations of period T given as per relation T = 2pi sqrt((I)/(mB_H)) Here I moment of inertia of bar magnet about its oscillation axis, magnetic moment of bar magnet and B_H horizontal component of earth's magnetic field at the place. An oscillation magnetometer is designed on this principle and is employed to compare magnetic moments of magnets and for determining earth's magnetic field at a place. The oscillation period of a given magnetic needle is 5 s at place A and 6 s at place B. The ratio of horizontal components of earth's magnetic field at A and B is |
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Answer» `5/6` |
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| 39316. |
Point like object is observed using microscope angle subtended by objective with object is 20^(@) If oil of refractive index 1.4 is kept between object and objective, numerical aperture will be ...... |
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Answer» `0.24` `=1.4 sin ((20)/(2))^(0)` `=1.4 sin 10^(@)` `=1.4xx0.1736=0.24` |
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| 39317. |
Distinguish between dispersion of white light and spectrum of white light. |
| Answer» Solution :DISPERSION is a PHENOMENON of SPLITTING of a WHITE ray of LIGHT into it.s constituent colors. | |
| 39318. |
A bullet is fired from gun. The force on bullet is, F = 600-2xx10^(5)t newton. The force reduces to zero just when bullet leaves barrel. Find the impulse imparted to bullet. |
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Answer» Solution :`F=600-2xx10^(5)t` F becomes zero as soon as the bullet LEAVES the barrel. `0=600-2xx10^(5)t , 600 = 2xx10^(5)t` `t=3XX10^(-3)s` Impulse `= int_(0)^(1) Fdt` `= int_(0)^(t)(600-2xx10^(5)t)dt=[600t-cancel(2)xx10^(5)(t^(2))/(cancel(2))]^(3xx10^(-3)` `=600xx3xx10^(-3)-10^(5)xx9xx10^(-6)=0.9 Ns` |
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| 39319. |
Explain construction of meter bridge used in laboratory. |
Answer» Solution : `rArr` Wire having uniform cross-section and uniform specific resistance (RESISTIVITY) and length of 1 mis tied to two end of steel plate as shown in FIGURE. As shown in figure plate are of shape of RIGHT angle which consist of connector al their end POINTS. `rArr` Between two end A and C of wire, battery, switch and sometimes rheostat is connected for regulating current. `rArr` Galvanometer is connected between two terminal at MIDDLE of metal plate. `rArr` Second end of galvanometer connected with jockey key. This can be moved on wire of meter bridge by keeping in contact with the wire. |
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| 39320. |
A body is moved along a straight line by a machine delivering constant power, the distance moved by the body in time in proportional to : |
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Answer» `t^(3//2)` `:. ML^(2)T^(-3)`= constant As M is constant `:. L^(2)T^(-3)const. implies L^(2) propT^(3) LpropT^(3//2)` or distance `PROP` `("time")^(3//2)` |
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| 39321. |
Where should an object be placed front of a concave mirror of the focal length f so that the image to be of the same size as the object ? |
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Answer» `R//2` |
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| 39322. |
What is displacement current? Give the expression for it |
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Answer» Solution :Current that RESULT due to the time rate of CHANGE of electric flux is called DISPLACEMENTCURRENT. Displacement current `I_(d)= epsi_(0) (d phi_(E ))/(DT)` `epsi_(0)`= Permittivity of free space, `(d phi_(E ))/(dt)`= rate of change of electric flux |
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| 39323. |
