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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.
| 7851. |
In Fig. 27-45, assume that epsi= 3.0 V, r = 100 Omega, R_(1)=250 22 and R_2= 300 Omega. If the voltmeter resistance RV is 5.0 kOmega. what percent error does it introduce into the measurement of the potential difference across R_1?. Ignore the presence of the ammeter. |
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Answer» |
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| 7852. |
If the error in the measurcmcnt of radius of a sphere is 2% , then the error in the determinationov volume of the sphere will be (2008) |
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Answer» 0.02 |
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| 7853. |
Statement-1 : The formula connecting u, v and f for a spherical mirror is valid only for mirrors whose sizes are very small compared to their radii of curvature. Statement-2 : Laws of reflection are strictly valid for plane surfaces, but not for large spherical surfaces. |
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Answer» statement-1 and 2 are true and statement-2 is a CORRECT explanation for statement-3 |
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| 7854. |
A long solenoid has 800 turns per meter length of solenoid. What is the magnetic induction at the end of the solenoid if it carries a current of 2.5A. |
| Answer» Solution :`B(mu_0nI)/2=(4pixx10^-7 XX 800 xx 2.5)/2=4pi xx 10^-4=1.256 xx 10^-3 (Wb)/m^2` | |
| 7855. |
Two progressive waves of equations y_1 = 6 sin (100 pit) and y_2 = 8 sin (100 pit + pi) super does at a point. The intensity of the resultant wave at the point is proportional to : |
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Answer» 28 |
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| 7856. |
The 300 turn primary of a tansformer has resistance 0.82Omega and the resistance of its secondary of 1200 turns is 6.2Omega. Find the voltage across the primary if the poweroutput from the secondary at 1600V is 32 k W. Calculate the power losses in both coils when the transformer effciency is 80%. |
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Answer» Solution :GIVEN : EFFICIENCY `eta=80%` `N_(p)=300,R_(p)=0.82Omega,R_(s)=6.2Omega,N_(s)=1200V,` Output power `=32kW,V_(s)=1600V.` To find : Power loss in the PRIMARY `=I_(p)^(2)R_(p)` Power loss in the sencondary `=I_(s)^(2)R_(s)` `V_(p)=?` Solution : `eta=("Out put power")/(Inpur power")=(V_(s)I_(s))/(V_(p)I_(p))` `(80)/(100)=(32xx10^(-3))/("Inpur Power")` Input Power `=(32xx10^(-3)xx10^(-3))/(80)=40kW.` `(N_(s))/(N_(s))=(V_(s))/(V_(s))` `V_(p)=(V_(s)xxN_(p))/(N_(s))` `V_(p)=(1600xx300)/(1200)=400V` `V_(p)=400V.` Input power `=V_(p)I_(p)` `40xx10^(-3)=400xxI_(p)` `I_(p)=(40000)/(400)=100A` Output power `=V_(s)I_(s)` `32xx10^(3)=1600xxI_(s)` `I_(s)=(32000)/(1600)=20A` Power loss in the primary `=I_(p)^(2)xxR_(p)` `=(100)^(2)xx0.82=8200` = 8.2 k W. Power loss in the secondary `=I_(s)^(2)xxR_(s)` `=(20)^(2)xx6.2=2480` `=2.48kW.` |
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| 7857. |
In the question number 34, the kinetic energy (in MeV) of the proton beam produced by the accelerator is (radius of dees =60 cm) |
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Answer» 5 `v=rxx2 pi upsilon=0.6xx2xx3.14xx12xx10^(6)=4.5xx10^(7)"m s"^(-1)` `therefore""K=(1)/(2)MV^(2)=(1)/(2)XX(1.67xx10^(-27)(4.5xx10^(7))^(2))/(1.6xx10^(-13))` `="10.6 MEV"` |
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| 7858. |
A double slit experiment is performed with light of wavelength 500 nm. A thin film of thickness 2mum and refractive index 1.5 is introduced in the path of the upper beam. The location of the central maximum will ...... |
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Answer» remain unshifted. `Deltax= mut-t` `=(nu-1)t` `=(1.5-1)xx2xx10^(-6)` `=1mu m` Now `Deltax=(yd)/(D) y=(Deltax)/(d)` `y=(D)/(d)xx1 mu m` Width of FRINGE `W=(dlambda)/(d)` `=(D)/(d)xx500xx10^(-9)m` `=(D)/(d)xx0.5xx1 mum` `:.y=(D)/(d)xx1 mum` `=2xx(D)/(d)XX(1)/(2)xx1 mu =2W` Hence, the central fringe will shift upward by nearly two fringes. |
