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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.
| 8151. |
Figures (a) and (b) show the field lines of apositive and negative point charge respectively. (a). Gives the sign of the potential difference V_(P)-V_(Q) : V_(B)-V_(A) (b) Given the sign of the potential energy difference of a small negative charge between the points Q and P, A and b (c) Give the sign of the work done by the field in moving a small positive charge from Q to P. (d) Give the sign of the work done by external agencyin moving a small negative charge from B to A. (e) Does the kinetic energy of a small negative charge increase or decrease in going from B to A? |
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Answer» Solution :a. As `V alpha 1/r, V_(P) gt V_(Q)." Thus "(V_(P)-V_(Q))` is positive. Also `V_(B)` is less negative than `V_(A)`. Thus, `V_(B) gt V_(A) or (V_(B)-V_(A))` is positive. b. A small negative CHARGE will be ATTRACTED towards positive charge. The negative charge moves from higher potential energy to lower potential energy. Therefore the sign of potential energy difference of a small negative charge between Q and P is positive. Similarly. `(P.E.)_A gt (P.E.)_B` and hence sign of potential energy DIFFERENCES is positive. c. In moving a small positive charge from Q to P, work has to be done by an EXTERNAL agency against the electric field. Therefore, work done by the field is negative. d. In moving a small negative charge from B to A work has to be done by the external agency. It is positive. e. Due to force of repulsion on the negative charge, velocity decreases and hence the kinetic energy decreases in GOING from B to A. |
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| 8152. |
Calculate the capacitanceof a parallel plate condenser, withplate area A and distancebetween plates d, when filled with a dielectric whose dielectric constant varies as, K(x)=1+(beta x)/(in_(0)) "" 0 lt x lt (d)/(2), "" K(x)=1+(beta)/(in_(0))(d-x)"" (d)/(2) lt x lt d. For what value of beta would the capacity of the condenser be twice that when it is without any dielectric? |
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| 8153. |
The figure shows a uniform, 3.0 T magnetic field that is normal to the plane of a conducting, circular loop with a resistance of 1.5 Omega and a radius of 0.024 m. The magnetic field is directed out of the paper as shown . Note: The area of the non-circular portion of the wire is considered negligible compared to that of the circular loop. If the magnetic field is held constant at 3.0 T and the loop is pulled out of the region that contains the field in 0.2 s, at what rate is energy dissipated in R? |
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Answer» `1.8xx10^(-2)W` |
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| 8154. |
For a transistor, the value of alpha= 0.9, then the value of B is |
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Answer» 9 |
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| 8155. |
A chain AB of length lis located on a smooth horizontal table so that its fraction of length h hange freely with end B on the table. At a certain moment, the end A of the chain is set free. With what velocity with this end of the chain slip off the table? |
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Answer» Solution :Let m be the mass per unit length of the CHAIN. Let at any instant x be the length of hanging PART of chain and T the tension in chain. Then x mg-T= x ma ....... (1) and T=(l-x)ma...... (2) ADDING these equations, we get x mg = [x m + (l -x) m ]a = l ma ( or) ` a=(x)/(l) g , " As a " (dv)/(dt) implies (dv)/(dt)= (x)/(l) g ` Multiplying both sides by `(dx)/(dt) , ` we get ` (dx)/(dt) (dv)/(dt) = (x)/(l) g(dx)/(dt) or vdv = (x)/(l) g dx ` INTEGRATING both sides , we get ` int_0^(v) v dv = (g)/(l) int_h^(l) x dx ` `[(v^(2))/(2)]_0^(v)=(g)/(l)[(x^(2))/(2)]_(H)^(l)=(g)/(l)[(l^(2)-h^(2))/(2)] :. v=sqrt( {(g(l^(2)-h^(2)))/(l)})` |
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| 8156. |
The value of galvanometer constant (G) is equal to : |
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Answer» `2 PI N` |
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| 8158. |
A 4 diopter lens is used as a magnifier. Where should the object be placed to obtain maximum angular magnification for normal eye? |
