Saved Bookmarks
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.
| 37701. |
A magnetic field set up using Helmholtz coilsis uniform in a small region in a direction perpendicular to both the axis of the coils and the elctrostatic field. If the beam remains undeflected when the eletrostatic field is 9.0 xx 10^(+5) Vm^(-1)make a simple guess as to what the beam contains . Why is the answer not unique ? |
|
Answer» Solution :Since the particle remains undeflected `v=(E)/(B) = (9 XX 10^(-5)xx 100)/(75) = 12 xx 10^(15)= 1.2 xx 10^(-6) ms ^(-1)` `KE - 1/2 mv^2= EV "" therefore e/m = (v^2)/(2V)` `therefore e/m = ((1.2 xx 10^(+))^(2))/(2xx 15 xx 10^3) = ((1.2)^2 xx 10^(12))/(30 xx 10^3) = 4.8 xx 10^7 C kg^(-1)` For deuterium `e/m = (1.6 xx 10^(-19))/(2 xx 1.67 xx 10^(-27)) = 4.8 xx 10^7 C kg^(-1)` Hence the particleis deuterium . Othere particles are `He^(++), Li^(+++)` etc. The answer is not unique because only the RATIO of change of mass is determined . |
|
| 37702. |
Which physical quantities are represented by the product of moment of inertia and (i) angular velocity (ii) angular acceleration? |
| Answer» Solution :(i) ANGULAR momentum (ii) external TORQUE PRODUCING the angular ACCELERATION. | |
| 37703. |
Derive an expression for the energy stored in a parallel plate capacitor. On charginga parallelplate capacitor to a potential V, the spacing between the plates is halved, and a dielectric medium of in_(r)=10 is introduced between the plates, without disconnecting the d.c. source. Explain, using suitable expressions, how the (i) capacitance, (ii) electric field and (iii) electric field and (iii) energy density of the capacitor change. |
|
Answer» Solution :Energy stored in a parallel plate capacitor : Work is done in charging a capacitor. This work done is stored as its electrical POTENTIAL energy. Suppose a capacitor is charged with charge q shows that potential difference between its plates is `V=(q)/(C)`. Work done to increase the charge by an around DQ is dW=C dq `=(q)/(C)` dq Total work done to charge the capacitor from 0 to Q is `W=underset(0)overset(Q)int (q)/(C) dq=(1)/(C) [(q^(2))/(2)]_(0)^(Q) = (Q^(2))/(2C)` `:.` Energy of the capactior, `U=(1)/(2) (Q^(2))/(2C)=(1)/(2) QV [:' C=(Q)/(V)]` (i) Here `K=in_(r)=10 :' C=(in_(0)A)/(d) :' C_(1)=K(in_(0)A)/(d//2)=10xx2xx(in_(0)A)/(d) = 20C`. `:. ` Capacitance becomes 20 times. (ii) Electric field between the plates remains unchanged `(E_(1)=E)` because the p.d. across the plates remains unchanged. (iii) `U=(1)/(2) CV^(2)` `:. U_(1)=(1)/(2)C_(1)V^(2)=(1)/(2) (20C)V^(2)` `=20((1)/(2)CV^(2))=20U`. `:.` Energy stored increased 20 times. |
|
| 37704. |
The diameter of a stretched string is increased 3% keeping the other parameters constant then the velocity is x% decreases what is the value of x ? |
|
Answer» |
|
| 37705. |
The slits in Young's double-slit experiment have equal width's and the light source is placed symmetrically relative to the slits. The intensity of the central fringe's is I_(0).if one of the slits is closed, the intensity at the point will be |
|
Answer» `I_(0)` |
|
| 37706. |
If the angle of a prism is 60° and angle of minimum deviation is 40°, then the angle of refraction will be .... |
|
Answer» Solution :For MINIMUM DEVIATION, `r_1 = r_2 = R` and prism angle `A=r_1+r_2=r+r=2r` `therefore r=A/2=(60^@)/(2)=30^@` |