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| 7859. |
A particle ‘P is moving in a circle of radius "r" with uniform speed v. AB is the diameter of circle and 'C' is the centre. The angular velocity of P about A and C are in the ratio. |
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Answer» `1:1` |
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| 7860. |
निम्न में से अक्रिस्टलीय ठोस है |
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Answer» प्लास्टिक |
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| 7861. |
Compare the oscillations in an LC circuit are analogous to the oscillation of a block at the end of a spring. |
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Answer» Solution :The LC oscillation is similar to the mechanical oscillation of a block attached to a spring. The LOWER part of each figure in question no. 22 depicts the corresponding stage of a mechanical system. Differential EQUATION of LC oscillation, `(d^(2)q)/( dt^(2))+( q)/( LC) =0` or ` (d^(2)q)/(dt^(2)) + omega_(0)^(2)q = 0` for a block of mass m oscillating with frequency `omega_(0)` the equation is, `(d^(2)x)/(dt^(2))+omega_(0)^(2) = 0` or ` (d^(2) x)/( dt^(2)) + ( Kx)/( m ) = 0 ` Here `omega_(0) = sqrt((k)/( m))` and k is the spring constant. From F = kc in mechanics `k = ( F )/( x ) ` means the external force requires to produce unit extension or compression is known as force constant. Unit of force constant `NM^(-1)`. Corresponding equation of this equation in LC circuit is` V = (q)/( C )` `:. (1)/(C ) = ( V )/(q)` means for established unit charge,necessary electrical difference is `(1)/( C )`. In below table analogies between mechnical and electrical quantities are shown. `{:("Mechanical system","Electrical system"),("Mass m","Inductance L"),("Force constant k","Reciprocal capacitance "1//C),("Displacement x","Charge q"),("Velocity" v=(dx)/(dt),"Current" I = ( dq)/(dt)),("Mechanical energy" E = (1)/(2) kx^(2) + ( 1)/(2) m v^(2),"Electromagnetic energy" E = ( 1)/(2) (q^(2))/(C ) kx^(2) + ( 1)/(2) m v^(2) ):}` |
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| 7862. |
Thetotal number of basic Si unit are |
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Answer» 3 |
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| 7863. |
When a compass needle is pivoted so that it can rotate in vertical plane (dip needle) will it align horizontally ? |
| Answer» Solution :Generally it will not ALIGN horizontally . The DIP needle , when properly adjusted , SHOWS an angle with the horizontal called dip . | |
| 7864. |
AB is a uniform resistance wire of 1 m length. Thegalvanometer G shows no deflection when the length AC = 20 cm. The resistance R is equal to |
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Answer» `2Omega` |
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| 7865. |
There are three voltmeters of same range with resistance 10,000Omega,8000Omegaand6000Omega, Which of these, can measure voltage most galvanometer. accurately ? |
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Answer» `10,000Omega` |
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| 7866. |
In a projectile motion the velocity a) is always perpendicular to the acceleration b) is not always perpendicular to the acceleration c) is perpendicular to the acceleration for one instant only d) is perpendicular to the acceleration for two instants. |
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Answer» a & B are correct |
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| 7867. |
What happens to a light ray when it travels from air into water? Which angle is more angle of incidence or angle of refraction? |
| Answer» Solution :When LIGHT ray travels from RARER medium to DENSER medium it bends towards the normal. HENCE angle of incidence is greater than angle of refraction. | |
| 7868. |
