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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.
| 5751. |
In the previous example find the mass of the wire if the magnetic field is equal to the earth magnetic field (4 xx 10^(-5) T), current in the wire is 5 A and length of the wire is 1m. The wire remains suspended in mid air in equilibrium. (take g = 10 m//s^(2)) |
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Answer» |
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| 5752. |
Obtain the equation for resolving power of microscope. |
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Answer» Solution :(i) A microscope is used to see the details of the object under observation. (ii) The ability of microscope depends not only in magnifying the object but also in resolving TWO points on the object separated by a small distance `d_('min")` · (iii) Smaller the value of `d_("min")` better will be the resolving power of the microscope. (iv) The radius of central maxima is already derived as equation (1), ` N = (1)/(a+b) "" .....(1)` ` r_0 = (1.22 lamda f)/( a)` (v). In the place of focal length f we have the image distance v. If the DIFFERENCE between the two points on the object to be resolved is `d_("min")` then the magnification m is, ` m = (r_0)/(d_("min"))` ` d_("min") = (r_0)/(m) = (1.22 lamda v)/( am ) =(1.22 lamda v)/(a (v/u))= (1.22 lamda u)/(a) "" [ THEREFORE m = v/u]` ` d_("min") = (1.22 lamda f)/(a) "" [ therefore u= f ]` On the object SIDE, ` 2 tan beta ~~ 2 sin beta = a/f "" [ therefore a = f 2 sin beta]` ` d_("min") = (1.22 lamda)/(2 sin beta)` (vi) To reduce the value of `d_("min")`the optical path of the light is increased by immersing the objective of the microscope in to a bath containing oil of refractive index n. |
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| 5753. |
A battery has an open circuit potential difference 10 V betweenthe terminals. When loads 9Omega and 4 Omegaare connected one by oneacross the battery , the power in the load resistance is the same. The amount of heat approximately generated in one second in the load when a load of 5OmegaIs connected across the battery will be |
| Answer» ANSWER :D | |
| 5754. |
The image of an object placed between the principal focus and the centre of curvature of a concave mirror is |
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Answer» ERECT, MAGNIFIED and VIRTUAL |
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| 5755. |
A source contain two phosporous radio nuclides _15P^32, T_(1/2)- 14.3 days and _15P^33, T(1/2) = 25.3 days. Initially, 10% of the decay come from _15^33P. How long one must wait until 90% do so ? |
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Answer» 250 DAYS |
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| 5756. |
(a) Derive a relation between the internal resistance, emf and terminal potential difference of a cell from which current I is drawn. Draw V vs I graph for the cell and explain its significance. (b) A voltmeter of resistance 988 Omega is connected across a cell of emf 2 V and internal resistance 2 Omega. Find the potential difference across the voltmeter and also across the terminals of the cell. Estimate the percentage error in the reading of the voltmeter. |
Answer» Solution :(a) Consider the circuit shown.  By Kirchhoff.s rules we have `E - IR - RI = 0` `E-V - Ir = 0` `E= V + Ir` The V-I GRAPH is as shown.  Significance of Graph: To find emf and internal resistance of the cell. (b) The diagram is as shown:  `V= E-Ir` `998 XX I =2 -2I` I= 0.002A Therefore, `V= 0.002 xx 998 = 1.996V` PERCENTAGE error `=(2-1.996)/(2) xx 100 = 0.2`% |
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| 5757. |
A cylindrical tube of length L has inner radius a while outer b. What is the resistance of the tube between its inner and outer surfaces? [The resistivity of its material is rho] |
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Answer» `RHO(2PI L)LN((B)/(a))` |
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| 5758. |
Calculate the mobility of a free electron in an electric field of 10^(+2)N//C. |
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Answer» `10^(-4)m^(-2)V^(-1)s^(-1)` |
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| 5759. |
Whendoestheterminalvoltage ofacellbecome (i)greaterthan itsemf(ii) lessthan itsemf ? |
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Answer» SOLUTION :(i) Whenthecell isbringchargedterminalpotentialdifference(V )becomesgreaterthanemf(E ), V = E +Ir (II)WHENTHE cellsisdischarged, then` V LTE ` ` V =E -Ir ` |
