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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.
| 37101. |
A 42 kg slab rests on a frictionless floor A 9.7 kg block rests on the top of the slab as shown in the figure. The co-efficient of static friction between the block and the slab is 0.53, while the co-efficient of kinetic friction is 0.38. The 9.7 kg block is acted upon by a horizontal force of 110 N. What are the resulting accelerations of (a) the block, and (b) the slab? (Take g=9.8m//s^(2)) |
Answer» SOLUTION : (a) From FBD of m `F-mu_(k)mg=ma_(2)` `a_(2)=(F)/(m)-mu_(k)g=((110N))/((9.7kg))-(0.38)(9.8m//s^(2))` `=7.6m//s^(2)` (towards left) (B) From FBD of M `f=Ma_(1) rArr (mu_(k)mg)/(M)=a_(1)` `a_(1)=((0.38)(9.7kg)(9.8m//s^(2)))/((42kg))` `=0.86m//s^(2)` (towards left) |
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| 37102. |
What is the significance of Lenz's law ? |
| Answer» SOLUTION :Lenz.s LAW VERIFIES the law of CONSERVATION of ENERGY. | |
| 37103. |
A source contains two phosphorus radio nuclides " "_(15)^(32)P(T_(1/2) = 14.3 d) and " "_(15)^(33)P(T_(1/2) = 25.3 d). Initially, 10% of the decays come from " "_(15)^(33)P . How long one must wait until 90% do so ? |
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Answer» Solution :Here at t=0 if activity of `" "_(15)^(32)P` be `R_(0)` and `R_(0)^(.)` that of `" "_(15)^(33)P` then it is given that `R_(0)`=10% of (`R_(0) + R_(0)^(.))` i.e., `R_(0)^(.) =9 R_(0) or R_(0)/R_(0)^(.)= 1/9` Let after a time of t days, new activities of `" "_(15)^(32)P` and `" "_(15)^(33)P` be R and R. such that R=9 R.. But from RADIOACTIVE decay law `R= R_(0)e^(-lambdat) and `R. = R_(0)^(.)e^(-LAMBDA^(.)t). `therefore R/(R.)=R_(0)/R_(0)^(.)e^(-lambdat)/e^(-lambda^(.)t)=R_(0)/R_(0)^(.)e^((lambda^(.)-lambda)t)=R/R_(0)^(.).e^(-(0.693/T_(1/2)^(.)-0.693/T_(1/2))t)` Substituting the values, we have `9=1/9.e((0.693/14.3-0.693/25.3)t) or 81=e^((0.693/14.3-0.693/25.3)t) or log_(e)81=(0.693(25.3-14.3))/(14.3xx25.3).t IMPLIES t=(log_(e)81xx14.3xx25.3)/(0.693xx11)=(2.303log_(10)81xx14.3xx25.3)/(0.693xx11)=204.2d`. |
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| 37104. |
, (A) and (B) are - |
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Answer» Chain ISOMER  Position of `-OH` group is change in above |
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| 37105. |
ML^0T^-2 is the dimensional formula of |
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Answer» MOMENT of inertia |
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| 37106. |
A particle moving with a constant acceleration describes in the last second of its motion 9//25^(th) of the whole distance .If it starts from rest,how long us the particle in motion and through what distance does it moves if it describes 6 cm in the first sec.? |
| Answer» Answer :A | |
| 37107. |
The power factor of A.C. circuit of resistance 12Omegaand impedance 15Omega is …… |
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Answer» 1.25 |
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| 37108. |
The two lenses of a compound microscope are of focal lengths 2 cm and 5 cm. If an object is played at a distance of 2.1 cm from the objective of focal length 2 cm the final image forms at the least distance of distinct vision of a normal eye. Find (i) the image distance of the image formed by the objective (ii) the magnification of the objective (iii) the distance between the objective and eyepiece and (iv) the magnifying power of the microscope. |
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Answer» |
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| 37109. |
Can there be a potential difference between two adjacent conductors carrying the same the charge ? |
| Answer» Solution :Yes, Since V =Q/C. If CONDUCTORS are different, there capacitances will also be different. i.e., V GETS CHANGE for the same value of charge Q. | |
| 37110. |
An a.c. voltage is applied to a pure inductor L and drives a current in the inductor. The currentn in the inductor would be |
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Answer» ahead of the voltage by `pi/2` |
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| 37111. |
A toroid has a core (non-ferromagnetic) of inner radius 25 cm and outer radius 26 cm, around which 3500 turns of a wire are found. If the current in the wire is 11 A, what is the magnetic field (i) outside the toroid, (ii) inside the core of the toroid, and (iii) in the empty space surrounded by the toroid. |