In a modified Young's double-slit experiment, source S is kept in front of slit S_(1) as shown in the following figure. Find the phase difference at a point that is equidistant from slits S_(1) and S_(2) and point P that is in front of slit S_(1) in the following situations: If a liquid is filled between the slit and the source S, what is the phase difference? |
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Answer» `(2PI)/(lambda)[nsqrt(d^(2)+x_(0)^(2))-x_(0)-(d^(2))/(2D)]` |
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| 7869. |
An electron hole pari is formed when light of maximum wavelength6000 Å is incident on the semiconductor. What is the band gap energy of the semiconductor ? (h=6.62xx10^(-34)Js) |
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Answer» `2.07xx10^(-19)J` Band GAP energy `E=(hc)/(LAMBDA)` `=(6.62xx10^(-34)xx3xx10^(8))/(6xx10^(-7))` `=3.31xx10^(-19)J` |
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| 7870. |
To determine the volume expansion coefficient of kororone, one end of a U-tube filled with it was held at 10^@C,and the other at 80^@C. The level of liquid in one tube was 280 mm and in the other 300 mm. Find the coefficient. |
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Answer» `(h_1)/(1 + beta t_1) = (h_2)/(1 + beta t_2) , beta = (h_2 - h_1)/(h_1 t_2 - h_2 t_1)` . 
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| 7871. |
A Capacitor of capacitance C_1=1 mu F can withstand a maximum voltage V_1=6.0 KV while another capacitor of capacitance C_2=2.0 muF withstands the maximum voltage V_2=4.0 KV . What maximum will the system of these two capacitance withstand when connected in series as shown below . |
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Answer» 5.0 KV |
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| 7872. |
The capacity of a parallel plate condenser with air as dielectric is 2 mu F. The space between the plates is filled with dielectric slab with K=5. It is charged to a potential of 200V and disconnected from cell. Work done in removing the slab from the condenser completely |
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Answer» `0.8J` |
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| 7873. |
In what way is diffraction from each slit related to the interference pattern in a double-slit experiment? |
| Answer» SOLUTION :The intensity of interference fringes in a double-SLIT experiment is MODULATED by the DIFFRACTION pattern of each slit. | |
| 7874. |
An X-ray tube is operated at a potential of 42,000V. The maximum frequency of the X-ray radiation produced is approximately |
| Answer» Answer :A | |
| 7875. |
The given figure shows an equilateral triangle ABC. A positive point charge |q is located at each of the three vertical A, B and C. each side of the triangle is of length a. A point charge Q (that may be positive or negative) is placed at the mid-point between B and C. Is it possible to choose the value of Q (that is non-zero) such that the force on Q is zero ? Explain why or why not. |
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Answer» Yes, because the FORCES on Q are vector and three vectors can add to ZERO |
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| 7876. |
An ideal diode and a 5 fl resistor are connected in series with a 15 V power supply as shown in figure below. Calculate the current that flows through the diode. |
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Answer» SOLUTION :The diode is forward biased and it is an ideal one. Hence, it ACTS like a closed switch with no barrier VOLTAGE. Therefore, current that flows through the diode can be calculated using Ohm’s LAW. `V=IR` `I=V/R` `15/5=3A` |
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| 7877. |
The magnitude of the electrostatic force between point charges q_(1) = 26.0 mu C and q_(2) = 47.0 mu C is initially 5.70 N. The separation is the changed such that the force magnitude is then 0.570 N. (a) What is the ratio of the new separation to the initial separation ? (b) What is the new separation ? |
| Answer» SOLUTION :(a) 3.16, (B) 4.39 m | |
| 7878. |
