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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.
| 39801. |
A force of 5 N making an angle with the horizontal acting on an object displaces it by 0.4 m along the horizontal direction. If the object gains kinetic energy of 1 J, the horizontal component of the force is : |
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Answer» 1.5 N |
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| 39802. |
What will happen when Ge and Si is doped with 3^(rd.)group element? What semiconductor is now formed? What is a hole ? |
| Answer» Solution :During doping covalent BOND are formed between the ATOMS of third group ELEMENT and the atoms of Ge and Si.In one of the convalent bonds of the impurity ATOM,a vacancy of electron remains unfilled.As a result P type semiconductor is formed.The DEFICIENCY of electron is generally known as hole. | |
| 39803. |
A parallel-plate capacitor consists of two square plates of effective area A, length L and plate separation d. If one of the plates is turned through a small angle theta, what is the change in the capacitance of the capacitor? |
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| 39804. |
An electrical technicanrequiresa capacitance fo 2 muF in a circuit across a potential difference of 1kV. A large number 1muF capacitor are available to him each of which can withstand a potential difference of not more than 400V. Suggest a possible arrangement that requires the minimum number of capacitors. |
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Answer» Solution :`C_("eff")=2muF= 2 xx 10^(-6)F ""V=1kV=10^(3)V""C=1muF=10^(-6)F," rate 400V"` Total change, `Q=C_("eff") V=2 xx 10^(-6) xx 10^(3)=2 xx 10^(3) C` Let there be .n. capacitors in series and m such rows in the circuit. Effective capacity, `C_("eff")=C/n xx m` `2 xx 10^(-6) =(10^(-6)m)/(n)"m=2N"` Also `(1000)/(n) LT 400 "i.e,"=(10)/n lt 4` `5/n lt 2 or 2n lt 5, n gt 2.5 i.e, " 3Hence m=6"` |
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| 39805. |
Choose the correct alternative from the clues given at the end of the each statement: The size of the atom in Thomson’s model is .......... the atomic size in Rutherford’s model. (much greater than/no different from/much less than.) |
| Answer» SOLUTION :No DIFFERENT from | |
| 39806. |
A plano-convex lens of refractive index 1.5 and radius of curvature 30 cm is silvered at the curved surface. Now, this lens has been used to form the image of an object. At what distance from this lens, an object be placed in order to have a real image of the size of the object. |
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Answer» 20 cm  `therefore(1)/(f_e)=-(n/f_1""-m/f_m)` .....(i) where `f_m` = focal length of mirror after silvered. `implies` From lens maker.s formula, `(1)/(f_1)=((n_e-n_1)/(n_1))((1)/(R_1)-(1)/(R_2))` `(1)/(f_1)=((1.5-1.0)/(1.0))((1)/(infty)-(1)/(-30))` `=1/2(+(1)/(30))` `=(1)/(60)` `therefore f_1=60` cm From EQUATION (1),`(1)/(f_e)=-((n)/(f_1)-(m)/(f_m))` [No. of lens =2,no. of nirrror=1] `(1)/(f_e)=-((2)/(60)-(1)/((-15)))` `(1)/(f_e)=-((1)/(30)+(1)/(15))` `(1)/(f_e)=-(3)/(30)=-(1)/(10)` `therefore f_e=-10` cm `therefore` MAGNITUDE `f_e=10` cm If image is to be obtained of same size of OBJEC then it should be placed at centre of curvature `therefore` Object distance `u = 2f_e = 2 xx 10 = 20` cm |
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| 39807. |
Height of the body from the ground can be calculated by using the formula h=-"gt"+(1//2)"gt"^(2) in a)A body projected vertically with velocity 'u' from the top of tower,reaches the ground in 't' sec. b)A body dropped from a balloon moving up with uniform velocity,reaches the grounf in 't' sec c)A body dropped from a helicopter moving up with uniform velocity ,reaches the ground in 't' sec d)A body projected vertically from the ground reaches the ground in 't' sec. |
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Answer» a,B and c CORRECT |
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| 39808. |
How do you differentiate between an ohmic and a non-ohmic resistor ? |
| Answer» Solution :When temperature and other physical CONDITIONS are kept fixed and if the resistance of a resistor remains constant for any potential difference applied then it is an ohmic resistance. If the resistance changes with applied potential difference then it is non-ohmic resistor. CONDUCTORS are EXAMPLE of ohmic resistor. DIODE is non-ohmic resistor. | |