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| 8159. |
In Fig. 27-9, what is the potential difference V._d -V_c between points d and c if epsi_(1)= 4.0 V. epsi_(2) - 1.0 V, R_1= R_2= 10Omega, and R_3= 8.0 Omega, and the battery is ideal? |
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| 8160. |
The pressure variation in a sound wave is given by DeltaP = 12 sin (8.18 x - 2700 t + pi//4) N//m^2.Find the displacement amplitude. |
| Answer» SOLUTION :`1.04 XX 10^(-5) m` | |
| 8161. |
In vernir callipers, m divisions of main scale with (m+1) divisions of vernier scale. If each division of main scale is d units, the least countof the instrument is |
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Answer» `d//(m-1)` |
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| 8162. |
What are the hazards of working in the glass bangles industry? |
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Answer» POOR health |
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| 8163. |
A particle moves in X-Y plane under the action of forces F such that the values of linear momentum 'p' at any times is p_(x) = 2 cos t and p_(y) = 2 sin t . The angle between vec(F) and at the time t will be: |
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Answer» Solution :Here `VEC(p)=vec(p_(x))hati + vec(p_(y))hatj=2costhati + 2sinthatj` |
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| 8164. |
“Magnetic mirror" is the term for the region of the magnetic field in which there is an intense concentration of lines of induction (Fig. 28.19a). Suppose a charged particle approaches a magnetic mirror, as shown in the figure. What will happen to it? |
Answer» Solution :As was shown above (see Problem 28.5), the particle will move along a helix winding around the lines of induction. Let us resolve  the velocity vector into two components: the LATERAL `v_(BOT)` (the orbital velocity) and the longitudinal (the drift velocity). Since the orbital moment of the CIRCULATING charge is directed against llice field, the magnetic forces Lend to push the charge OUL into the region of the weak field (compare with 37.4 and 4.10, where the dipole and Taxele moments are arranged it direction of the field). We see that as the charge approaches the magnetic MIRROR, its drift velocity decreases, becoming zero in the case of a sufficientiy high field gradient (fig. 28.19h). From this moment it begins to drilt in the opposite direction towards the weaker field (Fig. 28. 19c). 
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| 8165. |
If the maximum couple acting on a magnet in a field of induction 0.2T is 10Nm, what is its magnetic s moment? |
| Answer» SOLUTION :`50Am^(2)` | |
| 8166. |
A satellite is currently orbiting Earth in a circular orbit of radius R, its kinetic energy is K_(1), if the satellite is moved and enters a new circular orbit of radius 2R, what will be its kinetic energy? |
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Answer» `(K_(1))/(4)` |
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| 8167. |
What are alpha particles ? In the reaction ""_(z)X^(A) rarr Y + alpha-particle, give the atomic number and mass number of Y. |
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Answer» `""_(z)X^(A) RARR Y + alpha`-particle When `alpha(""_(2)He^(4))` is emitted, ATOMIC number decreases by 2 and mass number 4, then Y is written as `""_(z-2)Y^(A-4)` `""_(z)X^(A) rarr ""_(z-2)Y^(A-4)rarr ""_(2)He^(4)` |
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| 8168. |
Wavelength of K_(a). X-ray of an element Aislambda_(1) is and wavelength of K_(a) X-ray element B is lambda_(2) (lambda_(1))/(lambda_(2))is equal to(1)/(4) and Z_(1) and Z _(2)are the atomic numbers of element A and B respectively. Then relation between Z_(1) and Z_(2) are given by |
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Answer» `2Z_(2)-Z_(1)=1` |
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| 8169. |
What is the magnitude of the centripetal acceleration in above problem? |
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| 8170. |
What one means by view finder? |
| Answer» SOLUTION :The astronomical TELESCOPE has a large magnifying power but a small field of view. Hence, a small telescope with wide field of view and small magnifying power is fitted parallel to it to FACILITATE the focusing of DISTANT OBJECTS. This small telescope is called view finder. | |