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Answer» Solution :As per DATA GIVEN, MEAN radius of toroid `R = (25 + 26)/(2) = 25.5 cm = 0.255 m` Number of turns per unit length, `n = N/(2 pi R) = (350)/((2 pi) xx (0.255)) m^(-1)` (i) Magnetic field outside the toroid = 0 (ii) Magnetic field inside the core is `B = mu_0 I = (4 pi xx 10^(-7) xx 350 xx 11)/(2pi xx (0.255)) = 3.0 xx 10^(-2) T` (iii) Magnetic field in the empty space surrounded by the toroid is zero. |
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| 37112. |
A truck accelerates from speed v to 2 v. Work done in during this is |
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Answer» three times as the work done in accelerating it from REST to v. `:.(W_1)/(W_2)=(1/2m((2v)^2-v^2))/(1/2m(v^2-0))` =`(4v^2-v^2)/(v^2)=3` So`W_1=3W_2` |
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| 37113. |
A ray of light falls on a transparent glass slab of refractive index 1.53. If the refracted ray is perpendicular to the refracted ray, thenthe angle of incidence is approximately to : |
| Answer» ANSWER :D | |
| 37114. |
Net dipole moment per unit volume is called _____ |
| Answer» SOLUTION :MAGNETISATION | |
| 37115. |
Draw the circuit symbol of p-n- p transistor. |
Answer» SOLUTION :
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| 37116. |
Length of a string tied to two rigid supports is 40 cm. Maximum wavelength in cm of a stationary wave produced on it is |
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Answer» 20cm |
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| 37117. |
Statement A : Shock absorbers reduced the magnitude of change in momentum.Statement B : Shock absorbers increase the time of action of impulsive force |
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Answer» A & B are TRUE |
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| 37118. |
A pure inductor of 25.0 mH is connected to a sourceof 220V . Find the inductivereactance and rms current in the circuit if the frequency of the source is 50 Hz . |
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Answer» SOLUTION :`X_L= 2pi V L = 2 xx 3.14 xx 50 xx 25 xx 10^(-3)W= 7.8 Omega` The rms current in the circuit is `I = (V)/(X_L) = (220V)/(7.85 Omega) = 28A` |
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| 37119. |
Define magnetic inclination (I). |
| Answer» Solution :The angle SUBTENDED by the earth.s total MAGNETIC field `vec(B)` with the horizontal direction in the magnetic meridina is CALLED dip or magnetic INCLINATION (I) at the point. | |
| 37120. |
Charge on a body which carries 200 electrons is |
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Answer» `-3.2xx10^(-18)C` |
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| 37121. |
A ray of light falls on a transparent sphere with centre at C as shown in figure. The ray emerges from the sphere parallel to line AB. The refractive index of the sphere is |
| Answer» ANSWER :A | |
| 37122. |
A force of 16 N acts on a ball of mass 80 gm for one micro-second. Calculate the acceleration and impulse |
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Answer» `200 m / s^2`, `16x10^-6` N / SEC |
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| 37123. |
A perfect gas at 27^(@)C is heated at constant pressure to 327^(@)C. If original volume of gas 27^(@)C is V then volume at 327^(@)C is : |
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Answer» V |
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| 37124. |
If both the speed of a body and radius of it's circular path are doubled, What will happen to the centripetal force ? |
| Answer» SOLUTION :The CENTRIPETAL FORCE will be DOUBLED. | |
| 37125. |
A glass sphere (mu=1.5) of radius 20 cm has a small air bubble 4 cm below its centre. The sphere is viewed from outside and along a vertical line through the bubble. The apparent depth of the bubble below the surface of sphere is (in cm) |
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Answer» 13.33 |
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| 37126. |
Which of the following behaves as paramagnetic substance ? |
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Answer» Nickel |
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| 37127. |
Consider a thin square sheet ofside L and thickness t, made of a material of resistivity rho. The resistance between two opposite faces, shown by the shaded areas in the figure is |
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Answer» directly PROPORTIONAL to L Here, `l=L , A=Lt` `THEREFORE R=(rhoL)/(Lt) = rho/t` Hence, the resistance between two opposite FACES , shown by the SHADED areas in the figure is independent of L. 