| 39809. |
Minimum and maximum deviation in a prism A 60^(@) prism has a refractive index of 1.5. (a) Calculate the angle of incidence for minimum deviation. |
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Answer» Solution :As discussed in the Section : Angle of Minimum Deviation, the angle of incidence will be EQUAL to the angle of emergence . Calculation : As the prism is in the position of minimum deviation , `r-A//2=60^(@)//2-30^(@)`, so that at either face `sini=1.5sin30^(@)=0.75` `i=sin^(-1)(0.75)=49^(@)` (B) Calculate the angle of emergence of light at maximum deviation As we have descussed in the Section : Angle of maximum Deviation, the angle of incidence will be equal to `90^(@)`. Calculation :For maximum deviation `i_(1)=90^(@)` so that `r_(1)=theta_(C )=sin^(-1)(2//3)=42^(@)`. But as in a prism. `r_(1)+r_(2)=A` Therefore, `r_(2)=A-r_(1)=60^(@)-42^(@)=18^(@)` Now applying Snell.s law at the second face, `nsinr_(2)=sini_(2)`, `(3)/(2)sin18^(@)=sini_(2)` that is `i_(2)=sin^(-1)(1.5xx0.31)=sin^(-1)(0.465)=28^(@)` |
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| 39810. |
The displacement of the particle at x = 0 of a streached string carrying a wave in the positive xdirection is given by f(t)= A sin (t/T). The wave speed is V. Write the wave equation |
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Answer» `F(X,t)= A SIN (t/T -x/V)` |
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| 39811. |
The instantaneous velocity of a partical at any instant is equal to timederivative of its positionvector and the instancous vector.Which of the followingoptions are wrong ? |
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Answer» The instantancousvelocitydepends on theinstantaneouspositionvector. hence,velocityis independentof instaneousvalue of the displacementand acceleration does not depend uponthe instantaneousvalue of velocity. Hence,a,c,d, are incorrect |
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| 39812. |
What is radioactivity ? |
| Answer» SOLUTION :The SPONTANEOUS emission of radiations`(alpha,beta and GAMMA rays)` from radioactive substaneous iscalled RADIOACTIVITY. | |
| 39813. |
If the depth of a lake is 10 m, and the temperature of surroundings is -10^@C and that of water at the bottom of the lake is 5^@C. The thermal conductivity of the Ice and water are K_f and K_w respectively. The depth of ice will be |
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Answer» `(5K_w)/(K_w+K_l)` |
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| 39814. |
The phenomenon of polarisation of electromagnetic waveprovesthat theelectromagnetic waves are |
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Answer» LONGITUDINAL |
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| 39815. |
If F = 2/(sin theta + cos theta) then the minimum value of F out of the following option is |
| Answer» ANSWER :D | |
| 39816. |
In the Young's experiment, when a plate of thickness lambda and refractive index of 1.5 is introduced in path of beam the intensity at the position where central maximum occurred previously remained unchanged. The minimum thickness of glass plate is ...... I |
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Answer» `2lambda` Here `m=1 and n=1.5` `d=(lambda)/(1.5-1)=(lambda)/(0.5)=2lambda` `:.d=2lambda` |
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| 39817. |
Calculate the frequency of beats produced in air when two sources of sound are activated, one emitting a wavelength of 32 cm and the other of 32.2cm. The speed of sound in air is 350 m/s. |
| Answer» SOLUTION :7 PER SECOND | |
| 39818. |
When light ray beam is used to determine thi object position, light ...... will give the maximum accuracy. |
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Answer» is converging |
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| 39819. |
A tuning fork vibrating with sonometer, having tension in the wire as T , Produces 4 beats per sec. The beat frequency does not change when tension in the wire changes to 1.21 T . Find the frequency of the tuning fork (assuming that the sonometer wire is vibrating in the same mode in both cases ) |
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Answer» 80 Hz |
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| 39820. |
The peak voltage of an a.c. supply is 300 V. What is r.m.s. voltage ? |
| Answer» SOLUTION :`E_(RMS)= E_0/(SQRT2) = 300/(sqrt2 )V = 212.1 V`. | |
| 39821. |