| 8171. |
The relative permeability is represented by mu_(r) and the susceptibility is denoted by x for a magnetic substance.Then for a paramagnetic substance |
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Answer» `mu_(r) lt and X lt 0` |
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| 8172. |
A solid sphere of mass 1 kg rolls on a table with linear speed 2 m/s. Find its total kinetic energy. |
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Answer» Solution :DATA : M = 1 kg, v = 2 m/s The total kinetic energy of a rolling body, ` E = 1/2 Mv^(2) (1 + K^(2)/R^(2))` For a solid sphere, `k^(2) = 2/5 R^(2)` ` :. E = 1/2 Mv^(2)(1 + (2/5 R^(2))/R^(2)) = 1/2 Mv^(2) (1 + 2/5)` ` = 1/2 xx 7/5 Mv^(2) = 7/10 Mv^(2)` ` = 7/10 xx 1 xx 2^(2)` ` = (7 xx 4 )/10 = 2. 8 ` J |
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| 8173. |
Uncertainty in frequency of laser is 2000 Hz.Uncertainty in its periodic time will be……. |
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Answer» `5xx10^(-31)S` Now E=hf`therefore DeltaE=hDeltaf` `therefore hDeltaf=(h)/(2piDeltaT)` `therefore T=(h)/(2pihDeltaf)=(1)/(2piDeltaf)` `therefore DeltaT=(1)/(2xx3.14xx2000)=0.079xx10^(-3)` `therefore DeltaT~~7.9xx10^(-5)s=8xx10^(-5)S` |
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| 8174. |
The electrostatic potential on the surface of a charged conducting sphere is 100V. Twostatements arc made in this regard : S_(1) : At any point inside the sphere, electric intensity is zero. S_(2) : At any point inside the sphere, the ' electrostatic potential is 100V . Which of the following is a correct statement ? |
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Answer» `S_1 ` istruebut ` s_2` isfalse `E=-(dV)/(dr)` If electric field intensity E=0 then `(dV)/(dr)=0` `:. V` = constant Hence, uniform electric potential 100 V will be at any POINT inside the sphere when E = 0 in charged sphere. |
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| 8175. |
Ifa skingdiverfillshis lungsto full capacityof 5.5 Lwhen12belowthe surfaceto whatvolumewouldhis lungs expandIfhe quciky roseto the surface ? Is thisadvisable? |
| Answer» SOLUTION :11.9L , No | |
| 8176. |
A coil carries a current and experience a torque due to a magnetic field. The value of the torque is 80% of the maximum possible torque. The angle between the magnetic field and the normal to the plane of the coil is |
| Answer» ANSWER :D | |
| 8177. |
Estimate the distance for which ray optics is good approximation foran aperture of 4mm and wavelength 400 nm. |
| Answer» Solution :Expressionfor Fresnel distance is `z_(F)=(a^(2))/(LAMDA)=((4xx10^(-3))^(2))/(400xx10^(-9))=40M` | |
| 8178. |
A monochromatic light of wavelengthlambda is incident on a isolated metallic sphere of radius a. The threshol wavelength islambda_0 which is larger thanlambda. Find the number of photoelectrons emitted before the emission of photoelectrons will shop. |
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Answer» Solution : As the metallic sphere is isolated, it become POSITIVELY charged when electrons are ejected from it. There is an extra attractive force on the photoelectrons . If the potential of the sphere is raised to V, the electron should have a minimum energy `( varphi + eV)` to be able to COMES out. Thus, emission of photoelectrons will stop when ` (hc/ (lambda))= varphi + eV` ` = hc/ (lambda_0) + eV` or, `V= hc/e ((1/(lambda))-(1/(lambda_0))).` ` The charge on the sphere needed to take its potential ` to V is ` ` Q= (4 pi (epsilon_0)a) V.` ` The number of electrons emitted is, therefore, ` ` n= (Q/e)= (4 pi (epsilon_0)a V/ e)` ` (4 pi (epsilon _0) AHC /e^2)( 1/ lambda- (1/ lambda_0)).` |
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| 8179. |
The magnetic field in a plane electromagnetic wave is given by: By=12 xx 10^(-8) sin (1.20 xx 10^7z+3.60 xx 10^(15)t)T. Calculate the Speed. of the wave |
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Answer» speed `=3 XX 10^(8) m//s`. |
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| 8180. |
In a series LCR alternating current circuit under resonancecondition voltage across is exactly equal to the volt age across …………… |
| Answer» SOLUTION :INDUCTOR, CAPACITOR | |