There are two wires P and Q made of the same material and both have the same mass . The radius of wire P is half the radius of the wire Q. the resistance of P is 24 Omega. Find the resistance of Q. |
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Answer» Solution :Let m be the mass and d the density of MATERIAL. Let r be the RADIUS and l be the LENGTH. `m= (pi r^(2) l) d rArr l= (m)/(pi r^(2) d)` …(i) We know that resistance is given by the following relation. `R= rho (l)/(A) rArr R= rho (l)/(pi r^(2))` ...(ii) Substituting l from equation (i) in equation (ii) we get `R= rho (m)/(pi^(2) r^(4) d)` ...(III) In equation (iii) it is only the radius which is different for P and Q and rest of the parameters are same. Hence their resistances are INVERSELY proportional to fourth power of the radius of wire `(R_(P))/(R_(Q)) = ((r_(Q))/(r_(P)))^(4) rArr (R_(P))/(R_(Q)) = ((2r_(P))/(r_(P)))^(4) = 16` `rArr (R_(P))/(R_(Q)) = 16 rArr R_(Q) = (R_(P))/(16) rArr R_(Q) = (24)/(16) = 1.5 Omega` |
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| 39324. |
Digital signals |
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Answer» do not provide continuous set of values |
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| 39325. |
Heat is produced at a rate of 250 J per second in an X-ray tube when potential difference of 20 kV is applied between filament and target metal. How much current flows through the tube ? Assume only a small fraction of kinetic energy of electrons is converted into X-rays. |
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Answer» SOLUTION :If we assume that entire kinetic of ELECTRONS is GETTING CONVERTED into heat then : P=V I `RARR` I=P/V =250/20000=0.0125 A |
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| 39326. |
In A.C. circuit, when E = E_0 sin omegat supply applied, current is obtained in circuitI=I_0 sin (omegat-pi/2) , so power consumed is circuit will be …… |
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Answer» <P>`P=(E_0I_0)/SQRT2` `therefore` Real power `P = E_"RMS" I_"rms" cosdelta` `=E_(rms)I_(rms)xx "cos"pi/2` `=E I xx 0 [ because E_(rms) = E, I_(rms)=I]` `therefore` P=0 |
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| 39327. |
What is isotone? Give an example. |
| Answer» SOLUTION :ISOTONES are the atoms of different elements having same number of neutrons. `""_5^12B and ""_6^13B` are examples of isotones which 7 neutrons. | |
| 39328. |
Photons of energy 5 eV and incident on the surface of metal M_1 and electrons with kinetic energy E_(M_1)and of energy 5.75 eV are incident on surface of metal M_2 , the electrons with kinetic energy E_(M_2)=(E_(M_1)+2eV) and de Broglie wavelength lamda_(M_2)=(lamdaM_(1))/3 are ejected.If phi_(M_1) and phi_(M_2) are work function of metal surface , M_1 and M_2 respectively, then choose the correct option /s. {:(,"List -I",,"List - II"),(P,E_(M_1),1,2.25eV),(Q,E_(M_2),2,0.25eV),(R,phi_(M_1),3,4.75eV),(S,phi_(M_2),4,3.50eV):} |
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Answer» `{:(P,Q,R,S),(3,2,4,1):}` `E_(M_2)=5.75eV-phi_(M_2)""... (ii)` Given, `E_(M_2)=E_(M_1)-2eV ""...(III)` Also given , `lamda_(M_1)=3lamda_(M_2)"":.LAMDA=K/sqrtE` `E=K/lamda^2` `K/sqrtE_(M_1)=3K/sqrt(E_(M_2))impliesE_(M_2)=9E_(M_1)""...(iv)` From equations (iv) and (iii) `8E_(M_(1))=2eV` `E_(M_(1))=0.25eV` `E_(M_(2))=2.25eV` `phi_(M_(1))=4.75eV` `phi_(M_2)=3.50 eV` |
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| 39329. |
(a) Estimate the speed with which electrons emitted from a heated emitter of an evacuatedtube impinge on the collector maintained at a potential difference of 500 V with respect to the emitter.Ignore the small initial speeds of the electrons.The specific charge of the electron ,i.e., its (e)/(m) is given to be 1.76xx10^(11)Ckg^(-1) (b) Use the same formula you employ in (a)to obtain electron speed for an collector potential of 10 MV.Do you see what is wrong ?in what way is the formula to be modified? |