A slab of material of dielectric constant K has the same area as the plates of a parallel-platecapacitor but has a thickness ((3)/(4)) d where d is the separation of the plates. How Is the capacitance changed when the slab ls inserted between the plates ? |
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Answer» Solution :Let `E_(0) = (V_(0))/(d) be the ELECTRIC field between the plates when there is vacuum (air) and here the potential difference is V 0. If the dielectric is now inserted, the electric field in the dielectric will be`E = (E_(0))/(K)`where K is dielectric constant The potential difference will then be `V= (V_(0))/(4) +(3)/(4) .(V_(0))/(K)` `= (E_(0)d)/(4) +(3)/(d) .(E_(0)d)/(K)` `=E_(0)d[(1)/(4)+(3)/(4K)]` `= V_(0)[(K+3)/(4K)]` Hence, the potential difference decreases by the factor `(K+3)/(4K)`while the free charge `Q_(0)`on the plates remains unchanged. The capacitance thus increase, `C = (Q_(0))/(V) = (4K)/(K+3) .(Q_(0))/(V_(0))` `:. C = (4K)/(K+3).C_(0)[ because (Q_(0))/(V_(0))=C_(0)]` |
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| 7879. |
The given figure shows an equilateral triangle ABC. A positive point charge |q is located at each of the three vertical A, B and C. each side of the triangle is of length a. A point charge Q (that may be positive or negative) is placed at the mid-point between B and C. Determine an expression for the magnitude and sign of Q so that the net force on the charge at A is zero newtons. |
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Answer» `Q = +q((3sqrt(3))/(4))` |
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| 7880. |
What is astronomical unit? |
| Answer» SOLUTION :It is the MEAN DISTANCE of EARTH from SUN `=1.496xx10^11m` | |
| 7881. |
A direct currenty I flowsin a long straightconductorwhosecross-sectionalhas the fromof a the half-ring of radius R. The samecurrent flows in the oppositedirectionalonga thin condocutorlocatedon the "axis"of the first conductor (point O in FIg). Find teh magneticinteractionforce betweenthem reducedto a unitofthier length. |
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Answer» Solution :The MAGNITUDE FIELD due to the conductorwith semicircular cross section is `B = (mu_(0) I)/(pi^(2) R)` Then `(del F)/(del l) = BI = (mu_(0) l^(2))/(pi^(2) R)` |
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| 7882. |
A plane mirror M is arranged parallel to a wall W at a distance /from it: The light produced by a point.source S kept on the wall is reflected by the mirror and produces a patch of light on the wall. The mirror moves with velocity v towards the wall. Which of the following statement(s) is/are correct? |
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Answer» The PATCH of LIGHT will move with speed v on the WALL |
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| 7883. |
Obtain an expression for the torque on a current carrying loop placed in a uniform magnetic field. Describe the construction and working of a moving coil galvanometer. |
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Answer» Solution :Torque acting on a coil carrying a CURRENT kept in a uniform magnetic field : Let a reactangular current loop ABCD of length l =AB =CD and width b=AD=BC carrying a current ''I'' be suspended in magnetic field of flux density B. The normal ON drawn to the plane of the coil makes an angle `'theta'` with the magnetic field B.  Force on arm `AD= i barbxxbarB` acting upwards along the axis of suspension Force on arm `BC=i barbxxbarB` acting downwards along the axis of suspension Hence these two forces cancel.  