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| 5760. |
A block released fromrestfromthe top of a smooth inclined planeof angletheta_(1)reaches the bottom int_(1) . The same block relased from rest from the top of another smooth inclined planeof angletheta_(2) , reachesthe bottom in timet_(2). If thetwo inclined planes havethe same height, the relation betweent_(1) and t_(2)is |
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Answer» ` t_(2)/ t_(1) =(sin theta_(1))/(sin theta_(2))^(1/2)` `l_(1) = h/(sin theta_(1))and l_(2) = h/(sin theta_(2))` ACCELERATIONS of the BLOCK down the two planesare` a_(1) =g sin theta_(1) and a_(2) = g sin theta_(2)` As ` l_(1) = 1/2 a_(1)t_(1)^(2) and l_(2) a_(2)t_(2)^(2)` ` l_(1)/l_(2) = (a_(1)t_(1)^(2))/(a_(2)t_(2)^(2)) or t_(2)^(2)/(t_(1)^(2)) = (a_(1)l^(2))/(a_(2)l^(2)) = (sin theta_(1)/( g sin theta_(2)) xx ( sin theta_(1))/(sin theta_(2))` ` t_(2)/t_(1) = (sin theta_(1))/( sin theta_(2))` |
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| 5761. |
A straight wire is first bent into a circle of radius 'r' and then into a square of side 'a' each of 1 turn. If the currents flowing through them are in the ratio 2:3, find the ratio between their effective magnetic moments? |
| Answer» SOLUTION :`8//3 PI` | |
| 5762. |
A force of 200N is required to move a body with a velocity of 20 ms. The power developed is : |
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Answer» 100 walls P=4000 WAIT |
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| 5763. |
For the equation FpropA^(a)v^(b)d^(c), where F is the force, A is the area,v is the velocity and d is the density, the values of a,b and c are respectively : |
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Answer» `1,2,1` `[MLT^(-2)]=[L^(2a)][L^(b)T^(-b)][M^(c)L^(-3c)]` `=M^(c)L^(2a+c-3c)T^(-b)` `:.c=1|{:(2a+b-3c=1),(2a=-2+3+1),(a=1):}|{:(-b=-2),(b=2):}` Hence `FxxA^(1)V^(2)d^(1)` |
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| 5764. |
A terrestrial telescope is made by introducing an erecting lens of focal length f between the objective and eyepiece of the lenses of an astronomical telescope. This causes the length of telescope tube to increase by an amount equal to : |
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Answer» f |
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| 5765. |
At certain pressure and 127^(@)C temperature the mean kinetic energy of hydrogen molecules is 8 xx 10^(-12)J. (Mass of hydrogen atom =1.7 xx 10^(-27)kg & atomic weight of nitrogen =14). Then Root mean square speed of hydrogen molecule at 27^(@)C is |
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Answer» `0.88 xx 10^(4) "m/sec"` |
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| 5766. |
At certain pressure and 127^(@)C temperature the mean kinetic energy of hydrogen molecules is 8 xx 10^(-12)J. (Mass of hydrogen atom =1.7 xx 10^(-27)kg & atomic weight of nitrogen =14). Then The rms speed of nitrogen molecules at 27^@C is |
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Answer» `5.02 XX 10^(2)" m/sec"` |
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| 5767. |
A uniform rod of length 4m and mass 2sqrt(2)kg revolves with constant angular velocity omega what a vertical axis through a smooth joint A at one extermly of the rod so that it describes a cone of semi vertical angle 53^(@) as shown in the figure. Choose the correct statement (s). (g=10m//s^(2)) |
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Answer» The value of angular velocity `OMEGA` is `2.5rad//s` |
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| 5768. |
(A): Total induced emf in a loop is not confined to any particular point but it is distributed around the loop. (R): In general when there is no change in magnetic flux, no induced emf is produced |
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Answer» Both A and R are TRUE and R is the CORRECT explanation |
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| 5769. |
Find the size of image formed in the situation shown in figure. |
| Answer» SOLUTION :1.2 CM, APPROX | |
| 5770. |
An electron and proton are possessing the same amount of kinetic energy. Which of the two have greater wavelength ? |
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Answer» SOLUTION :`E_e=1/2m_enu_e^2` and `E_p=1/2 m_pnu_p^2` `rArr m_enu_e=SQRT(2E_em_e) and m_pv_p =sqrt((2E_p" "m_p)` But `E_e=E_p rArr (lambda_e)/(lambda_p)=sqrt(m_p)/(m_e)GT 1` `THEREFORE lambda_e gt lambda_p`. |