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Answer» Solution :Here `V=500 V,(e )/(m)=1.76xx10^(10)(C )/(kg)` `m_(0)=9.1xx10^(-31)` kg (a) Kinetic energy of electron, `(1)/(2)mv^(2)=eV` `therefore v=sqrt((2eV)/(m))`…….(1) `therefore=sqrt((2xx1.76xx10^(11)xx500)/(1))` `therefore v=sqrt(1.76xx10^(14))` `therefore v=1.3266xx10^(7)` `therefore v~~1.33xx10^(7)m//s` (b)`V.=10 MV=10xx10^(6)=10^(7)V` `therefore` From equation (1), `v.=sqrt((2eV)/(m))=sqrt(2xx1.76xx10^(11)xx10^(7))` `=sqrt(3.52xx10^(10))` `=1.876xx10^(9)` `~~1.88xx10^(9)m//s` Such speed is not possible.Because speed of light is `|c=3xx10^(8)ms^(-1)|`.Any particle can not have velocity more than speed of light .So above equation can only be used for speed `VLTC` `therefore K=mc^(2)-m_(0)c^(2)=(m-m_(0))c^(2)` Where `m_(0)`=rest mass of particle and `m=(m_(0))/(sqrt(1-v^(2)/c^(2)))` `therefore eV=(m_(0)c^(2))/(sqrt(1-(v^(2))/(c^(2))))-m_(0)c^(2)[because K=eV]` `therefore (eV)/(m_(0)c^(2))=(1)/(sqrt(1-(v^(2))/(c^(2))))-1` `therefore(eV)/(m_(0)C^(2))+1=(1)/(sqrt(1-v^(2)/(c^(2))))` `therefore=(1.6xx10^(-19)xx10^(7))/(9.1xx10^(-31xx9xx10^(16)))+1=(1)/sqrt(1-(v^(2))/(c^(2)))` `19.54+1=(1)/(sqrt(1-(v^(2))/(c^(2)))` `1-(v^(2))/(c^(2))=(1)/((20.54)^(2))` `therefore1-(1)/((20.54)^(2))=(v^(2))/(c^(2))` `therefore1-0.00237=(v^(2))/(c^(2))` `0.99763=(v^(2))/(c^(2))` `therefore sqrt(0.99763)=(v)/(c )` `therefore` v=0.999c |
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| 39330. |
The mechanism of beta emission can be stated as a neutron decay. Explain. |
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Answer» SOLUTION :A neutron decays into a proton `(""_(1)^(1)H)` and an electron `(""_(-1)^(0)e). beta`- EMISSION is the emission of ELECTRONS from the nucleus. `""_(0)^(1)N to ""_(1)^(1)H + (-1)^(0)e + "energy"` Actually there is an emission of a third particle called anti-neutrino `(barupsilon)` in the p-emission process, so that both momentum and energy to be conserved. `:. ""_(0)^(1)n to ""_(1)^(1)H + ""(-1)^(0)e + barupsilon + "energy"`. |
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| 39331. |
A horizontal electrical power line carries a current of 90A from east to west direction. What is the magnitude and direction of magnetic field produced by the power line at a point 1.5 m below it? |
| Answer» SOLUTION :`1.2 xx 10^(-5)` T South WARD | |
| 39332. |
Under the action of a constant force, A particle is experiencing a constant acceleration. The power is |
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Answer» POSITIVE constant |
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| 39333. |
Two blocks A and B having masses 5 kg and 10 kg respectively connected through a massless string which is passing over a pulley (disc) of mass 2 kg and radius 25 cm as shown in figure If system is. released from rest. then angular acceleration of the pulley is(String does not slip over the pulley horizontal surface is smooth and g = 10ms ^-2 (AAK_TST_02_NEET_PHY_E02_011_Q01) |
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Answer» |
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| 39334. |
Two parallel long wires carry currents 18A and 3A. When the currents are in the same direction, the magnetic field at a point midway between the wire is B_1 If the direction of i_2 is reversed, the field becomes B_2 Then the value of B_1,//B_2 is |
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Answer» `5:7 ` |
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| 39335. |
An electric dipole made up of a positive and negative charge, each of 1 mu Cseparated by a distance of 2cm is placed in an electric filed of 10^5 N//C, then the work done in rotating the dipole from the position of stable equilibrium through an angle of 180^@ is |
