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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.
| 40401. |
A rectangular coil of length 0.12 m and width 0.1 m having 5Q turns of wire is suspended vertically in a uniform magnetic field of strcnglh 0.2 Weber/m^(2). The coil carries a current of 2 A. If the plane of the coil is inclined at an angle of 30° with the direction of the field, the torque required to keep the coil in stable equilibrium will be |
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Answer» 0.12 Nm |
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| 40402. |
A solid non-conduction cylinder of radius R is charge such that volume charge density is proporation to r where r is distance from axis. The electric field E at a distance r(r lt R) well depend on r as. |
Answer» Solution : `E2pirl=(intalphar.2pil. DR)/(epsi_(0))` `EALPHA r^(2)` |
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| 40403. |
The velocity time graph of a body moving in a straight line is as follows. Find the magnitude of displacement of body and distance travelled by body during first 6 seconds. |
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| 40404. |
Six identical conducting plates of area A each are connected as shown. The separation d. between any two adjescent plates is same. epsilon_(0)is permittivity of free space between the plates. The effective capacitance between the terminals of battery is (3kepsilon_0A)/(2d)The value of .k. is |
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| 40405. |
charges of an electron in esu determined as4.806 xx 10^(-10)esu. Show that1C = 3xx 10 ^(9)esu |
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Answer» Solution :a . `1C =3xx10 ^(9) esu "" because lesu (1)/( 3xx10 ^(9) C ` NUMBER of electron in 1C =` 6.25 xx 10 ^(12) ` ` because ` Number of electron in 1 esu ` = (1)/( 3xx10 ^(9)) xx 6.25 xx 10 ^(18)- 208 xx10 ^(9) ` Number of electronsin 1 esu ` = (1)/( 3xx10 &(9) ) xx 6.25 xx10 ^(18) ` ` because ` charges esu is ` 6.25 xx10 ^(6) xx 4.806 xx10 ^(10) ` ` therefore 1C =6.25 xx10 ^(18)xx 4.806 xx 10 ^(10)= 3xx 10 ^(9) ` esu of charges. |
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| 40406. |
240 A.C. mains supply is given to one transformer which draws 0.7 A from it. This transformer is used to operate a bulb having power rating "24 V, 140 W". Then efficiency of this transformer is ...... |
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Answer» Solution :INPUT POWER `P_1=VI=240xx0.7` `therefore` input power `P_1`= 168 W output power `P_2`= 140 W `therefore` EFFICIENCY of transformer `=("Output power" (P_2))/("Input power" (P_1))=140/168xx100` = 80% |
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| 40407. |
A radioactive sample decays to form a stable nuclide. A graph showing variation of rate of disintegration ((dN)/(dt)) of radioactive sample with time (t) is |
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Answer»
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| 40408. |
A metallic conductor at 10^@ C connected in the left gap of meter bridge gives balancing length 40 cm. When the conductor is at 60°C, the balancing point shifts by ---cm, (temperature coefficient of resistance of the material of the wire is (1/220)/""^@C) |
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Answer» `4.8` |
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| 40409. |
What did the British do with the body of Bhagat Singh? |
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Answer» CHOPPED it |
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| 40410. |
Assertion :In He-Ne laser, population inversion takes place between energy levels of neon atoms. Reason: The base to emitter region is forward biased. |
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Answer» If both the Asseration and Reason are TRUE and reason explains the ASSERTION : |
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| 40411. |
