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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.
| 40151. |
A solenoid pi m long and 5.0 cm in diameter has two layers of windings 1000 turns each and carries a current of 5 A .The magnetic induction at its centre along the axis : |
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Answer» `4xx10^3 T` |
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| 40152. |
An object falling through a static water column is observed to have an acceleration given by the relation ang-bu where g is acceleration due to gravity and 'b' is a constant. After sufficiently long time of its release, it is observed to fall with a constant speed. What must be the value of its constant speed ? |
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Answer» v=g/b After a long intervel the viscous restisance become constant and BODY BEGINS to fall with a consatnt SPEED. As at that moment a=0 `:.` 0=g-b.v `:.` bv=g or v=g//b |
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| 40153. |
अन्तः केंद्रित इकाई कोशिका की भुजा a तथा गोले की त्रिज्या r में सम्बन्ध क्या |
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Answer» `a=2r` |
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| 40154. |
Mention any two differences between primary rainbow and secondary rainbow. |
Answer» SOLUTION :
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| 40155. |
The resolving power of a microscope is basically determined by the |
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Answer» speed of the light used |
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| 40156. |
A series LCR circuit is connected to a 220 V variable frequency supply. If L = 20 mH, C = [(800)/(pi^(2))]muF and R = 110 Omega, (a) Find the frequency of the source for which the average power absorbed by the circuit is maximum. (b) Calculate the value of maximum current amplitude. |
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Answer» Solution :Given, `L = 20 mH = 20 XX 10^(-3)H` `C = (800)/(pi^(2)) = muF = (800)/(pi^(2)) xx 10^(-6)F` `R = 110 Omega` (a) `v = (1)/(2pi sqrt(LC))` `= (1)/(2pi sqrt(20 xx 10^(-3) xx (800 xx 10^(-6)))/(pi^(2)))` `= (1)/(2sqrt(16 xx 10^(-6))) = (1)/(8 xx 10^(-3))` `= (1000)/(8) = 125 Hz` (b) `I_(max) = (E)/(Z_(min)) = (E )/(R )` `= (220)/(110) = 2A` |
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| 40157. |
In Young's experiment, fringe width was found to be 0.8 mm. If whole apparatus is immersed in water of refractive Index, mu = 4/3, new fringe width is |
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Answer» 0.6 |
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| 40158. |
The V-Igraph is given for two conductors of same area and length. If sigma_(1)and sigma_(2) are the cnductivities of the conductors 1 and 2 respectively, (sigma_(1))/(sigma_(2))= |
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Answer» Solution :`sigmaprop1/R"When l and A=same"` `Rprop tan theta,sigmaprop(1)/(tan theta)` `(sigma_(1))/(sigma_(2))=(TAN30^(@))/(TAN60^(@))=1/3` |
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| 40159. |
A nucleus of mass 218 amu is in free state decays to emit an alpha-particle. Kinetic energy of alpha-particle emitted is 6.7Mev. The recoil energy in (MeV) emitted by the daughter nucleus is |
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Answer» `1.0` |
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| 40160. |
A fish is at a depth of 12 cm is water is viewed by ann observer on the bank of a lake. To what height the image of the fish is raised in cm? (mu_(w)=4//3) |
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| 40161. |
Calculate the capacitance of a sherial capacitor consisting of two concentric spherical conductros of radii a and b(bgta), when the outer one is charged and the inner is earthed. The space between the spheres is filled by insulating material of relative permittivity epsilon_(r). |
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Answer» |
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| 40162. |
A cylindrical bar magnet is kept along the axis of a circular coil. If the magnet is rotated along its axis, then |
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Answer» A CURRENT will be INDUCED in the COIL |
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| 40163. |
The time period when current in inductor loop grows to a maximum value or when current decays to zero are called .............. of the circuit. |
| Answer» SOLUTION :TRANSIENT STATE | |
| 40164. |
What is the ratio of radii of the orbits corresponding to first excited state and the ground state in a hydrogen atom. |
| Answer» Solution :Radius r PROP `n^2` for GROUND STATE n=1 and for the first EXCITED state n=2 ,`r^2//r^1=2^2//1^2`=4:1 | |
| 40165. |