| 8181. |
When an object is placed normallyon the principal axis of a spherial mirror at a distance 'a' from its pole, its image is formed at a distance 'v' from the pole of mirror such that (1)/(v)+(1)/(u)=(1)/(f), where f = focal length of given mirror. The relation is called mirror formula and is true for all types of mirrors under all conditions. However, values of u, v and f must be put with proper signs are per the sign convention followed. If 'h' be the height of the linear object and h' the height of image, then ratio (h')/(h) is called linear magnification or lateral magnification and its value m is given as : m=(h')/(h)=-(v)/(u) What is the surface area of image of linear object formed by the given concave mirror? |
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Answer» SOLUTION :`because"Linear magnification m"=(h.)/(h)=-(v)/(u)` `therefore"Areal magnification "=(A.)/(A)=m^(2)=(v^(2))/(u^(2))` `rArr""A.=(v^(2))/(u^(2))xxA=((60)/(30))^(2)xx6cm^(2)=24cm^(2)` |
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| 8182. |
An iron bar of length 10 cm and diameter2cm is placed in a magnetic field of intensity 1000Am^(-1) with its length parallel to the direction of the field. Determine the magnetic moment produced in the bar if permeability of its material is 6.3 xx 10^(-4) TmA^(-1). |
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Answer» SOLUTION :Here `H=1000 Am^(-1) , MU = 6.3 xx 10^(-4) TmA^(-1)` , `l=10 cm , d=2 cm ` Radius of the iron bar = 1CM = `10^(-2)m` we known that , `mu= mu_0 ( 1 + chi_m)` `implies chi_m = (mu)/(mu_0) -1 = (6.3 xx 10^(-4))/(4pi xx 10^(-7))-1` `=500.6` Intensity of magnetisation , ` I = CHIH = 500 .6 xx 1000` `=5 xx 10^(5) Am^(-1)` Magnetic moment , ` M=IV = 5 xx 10^(5) xx pi xx (10^(-2))^(2) xx 0.1 = 5 pi Am^(2)` |
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| 8183. |
When an object is placed normallyon the principal axis of a spherial mirror at a distance 'a' from its pole, its image is formed at a distance 'v' from the pole of mirror such that (1)/(v)+(1)/(u)=(1)/(f), where f = focal length of given mirror. The relation is called mirror formula and is true for all types of mirrors under all conditions. However, values of u, v and f must be put with proper signs are per the sign convention followed. If 'h' be the height of the linear object and h' the height of image, then ratio (h')/(h) is called linear magnification or lateral magnification and its value m is given as : m=(h')/(h)=-(v)/(u) A small match box lies along the principal axis of a concave mirror and one end of match box is at centre of curvature of the mirror. Draw a ray diagram to show its image. Is its magnetification uniform? |
| Answer» Solution :Ray diagram Fig. 9.13 is SHOWN here. A. B. is the image of the match box AB. Its MAGNIFICATION is non - UNIFORM i.e., the image is distored. | |
| 8184. |
Give information about electromagnets. |
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Answer» Solution :It is known as the ELECTROMAGNET that only have magnetic property until the electric current from the ferromagnetic material is passed. Core of electromagnets are made of ferromagnetic MATERIALS. These materials have high permeability and low retentivity. So soft iron is a SUITABLE material for electromagnets. On PLACING a soft iron rod in a solenoid and passing a current, the magnetism of the solenoid increased by thousand fold. When we switch off the solenoid current, the magnetism is switched off because the soft iron core has a low retentivity.  In transformer core and telephone diaphragms, the material goes through an ac cycle of magnetisation for a long period. The hysteresis curve of such materials must be NARROW, so the energy dissipated and the heating will be small. The material must have high resistivity to lower eddy current losses. Uses of electromagnets : Electromagnets are used in electric bells, loudspeakers and telephone diaphragms. Giant electromagnets are used in cranes to lift machinery and bulk quantities of iron and steel. |
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| 8185. |