An ideal battery of emf 12 V is connected across an electric network shown here. Now answer the following questions:The reading of ammeter A is |
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Answer» 3A |
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| 40412. |
If the luminous intensity of a 100 W unidirectional bulb is 100 candela, then total luminous flux emitted from the bulb is |
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Answer» 861 LUMEN |
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| 40413. |
Explain the equivalent resistance of a series and parallel resistor network. |
Answer» Solution :Resistors in series: When two or more resistors are connected end to end, they are said to be in series. The resistors could be simple resistors or BULBS or heating elements or other DEVICES. Fig. (a) shows three resistors Rp R2 and R3 connected in series.  The amount of charge passing through resistor R] must also pass through resistors `R_(2)` and `R_(3)`since the charges cannot accumulate anywhere in the circuit. Due to this reason, the current I passing through all the three resistors is the same. According to Ohm’s law, if same current pass through different resistors of different values, then the potential difference ACROSS each resistor must be different. Let `V_(1), V_(2)`and `V_(3)`be the potential difference (voltage) across each of the resistors`R_(1),R_(2)`and `R_(3)`respectively, then we can WRITE `V_(1) = IR_(1),V_(2) = IR_(2)`and `V_(3) = IR_(3)`. But the total voltage V is equal to the sum of voltages across each resistor. `V = V_(1) + V_(2) + V_(3)` `= IR_(1) + IR_(2) + IR_(3)` `V = I(R_(1) + R_(2) + R_(3))` `V=I.R_(3)` Where `R_(S)` is the equivalent resistance. `R_(S) = R_(1) +R_(2) + R_(3)` When SEVERAL resistances are connected in series, the total or equivalent resistance is the sum of the individual resistances as shown in fig. |
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| 40414. |
The line that draws power supply to your house from street has |
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Answer» ZERO average current. The line (cable) is having some resistance so power factor, cos `phi=R/Z ne 0` Hence, `phi ne pi/2` `therefore phi lt pi/2` Means, phase difference lies between 0 and `pi/2` |
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| 40415. |
Threshold wavelength of metal is 30 Å If radiation of 2000Å Is incident metal surface then …… |
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Answer» positron will be EMITTED. |
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| 40416. |
The correct match is |
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Answer» a-f, B-e, c-d |
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| 40417. |
The correct match is |
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Answer» a-h , B-g , c-f , d-e |
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| 40418. |
(a) Describe briefly the functions of the three segments of n-p-n transistor. (b) Draw the circuit arrangment for studying the output characterstics of n-p-n transistor in CE configuration. Explain how the output characterstics is obtained. |
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Answer» Solution :(a) A junction transistor is obtained by growing a thin layer of ONE type semi-conductor in between two thick layers of other similar type semi-conductors. Three layers and functions Emitter( E) : The left hand thick layer of the transistor which is heavily doped and moderate in size is called emitter. If emits majority charge carriers.  Output Characteristics : The variation fo the collector. Current `(I_(C))` with the collector emitter voltage `(V_(CE))` keeping the base current `(I_(b))` constant is called output characteristics. (i) Baxe (B) : It is a central thick layer of the transistor which is lightly DEPED. Correlates INTERACTION between emitter and collector. (iii) Collector ( C) : The right hand hand side thick layer of the transistor which is moderately deped is called collector. It collect the majority charge carriers. 