|
| 37707. |
A particle having mass m and charge q is , rest. On applying a uniform electric field Eit, it starts moving. What is its kinetic energ when it travels a distance y in the direction force ? |
|
Answer» `qE^(2)y` `DELTAK = W` `:. K - K_(0)= "Force" xx ` DISPLACEMENT `:. K -0 = F xxy "" [ because K_(0)=0]` `:. K = EQY"" [ because F= Eq]` |
|
| 37708. |
Eight equal drops of water are falling through air with a steady velocity of 10 cms^(-1).If the drops combine to form a single drop big in size,then the terminal velocity of this big drop is |
|
Answer» 20 cm/s `therefore4/3piR^(3)=8 4/3pir^(3)` `rArrR=2r` Since TERMINAL velocity, v`=2/(9eta).r^(2)[p-sigma]g` `rArrvpropr^(2)` `therefore` terminal velocity becomes 4 times as RADIUS becomes DOUBLE. `therefore` New terminal velocity `= 4 xx 10` = 40 emfs. Thus the CORRECT choice is (c). |
|
| 37709. |
In an apparatus, the electric field was found to oscillate with an amplitude of 18 V // m. The magnitude of the oscillating magnetic field will be : |
|
Answer» `4 xx 10^(-6) T` |
|
| 37710. |
A manstanding. Observesrainfallingwithvelocityof20m//satanangleof 30^@withthe verticalfind outvelocityof manso thatrainappearsto fallvertically |
|
Answer» Solution :` vecV_(R)=- 10hati- 10sqrt(3) hatj` ` vecV_m=-v_xhati` ` vecV_(RM)=- (10- v_x) HATI - 10 sqrt(3) hatj ` Anglewiththevertical`=30^@` ` impliestan30^@=( 10 -v_s )/( - 10sqrt(3)) impliesv_x= 20 m//s` 
|
|
| 37711. |
The wavelength of infrared rays is of the order of: |
|
Answer» `5 xx 10^(-7) m` |
|
| 37712. |
What is the thin prism? |
| Answer» Solution :A PRISM of very SMALL angle ( around `5^(@)` to `10^(@)` ] is CALLED a THIN prism. | |
| 37713. |
Obtain the answers (a) to (b) in Exercise 7.13 if the circuit is connected to a high frequency supply (240 V, 10 kHz). Hence, explain the statement that at very high frequency, an inductor in a circuit nearly amounts to an open circuit. How does an inductor behave in a d.c. circuit after the steady state ? |
|
Answer» Solution :Here `L = 0.5 H, R = 100 Omega, V_(rms) = 240 V` and v= 10 kHz `=10^(4)` Hz `therefore omega = 2PI v = 2 xx 3.14 xx 10^(4)= 6.28 xx 10^(4)s^(-1)` (a) `therefore` Impedance `Z = sqrt(R^(2) + (Lomega)^(2)) = sqrt((100)^(2) + (0.5 xx 6.28 xx 10^(4))^(2)) = 3.14 xx 10^(4) Omega` `therefore I_(m) = V_(m)/Z =(sqrt(2)V_(rms))/Z = sqrt(2) xx 240 = 1.1 xx 10^(-2) A` (B) `phi = tan^(-1) (OLomega)/R = tan^(-1) (0.5 xx 6.28 xx 10^(4))/100 = tan^(-1)( 31.4) = pi/2` rad `therefore t = phi/ (2pi v) =pi/(2 xx 2pi xx 6.28 xx 10^(4)) = 4.0 xx 10^(-6)` s. From these calculations, it is clear that for a high frequency, current is very very small i.e., at high FREQUENCIES the inductor behaves as if the circuit is an open circuit or its impedance is extremely high. An inductor OFFERS zero resistance in a d.c. circuit after the steady state is reached. |
|
| 37714. |
What is the shape of a wavefront at a large distance away from a point source? |
| Answer» SOLUTION :PLANE WAVEFRONT | |
| 37715. |
Consider the possible vibration modes in the following linear molecules: (a) CO_(2) (O-C-O): (b) C_(2)H_(2)(H-C-C-H). |
|