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| 5771. |
What are Isogonic, Agonic, Isoclinic lines. |
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Answer» SOLUTION :(i) Isogonic are lines of JOINING place of equal DECLINATION. (II) Are place of ZERO declination. (iii) Lines are place of equal dip. |
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| 5772. |
If the horizontal velocity given to the satellite is greater than escape velocity, then the satellite moves |
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Answer» CIRCULAR path |
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| 5773. |
At a given time there are 25% undecayed nuclei in a sample. After 10 seconds number of undecayed nuclei reduces to 12-5%. Then mean life of the nuclei will be about: |
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Answer» 10 sec `N_(1)/N_(0)=1/4 or t_(1)=2T` `N_(2)=1/8 or t_(2)=3T` `T=t_(2)-t_(1)=10s"=HALF LIFE"` `T_("aver.")=(T)/(0.693)=14.43sec` |
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| 5774. |
A person cannot see objects clearly beyond 50.cm. The power of lens to correct the vision |
| Answer» ANSWER :C | |
| 5775. |
Consider a charged particle moving with constant velocity inside a region of space where electric and magnetic field both are present. How can the fields and velocity be directed to achieve this state? |
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Answer» Solution : LET us assume it as one positive charge and for the negative charge, directions of fields will be + just opposite. Electric field applies force on the positive charge in its direction. And direction of magnetic force is governed by right hand rule as described in `vecF = q vecv xx vecB` To make net force zero we require `vecv xx vecB`opposite to `vecE` . We know that the direction of cross product is always perpendicular to the common plane of those TWO vectors in cross product. So we can say that both `vecv "and " vecB`must be perpendicular to `vecE`. You can now visualise the whole situation as follows: Draw a plane perpendicular to electric field intensity vector. Velocity vector and magnetic field R V intensity vector both must lie on this plane. Angle between velocity vector and magnetic field intensity vector can be other than `90^(@)`also. Let this angle be `theta`. Now orientation of velocity vector and magnetic field intensity vector on this plane must be such that cross product is ANTIPARALLEL to electric field, which is already perpendicular to this plane. For net force to become zero, we can write as follows: `qE = QVB sin theta rArr E = B v sin theta` if angle `theta = 90^(@), " then " E = Bv`. 
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| 5776. |
Poisson's ratio of a certain material is 0.2. If a longitudinal strain is 4xx10^(-3) is produced in a uniform rod or this material, the percentage change in its volume will be |
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Answer» 0.0024 |
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| 5777. |
One requires 11eV of energy to dissociate a carbon monoxide molecule into carbon and oxygen atoms. The minimum frequency of the appropriate electromagnetic radiation to achieve the dissociation lies in |
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Answer» VISIBLE REGION `therefore v=(E )/(h)=(11eV)/(6.62xx10^(-34)JS)` `therefore v=(11xx1.6xx10^(-19)J)/(6.62xx10^(-34)Js)` `therefore v=2.6586xx10^(15)Hz` which is in ultraviolet region. |
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| 5778. |
A heater coil is cut into two equal parts and only one part is now used in the heater . What is the percentage of increase or decrease in the rate f productu=ion of heat ? |
| Answer» Solution :In both case ,the potential differnces between the two ends of the coil (V) are same . So , rate of PRODUCTION of HEAT `prop(V^(2))/(R)` . When the oil is cut into two equal parts , the value ofR becomes half . So ,rate of production of heatbecomes double , i.e ., this rate increases by 100 % | |
| 5779. |
A binary star consists of two stars A (mass 2.2 M_(s)) and B (mass 11 M_(s)), where M_(s) is the mass of the sun. They are separated by a distance d and are rotating about their centre of mass, which is stationary. The ratio of the total angular momentum of the binary star to the angular momentum of star B about the centre of mass is.................... . |