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| 40419. |
What is minimum deviation ? |
| Answer» SOLUTION :The LEAST POSSIBLE angle of deviation of LIGHT ray REFRACTED through a prism is called angle of minimum deviation. | |
| 40420. |
What did the author see when his head popped out of the water? |
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Answer» The SHIP was sinking |
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| 40421. |
Force on current carrying loop (Radius = R) in uniform magnetic (B) field which is at an angle 30” with the normal will be :- |
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Answer» zero |
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| 40422. |
In a parallel plate capacitor the potential difference of 100 V is manitianed between the plates. If distance between the plates be 5mm, what will be the electric field at points A and B ? |
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Answer» Solution :ELECTRIC field at both paints A and B is same having a value `E = V/d = (100 V)/(5mm) = (100 V)/(5 xx 10^(-3)m) = 2 xx 10^4 NC^(-1)` |
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| 40423. |
Choose correct statement from the following. A)Large mass number nuclei undergo fission B) Low mass number nuclei undergo fusion C)For heavy nuclei the decrease in binding energy per nucleon shows the contribution of the increasing coulomb repulsion. |
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Answer» A, B are CORRECT |
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| 40424. |
When a light wave goes from air into water, the quality that remains unchanged is its |
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Answer» speed. |
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| 40425. |
The half-life of a radioactive substance is 20 minutes. The time taken between 50% decay and 87.5% decay of the substance will be |
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Answer» (a)20 minutes |
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| 40426. |
Why do metals have a large number of free electrons ? |
| Answer» Solution :In metals, the ELECTRONS in the OUTER most shells are LOOSELY bound to the nucleus. Even at room temperature, there are a large number of free electrons which are moving INSIDE the metal in a random manner. | |
| 40427. |
A npn transistor is connected in common emitter configuration in a given amplifier. A load resistance of 800 Omega is connected in the collector circuit and the voltage drop across it is 0.8 V. If the current amplification factor the voltage gain and the power gain of the amplifies will respectively be: |
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Answer» `4,3.84` GIVEN current gain `=ALPHA=0.96` Voltage gain for common base CONFIGURATION `A_(v)=alpha.(R_(L))/(R_(p))=0.96xx800/192=4` POWER gain for common base configuration `P_(v)=A_(v)alpha=4xx0.96=3.84` |
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| 40428. |
A battery of emf 12 V and internal resistance 2 Omega is connected to a 4 Omega resistor as shown in the Fig. Show that a voltmeter when placed across the cell and across the resistor, in turn, gives the same reading. |
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Answer» SOLUTION :Here emf of battery `EPSI` =12 V, internal resistance R=2 `Omega` and EXTERNAL resistance R=4`Omega` CIRCUIT current `I = (epsi)/(r + R) = (12)/(2 + 4) = 2A` ` therefore ` Voltmeter reading when placed across the cell `V_1 = epsi - Ir = 12 - 2 xx 2 = 8V` and voltmeter reading when connected across the resistor `V_2 = IR = 2 xx 4 = 8V` Obviously `V_1 = V_2 = 8V` |
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| 40429. |
A : For a real gas internal energy depends on its temperature as well as volume also. R:For a gas real gas interatomic potential energy depends on volume and kinetic energy depends on temperature. |
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Answer» If bothAssertion & Reason are true and the reason is the correct explanation of the ASSERTION, then Mark (1) |
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| 40430. |
Three perfect gases at absolute temperatures T_(1), T_(2) and T_(3) are mixed. The masses of molecules are m_(1), m_(2) and m_(3) and the number of molecules are n_(1), n_(2) and n_(3) respectively. Assuming no less of energy, the final temperature of the mixture is: |
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Answer» `(n_(1)T_(1)+n_(2)T_(2)+n_(3)T_(3))/(n_(1)+n_(2)+n_(3))` `(F)/(2)(n_(1)+n_(2)+n_(3))KT =(f)/(2)n_(1)kT_(1)+(f)/(2)kT_(2)+(f)/(2)n_(3)kT_(3)` `therefore T=(n_(1)T_(1)+n_(2)T_(2)+n_(3)T_(3))/(n_(1)+n_(2)+n_(3))` Correct choice : (a). |
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| 40431. |
A screen is placed 90cm from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by 20cm. Determine the focal length of the lens. |