Answer» Solution :(a) `CO_(2)(O-C-O)` The molecule has `9` degrees of freedom `3` for each atom. This means that it can have up nine frequencies, `3` degree of freedom correspond to rigid translation, the frequency ASSOCIATED with this is zero as the potential energy of the system can not change under rigid translation. The `P.E` will not change under ROTATIONS about axes passing through the `C` atoms and PERPENDICULAR to the `O-C-O` line. Thus there can be be at most four non zero frequencies. we must look for modes different from the above. One mode is  ANOTHER mode is  These are only collinear modes. A third mode is doubly degenerate:  (vibration in & `_|_` to the plane of paper). (b) `C_(2)H_(2) (H-C-C-C-H)` There are `4xx3-3-2=7` different vibrations. There are three colliner modes.    TWO other doubly degenerate frequencies are   together with their counterparts in the plane `_|_` to the paper. |
|
| 37716. |
A Zener of power rating 1 W is to be used as a voltage regulator. If zener has a breakdown of 5V and it has to regulate voltage which fluctuated between 3V and 7V, what should be the value of R_(s) for safe operation (See figure)? |
|
Answer» Solution :`P=V_(Z)I_(Z)` `THEREFORE (P)_("max.")=V_(Z)(I_(Z))_("max.")""(because V_(Z)=` CONSTANT) `therefore 1 = 5xx(I_(Z))_("max")` `therefore (I_(Z))_("max".)=(1)/(5)=0.2A` Here `v_(i)=I_(Z)=R_(S)+V_(Z)` `therefore (V_(i))_("max".)=(I_(Z))_("max.")R_(S)+V_(Z)` `therefore 7=0.2R_(S)+5` `therefore R_(S)=(7-5)/(0.2)=(2)/(0.2)=10Omega` |
|
| 37717. |
Wavelength of a microwave is 1 mm and that of radio wave 150 m. What is the ratio of their speeds in free space ? |
|
Answer» `1:1` |
|
| 37718. |
The input characteristics of a transistor in CE mode is the graph obtanied by plotting |
|
Answer» `I_(B)` against `I_(C )` at constant `V_(BE)` |
|
| 37719. |
(a) When an unpolarised light of intensity I_(0) is passed through a polaroid, what is the intensity of the linearly polarised light ? Does it depend on the orientation of the polaroid ? Explain your answer. (b) A plane polarised beam of light is passed through a polaroid. show graphically the variation of the intensity of the transmitted light with angle of rotation of the polaroid in complete one rotation. |
| Answer» Solution :(a) When an UNPOLARISED LIGHT of INTENSITY `I_(0)` is passed through a polaroid, the intensity of linearly polarised light obtained on other side of polaroid is `I=I_(0)//2`. It does not DEPEND on the orientation of the polaroid. If the polaroid is rotated, the direction of orientation of electric vector of transmitted light changes but intensity of light remains UNCHANGED. | |
| 37720. |
The difference in speed of a ray of light in glass and water is 2.5 xx 10^7 m//s. If the R.I. of glass and water are 3/2 and 4/3 respectively, then the velocity of light in air is : |
|
Answer» `2 XX 10^8 m//s` |
|
| 37721. |
If the cohesive force is less than the adhesive force, then the liquid surface will be |
|
Answer» concave |
|
| 37722. |
A circular disc is rolling in a horizontal plane. its total K.E. is 300J . What is its rotational K.E.? |
|
Answer» 50 J |
|
| 37723. |
When a charged particle enters a uniform magnetic field its kinetic energy |
|
Answer» REMAINS CONSTANT |
|
| 37724. |
True/False Type Questions Q A large-angle prism deviates a light ray more. |
|
Answer» more |
|
| 37725. |
You are to compare two resistance that are in the ratio 1:2. The wire of metre bridge given is of length 99 cm only. With full calculations explain how the error is minimized by taking additional readings with resistance interchanged. |