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| 37128. |
observer S detects two flashes of light. A big flash occurs at x_(1)=1000 m and, slightlylater, a small flash occurs at x_(2)=480m. The time interval between the flashes is Deltat=t_(2)-t_(1). What is the smallest value of Deltat for which observer S' will determine that the two flashes occur at the same x' coordinate? |
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Answer» |
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| 37129. |
The electric intensity ( E ), current density (J) and specific resistance (P) are related to each other |
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Answer» `E= J/rho` |
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| 37130. |
Suppose you are given a chance to repeat the alpha- particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below 14 K.) What results do you expect? |
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Answer» Solution :The NUCLEUS of a hydrogen atom does not contain neutrons it contains only one proton and its mass is `1.67 xx 10^(-27) kg` and hence mass of hydrogen nucleus is `1.67 xx 10^(-27 )` kg. The a-particle nucleus is MADE up of 2 protons and 2 neutrons hence mass of nucleus of - particle is `= 4 xx 1.67xx 10^(-27) kg` `=6.68xx10^(-27)kg` Since the mass of the scattered a-particle is greater than the mass of nucleus of hydrogen atom, so it cannot be thrown back in head on collision but it will pass through the atom of hydrogen without deviating from its PATH. This is the case where a football COLLIDES with a tennis ball that is at REST. |
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| 37131. |
Solve Problem 25.7 assuming that the plates remain connected to the voltage source. |
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Answer» Solution :In this case, too, the capacitance of the capacitor increases as the LIQUID rises, but the energy of the ELECTRIC field is not conserved but increases. Besides, the potential energy of the rising water increases as well. Is this not in contradiction with the LAW of conservation of energy? Of course not. The power supply PERFORMS work to raise the liquid, the increase in the energy of the system being equal to the work of the power supply in displacing the charges to the capacitor plates: `W_(p.s)=DeltaW_(el)+Delta_("pot")` But `W_(p.s)=Deltaq. varphi=(C-C_(0))varphi^(2)` Substituting, we obtain the result sought after some simple transformations. |
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| 37132. |
Find the magnetic induction at point O, if the current carrying wire is in the shape shown in the figure. |
| Answer» SOLUTION :`(mu_0 i)/(4 PI R) [ 3/2 pi + 1]` | |
| 37133. |
A vertical wire carries a current of 17 A. If the neutral point is observed at a distance of 10 cm from the wire, calculate the horizontal compoinent of earth's magnetic field . |
| Answer» SOLUTION :`0.34 XX 10^4 T` | |
| 37134. |
A terrestrial telescope has an objective of focal length 180cm and an eyepiece of focal length 5.0cm. The erecting lens has focal length of 3.5cm. What is the separation betwee the objective and the eyepiece?What is the magnifying power of the telescope? |
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Answer» Solution :The image formed by the objective FALLS at 2F of the erecting lens which forms an ERECT image of the same size. The separation between the objective and the EYEPIECE = `f_0+f_e+4xxfocal length of the erecting lens `180+5+4xx3.5` = 199cm. Magnifying power of the TELESCOPE `f_o/f_e` = 180/5 = 36 |
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| 37135. |
State the law which relates to the generation of induced emf in a conductor being moved in a magnetic field. Apply this law to obtain an expression for the induced emf when one .rod. of a rectangular conductor is free to move in a uniform, time independent and.normal. manetic field. Apply the concept of the Lorentz (magnetic) force acting on a moving charge to justify the expression obtained. |
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Answer» Solution :Farday.s law of electromagnetic induction : It states that the induced emf is equal to the rate of change of magnetic flux. That `E = - (d phi)/(dt)`...(i) But `E = vec(oint)vec(E)*vec(dl)` and `phi= oint_(S) vec(B)*vec(dS)`...(ii) Hence the above relation may be written as `oint vec(E) * vec(dl) = oint_(S) (d vec(B))/(dt) d vec(S)` ...(iii) Induced emf by Changing the Area A. Suppose a uniform magnetic field B perpendicular to the plane of the paper and directed outwards [shown by dots in Fig.] is confined in region PQRS. Leta rectangular loop of wire KLMN be held partially in the magnetic field in the plane of the paper. Let `KL=l`. Suppose `x` is the portion of the loop in the field at any instant. t. When the loop is moved with a velocity `v` in the plane of the paper, area of the loop associated with the magnetic field changes. An induced emf is set up in the wire. This is detected by deflection in the galvanometer G CONNECTED in the loop.  