When an object is placed normallyon the principal axis of a spherial mirror at a distance 'a' from its pole, its image is formed at a distance 'v' from the pole of mirror such that (1)/(v)+(1)/(u)=(1)/(f), where f = focal length of given mirror. The relation is called mirror formula and is true for all types of mirrors under all conditions. However, values of u, v and f must be put with proper signs are per the sign convention followed. If 'h' be the height of the linear object and h' the height of image, then ratio (h')/(h) is called linear magnification or lateral magnification and its value m is given as : m=(h')/(h)=-(v)/(u) (d) Define longitudinal magnification. |
| Answer» Solution :LONGITUDINAL MAGNIFICATION is defined as the ratio of length of image to the length of an OBJECT held along the principal axis of the mirror. | |
| 8186. |
When an object is placed normallyon the principal axis of a spherial mirror at a distance 'a' from its pole, its image is formed at a distance 'v' from the pole of mirror such that (1)/(v)+(1)/(u)=(1)/(f), where f = focal length of given mirror. The relation is called mirror formula and is true for all types of mirrors under all conditions. However, values of u, v and f must be put with proper signs are per the sign convention followed. If 'h' be the height of the linear object and h' the height of image, then ratio (h')/(h) is called linear magnification or lateral magnification and its value m is given as : m=(h')/(h)=-(v)/(u) (e) Find an expression for longitudinal magnification of a small object placed linearly along the axis of a spherical mirror. |
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Answer» Solution :Let linear size of object be du andlinear size of its image be dv. As per mirror FORMULA, we have `(1)/(v)+(1)/(u)=(1)/(f)` On differentiation, we GET `(-dv)/(v^(2))-(du)/(u^(2))=0` `rArr""(dv)/(v^(2))=-(du)/(u^(2))` `therefore"LONGITUDINAL magnification "=(dv)/(dy)=-(v^(2))/(u^(2))` 
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| 8187. |
The variation of frequency of carrier wave with respect to the amplitude of the modulating signal is called_____________ |
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Answer» AMPLITUDE MODULATION |
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| 8188. |
A 50 kW carrier is to be modulated to a level of80%. What is the total side band power in each case. |
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Answer» <P> SOLUTION :`P_("SIDE BAND")=1/2 m^2 P_("CARRIER")``=1/2(0.8)^2xx50=16kW` |
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| 8189. |
Momemtum of a photon with wavelength lambda is______(i) hlambda(II)h/lambda (III)lambda/h (IV)h+lambda |
| Answer» SOLUTION :`h/lambda` | |
| 8190. |
A gravitational field due to a mass distribution is E-K//r^(3) in the X-direction. (K is a constant). Taking the gravitational potential to be zero at infinity, its value at a distance x is : |
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Answer» K/x |
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| 8191. |
Wavelength of H_alpha line in balmer series is approximately_____A.U. |
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Answer» 6000 |
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| 8192. |
When an object is placed normallyon the principal axis of a spherial mirror at a distance 'a' from its pole, its image is formed at a distance 'v' from the pole of mirror such that (1)/(v)+(1)/(u)=(1)/(f), where f = focal length of given mirror. The relation is called mirror formula and is true for all types of mirrors under all conditions. However, values of u, v and f must be put with proper signs are per the sign convention followed. If 'h' be the height of the linear object and h' the height of image, then ratio (h')/(h) is called linear magnification or lateral magnification and its value m is given as : m=(h')/(h)=-(v)/(u) (a) A planer object of length 3 cm and breadth 2 cm is held normal to the principal axis of a concave mirror of focal length 20 cm at a distance of 30 cm from its pole. Where is its image formed? |
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Answer» Solution :(a) Here `u=-30cm, f=-20cm` and area of PLANAR object `A=lb=3xx2=6cm^(2)`. `THEREFORE""(1)/(v)=(1)/(f)-(1)/(u)=(1)/((-20))-(1)/((-30))=-(1)/(60)` `RARR""v=-60cm` thus, image is formed in front of mirror at a distance of 60 cm from the POLE. |
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| 8193. |