|
Answer» Solution :`R/2 = 1/2 = l/(99-l) ` GIVES `l= 33 CM` But we read the lengths as `l=33` cm and 100-33 = 67 cm and WRITE `R/S = (33)/(67) = 0.4925` Now , then we interchange the resistances, we shouldhave `R/S = 2/1 = l/(99-l) ` which gives `l= 66` cm - But we read lengths as `l= 66` cm and 100 - 66 = 32 cm and write `R/S = (66)/(34) ` or R/S = (34)/(66)=0.5152` is 0.5038 and is very close to `1/2` |
|
| 37726. |
SI Unit Of Mass Is |
|
Answer» Meter |
|
| 37727. |
Two batteries of emf e_(1) and e_(2) having internal resistance r_(1) and r_(2) respectively are connected in series to an external resistance R. Both the batteries are getting discharged. The above described combination of these two batteries has to produce aweaker current than when any one of the batteries is connected to the same resistor. For this requirement to be fulfilled |
|
Answer» <P>`(epsi_(2))/(epsi_(1))"must not lie between" (r_(2))/(r_(1)+R)and(r_(1))/(r_(2)+R)` `R'=(V^(2))/(P)=((100)^(2))/(1000)gt10Omega` If the heater has to operature with a power P'=62.5W, the voltage V' across its coil should be `V'=(P'R')^(1//2)=(6.25xx10)^(1//2)=25VA` Thus, out of 100V, a valtage drop of 25V occurs across the heater and the rest 100-25=75V occurs across the `10Omega` resistor. Therefore CURRENT in the cicruit is `I=(75)/(10)=7.5A` Now, current throght the heater =`(V')/(R)=(25)/(10)=2.5A` Therefore, current throgh reistor R=7.5-2.5=5.0A `"Here",R=(V')/(5.0A)=(25V)/(5.0)=5Omega` |
|
| 37728. |
What is scattering of light ? And on which doe the scattering depend ? |
|
Answer» Solution : As sunlight travels through the earth.s atmosphere it gets scattered (changes it: direction) by the atmospheric particles. LIGHT o shorter wavelengths is scattered much more than light of longer wavelengths. The amount of scattering is inversely proportional to the fourth power of the wavelength. This is known as Rayleigh scattering Hence, the BLUISH colour predominates in a clear sky since blue has a shorter wavelength than red and is scattered much more strongly. In fact, violet gets scattered EVEN more than blue, having a shorter wavelength. Large particles like dust and WATER droplets, present in the atmosphere behave differently. The relevant quantity here is the relative size of the wavelength of light `lambda` and the scatterer (of typical size SAY a). For a `lt lt lambda`, one has Rayleigh scattering which is proportional to `(1)/(lambda^4)` If a `gt gt lambda` then it is called geo-metric scattering. If a = `lambda`, then it is called mie scattering. |
|
| 37729. |
Pressure inside two soap bubbles is 1.01 and 1.02 atmosphere. Ratio between their volume is : |
|
Answer» `102:101` |
|
| 37730. |
Two sinusoidal waves with the same amplitude and wavelength travel through each other along a string that is stretched along an x axis. Their resultant wave is shown twice in Fig. 16-53, as the antinode A travels from an extreme upward displacement to an extreme downward displacement in 6.0 ms. The tick marks along the axis are separated by 15 cm, height H is 1.20 cm. Let the equation for one of the two waves be of the form y(x, t) =y_(m) sin(kx + omega). In the equation for the other wave, what are (a) y_(m) (b) k, (c) omega, and (d) the sign in front of omega? |
|
Answer» |