In an interference arrangementsimilarto Young.sdouble - slit experiment , the slitsS_(1) ans S_(2)are illuminatedwith coherent microwavesources . Eachof frequency 10^(6) Hz . The sources are synchronized to have zero phase difference and the slits are separated by a distanced = 150.0 m . The intensity I(theta) is measured at a large distance as a function of theta , where thetais defined as shown . If I_(0) is the maximum intensity , then (theta) for 0 le theta le 90^(@) is given by |
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Answer» `I(theta) = I_(0)//2 ` for `theta = 30^(@)` |
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| 40166. |
The height and width of each step of a staircase are 20cm and 30cm respectively. A ball rolls off the top of a stair with horizontal velocity v and hits the fifth step. The magnitude of vis (g=10 ms^(-2)) |
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Answer» `1.5 SQRT(5) ms^(-1)` |
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| 40167. |
Use the expression vecF = q(vecv xx vecB) to define the SI unit of magnetic field. |
| Answer» SOLUTION :As per expression `vecF = Q(vecv xx vecB)`, we have `F = q v B` where `theta` is the angle between the directions of v and B . From this expression we define MAGNETIC field B in space at a given point as the maximum value of FORCE acting on a unit positive charge moving with unit velocity there. | |
| 40168. |
Refraction of ray passing through tank filled with water An observer is viewing along the line shown in Fig. 34-4. When there is no water filled in the tank, he can see only wall AB and no other part of the base of the tank. Then water is filled in till height h and he is just able to see the point E as shown in Fig. 34-3. Find the angle of refraction into air (phi) and angle of incidence from water (phi). Also find the depth to which water has been filled. |
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Answer» Solution :Here, we can SEE that refraction of ray enables the observer to see E. So, we can apply Snell.s law and geometry. Calculation : From Fig. 34-3 , we can see that the normal and wall AB are parallel to each other. So, angle of refraction into air is `tantheta=((160)/(3))/(40)impliestheta=53^(@)` From Snell.s law, we can say that `1xxsin theta=(4)/(3)sinphiimpliesphi=37^(@)` From Fig. 34-3, we can say that `tan53^(@)=(x)/(40-h)` and `tan37^(@)=(30-x)/(h)`. Solving, we get `h=40cm`.  Figure 34-3 when the water is filled in a TANK, we can see the hidden part of the tank as well with the help of refracted RAYS. |
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| 40169. |
What did Noodle suggest about the books? |
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Answer» That they are the COMMUNICATION devices |
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| 40170. |
de-Broglie wave length is expressed as: |
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Answer» `(2.3)/(SQRT(V))Å` |
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| 40171. |
Spiders and small insects move easily on the surface of water without sinking because - |
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Answer» They can swim on water |
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| 40172. |
(a) For what kinetic energy of a neutron will the associated de Broglie wavelength be 1.40 xx 10^(-10) m? (b) Also find the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of (3/2) k T at 300 K. |
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Answer» Solution :(a) `6.686xx10^(-21)J=4.174xx10^(-2)eV` (B) 0.145 nm |
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| 40173. |
How old was Evelyn when she went to Royal Music academy? |
| Answer» Answer :D | |
| 40174. |
Ozone layer in atmosphere is useful because it : |
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Answer» stops ULTRAVIOLET radiation |
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| 40175. |
If a bar magnet of magnetic moment 80 units be cut into two halves of equal lengths, the magnetic moment of each half is, |
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Answer» 40 units |
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| 40176. |
A current of i amperes is flowing through each of,the bent wires as shown. Find the magnitude and direction of magnetic field at O. |
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Answer» Solution :`B(theta)= (mu_0I)/(4piR)theta,,B_("net")=(mu_0I)/(4pi)[(pi//2)/(R) + (3pi//2)/(R^l)] = (mu_0I)/(8)[1/R + 3/(R^l)]` (NORMALLY into the PAGE) |
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| 40177. |
Two cells with same e.m.f. .E. and different internal resistances r_(1) and r_(2) are connected in series to an external resistance .R.. The value of R so that the p.d. across the first cell be zero is |
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Answer» `SQRT(r_(1)r_(2))` |
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| 40178. |
The potential difference between the two ends of the three mponents of L-C-R series A.C. circuit are V_L,V_C and V_R respectively. Then voltage of A.C. source is …….. |