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Answer» `2 xx 10^(-3)` Joule |
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| 39336. |
In young double slitexperiment the n^(th) red brightbandcoincides with (n+1)^(th)blue bright band . Ifthe wavelengthof redand bluelights are 7500 A^(0) and 5000 A^(0), the value of .n. is |
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Answer» `5762 A^(@)` |
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| 39337. |
For pair production i.e. for the production of electron and positron, the incident photon must have a minimum frequency of the order of |
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Answer» `10^(18) s^(-1)` |
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| 39338. |
A volume occupied by a saturated vapour is reduced isothermally n-fold. Find what fraction eta of the final volume is occupied by the liquid phase if the specific volumes of the saturated vapour and the liquid phase differ by N times (N gt n). Solve the same problem under the condition that the final volume of the substance corresponds to the midpoint of a horizontal portion of the isothermal line in the diagram p, V. |
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Answer» Solution :We let `V'_l` = specific volume of liquid. `V'_v = N V'_l` = specific volume of vapour. Let `V` = original volume of the vapour. Then `M (PV)/(RT) = m_l + m_v (V)/(N V'_l)` or `(V)/(n) = (m_l + N m_v) V'_l` So `(N - 1)m_l V'_l = V (1 - (1)/(n)) = (V)/(n)(n - 1)` or `eta = (m_l V'_l)/(V//n) =(n-1)/(N - 1)` In the case when the final volume of the substance CORRESPONDS to the midpoint of a HORIZONTAL portion of the isothermal LINE in the `p, v` diagram, the final volume must be `(1 + N)(V'_l)/(2)` per unit mass of the substance. Of this the volume of the liquid is `V'_1//2` per unit total mass of the substance. Thus `eta = (1)/(1 + N)`. |
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| 39339. |
A parallel plate capacitor with air as dielectric between the plates has a separation of 5mm and plate area 50 cm^(2) is connected to a 200V source. Calculate a. the charge on the plates. (b) the charge on the plates if the space between the plates is filled with a dielectric K=5 |
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Answer» SOLUTION :DATA supplied, `C=(epsi_(0) KA)/(d) and q=CV ""epsi_(0)=8.854 xx 10^(-2) C^(2)//Nm^(2)` K=1 for air `A=50 cm^(2)=50 xx 10^(-1) m^(2)""d=5mm=5 xx 10^(-3)m` V=200V `a, therefore q=(epsi_(0)KV)/(d)=(8.854 xx 10^(-12) xx 1 xx 50 xx 10^(-4) xx 200)/(5 xx 10^(-3))=1.77 xx 10^(-9)C=1.77nC` (b), `q=(epsi_(0) K.A.V)/(d)=(8.854 xx 10^(-12) xx 5 xx 50 xx 10^(-4) xx 200)/(5 xx 10^(-3))""q=8.85nC` |
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| 39340. |
Magnetic field lines can be entirely confined within the core of a toroid but not within a straight solenoid. |
| Answer» SOLUTION :TRUE - The core of the toroid is a closed SURFACE and so, as PER Guass theorem, FLUX of field `vecB` must be zero. | |
| 39341. |
From ideal instruments, current measured is I = 10.0 amp, potential difference measured is V = 100.0 volt, length of wire is 31.4 cm, and diameter of wire is 2.00 mm, the resistivity of wire will be (in correct significant figure) (pi = 3.14) |
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Answer» `1.00 xx 10^(-4) Omega - m` `rho = 1.00 xx 10^(-4) Omega - m` |
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| 39342. |
A certain gas laser can emit light at wavelength 550 nm, which involves populationinversionbetween ground state and an excited state. At room temperature, how many moles of neon are needed to put 12 atoms in that excited state by thermal agitation? |
| Answer» SOLUTION :`1.5xx10^(15)` MOL | |
| 39343. |