|
| 37731. |
The resistance of a 10 m long potentiometer wire is 20 Omega . It is connected in series with a 3 V battery and 10 Omegaresistor. The potential difference between two points separated by distance 30 cm is equal to ....... |
|
Answer» Solution :0.06 V `L rho = 20 OMEGA , L = 10 ` m `therefore rho - 2 (Omega)/(m)` Now, `V_(l) = I rho `l `therefore I =(E rho l )/(I rho + R + r)= (3 XX 2 xx 0.3)/(20 + 10 + 0) = (1.8)/(30)` `therefore V_(l) = 0.06 ` V |
|
| 37732. |
Figure 27-12 shows a circuit whose elements have the following values: epsi_(1) = 3.0 V, epsi_(2)= 6.0 V, R_1 =2.0 Omega R_2= 2.0Omega. The three batteries are ideal batteries. Find the magnitude and direction of the current in each of the three branches |
|
Answer» Solution :It is not worthwhile to try to simplify this circuit, because no two resistors are in parallel, and the resistors that are in series (those in the right branch or those in the left branch) present no problem. So, our plan is to apply the junction and loop rules. Junction rule: Using arbitrarily chosen directions for the CURRENTS as SHOWN in Fig. 27-12, we apply the junction rule at point a by writing `i_(3)=i_(1)+i_(2)` An application of the junction rule al junction b gives only the same equation, so we next apply the loop rule to any two of the three loops of the circuit. Left-hand loop: We first arbitrarily choose the left-hand loop, arbitrarily start at point b, and arbitrarily traverse the loop in the clockwise direction, obtaining `-i_(1) R_(1)+epsi_(1)-i_(1)R_(1)-(i_(1)+i_(2)) R_(2)-epsi_(2)=0`  Figure 27-12 A maltiloop circuit with three ideal batteries and five resistances. where we have used `(i_(1)+i_(2))` instead of `i_(3)` in the middle branch. Substituting the given data and simplifying yeild `i_(1) (8.0 OMEGA) +i_(2) (4.0 Omega)=-3.0V` Right-hand loop: For our second application of the loop rule, we arbitrarily choose to traverse the right-hand loop counterclockwise from point b, finding `-i_(2)R_(1)+epsi_(2)-i_(2)R_(1)-(i_(1)+i_(2)) R_(2)-epsi_(2)=0` Substituting the given data and simplifying YIELD `i_(1) (4.0 Omega) +i_(2) (8.0 Omega) =0` Combining equations: We now have system of two equations (Eqs. 27-27 and 27-28) in two unknowns `(i_(1), and i_(2))` to solve either "by hand” (which is easy enough here) or with a "math package." (given in Appendix E.) We find `i_(1)=-0.50A` (The minus sign signals that our arbitrary choice of direction for `i_(1)` in Fig. 27-12 is wrong, but we must wait to correct it.) Substituting `i_(1)=-0.50 A` into Eq. 27-28 and solving for `i_(2),` then give us `i_(2)= 0.25 A.` With Eq. 27-26 we then find that `i_3=i_(1)+i_(2)=-0.50A+0.25 A` The positive answer we obtained for `i_2` signals that our choice of direction for that current is correct. However. the negative answers fort, and indicate that cour choices for those currents are wrong. Thus, as a last ster here, we correct the answers by reversing the arrows for and in Fig, 27-12 and then writing. 20.50 and `i_(1)=0.50 A and i_(3)=0.25 A` Caution: Always make any such correction as the last step and not before valeulaling at the currents. |
|
| 37733. |
Alongmagnet is cut into two parts in such a way that the ratio of hteir lengths is 2:1 the ratioof polestrengths of both the sectionis |
| Answer» ANSWER :A | |
| 37734. |