To a germanium crystal equal number of aluminum arid indium are added. Then, |
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Answer» it remainsas intrinsic SEMICONDUCTOR |
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| 39822. |
A small source of sound moves on a circle as shown in the figure and an observer is sitting on O. Let n_(1), n_(2)and nz be the frequencies heard when the source is at A, B and C respectively, then |
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Answer» `n_(1) gt n_(2) gt n_(3)`  `THEREFORE n lt n`. At B, source MOVE towards O. `therefore n_(2) gt n` At C source is at PERPENDICULAR to O. `therefore n_(3) =`n HENCE `n_(2) gt n_(3) gt n_(1)` |
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| 39823. |
What effect will be noticed when a source of alpha-particles is introduced in a changed gold leaf electroscope ? |
| Answer» Solution :The leaves will collapse very fast. `ALPHA` - PARTICLE ionises the dry air in the ELECTROSCOPE, making it a good conductor and the CHARGE from the leaves flow to earth through this conducting air. | |
| 39824. |
A thin spherical shell radius of r has a charge 2 uniformly distributed on it. At the centre of the shell, a negative point charge -q is placed. If the shell is cut into two identical hemi spheres, still equilibrium is maintained. Then find the relation between Q and q? |
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Answer» Solution :Here the outward electric PRESSURE at EVERY point on the shell due to its own charge is `P_1 = (sigma^2)/(2 in_0) ((Q)/(4pi r^2))^2 , P_1 = (Q^2)/(32 PI^2 in_0 r^4)` Due to -q, the electric field on the surface of the shell is `E = (1/(4 pi in_0) q/(r^2))`. This electric field pulls every point of the shell in inward direction. The inward pressure on the surface of the shell due to the negative charge is `P_2 = sigma E` `= ((Q)/(4 pi r^2)) (1/(4 pi in_0) q/(r^2)) = (QQ)/(16 pi^2 in_0 r^4)` For equilibrium of the hemispheircal shells `P_2 ge P_1` or `(Qq)/(16 pi in_0 r^4) ge (Q^2)/(32 pi^2 in_0 r^4) "" q ge (Q)/(2)`. |
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| 39825. |
An electron is moving with a^4 velocity 5 xx 10^(6) ms^(-1) hat i in a uniform electric field of 5 xx 10^(7) Vm^(-1) j. Find the magnitude and direction of minimum uniform magnetic field in tesla that will cause the electron to move underviated along is original path. |
| Answer» SOLUTION :`10 HATK` | |
| 39826. |
The horizontal component of the Earth's magnetic field is 16 A/m. Calculate the dimensions of the Helmholtz coils designed to compensate the Earth's magnetic field, if the current in a coil is 200 mA and the number of turns in each winding is 50. |
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| 39827. |
Calculate the binding energy per nucleon of ._20^40Ca. Given that mass of ._20^40Ca nucleus = 20 39.962589 u. mass of a proton = 1.007825 u , mass of neutron = 1.008665 u and luis equivalent to 931 Mev. |
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Answer» 4.55 Mev |
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| 39828. |
The total magnification produced by a compound microscope is 20. The magnification produced by the eye piece is 5. The microscope is focussed on a certain object. The distance between the objective and eyepiece is observed to be 14cm. If the least distance of distinct vision is 20cm, calculate the focal length of the objective and the eye piece. |
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Answer» SOLUTION :Given , M = 20, `m_(e ) = 5, D = 20cm`. `m_(e ) = ( D )/( u_(e )) ` ( For eyepiece ) ,br> `5 = ( 20) /( u_(e ))` `u _(e ) = 4 cm` `u _(e ) = - 4cm, V _(e) =D= - 20cm` `(1)/(f_(e)) = (1)/( v_(e)) - (1)/( u_(e ))` `= ( 1)/( -20) +(1)/( 4)` `= ( -1+5)/(20) = ( 4)/( 20)` Hence, FOCAL length of eyepiece, `f_(e ) = 5 cm` `M = m_(0) xx m_(e )` `m_(0) = ( M)/( m_(e )) = ( 20)/( 5) = 4` `v_(0) = 14-4 =10cm` For objective lens, `M_(0) = - ( v_(0))/( u_(0))` `u= - (10)/( u )( :. M_(0)=4)` `rArr u_(0) = - (10)/( u ) = - 2.5 cm` Now, `(1)/(f_(0)) = ( 1)/(v_(0)) - ( 1)/(u_(0))` `= ( 1)/( 10) - (10)/( - 2.5)` `= (1)/(10) + (10)/( 2.5)` `= (1)/(10) + ( 10)/( 25)` `= (1)/(2)` `:.``f_(0) = 2 cm` Hence, focal length of objective `f_(0) = 2.0 cm`. |
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| 39829. |
At the magnetic north poleof the earth the value of the horizontal componentof earth's magneticfield and angle of dip are respectively |
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Answer» ZERO MAXIMUM |