Force on arm AB= ilB acting perpendicular to the plane as shown. Force on arm CD=ilB acting perpendicular to the plane as shown. These two forces constitute a couple on the coil. Moment of the couple = (Force) `xx` (Perpendicular distance between the forces) `= ilB (PQ sin theta)` Torque `=ilB b sing theta`. But `lxxb`= Area of coil `therefore" Torque "=iAB sin theta` If the loop has 'n' turns the troque on the coil `tau=n i AB sin theta` If `'phi'` is the deflection of the coil, that angle between the plane of the coil and magnetic field B `tau= n i AB cos phi` Moving coil galvanometer : Principle : When a current carrying coil is placed in the uniform magnetic field, it experiences a torque. Construction : i) It consists of a coil wound on a non metallic frame. ii) A rectangular coil is suspended between two concave shaped magnetic poles with the help of PHOSPHOUR Bronzee wire. iii) The lower portion of the coil is connected to a spring. iv) A small plane mirror M is fixed to the phosphour Bronze wire to measure the deflection of the coil. v) A small SOFT iron cylinder is placed with in the coil without touching the coil. The soft iron cylinder increases the induction field strength. vi) The concave shaped magnetic poles render the field radial. So maximum torque acting on it. vii) The whole of the apparatus is kept inside a brass case provided with a glass window.  Theory : Consider a rectangular coil of length l and breadth b and carrying current i suspended in the induction field strength B. Deflecting torque `(tau)=" B i A N" rarr (5)` where A= Area of the coil N = Total number of turns. The restoring torque developed in the suspension `=C theta rarr(2)` Where C is the couple PER unit twist and `theta` is the deflection made by the coil. When the coil is in equilibrium position Deflecting torque = Restoring torque `"B i A N "= C theta` `=((C)/("BAN"))theta` where `K=(C)/("BAN")=` Galvanometer constant. `i=K theta rarr (3)` `i prop theta` Thus delection of the coil is directly proportional to the current flowing through it. The deflection in the coil is measured using lamp and scale arrangement. |
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| 7884. |
The distance travelled by a particle performing S.H.M. from its mean position in 2 second is equal to sqrt3/2A. Then what is its period ? |
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Answer» SOLUTION :`x=Asin omegat` `sqrt(3A)/2=A SIN omegatsqrt3/2=sin((2PI)/T)` `thereforesin(pi/3)=sin((2pi)/2T)` `pi/3=(2pi)/Txxt=(2pi)/Txx2` `T=4xx3=12sec.` |
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| 7885. |
Nano wires are used in ............. |
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Answer» TRANSISTORS |
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| 7886. |
The objective and eye piece of an astronomical telescope are double convexlenses with refractive index 1.5. When the telescope is adjusted to infinity the seperation between the lenses is 16 cm. If the space between the lenses is now filled with water and again telescope is adjusted for infinity, Then the present separation between the lenses is |
| Answer» Answer :4 | |
| 7887. |
A 800 turn coil of effective area 0.05 m^2 is kept perpendicular to a magnetic field 5 xx 10^(-5) T. When the plane of the coil is rotated by 90^@around any of its coplanar axis in 0.1 s, the emf induced in the coil will be |
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Answer» 0.02 V Induced emf , `E=(-N phi_2-phi_1)/(Deltat) =(-N(AB cos 90^@ -AB cos 0^@)/(Deltat)` `=-(N(0-AB))/(Deltat)=(NAB)/(Deltat)` `=(800xx0.05xx5xx10^(-5))/0.1` `=2xx10^(-2)` V =0.02 V |
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| 7888. |
Find the refreactive index of metallic sodium for electrons with kinetic energy T=135 eV. Only one free electron is assumed to correspond to each sodium atom. |
Answer» Solution : The potential energy inside the metal is `-U_(0)` for the ELECTRON and it related to the work function `A` by `U_(0)=E_(F)+A` If `T` is the `K.E` of electron outside the metal, its `K.E` inside the metal will be `(E+U_(0))`. On entering the metal electron canot experience any tangential force so the tangential componet of mometum is unchanged. Then `(2mT)SIN alpha=sqrt(2m(T+U_(0))) sin beta` Hence `(sin alpha)/(sin beta)=sqrt(1+(U_(0))/(T))=n` by DEFINATION ofrefractive index. In sodium with ONE free electron PER `Na` atom `n=2.54xx10^(22) per c.c` `E_(F)= 3.15eV` `A= 2.27eV`(from table) `U_(0)= 5.42eV` `n=1.02` |
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| 7889. |