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Answer» `V_L+V_C+V_R` |
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| 40179. |
Obtain the formulafor the effective capacitance of the parallel combinationof different n capacitors. |
Answer» Solution :FIGURE shows n capacitors of capacitance `C_(1),C_(2), cdots , C_(n)` are arranged in parallel.  Here potential difference across each capacitor is same but the plate charges are different to each capacitor. Suppose, `Q_(1),Q_(2),cdots, Q_(n)` are the charges on the capacitors `C_(1),C_(2), cdots, C_(n)` respectively. The total charge `Q= Q_(1)+Q_(@)+cdots, Q_(n)` But `Q_(1)=C_(1)V_(1), Q_(2)=C_(2)V_(2),cdots, Q_(n)=C_(n)V` `:. Q=C_(1)V+C_(2)V+cdos+C_(n)V` `:. (Q)/(V)=C_(1)+C_(2)+cdots C_(n)` If `(Q)/(V)` is the EFFECTIVE capacitance of combination then `C=C_(1)+C_(2)+ cdotsC_(n)` The effective capacitance of n numbers of capacitors in parallel is equal to the ALGEBRAIC sum of the individual capacitance of capacitors. The effective capacitance of the capacitors in parallel is more than capacitance of any one of the parallel connected capacitors. |
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| 40180. |
A slab of material of dielectric constant K hasthe same area as the plates of a parallel capacitor, but hasa thickness ((3)/(4) d), where d is the separation of the plates. How is the capacitancechangedwhen the slab is inserted between the plates |
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Answer» Solution : ` V = E_o ((d)/(4))+(E_o)/(K)((3d)/(4)) = E_o d ((K+3)/(4K)) ` ` V = V_o (( K + 3 )/(4K)) ` ` C = (Q_o)/(V)= (4K)/(K + 3) (Q_o)/(V_o) = (4 K )/(K + 3 ) CO` |
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| 40181. |
State the methods of connection of p-n junction. |
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Answer» Solution :The p-n junction can be BIASED in TWO WAYS. (i) FORWARD bias (ii) Reverse bias |
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| 40182. |
In one dimensional motion a 1 kg body experiences a force which is function of time and is F= 2 r in the direction of motion. The work done by the force in first 4 sec. is : |
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Answer» 16 J |
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| 40183. |
Which direction would a compass needle point to if located on the magnetic north or south pole ? |
| Answer» Solution :If it is capable of rotating in a vertical plane, it will ALIGN vertical, PERPENDICULAR to the surface. If it is free to ROTATE in a horizontal plane, it may stay in any direction because `B_H` = 0 | |
| 40184. |
A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3 Omega. L = 25.48mH. And C = 796mu F. The impedance of the circuit |
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Answer» <P> Solution :(a) To find the impedance of the circuit, we first calculate `X_(L) and X_(C )`.`X_(L)=2pi vL` `=2xx3.14xx50xx25.48xx10^(-3)Omega=8Omega` `X_(C )=(1)/(2pi vC)` `=(1)/(2xx3.14xx50xx796xx10^(-6))=4Omega` Therefore, `Z=sqrt(R^(2)+(X_(L)-X_(C ))^(2))=sqrt(3^(2)+(8-4)^(2))` `=5Omega` (b) Phase difference, `phi="tan"^(-1) (X_(C )-X_(L))/(R )` `=tan^(-1)((4-8)/(3))= -53.1^(@)` Since `phi` is NEGATIVE, the current in the circuit LAGS the VOLTAGE across the source. (c) The power dissipated in the circuit is `P=I^(2)R` Now, `I= (i_(m))/(sqrt(2))=(1)/(sqrt(2)) ((283)/(5))=40A` Therefore, `P=(40A)^(2)xx3Omega=4800W` (d) Power factor `=cosphi=cos(-53.1^(@))=0.6` |
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| 40185. |
According to Einstein's photoelectric equation graph of kinetic energy of emitted photo electrons from metal versus frequency of incident radiation is linear .Its slope…… |
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Answer» depends on TYPE of metal used |
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| 40186. |
The apparent depth of ater in water in cylindrical water tank of diameter 2R cm is reducing at the rate of x cm//min when is being drined out a constant rate. The amount of water drained in cc//min is (n_1=refractive index of air, n_2=refractive index of water) |
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Answer» `(xpiR^2n_1)/n_2` `:. d=n_2/n_1 d_(app)` `-d/(DT) =n_2/n_1(-d/(dt) d_(app))=(n_2/n_1)X` `(DV)/dt=A[-d/(dt)(d)]=(xpiR^2n_2)/n_1` |
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| 40187. |
A rod ABC shown in the figure is made up of two parts . The part AB is non - conducting while the part BC is conducting . The rod rotates with constant angular speed omega about an axis perpendicular to its length and passing through the end A. The axis carries a constant current I. The emf iduced across the ends A and C is |
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Answer» Zero |
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| 40188. |
