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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Choose the correct stateme |
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Answer» TEMPERATURE is a SCALAR but temperature gradient is a vector |
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| 2. |
The energy equivalent of one atomic mass unit is : |
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Answer» `1.6xx10^(-19)J` |
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| 3. |
In a given process on an ideal gas, dW = 0 " and " dQ lt 0. Then for the gas |
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Answer» the temperature will DECREASE But as per question, `dQ lt 1` ` :. dU lt 1` ` :. ` The temperature will decrease. |
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| 4. |
When two resistances (50)/(3)Omega and unknown resistance are connected in the left and right gaps of a meter bridge, the balance length is 70cms. When resistance 'x' is connected in parallel to (50)/(3)Omega the balance length becomes 35cms. The value of 'x' is |
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Answer» `2OMEGA` |
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| 5. |
Match the situation given in column A with the possible curves in column B |
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Answer» <P> |
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| 6. |
The neutron was discovered by |
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Answer» MARIE Curie |
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| 8. |
A thermometer liquid which can be used to measure temperature between -40^@C to 40^@C is |
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Answer» a)Water |
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| 9. |
a. What is positron ? "" b. What is its charge ? |
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Answer» SOLUTION :a. A POSITRON is the ANTIPARTICLE of ELECTRON . B. Same as that of electron but +ve. |
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| 10. |
In a refrigerator, the low temperature coil of evaporator is at -23^@C and the compressed gas in the condenser has a temperature of 77^@C. The amount of electrical energy spent in freezing 1 kg of water at 0^@C is |
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Answer» `134400 J` |
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| 11. |
A circuit contains an ammeter, a battery of 30 V and a resistance 40.8 ohm all connected in series. If the ammeter has a coil of resistance 480 ohm and a shunt of 20 ohm, the reading in the ammeter will be |
| Answer» ANSWER :B | |
| 12. |
One kilowatt hour equals: |
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Answer» `36xx10^(5)J` |
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| 13. |
An inductor and a bulb are connected in series to an AC source of 220V, 50Hz. 7C A current of 11A flows in the circuit and phase angle between voltage and current is (pi)/(4)radians. Calculate the impedance and inductance of the circuit |
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Answer» Solution :Here ,`V=220 V , f=50 HZ I=11 A , phi= pi //4.Z =? L=?` `I_(rms) = 0.7071 I_0 ` ` I_0=(I_(rms))/(0.707 ) = (11)/(0.707 ) = 15.55 A` `V_(rms)=0.707 V_0` `V_0 =(V_(rms))/( 0.707 )= (220 )/( 0.707 ) = 311.17 V ` indedence`Z=(V_0)/(I_C) = (311.17)/( 15.55) = 20.01Omega ` `cosphi = (R )/(Z)` ` R=Z COS phi= 20.01cos ((pi)/(4))` ` R= 20.01 xx (1)/(sqrt(2)) = (20.01)/(1.414)= 14.151 OMEGA ` ` Z= sqrt(R^2+ ( omega L)^2)` ` Z^2= R^2 + (2 pi fL)^2 "" (:.omega= 2 pi f )` `(Z^2 -R^2)/( 4 pi ^2f^2) = L^2impliesL^2=((20.01)^2 - (14.151)^2)/( 4 xx (3.14 )^2 xx (50)^2)` `L^2 = ( 400.40- 200.25 ) /( 98596 )` `= (200.15 )/( 98596 )= 0.00202H` `L= 0.0449H= 44.9mH .` |
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| 14. |
An equilateral triangular plate of mass 'm' is tied at the three vertices by threads S_(1), S_(2) and S_(3) respectively, keeping the plate in equilibrium such that the plane of plate is kept horizontal and three blocks of mass 'm' each, are connected to the corners of the plate as shown in figure. Side of the triangular plate is l. At t= 0, string 'S_(1)' is burnt, then the tension in the sring S_(3) , just afterward , is |
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Answer» `(4)/(3) MG` |
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| 15. |
In a rocket of mass 1000 kg, fuel is consumed at the rate of 40 kg/s. The velocity of the gases ejected from rocket is 5x10^4 m / s. The thrust on the rocket is |
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Answer» `2 x 10^3 N` |
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| 16. |
Estimate the distance for which ray optics is good approximation for an aperture of 4 mm and wavelength 400 nm |
| Answer» ANSWER :C | |
| 17. |
The binding energy per nucleon is maximum in the case of |