To calculate the emf induced, suppose in a small time `Delta t`, the loop is moved out of magnetic field through a small distance `Delta x`. `:.` Decrease in area of the loop `= - l Delta x` Decrease in magnetic flux linked with the loop `d phi = -Bl Delta x` As induced emf, `epsilon = - d phi // dt` `:. epsilon = (B l Delta x)/(Delta t) = Blv` i.e., `epsilon = Blv` It is also possible to explain the motional emf expression by invorking the Lorentz force acting on the free CHARGE CARRIERS of conductor PQ. Consider any arbitrary charge q in the conductor PQ. When the rod moves with a speed v, the charge will also be moving with speed v in the magnetic magnitude, and its direction is towards Q. All charges EXPERIENCE the same force, in magnitude and direction, irrespective of their positionin the rod PQ. The work DONE in moving the charge from P to Q is. `W = q v Bl` Since emf is the work done per unit charge. `epsilon = W/q` `=Blv` This equation gives emf induced across the rod PQ. |
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| 37136. |
Which one of the following involves analog quantities? |
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Answer» NUMBER of ATOMS in a sample of material |
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| 37137. |
Calculate the magnetic induction at a point on the axial line of a bar magnet. |
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Answer» Solution :Magnetic field at a point along the axial line of the magnetic dipole (bar magnet): Consider a bar magnet NS. Let N be the North Pole and S be the south pole of the bar magnet, each of pole strength `q_(m)` and separated by a distance of 2l. The magnetic field at a point C (lies along the axis of the magnet) at a distance from the geometrical CENTER O of the bar magnet can be computed by keeping unit north pole `(q_(MC)"= 1 Am")` at C. The force experienced by the unit north pole at C due to pole strength can be computed using Coulomb.s law of magnetism as follows: The force of repulsion between north of the bar magnet and unit north pole at point C (in free space) is `vecF_(N) =(mu_(0))/(4pi) (q_(m))/((r-1)) hati ............(1)` where r-l is the distance between north pole of the bar magnet and unit north pole at C. The force of altration between South Pole of the bar magnet and unit North Pole at point C (in free space) is `veF_(S) (q_(m))/((r+l)^(2)) hatl ...........(2)` where r+1 is the distance between south pole of the bar magnet and unit north pole at C.  From equation (1) and (2), the net force at point C is `vecF=vecF_(N)+vecF_(S)`. From definition, this net force is the magnetic field due ot magnetic dipole at a point `C(vecF=vecB)` `vecB=(mu_(0))/(4pi (r-l)^(2))hati+((mu_(0))/(4pi) (q_(m))/((4+l)^(2))hati)=(mu_(0)q_(m))/(4pi) ((1)/((r-l)^(2))-(1)/((r+l)^(2)))hati rArr vecB=(mu_(0)2R)/(4pi) ((q_(m).(2l))/((r^(2)-l^(2))^(2)))hati..........(3)` Since, magnitude of magnetic dipole MOMENT is `|vecp_(m)|=p_(m)=q_(m).2l` the magnetic field point C equation (3) can be written as `vecB_("axial")=mu_(0)/(4pi) ((2rp_(m))/((r^(2)-l^(2))^(2)))hati ...........(4)` If the distace between two poles in a bar magnetic are small (looks like short magnet) compared to the distance geometrical centre O of hte bar magnet and the location of point C i.e, `r gt gt 1" then "(r^(2)-l^(2))^(2) approx r^(4) .............(5)` Therefore, using equation (5) in equation (4), we get `vecB_("axial")=mu_(0)/(4pi) ((2rp_(m))/(r^(3)))hati=mu_(0)/(4pi) 2/r^(3) vecp_(m)` Where `vecp_(m)=p_(m) hati. ............(6)` |
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| 37138. |
A particle is vibrating in simple harmonic motion with an amplitude of 4 cm. At what displacement from the equilibrium position is its energy half potential and half kinetic? |
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Answer» `SQRT(2)cm` |
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| 37139. |
If an inflated tyer bursts, the air escaping out is cooled, why ? |
| Answer» Solution :When the tyre BURSTS, there is adiabatic expansion, the air because the pressure of air inside is sufficiently GREATER than ATMOSPHERIC pressure. In expansion, the air does some work against sorrundings and so its internal energy decreases.This causes a fall in TEMPERATURE. | |
| 37140. |
Which is the most potent and versatile art form? |
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Answer» Cinema |
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| 37141. |
Which of the following series fall in the visible range of electromagnetic spectrum of a hydrogen atom ? |
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Answer» Brackett SERIES |
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| 37142. |