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Answer» SOLUTION :Here distance of COM w.r.t. A `=((11M_(S))/(11M_(S)+2.2M_(S)))xxd=(5D)/(6)` `L_(A)=2.2M_(s)omega(5d//6)^(2)=55M_(s)omega((d^(2))/(36))` `L_(B)=11M_(s)omega(d//6)^(2)=11M_(s)omega((d^(2))/(36))` and `L_("total")=L_(A)+L_(B)=(11)/(6)M_(s)omegad^(2)` `therefore (L_("total"))/(L_(B))=((11)/(6)M_(S)omegad^(2))/(11M_(S)omega(d//6)^(2))=6` |
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| 5780. |
What are energy levels in a hydrogen atom ? |
| Answer» Solution :The ENERGIES of the electrons in an atom can have only CERTAIN VALUES. These values are called ENERGY levels of the atoms. | |
| 5781. |
A man standing in a swimming pool looks at a stone lying at the bottom. The depth of the swimming pool is h. At what distance from the surface of water is the image of the stone formed. Line of vision is normal. The refractive index of water is mu: |
| Answer» ANSWER :D | |
| 5782. |
In a moving -coil instrument, the coil is is suspended in a radial magnetilc field instead of a uniform magnetic field. This is done to |
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Answer» increase the sensitivity of the instrument |
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| 5783. |
The figure shows a snap photograph of a vibrating string at t = 0. The particle P is observed moving up with velocity 20 sqrt3 cm/s. The tangent at P makes an angle 60^@ with x-axis. Find the direction in which the wave is moving. |
| Answer» SOLUTION :NEGATIVE X | |
| 5784. |
An alternating voltage of 141.4 V (rms) is applied to a vacuum diode as shown in the figure. The maximum potential difference across the condenser will be |
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Answer» 100 V |
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| 5785. |
The velocity components of a particle moving in the xy plane of the reference frame K are equal to v_x and v_y. Find the velocity v^' of this particle in the frame K in the positive direction of its x axis. |
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Answer» Solution :By the VELOCITY addition FORMULA `v_x'=(v_x-V)/(1-(Vv_x)/(c^2)), v_y'=(v_ysqrt(1-V^2//c^2))/(1-(v_xV)/(c^2))` and `v^'=sqrt(v_x'^2+v_y'^2)=(sqrt((v_x-V)^2+v_y^2(1-V^2//c^2)))/(1-(v_xV)/(c^2))` |
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| 5786. |
The figure shows a snap photograph of a vibrating string at t = 0. The particle P is observed moving up with velocity 20 sqrt3 cm/s. The tangent at P makes an angle 60^@ with x-axis. Find the total energy carried by the wave per cycle of the string. Assume that the mass per unit length of the string = 50 g/m. |
| Answer» SOLUTION :`1.6 XX 10^(-5)J` | |
| 5787. |
A : More accurate formula for the Doppler effect which is valid when the speeds close to that of light, requires the used of Einsteins special theory or relativity. R : Doppler effect is the basis for the measurements of the raidal velocities of distant galaxies. |
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Answer» Both A and R are true and R is the CORRECT explanation of A |
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| 5788. |
The figure shows a snap photograph of a vibrating string at t = 0. The particle P is observed moving up with velocity 20 sqrt3 cm/s. The tangent at P makes an angle 60^@ with x-axis. Write the equation of the wave. |
| Answer» SOLUTION :`x = 0.4 sin [10 pi t -pi/2 XX + (3pi)/4]` | |
| 5789. |
A 20 mu F capacitor is being charged by some constant voltage source. When p.d. across capacitor changes with time at a rate 3 V/s, find conduction current in the wires joining battery with capacitorand displacement current in the space between the plates. |
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Answer» 0, 0 `V=(Q)/(C )` `THEREFORE Q=VC` `therefore (dQ)/(dt)=C(dV)/(dt) ""(because " C = constant")` `therefore i_(C )=C (dV)/(dt)` `=20xx10^(-6)xx3` `= 60xx10^(-6)A` `i_(C )=60mu A` `rArr i_(d) = 60 mu A "" (because i_(C )= i_(d))` |
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| 5790. |
If 'G''c' and 'h' are usual constants of physics, the unit of time is expressed as : |
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Answer» `sqrt((HC)/G)` `M^(0)L^(0)T^(1)=(M^(-1)L^(3)T^(-2))^(x)XX(L^(1)T^(-1)]^(y)xx[ML^(2)T^(-1)]^(z)` `=M^(-x-y)L^(3x+y+2z)T^(-2x-y-z)` `-x+z=0 3x+y+2z=0-2x-y-z=1` `x=z3z+2z=-y-2z-z=1+y` `5z=-y-3z=y+1` `5z=3z+1y=-3z-1` `x=1//2z=1` or `z=1//2y=-5//2` or `T=G^(1//2)c^(-5//2)h^(1//2)` `sqrt((hG)/(c^(5)))` Hence correct choice is `(b)`. |
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| 5791. |