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Answer» `_92U^235` |
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| 18. |
A convex and a lens separated by distane d are then put in contact. The focal length of the combination |
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Answer» Becomes 0 `(1)/(F)=(1)/(f_(1))+(1)/(f_(2))-(d)/(f_(1)f_(2))` As `f_(2)` is negative, `(1)/(F_(1)) = (1)/(f_(1)) - (1)/(f_(2)) + (d)/(f_(1)f_(2))` When put TOGETHER, `(1)/(F_(2)) = (1)/(f_(1)) - (1)/(f_(2))` `therefore F_(1) = (f_(1)f_(2))/(f_(2) - f_(1) + d) ,F_(2) = (f_(1)f_(2))/(f_(2) - f_(1))` Therefore the focal lenght for this combination, become LARGER. If `f_(1) = f_(2)`, this combination GIVES infinity in contact and not zero. Therefore for the option given different focal lenght are assumed. |
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| 19. |
findoutmotionof treeboyas seenbu oldman . |
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Answer» SOLUTION :withrespect tooldman: ` V_("BOY")= 18m//s( to )` ` T_("tree") = 2 m//s( to )` ` V_(" Bird") = 18m//s( to )and 12m//s( UARR)` |
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| 20. |
A particle is moving three times as Fast as an electron .The ratio of the de-Broglie wavelength of the particle to that of the electron is 1.813xx10^(-4). Calculate the particle's mass and indentify the particle. |
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Answer» Solution :Suppose ,the speed of ELECTRON is `v_(e)` and speed of PARTICLE is v.Mass of electron `m_(e)` and mass of particle is m and de-Broglie wavelength of electron `lambda_(e)` and de-Broglie wavelength of particle is `lambda`. Now de-Broglie wavelength of particle, `lambda=(H)/(p)=(h)/(mv)` [`thereforep=mv`] `therefore m=(h)/(lambdav)` `therefore` For electron `m_(e)=(h)/(lambda_(e)v_(e))` For particle `m=(h)/(lambdav)` `therefore (m)/(m_(e))=(lambda_(e))/(lambda)XX(v_(e))/(v)` `therefore m=(m_(e))xx(lambda_(e))/(lambda)xx(v_(e))/(v)` `=9.11xx10^(-31)xx(1)/(1.813xx10^(-4))xx(1)/(3)` `therefore m=1.675xx10^(-27)kg` `therefore` This particle can be proton or NEUTRON. |
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| 21. |
The absolute temperature of a gas is increased three times. The root mean square velocity of the gas molecules will be |
| Answer» Answer :D | |
| 22. |
A bob of mass m, suspended by a string of length l_(1) is given a minimum velocity required to complete a full circle inthe vertical plane. At the highest point, it collides elasticallywith another bob of mass 2m suspended by a string of length l_(2), which is initially at rest. Both strings are massless and inextensible. If the second bob, after collision acquires the minimum speed required to complete a full circle in the vertical plane, the ratio l_(2)//l_(1) is (5//4)n. The value of n is |
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Answer» 3 |
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| 23. |
A condenser of capacity 16 mF charged to a potential of 20 V is connected to a condenser of capacity C charged to a potential of 10 V as shown in the figure. Ifthe common potential is 14 V. the capacity C is equal to |
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Answer» 18 mF |
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| 24. |
In terms of Rydberg constant R, the wave number of the first Balmer line is |
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Answer» R |
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| 25. |
Can we interchange emitter and collector of a transistor ? |
| Answer» Solution :No. we cannot interchange EMITTER and collector due to two reasons.(i) The doping level of emitter is higher than that of collector.(II) The CONTACT area of emitter-collector JUNCTION is larger than that of emitter base junction | |
| 26. |
If the Earth suddenly shrinks so as reduce its volume, mass remaining unchanged, what will be the effect on the duration of the day ? |
| Answer» Solution :If the Earth SUDDENLY shrinks, mass remaining CONSTANT, the moment of INERTIA of the Earth will descrease, and consequently the angular speed of ROTATION `omega` about its axis will increase. SINCE period `Tprop1/omega`. The duration of the day T will decrease. | |
| 27. |
The process of separating radio signal from the modulated wave is known as |
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Answer» SUPERIMPOSITION |
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| 28. |
Which of the following is not an application of eddy currents |
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Answer» Induction FURNACE |
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| 29. |
Let an electron requires 5xx10^(-19) joule energy to just escape from the irradiated metal.If photoelectron us emited after 10^(-9) s of the incident light,calculate the rate of absorption of energy .If this process is considered clasically.The light energy is assumed to be continously distributed over the wave front.Now,the electron can only absorb the light incident withina small area,say 10^(-19)m^(2).What is the intensity of illumination in order to see the photoelectric effect? |