State Bohr's quantum condition for stationary orbits in terms of de-Broglie wavelength. |
| Answer» Solution :TOTAL length of an orbital PATH of electron in nth state of atom should be n times (where n is an INTEGER), the value of de-Broglie WAVELENGTH `lambda` . Thus`2pi r_(n) = n lambda_(de)` | |
| 7890. |
The time in seconds required to produce a potential difference of 20V across a capacitor of 1000mu F when it is charged at the steady rate of 2004 mu C/s is |
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Answer» 50 |
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| 7891. |
In the relation y=rsin (omegat-kx) the dimensions of (omega)/(k) are : |
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Answer» `[M^(0)L^(0)T^(0)]` Here `omegat=angle:.omega=(1)/(T)=T^(-1)` SIMILARLY `kx`=angle`impliesk=(1)/(L)=L^(-1)` `:.(omega)/(k)=(T^(-1))/(L^(-1))=LT^(-1)` or Simply `(omega)/(k)` REPRESENT WAVE velocity as `(omega)/(k)=(2piv)/(2pi//lambda)=vlambda` Hence correct choice is `(b)`. |
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| 7892. |
Find de-Broglie wavelengthcorresponding to the root-mean square velocity of hydrogen molecules at room temperature (20^@C). |
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Answer» <P> |
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| 7893. |
Plot a graph of resistivity of a semiconductor as a function of absolute temperature. |
Answer» SOLUTION :
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| 7894. |
The earth's magnetic field at a given point is 0.5xx10^(-5) "Wb m"^(-2) . This field is to be annulled by magnetic induction at the centre of a circular loop of radius 5 cm. The current required to be flown in the loop is nearly |
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Answer» 0.2 A `therefore I=(2B_H R)/mu_0 =(2xx0.5xx10^(-5)xx0.05)/(4PIXX10^(-7))`= 0.3978 `approx` 0.4 A |
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| 7895. |
A point charge q is placed at origin. Let E_A, E_B " and " E_c be the electric fields at three points A (1,2,3) , B (1,1,-1) and C (2,2,2) respectively due to the charge q. Then , the relation between them is 1. E_A bot E_B ""2.E_A || E_c 3. |E_B|=4|E_c| "" 4.|E_B|=8|E_c| |
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Answer» 1,4 are correct `E_(A)= { (KQ)/((1^2 + 2^2+3^2)^(3//2))} ( hat I + 2 hat j + 3 hat k)` `E_(B)= (kq)/((1^2+1^2+(-1)^2)^(3//2)).(hati+hatj-HATK)` `E_( C)= (kq)/((2^2+2^2+2^2)^(3//2)).(2 hat i+2 hatj+2 hatk)` `E_(A)= (kq)/(14^(3//2))(hati+2hatj+3hatk)` `E_(B)= (kq)/(3^(3//2))(hati+hatj-hatk)` `E_(C)= (kq)/(12^(3//2))(2hati+hatj+2hatk)` As, `E_(A)_|_E_(B)=0 and E_B =4 |E_C|` |
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| 7896. |
A cup of tea cools from 65.5^@C" to " 62.5^@C in 1minute in a room at 22.5^@CHow long will it take to cool from 46.5^@C " to " 40.5^@CC in the same |
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Answer» 4 minutes (I) `3/1=k(64-22.5)` `3/1=k(41.5)""...(1)` (II) `6/t=k(43.5-22.5)` `6/t=k(21)" "...(2)` `((1))/((2))rArr3/1xxt/6=41.5/21` t = 4 MIN |
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| 7897. |
If the displacement of a particle in S.H.M. is y=0.5 sin(4pit+phi), The frequency is |
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Answer» 1 Hz |
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| 7898. |
भारत कौन- से देश के बाद सबसे बड़ा चावल का उत्पादक देश है? |
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Answer» बांग्लादेश |
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| 7899. |
If the surface is a perfect reflector. The change in momentum of the wave after falling on the surface is : |
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Answer» Solution :When the WAVE falls on perfect reflector then its momentum is reversed. so change iin momentum `DELTA p = p_(2) - p_(1) = - p -= - 2p` . So CORRECT choice is d. |
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| 7900. |
Centripetal force needed for an electron to revolve in itsstableorbitaround the nucleus is provided by________. |
| Answer» SOLUTION :ELECTROSTATIC FORCE | |