A : There is no specific important physical difference between interference and diffraction. R : When there are only few sources (say two), the result is usually called interference, but if there is a large number of sources, the result is diffraction. |
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Answer» Both A and R are true and R is the CORRECT EXPLANATION of A |
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| 5792. |
A weight of mass 1 kg attached to a spring with a force constant of 20 N/m is able tooscillate on a horizontal steel rod Fig. The initial displacement from the position of equilibrium is 30 cm. Find how many swings the weight will make before stopping completely. One swing is the movement from maximum displacement to the equilibrium position (or back). For numerical calculation put g=10m//s^(2) and coefficient of friction mu=0.05. |
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Answer» `(1)/(2)kA_(0)^(2)-mumgA_(0)=(1)/(2)kA_(1)^(2)+mumgA_(1)` from which it follows that `A_(1)=A_(0)-(2mumg)/(k)` The same will be true for all the subsequent oscillations, i.e. the AMPLITUDES will form an arithmetical progression: `A_(n)=A_(0)-(2mumg)/(k)` The pendulum will stop when its amplitude becomes zero. Substituting `A_(n)=0` given `n=(kA_(0))/(2mumg)` Since all the amplitudes except the initial one are passed through twice the numberof swings is `N=2n-1=(kA_(0))/(mumg)-1` As may be seen, due to friction the mechanical energy rather quickly transforms into INTERNAL energy, and the oscillations cease. |
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| 5793. |
A convex lens is placed in contact with a plane mirror. A point object at a distance of 20 cm on the axis of this combination has its image coinciding with itself. What is the focal length of the lens ? |
| Answer» SOLUTION :The FOCAL length of the GIVEN convex lens is 20 cm. | |
| 5794. |
A conducting rod of mass m and length L is connected by two identical springs as shown in the figure. Initially the system is in equilibrium. A uniform magnetic field of magnitude B directed pependicular to the plane of the paper outwards also exists in the region. If a current l is switched on that passes from P to Q through the rod. Further maximum elongation in the spring is ["Given": |mg| = |BIL|] |
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Answer» `(BIL)/(K)` 
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| 5795. |
The ptoential gradient is a |
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Answer» VECTOR QUANTIY |
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| 5796. |
A small object of height 0.5 cm is placed in front of a convex surface of glass. (mu=1.5) of radius of curvature 10 cm on its principal axis a distance of 30, cm from the pole. Find the height, nature and position of image. |
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Answer» 1 cm real, inverted, magnified, formed in the GLASS |
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| 5797. |
Suppose that the electric field part of an electromagnetic wave in vacuum is vec(E )={(3.1 N//C)cos [(1.8 rad//m)y+(5.4xx10^(6)rad//s)]}hat(i)What is the amplitude of the magnetic field part of the wave ? |
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Answer» Solution :ACCORDING to FORMULA, `c=(E_(0))/(B_(0))` `therefore B_(0)=(E_(0))/(c )` `=(3.1)/(3xx10^(8))` `therefore B_(0)=1.033xx10^(-8)T ~~ 10.3 NT ""`…(7) |
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| 5798. |
Two short magnets of dipole moment M each are placed crossed at 90^(@) (see adjacent figure). The magnetic field at a distance from the centre on the bisector of the right angle is |
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Answer» `(mu_(0) 2M)/(4 PIR^(3))` |
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| 5799. |
Suppose that the electric field part of an electromagnetic wave in vacuum is vec(E )={(3.1 N//C)cos [(1.8 rad//m)y+(5.4xx10^(6)rad//s)]}hat(i)What is the frequency v ? |
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Answer» Solution :From equation (3), `omega = 5.4xx10^(6)(rad)/(s)` `therefore 2pi V=5.4xx10^(6)(rad)/(s)` `therefore v=(5.4xx10^(6))/(2xx3.14)=8.599xx10^(5)Hz` `~~86 " MHZ"`…(6) |
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| 5800. |
The volume occupied-by an atom is greater than the volume of the nucleus by a factor of about |
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Answer» `10^(1)` |
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