Answer» Solution :`rArr` LET, image of object shown at O for combination of lenses `L_1` and `L_2` obtained at I.  `rArr` Here, clearly, `x+ 20 + x = 90 cm ` `therefore2 x = 70 ` `thereforex = 35 cm ` `rArr`x = 35 cm `rArr ` If lens is at `L_1`, thenObject distance u = -x = - 35 cm Image distance v = 20+ x = 20 + 35 = 55 cm `therefore` From lens formula `1/f = 1/v -1/u = 1/55- (1)/(-35) = (1)/(55) + (1)/(35)` `therefore 1/f = (7 + 11)/(385) = (18)/(385)` `therefore f = (385)/(18) = 21.4 cm ` Second METHOD : Distance between object and image D = 90 cm Distance between two lenses d = 20 cm Now `f= (D^2 - d^2)/(4D)` (this has to be remembered) `thereforef= ((90)^2 - (20)^2)/(4 xx 90) = (8100-400)/(360)` `therefore f = (7700)/(360) = 21.38 cm ` `thereforef ~~ 21.4 cm ` |
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| 40432. |
An ideal battery of emf 12 V is connected across an electric network shown here. Now answer the following questions:The equivalent resistance of the network is : |
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Answer» `2 OMEGA ` |
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| 40433. |
(a) an electrostatic field line is a contiuous curve that isfieldline cannot have sudden breaks why not (b) explain why two field linesnever cross each other at any point |
| Answer» Solution :An electrostatic field line is a continuous CURVE because a charge experiences a continuous force when TRACED in an electrostatic field . Thefield linecannot have sudden breaks because the charge MOVES continously anddoes not JUMP from ONE point to the other. | |
| 40434. |
निम्नलिखित में से किसके द्वारा अनंत पर स्थित किसी बिंब का प्रतिबिंब अत्यधिक छोटा बनेगा? |
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Answer» केवल अवतल दर्पण |
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| 40435. |
When the planet is at farthest distance from the sun it is called perihilion.Is it true? |
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Answer» |
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| 40436. |
भारत के तमिलनाडु तट पर स्थित मैदान को किस नाम से जाना जाता है? |
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Answer» उत्तरी सरकार तट |
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| 40437. |
Two charges 3 xx 10^(-8) C are located 15 cm apart. At what point on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero. |
Answer» SOLUTION :LET us take the origin O at the location of the positive charge. The line joining the two charges is taken to be the x-axis, the negative charge is taken to be on the right side of the origin.  Let P be the required point on the x-axis where the potential is zero. If x is the x-coordinate of P, obviously x must be positive. (There is no possibility of potentials due to the two charges adding up to zero for `x lt 0`). If x lies between O and A, we have `1/(4piepsilon_(0))[(3 xx 10^(-6))/(x xx 10^(-2)) - (2 xx 10^(-8))/((15-x) xx 10^(-2)) ]=0` where x is in cm. That is, `3/x - 2/(15-x) =0` If x lies on the extended line OA, the required condition is `(3/x - 2/(x-15))=0` which gives x = 45 cm THUS, electric potential is zero at 9 cm and 45 cm away from the positive charge on the side of the negative charge. Note that the formula for potential used in the calculation required choosing potential to be zero at infinity. |
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| 40438. |
Find the mathematical and the thermodynamical probabilities of the five possible distributions of four balls in two halves of a vessel (Fig.), assuming them to be distinguishable. |
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Answer» To obtain the mathematical probabilities, divide the values obtainted by the total NUMBER of cases. Use the data OBTAINED to compile Table 18.4b, which shows the thermodynamic and mathematical probabilities of interest to us. |
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| 40439. |
An ideal battery of emf 12 V is connected across an electric network shown here. Now answer the following questions:Net current drawn by the circuit from the battery is: |
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Answer» 6A |
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| 40440. |
A water molecule has an electric dipole moment of 6.3xx10^(-30) cm . A sample contains 10^(22) watermolecules with all the dipole moments aligned parallel to the external electric field of magnitude 3xx10^(5)NC^(-1) . How much work is required to rotate all the water molecules from theta=0^(@) "to" 90^(@) ? |