If a electric field of magnitude 570 N C^(-1), is applied in the copper wire. find the acceleration experienced by the electron. |
|
Answer» SOLUTION :`E=570NC^(-1),r=1.6xx10^(-19)`C, `m=9.11xx10^(-31)kg` and a=? F=ma=eE `impliesa=(eE)/m=(570xx1.6xx10^(-19))/(9.11xx10^(-31))=1.001xx10^(14)ms^(-2)` |
|
| 37735. |
(A) : The microscopic roots of magnetism can be traced back to intrinsic spin of electrons. (R) : The spinning electron is equivalent to a magnetic dipole. |
|
Answer» Both 'A' and 'R' are true and 'R' is the correct explanation of 'A'. |
|
| 37736. |
Consider the earth as a short magnet with its centre coinciding with the centre of earth . Show that the angle of dip phi is related to magnetic latitudelambdathrough the relationtan phi = 2 tan lambda |
Answer» Solution : consider the SITUATION as shown in the figure . For DIPOLE , at position ` (r , theta )` we have `B_r = (mu_0)/(4pi)(2 M COS theta )/(r^(3)) and B_0 = (mu_0)/( 4pi)(Msin theta )/(r^(3))` .......... (1) and as ` tan phi = (B_V)/(B_H) = -(B_r)/(B_(theta))`, so in the LIGHT of eq (1) , ` tan phi =-2 cot theta ` But from figure ` theta = 90 + lambda ` so , ` tan phi = -2 cot (90 + lambda )` , i.e., ` than phi = 2 tan lambda`. |
|
| 37737. |
After switch is closed in the LC circuit shown in the figure, the charge on the capacitor is sometimes zero, but at such instants the current in the circuit is not zero. How is this behaviour possible ? |
|
Answer» |
|
| 37738. |
Two prisms A and B have dispersive powers of 0.012 and 0.018 respectively. The two prisms are incontact with each other. The prism 'A' produces a mean deviation of 1.2^@, the mean deviation produced by 'B' if the combination is achromatic is |
|
Answer» `3.6^@ ` |
|
| 37739. |
A load of mass M is attached to the free and of a uniform wire of mass m as shown in figure. If a transverse wave pulse is generated at the lower end of wire, find the ratio of velocity of wave at bottom end and at top end of wire: |
|
Answer» `M/(M+m)` |
|
| 37740. |
What does the poet mean by "clay"? |
|
Answer» He MEANS his body |
|
| 37741. |
In Young's experiment, the wavelength of monochromatic light used in 6000Å. The optical path difference between the rays from the two coherent sources are 0.0075 mm and 0.0015 mm at points P and Q, respectively, onthe screen and on opposite sides of the central bright band. How many bright and dark bands are observed between points P and Q? |
|
Answer» <P> Solution :Data: `Delta_(1) = 7.5 xx 10^(-6)m, Delta_(2) = 1.5 xx 10^(-6)m, lambda=6 xx 10^(-7)m`For point P: Let p be an integer such that `p(lambda/2) = Delta_(1)` `THEREFORE p = (2Delta_(1))/(lambda) = (2 xx 7.5 xx 10^(-6))/(6 xx 10^(-7)) = 150/6=25` `therefore` The path difference `Delta_(i)` is an odd integral multiple of `lambda/2, Delta_(1) = (2m-1)lambda/2`, where m is an integer. `therefore 2m-1=25 therefore m=13` `therefore` Point P is on the 13th dark band. For point Q: Let q be an integer such that `q lambda/2= Delta_(2)`, `therefore = (2Delta_(2))/(lambda) = (2 xx 1.5 xx 10^(-6))/(6 xx 10^(-7)) = 30/6 = 5` `therefore` the path difference `Delta_(2)` is an odd integral multiple of `lambda/2`, `Delta_(2) = (2n-1)lambda/2`, where n is an integer. `therefore 2n-1=5 therefore n=3` `therefore` Point Q is on the 3rd dark band. Betwen points P and Q, excluding the bonds at these points, the NUMBER of dark bonds `=12+2=14` and the number of BRIGHT bands (including the central bright band) `=12+2+1=15` |