An electric charge 10^(-3)mu C is placed at the origin (0,0) of x-y coordinate system. Two points A and B are situated at (sqrt2, sqrt2) and (2,0) respectively. The potential difference between the points A and B will be |
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Answer» zero `:. V_A = V_B = q/(4 PI epsi_0.r)` and hence `V_A - V_B = 0` |
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| 8194. |
What is the relation between polarizing angle and critical angle ? |
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Answer» SOLUTION :`n=tantheta` `n=frac{1}{sinc}`thereforetantheta=frac{1}{sin c}` where THETA is the polarizing ANGLE and c is the CRITICAL angle. |
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| 8195. |
A particle moves in x-y plane. The position vector of particle at any time t is vecr={(2t)hati+(2t^2)hatj} m. The rate of change of thetaat time t = 2 s (where theta is the angle which its velocity vector makes with positive x-axis) is |
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Answer» `2/17 "rad s"^(-1)` Comparing it with standard equation of position vector , `vecr=xhati+yhatj` , we get x=2t and `y=2t^2` `rArr v_x=(dx)/(dt)=2` and `v_y=(dy)/(dt)=4t` `THEREFORE tan THETA = v_y/v_x = (4t)/2=2t` Differentiating with respect to time we get , `(sec^2 theta ) (d theta)/(dt)=2` or `(1+tan^2 theta) (d theta)/(dt) = 2` or `(d theta)/(dt) =2/(1+4t^2)` at t=2 s , `((d theta)/(dt))=2/(1+4(2)^2)=2/17 "rad s"^(-1)` |
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| 8196. |
The radiations from the sun have maximum wavelength of 5760Å. The surface temperature of the sun is : (b=0.288xx10^(-2)"mK") : |
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Answer» 6000 K `rArrT=(0*288xx10^(-2))/(5760xx10^(-10))` `=5000K`. THUS CORRECT choice is (b). |
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| 8197. |
Obtain an expression for the average power supplied to a series RLC circuit. Discuss the average power when the series RLC circuit behaves as a pure resistive, inductive or capacitive circuit. |
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Answer» <P> Solution :For a series RLC circuit, instantaneous power supplied is given by`P=vi` i.e., `P=(v_(m)sinomegat)(i_(m)sin(omegat+phi))` `=(v_(m)i_(m))/(2)[cosphi-cos(2omegat+phi)]` The average power supplied, `barp=p_(av)=(v_(m)i_(m))/(2)cosphi` because `ltcos(2omegat)gt =0,v_(m)=sqrt(2)v_(RMS),i_(m)=sqrt(2)i_(rms)` i.e., `barp=v_(rms)i_(rms)cosphi` Power factor of an AC circuit, `cosphi=(R)/(Z)` so `barp=v_(rms)i_(rms)=(R)/(Z)` Case (i): For a series RLC to behave as a resistive circuit:- `X_(L)=X_(C)` so that `Z=Z_(min)=R,X_(L)-X_(C)=0` `therefore` Power factor `cosphi=(R)/(Z_(min))=1` Average power dissipation at resonance = `barp=v_(rms)i_(rms)` Case (ii): For a series RLC to behave as an inductive circuit:- `X_(L)gtX_(C)` i.e., `2pif_(H)Lgt(1)/(2pif_(H)C)` or `f_(H)gt(1)/(2pisqrt(LC))orf_(H)gtf_(0)`where, `f_(0)=(1)/(2pisqrt(LC))` Power factor `cosphi=(R)/(Z)=(R)/(sqrt(R^(2)+X_(L)^(2)))~~(R)/(X_(L))" for "X_(L)gtR`. Case (iii): For a series RLC to behave as a capacitive circuit :- `P.f=cosphi=(R)/(sqrt(R^(2)+X_(C)^(2)))" for "X_(C)gtX_(L)andf_(L)ltf_(0)` |
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| 8198. |
A magnetic needle of dipole moment 0.067 JT^(-1) oscillates with a period of 2/3 s in a uniform magnetic field of 100 G . Find the moment of inertia of the needle about the axis of oscillations. |
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| 8199. |
Lateral magnification of an object by spherical mirror is 0.3. If focal length of mirror is 30 cm, find type of mirror. |
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Answer» concave `m=(F)/(f-u)=(f-v)/(f)` (1) `m=(f)/(f-u)` `therefore 0.3 =(30)/(30-u)` `30-u=100` `therefore u=-70` cm `m=(f-v)/(f) `therefore 0.3=(30-v)/(30)` `therefore 9 =30-v` `therefore v=21` cm `therefore` IMAGE obtained behind the mirror, hence convex. |
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| 8200. |
A:When monochromatic beam of light is incident on metal surface ,it is dispersed in form of kinetic energy. R:Work function of metal depend on depth of metal. |
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Answer» Both assertion and reason are true and the reason is CORRECT explanation of the assertion. WORK function of metal DEPEND on type of metal. |
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