Using the Rydberg formula, calculate the wavelengths of the first four spectral lines in the Lyman series of the hydrogen spectrum. |
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Answer» SOLUTION :The Rydberg formula is `(hc)/(lambda_(l))=(me^(4))/(8epsi_(0)^(2)H^(2))(1)/(n_(f)^(2)-(1)/(n_(i)^(2)))` The wavelengths of the first four lines in the Lyman SERIES correspond to transitions from `n_(1) = 2, 3, 4, 5, "to" n_(f) =1` . We know that `(me^(4))/(8epsi_(0)^(2)h^(2))=13.6eV=21.76xx10^(-19)J` `:. lambda_(s)=(hc)/(21.76xx10^(-19))((1)/(1)-(1)/(n_(i)^(2)))m` `=(6.625xx10^(-34)xx3xx10^(8)xx n_(i)^(2))/(21.76xx10^(-19)xx(n_(i)^(2)-1))` `=(0.9134 n_(i)^(2))/((n_(i)^(2)-1))xx10^(-7)m` `=(913.4n_(i)^(2))/((n_(i)^(2)-1)Å)` Substituting `n_(i)=2,3,4,5` we get `lambda_(21)=1218Å`, `lambda_(31)=1028Å, lambda_(41)=974.3Å, and lambda_(51)=951.4019Å` |
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| 40189. |
Calculate the potential at a point P due to a charge of 4 × 10^(-7) C located 9 cm away. |
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Answer» <P> Solution :Here `Q = 4xx10^(-7)`Cr=9 cm `9XX10^(-2)` m Potential at a point P, `V=(KQ)/(r)` `=(9xx10^(9)xx4xx10^(-7))/(9xx10^(-2))` `=4xx10^(4)V` |
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| 40191. |
A boy of mass 40 kg is climbing on a vertical pole at a constant speed. If the coefficient of friction between his palms and the pole is 0.8 and g = 10 m//s^(2), the horizontal force that he is applying on the pole is |
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Answer» 300 N LET F be horizontal force that the BOY is APPLYING on the pole. The various forces are acting on the boy as shown in the figure. FRICTIONAL force,f= μN= mg `N=(mg)/(mu)=(40xx10)/(0.8)=500N` F=N=500N 
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| 40192. |
A thin convex lens has focal length f_(1) and a concave lens has focal length f_(2). They are kept in contact. Now, match the following two columns. |
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Answer» <P> `f_(1)and P_(1)` are POSITIVE, `f_(2) and P_(2)` are negative. If `ABS(f_(1))gtabs(f_(2))`, POWERS of system is negative If `abs(f_(1))ltabs(f_(2))`, powers of system is positive . |
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| 40193. |
A transmitting antenna is at a height of 20m and receiving antenna is at a height of 80m. The maximum distance between satisfactory communication is them for |
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Answer» 16 km |
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| 40194. |
Two capacitors of 10 PF and 20 PF are connected to 200 V and 100 V sources respectively. If they are connected by the wire, what is the common potential of the capacitors ? |
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Answer» 133.3 volt |
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| 40195. |
Light from a narrow source falls on a razor's edge at a distance of 20 cm from the source. Calculate the separation between the first and fourth maxima on a screen held 30 cm away from the edge. (Wavelength of light =6000 Å) |
| Answer» SOLUTION :`11.04xx10^(-4) m` | |
| 40196. |
(A): Motion of conduction electrons in electric field barE is the sum of motion due to random collisions and that due to barE. (R): The contribution to V_(d) of electron comes only by barE but not by random motion of electrons. |
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Answer» Both 'A' and 'R' are true and 'R' is the CORRECT explanation of 'A' |
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| 40197. |
Column 1 gives four different situations involving two blocks of mass m_(1) and m_(2) placed in different ways on smooth horizontal surface as shown. In each of the situations, horizontal forces F_(1) and F_(2) are applied on blocks of mass m_(1) and m_(2), respectively and also m_(2)F_(1) lt m_(1)F_(2). Match the statements in column I with the corresponding results in column II. |
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| 40198. |
Which of the following curves shown below can possibly represent electrostatic field lines? |
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Answer»
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| 40199. |
Two parallel wires carry current in the same direction. The wires |
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Answer» ATTRACT each other |
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| 40200. |
Radioactive isotopes are produced in a nuclear physics experiment at a constant rate dN/dt = R. An inductor of inductance 100 mH, a resistor of resistance 100Omega and a battery are connected to from a series circuit. The circuit is switched on at the instant the production of radioactive isotope starts. It is found that i/N remains constant in time where i is the current in the circuit at time t and N is the number of active nuclei at time t. Find the half-life of the isotope. |
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