A rod of lengthl and mass m rests ontwo smooth parallel conductorsshortedat oneendby an inductorL and open at theend . The circuits is in a unifromfieldBperpendcularin tothe plance. Theconductoris suddenlyimpartedaninitialvelocityV_(0) directiontothe right . Show that the motionis simple harmonic. Find its angular frequency and amplitude . |
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Answer» |
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| 39344. |
(i) Derive expression for drift velocity of electrons in a conductor. Hence deduceohm's law . (ii) A wire whose cross-sectional area is increasinglinealy from itsone end to the other, is connectedacrossa batteryof V volts. Which of the followingquantities remains constantin the wire ? (a)drift speed , (b)current density (c ) electric current , (d) electricfield. Justify your answer. |
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Answer» Solution :(i) Consider an electron of mass m and chargee moving insidea conductorwith a driftvelocity`V_(d)` when an electricfield E is applied. Force on electron= eE. ma is the mechanical force the electron. `{:( :.,ma=eE,rArr,a=(eE)/m),("Now",v=u+at,,),("Here",u=0,,a=(eE)/(m)),(,t=tau(" relaxation time"),:.,u_(d)=(eE)/(m)tau):}` Ohm's law : `{:( :.,I="ne"Av_(d),:.,I="ne"(AEE)/(m)tau),( :.,I=("ne"^(2)AV)/(ml)tau,:.,R=V/IrArr (ml)/(A" ne"^(2)tau)),( :., V=(ml)/(" ne"^(2)Atau)I,:.,V prop I):}` Here Resistance `R =(ml)/(" ne"^(2)Atau)`. ltbgt(II) Currentdensity remainsconstant in the WIRE whose cross sectionalarea is increasing linearly from its one end to the other as , Current density `J = (I)/(A)`, It is currentper unit AREA,i.e, doesn't depend on area ofcross sectional . Drift speed `v_(d) = 1/(A"ne")` Electric field `= J//A`, is electricalconductivity. |
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| 39345. |
An object 2.4 m in front of lens forms a sharp image on a film 12 cm behind the lens. A glass plate 1 cm thick, of refractive index 1.50 is interposed between lens and film with its plane faces parallel to film. At what distance (from lens) should object shifted to be in sharp focus on film ? |
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Answer» 5.6 m `therefore "Shift" = 1 (1 - (2)/(3)) = (1)/(3)` Now `v. = 12 - (1)/(3) = (35)/(3) cm` `therefore (21)/(240) = (3)/(35) - (1)/(u)` `(1)/(u) = (3)/(35) - (21)/(240) = (1)/(5) ((3)/(7) - (21)/(48))` `(5)/(u)=|(144-147)/(48xx7)|` |
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| 39346. |
The ratio of radii of nuclei _13Al^27 and _52Te^125is approximately |
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Answer» 0.57777777777778 |
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| 39347. |
Suppose the mass of electron decreases by 25%. How will it affect the Rydberg constant? |
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Answer» REMAINS unchanged |
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| 39348. |
Both the capacitors shown in figure are made of square plates of edge a. The separatiopn between the plates of the capacitors are d_1 and d_2as shown in the figure. A potential difference V is applied between the points a and b. An electron is projected between the plates of the upper capacitor along the central line. With what minimum speed should the electron be projected so that it does not collide with any plate ? Consider only the electric forces. |
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Answer» `[("VEA"^2)/(md_(2)(d_1-d_2))]^(1//2)` |
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| 39349. |
Define mobility. Mention its S.I. Unit |
| Answer» Solution :It is the RATIO of drift VELOCITY of FREE electrons to the applied electric FIELD. | |
| 39350. |
In alpha-particle scattering experiment it was found that about one alpha particle in 8000 deflects by more than 90^(@). |
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Answer» |
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