Consider three charged bodies P, Q and R. If P and Q repel each other and P attracts R, what is the nature of force between Q and R ? |
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Answer» <P>Attractive, because Q and R will CARRY opposite charges. |
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| 37143. |
When two progressive waves of intensity I_(1)and I_(2) , but slightly different frequencies superpose, the resultant intensity fluctuates between |
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Answer» `(SQRT(I_(1)) + sqrt(I_(2)))^(2)` and `(sqrt(I_(1)) - sqrt(I_(2)))^(2)` `I_("Resultant") = l_(1) + l_(2) + 2sqrt(I_(1)I_(2) cos phi)` Maximum value of intensity is at `phi =0` and MINIMUM value of intensity is at `phi = 180^(@)` `therefore` Maximum intensity is, `I_(rms) = I_(1) + I_(2) + 2sqrt(I_(1) + I_(2)) cos 0^(@) = (sqrt(I_(1)) + sqrt(I_(2))^(2))` and the minimum intensity is, `I_("MIN") = I_(1) + I_(2) +2sqrt(I_(1)I_(2)) cos 180^(@)` `(sqrt(I_(1) - sqrt(I_(2))^(2)))` |
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| 37144. |
The potential difference between points A and B in the given uniform electricl field is : |
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Answer» EA |
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| 37145. |
The magnetic moment of paramagnetic material is |
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Answer» infinity |
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| 37146. |
A point source S is placed at the bottom of different layers as shown in the figure. The refractive index of bottom most layer is mu_0 . The refractive index of any other upper layer is mu(n)= mu_0 -(mu_0)/(4n - 18) where n = 1, 2, ...... A ray of light with angle i slightly more than 30^@ starts from the source S. Total internal reflection takes place at the upper surface of a layer having n equal to |
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Answer» 3 |
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| 37147. |
N(lt100) molecule of a gas have velocities 1,2,3….N, km/s respectively. Then |
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Answer» rms SPEED and average speed of molecules are same `v_(av)=(1+2+...+N)/(N)=(N(N+1))/(2N)=(N+1)/(2)` rms speed, `v_(rms)=sqrt((1^(2)+2^(2)+3^(2)+...+N^(2))/(N))` `=sqrt((N(N+1)(2N+1))/(6N))=sqrt(((N+1)(2N+1))/(6))` `(v_(rms))/(v_(av))=sqrt(((N+1)(2N+1))/(6))xx(2)/((N+1))=(2)/(sqrt(6))=sqrt(((2N+1)/(N+1)))` |
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| 37148. |
Distance between the centres of two stars is 10a. Themasses of these stars are M and 16M and their radii a and 2a respectively. A body of mass m is fired straight from the surface of the larger star towards the smaller star. The minimum initial speed for the body to reach the surface of smaller star is |
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Answer» `2/3sqrt((Gm)/a)`  i.e., once the body reaches P, the gravitational pull of attraction due to M takes the lead to make m move towards it AUTOMATICALLY as the gravitational pull of attraction due to 16 M vanishes i.e., a minimum KE or velocity has to be imparted to m from surface of 16 M such that it is just able to OVERCOME the gravitational pull of 16M. By law of conservation of energy `rArr 1/2mv^2+[-(G(16M)M)/(2a)-(GMM)/(8a)]+[-(GMm)/(2a)-(G(16M)m)/(8a)]=0` `rArr 1/2mv^2-(GMm)/(8a)(45)rArr v=3/2sqrt((5GM)/a)` |
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| 37149. |
A solenoid of length l metre has self inductance L henry. If number of turns are doubled, its self inductance |
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Answer» Becomes 2 L HENRY `(L)/(L.)=((N)/(N.))^(2)=((N)/(2N))^(2)=(1)/(4)RARR L.=4 L`henry |
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| 37150. |
Two point charges +Q_(1) and -Q_(2) are placed at A and B respectively. A line of force emanates from Q_(1) at an angle theta with the line joining A and B. At what angle will it terminate at B ? |
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Answer» Solution :We know that number of lines of force emerge is proportional to magnitude of the CHARGE. The field lines emanating from `Q_(1)`, spread out equally in all directions. The number of field lines or FLUX through cone of HALF ANGLE `THETA` is `(Q_(1))/(4pi) 2pi (1- cos theta)`. Similarly the number of lines of force terminating on `-Q_(2)` at an angle `phi " is " (Q_(2))/(4pi) 2pi (1- cos phi)`. The total lines of force emanating from `Q_(1)` is equal to the total lines of force terminating on `Q_(2)` `rArr (Q_(1))/(4pi) 2pi (1- cos theta) = (Q_(2))/(4pi) 2pi (1- cos phi) or (Q_(1))/(2) (1- cos theta) = (Q_(2))/(2) (1- cos phi)` `Q_(1) sin^(2) theta//2 = Q_(2) sin^(2) phi//2` `sin phi//2= sqrt((Q_(1))/(Q_(2))) sin theta//2 rArr phi= 2 sin^(-1) {sqrt((Q_(1))/(Q_(2))) sin theta//2}` |
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