|
| 37742. |
A rifle was aimed at the vertical line on the target located precisely in the northern direction, and then fired. Assuming the air drag to be negligible, find how much off the line, and in what direction, will the bullet hit the target. The shot was fired in the horizontal direction at the latitude varphi=60^@, the bullet velocity v=900m//s, and the distance from the target equals s=1.0km. |
|
Answer» Solution :Define the axes as shown with z along the local vertical, x due EAST and y due north. (We assume we are in the northern hemisphere). Then the Coriolis FORCE has the componets. `vecF_(cor)=-2m(vecomegaxxvecv)` `=2momega[v_ycos theta-v_xsin theta)veci-v_xcos thetavecj+v_xcos thetaveck]=2momega(v_ycos theta-v_xsin theta)veci` since `v_x` is small when the direction in which the gun is fired is due north. Thus the equation of motion (neglecting centrifugal forces) are `overset(..)x=2momega(v_ysin varphi-v_xcos varphi)`, `overset(..)y=0` and `overset(..)z=-g` Integrating we GET `overset(.)y=V` (constant), `overset(.)z=-g t` and `overset(.)x=2omegavsin varphit+omegag t^2cos varphi` FINALLY, `x=omegavt^2sin varphi+1/3omega g t^3cos varphi` Now `vgt gtg t` in the present case. so, `x~~omegavsin varphi(s/v)^2=omega sin varphi(s^2)/(v)` `~~7cm` (to the east). 
|
|
| 37743. |
The electrostatic force between two charges Q_1 and Q_2 at separation r is given by F = (K.Q_1Q_2)/(r^2). Theconstant K |
|
Answer» DEPENDS on the SYSTEM of UNITS only |
|
| 37744. |
Assertion : the following circuit represents 'OR' gate Reason : for the above circuit Y=barX=barbar (A+B)+A+B |
|
Answer» If both assertion and REASON are true and the reason is the CORRECT EXPLANATION of the assertion 
|
|
| 37745. |
Two charges 2muC and -2muC are placed at points A and B is 6 cm apart:- What is the direction of the electric field at every point on the equipotential surface between them? |
| Answer» SOLUTION :NORMAL to the PLANE is the DIRECTION AB. | |
| 37746. |
The magnetic moment of magnet of mass 75 gm is 9 xx 10^(-7)A-m^(2). If the density of the material of magnet is 7.5 xx 10^(3)kg m^(-3), then find intensity of magnetisation ? |
| Answer» SOLUTION :`0.09Am^(-1)` | |
| 37747. |
What should be the value of F so that block of mass 1 kg remains in equilibrium. The co-efficient of friction between 1 kg and 4 kg block is 0.5 and except it all surfaces are smooth :- |
|
Answer» 220N |
|
| 37748. |
Two charges Q_1and Q_2 Coulombs are shown in fig. A third charge Q_3 coulomb is moved from point R to S along a circular path with P as centre. Change in potential energy is |
|
Answer» `(Q_1 Q_2 Q_3)/( 4 PI in_0)` |
|
| 37749. |
For ensuring dissipation of same energy in all three resistors (R_(1), R_(2), R_(3)) connected as shown in figure, their values must be related as |
|
Answer» `R_(1)=R_(2)=R_(3)` |
|
| 37750. |
A projectile at its highest point, breaks into two equal parts due to an internal explosion. One part moves vertically up at 30 m/s with respect to ground. Then the other part will move with speed if (Speed just before explosion is 20 m/s) : |
|
Answer» `20m//s`  ` vecP_(1)=vecP_(f)` `mxx20 hati= (m) /(2) xx30 HATJ+(m)/(2)XX VECV` ` vecv- 40hati- 30 hatj` ` | vecv | = 50` |
|