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| 39830. |
In an intrinsic semiconductor, the electron and hole concentrations are_____. |
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Answer» |
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| 39831. |
A thin spherical soap bubble has surface tension S . It's surface is charged with charge Q, its's volume is V it is found that when Q^(2)=npi varepsilon_(0) SV the excees pressure inside the bubble becomes zero find n. |
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Answer» `sigma=Q/(4pir^(2))` ……(2) `sigma^(2)/(2in_(0))+p-p_(0)=(4S)/r` …..(3) `p-p_(0)=0` ….(4) `Rightarrow1/(2 in _(0)).Q^(2)/(16pi^(2)r^(4))=(4s)/r` `RIGHTARROW 1/(2 in_(0)) . (npi in_(0)s)/(16pi^(2)r^(3)). 4/3 pir^(3)=4s` `n=96` 
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| 39832. |
A charged particle oscillates about its mean equilibrium position with a frequency of 10^(9)Hz. What is the frequency of the electromagnetic waves produced by the ocillator? |
| Answer» Solution :Frequency of electromagnetic waves = Frequency of ocillations of CHARGED PARTICLE `10^(9)HZ`. | |
| 39833. |
Figure 1.24 shows electric field lines corresponding to an electric field E. Figure suggests that. |
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Answer» `E_A lt E_B OT E_C` |
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| 39834. |
In the case of interference, the maximum and minimum intensities are in the ratio 16 : 9. Then |
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Answer» The MAXIMUM and MINIMUM AMPLITUDES will be in the ratio 9 : 5 |
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| 39835. |
What is the meaning of 'pleasant'? |
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Answer» Unpleasant |
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| 39836. |
Assertion : Corpuscular theory of light cannot explain change in velocity of light when it changes medium. Reason : According to corpuscular theory of light, speed of light is more in denser medium than in rarer medium. |
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Answer» If both ASSERTION and REASON are CORRECT and reason is a correct explanation of the assertion. |
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| 39837. |
What is noice? |
| Answer» Solution :When a signal is transmitted, the UNDESIRED signals get MIXED with it LEADING to DISTORTION of the signal. | |
| 39838. |
What is the function of a step up transformer ? |
| Answer» SOLUTION : To CONVERT a low alternating VOLTAGE INPUT into a HIGH alternating voltage output. | |
| 39839. |
A bullet on penetrating 30 cm into its target loses its velocity by 50%. What additional distanee will it penetrate into the target before it comes to rest? |
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Answer» 30 cm `a = -(3nu^(2))/(8(30xx10^(-2)))` For latter part of PENETRATION by EQUATION of motion `0-((nu)/(2))^(2)=2ax` `X= -(nu^(2))/(8A)` `= (nu^(2))/(8) ((8(30xx10^(-2)))/(-3nu^(2)))"" `(Using (i)) `= 10xx10^(-2)m = 10cm` |
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| 39840. |
(A) : Voltmeter is connected in parallel with the circuit (R) : Resistance of a volumeter is very large. |
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Answer» Both 'A' and 'R' are TRUE and 'R' is the correct explanation of 'A'. |
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| 39841. |
In working of atransistori, the emitter-base (EB) junction is forward biased while collector-base (CB) junction is reverse biased. Why ? |
| Answer» Solution :Only FORWARD biased emitter-BASE junction can send the-MAJORITY charge carriers from emitter to base and only reverse biased collectro can collect these majority earners from the base REGION. If the emitter is reverse biased, no charge carriers will flow towards the collectror and hence no CURRENT will flow through the transistor. | |
| 39842. |
A galvanometer coil has a resistance of 10 Omega and the meter shows full scale deflection for a current of 1 mA. The shunt resistance required to convert the galvanometer into an ammeter of range 0 - 100 m A is about |
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Answer» `10 OMEGA` `:.` Shunt resistance `r_s = (R_G CDOT I_g)/(I - I_g) = (10 Omega xx 1 mA)/((100 mA - 1 mA)) = 10/99 Omega = 0.1 Omega` (appx.) |
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| 39843. |
Rectification is the process of conversion of |