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Answer» Solution :When the water molecules are aligned in the direction of the electric field it has minimum POTENTIAL ENERGY. The work done to rotate the dipole from `theta = 0^(@) "to" 90^(@)` is equal to the potential energy DIFFERENCE between these two configurattions `W=DeltaU=U(90^(@))-U(0^(@))` As we know U = `-pE cos theta ` Next we calculate the work done to ratate one water molecule from `theta=0^(@)"to " 90^(@)` For one water molecule `W=-pE cos 90^(@)+pE cos 0^(@)=pE` `W=6.3xx10^(-30)xx3xx10^(5)=18.9xx10^(-25)` J For `10^(22)` water molecules the total work done is `W_("tot")=18.9xx10^(-25)xx10^(22)=18.9xx10^(-3)J` |
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| 40441. |
Three polarizing sheets are placed in a stack with the polarizing directions of the first and third perpendicular to each other. What angle should the polarizing direction of the middle sheet make with the polarizing direction of the first sheet to obtain maximum transmitted intensity when unpolarized light is incident on the stack ? |
| Answer» Answer :C | |
| 40442. |
Yellow light is used in single slit diffraction experiment with slit width 0.6 nm. If yellow light is replaced by X-rays then the ...... |
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Answer» no diffraction PATTERN will obtained. For YELLOW ligth `(lambda)/(d)=(6xx10^(-7))/(6xx10^(-10))=1000` `:.` Diffraction is possible but for X-ray `=(lambda)/(d)=(100Å)/(6xx10^(-10))=(100xx10^(-10))/(6xx10^(-10))` `=(lambda)/(2)=16.25` which is far less than the yellow light so it will not cause diffraction. So the diffraction pattern will not be obtained. |
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| 40443. |
Four vectors bara ,barb,barc, andbard are lying in the same plane. The vectors bara and barb are equal in magnitude and inclined to each other at an angle of 120^@. is the resultant of bara and barb. Further bara + barb + bard=0. If the angle between bara and bard is betaand the angle between bara and barc is a, find the correct relation between alphaand beta |
| Answer» SOLUTION :`ALPHA= 1/2beta ` | |
| 40444. |
A bomb of mass 9 kg explodes into two pieces of mass 3 kg and 6 kg. The velocity of 3 kg mass is 16 m/s. The Kinetic energy of 6 kg mass will be |
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Answer» 96 J |
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| 40445. |
A body floats with one third of it's volume outside water and 3/4 of it'svolume outside another liquid. The density of other liquid is |
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Answer» a)9/4 gm/cc |
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| 40446. |
What should one do if someone has done some harm to oneself? |
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Answer» One should think ill of them |
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| 40447. |
A smooth hoop lie as on a smooth horizontal table and is fixed. A particle is projected on the table form a point A on the inner circumference of the hoop at angle theta with radius vector . If e be the coefficient of restitution and the particle returns to the point of projection after two successive impacts. The final angle theta' made by velocity vector with radius of hoop is |
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Answer» `tan theta' = ( e^(2))/( ( 1+ e^(2))) tan theta ` |
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| 40448. |
Prove thatwhen a reflectingsurface of light by an angle theta, the reflected light will be tited by an angle 20. |
Answer» Solution :For the reflecting SURFACE AB. The incident ray IO and the reflected ray `OR_(1)` subtend angle I with the normal N as angle of incidence is EQUAL to angle of reflection. When the surface AB US tilted to A.B. by an angle `THETA`. Remember the position of incident ray IO remains unaltered. But, in the tilted system the angle of incidence is now `i+theta` and the angle of reflection is ALSO `i + theta`. NOw, `OR_(2)` is the reflected ray. The angle between `OR_(2) and OR_(1)` is `angleR_(1)OR_(2) = angle N.OR_(2) - angleNOR_(1)` 
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| 40449. |
A magnet place in the north pointing north position, balances the earth's magnetic field at a point, which is 27 cm from either pole. If it is broken into three pieces and one such piece is similarly placed, find the position of the neutral point. |
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Answer» |
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| 40450. |
In an A.C. circuit, voltage applied is V = 220 sin 100t. If the impedance is 110Omega and phase difference between current and voltage is 60^@, the power consumption is equal to ........ |
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Answer» Solution :Comparing EQUATION V=220 sin 100t with `V=V_m sin omegat`, `V_m`=220 `|Z|=110 Omega` `therefore I_m=V_m/"|Z|"=220/110`=2A `delta=60^@ =pi/3` rad `P=(V_mI_m)/2 COS delta =(220xx2)/2 cos 60^@` `=(220xx2)/2 xx1/2`=110 W |
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