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Answer» a.c into d.c |
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| 39844. |
A network of resistore is connected to a 16 V battery with internal resistance of 1Omega, as shown in (a) Compute the equivalent resistance of the network. (b) Obtain the current in each resistor.(c ) obtain the voltage drops V _(AB), V _(BC) and V _(CD) |
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Answer» Solution :(a) The network is a simple series and parallel combination of resistors. First the two `4Omega` resistors in parallel are equivalent to a resistor `= [(4xx 4) //(4 +4) Omegta = 2 alpha` In the same way. The `12 Omega and 6 Omega` RESISTORES in parallel are equivalent ot a resistor of ` [ (12 xx 6) //(12+ 6) ]Omega = 4 Omega` The equivalent resistance R of the networn is obtained by conmbining these resistore `(2 Omega and 4 Omega)` with `1 Omega` in series. that is. `R = 2 Omega + 4 Omega + 1Omega = 7 Omega` (b)The total CURRENT In in the circuit is `I = (in )/(R +r) = (16 v)/((7 +1) Omega) = 2A` Consider the restors between A and B. If `I _(1)` is the current in one of the `4Omega` resistors and ` I (2)` the current in the other. `I _(1) xx 4 = I _(2) xx 4` then is, ` I _(1) = I _(2),` which is otherwise obvious from the symmetry of the two arma. But `I _(1) + I _(2) = I =2 A.` THUS. `I _(1) = I _(2) =1A` that , is current in each `4Omega` resistor is 1A. Current in `1Omega` resistor between B and C would be 2 A. Now, consider the resistance between C and D. If `I _(3)` is the current in the `12Omega` resistor, and `I_(4)` in the `6 Omega` RSISTOR. `I _(3) xx 12 =I _(4) xx 6,i.e., I _(4) = 2I _(3)` But, `I_(3) + I _(4) =I =2A` Thus, ` I _(3) = ((2)/(3)) A, I _(4) = ((4)/(3)) A` that is, the current in the `12Omega` resistor is (2/3) A, whithe the current in the `6 Omega` resistor is (4/3) is `V _(AB) = I _(A) xx 4= 1A xx 4 Omega = 4V.` This can also be obtained by multtplying the total current between A and B by the equivalent resistance between A and B, that is. `V_(AB) =2A xx 2 Omega =4V` The voltage drop across BC is `V _(BC)=2 A xx 1 Omega =2 V` Finally, the voltage drop across CD is `V _(CD) =12 Omega xx I _(3) =12 Omega xx ((2)/(3)) A =8V.` This can alternately be obtained by multiplying total current between C and D by the equivalent resistance between C and D. that is `V _(CD) =2 A xx 4 Omega =8V` Note that the total voltage drop across AD is `4 V + 2V +8V =14V.` Thus, the terminal voltage of the bettery is 14 V, while its emf is 16 V. The loss of the voltege `(=2V)` is accounted for by the internal resistance `1 Omega` of the bettery `[2 A xx 1 Omega = 2V].` |
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| 39845. |
In several fungi and plants the bisexual condition is denoted by |
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Answer» Homothallic and monoecious |
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| 39846. |
Greater the height of a TV transmitting antenna. Explain. |
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Answer» Solution :`d=sqrt(2R)` In CASE, height of tower (H) is INCREASED. the DISTANCE (d) upto which TV coverage can be done will also increases. |
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| 39847. |
The following figures in the Column-II indicate the direction of motion of source and the observer match the corresponding graph in Column-I drawn between the apparent frequency and time V_(0)is the velocity of observer V_s the velocity ,of source, and n_0 the frequency of the source: |
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Answer» |
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| 39848. |
A 5^(0)C rise in the temperature is observed in a conductor by passing some current. When the current is doubled, then rise in temperature will be equal to |
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Answer» `5^(0)C` |
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| 39849. |
Needles N_1, N_2, N_3 are made of a ferromagnetic, a paramagnetic and a diamagnetic substance respectively. A magnet when brought close to them will ...... |
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Answer» ATTRACT all THREE of them |
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| 39850. |
In previous example, calculate the stress produced in brass plate |
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Answer» |
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