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Answer» SOLUTION :The rate of ABSORPTION of energy (power) is `p(e )/(t)=(5xx10^(-19))/(10^(-9)=5xx10^(10)(J)/(S)` From the definition of the INTENSITY of light, `I=("Energy")/("time"xx"area")=(5xx10^(-10))/(10^(-19))=5xx10^(9)(J)/(s.m^(2))` `(i.e. 500 billion ("Watt")/(m^(2)))` Since, practically it is IMPOSSIBLY high energy,which suggests that explanation of the photoelectric effect in classical term is not POSSIBLE. |
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| 30. |
MAGNETIC FIELD DUE TO CURRENT CARRYING SOLENOID |
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Answer» If both assertion and reason are true and reason is the correct explanation of assertion. The magnetic field inside the solenoid is uniform. |
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| 32. |
Answer the following questions: (b) As you have learnt in the text, the principle of linear superposition of wave displacement is basic to understanding intensity distributions in diffraction and interference patterns. What is the justification of this principle? |
| Answer» Solution :SUPERPOSITION PRINCIPLE follows from the linear character of the (differential) equation governing WAVE motion. If `y_(1) and y_(2)` are SOLUTIONS of the wave equation, so is any linear COMBINATION of `y_(1) and y_(2)` . When the amplitudes are large (e.g., high intensity laser beams) and non-linear effects are important, the situation is far more complicated and need not concern us here. | |
| 33. |
Electronicmagnetic wave have wide range of wavelengths starting from 10^-14m to 10^3m. The em wave of different wavelength are used for different purpose. The gamma-rays which have lowest wavelength are least energetic. Read the above the passage and answer the following questions: (i) What are more energetic waves, x-rays or ultra violet rays? (ii) Why are teh radio waves not used to detect fracture in the bones of a human body when they can deliver a message at large distances? (iii) What are the basic values displayed by the above study? |
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Answer» Solution :(i) X-rays have the wavelength range `10^-13m` to `3xx10^8m`, which is smaller than that of ULTRAVIOLET rays of wavelength range `6xx10^-9m` to `4xx10^-7m`, i.e., `lambda_X lt lambda_(UV)`. Since, energy `E=(hc)/lambda` or `Eprop 1/lambda` , so `(E_X)/(E_(UV))=(lambda_(UV))/(lambda_X) gt 1` or `E_X gt E_(UV)`. Thus X-rays are more energetic than ultraviolet rays. (ii) The wavelength of RADIOWAVES is of range 3.0m to `6xx10^2m`, which is very large as compared to the size of the molecules of our blood, FLESH, bones etc. As the energy of radiowaves is quite small so these radiowaves can not penetrate the blood,flesh and bones of our body. That is why we can not use the radiowaves to detect the FRACTURES in bones. (iii) Just as different em waves have different applications depending on their wavelength, in the same way, every person has different qualities, which make him suitable for different pusposes/fields. The aim of a TEACHER of manager is to identify the quality in different children/persons and put them to their best use accordingly. |
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| 34. |
Green light of wavelength 5460 A is incident on an air glass interface. If the refractive index of glass is 1.5, the wavelength of light in glass would be (C = 3 xx 10^8 ms^-1) |
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Answer» 3640 A |
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| 35. |
In absorption spectrum of Na the missing wavelength (s) are |
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Answer» 589 nm |
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| 36. |
A coil has 1000 turns and 500 cm^(2) as its area. It is placed at right angles to a magnetic field of 2xx10^(-5) Wb m^(-2). The coil is rotated through 180^(@) in 0.2s. Find the average emf induced in the coil. |
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Answer» |
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| 37. |
The ceiling of a hall is 25 m. What is the maximum distance at which a ball can be thrown inside the hall ? |
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Answer» 100m `R_(MAX)`= 4H =` 4 xx 25 `= 100 m. |
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| 38. |
Maximum kinetic energy of photoelectron emitted is independent of …… |
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Answer» FREQUENCY of INCIDENT radiation |
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| 39. |
Electric field is always directed ____________ to an equipotential surface. |
| Answer» SOLUTION :PERPENDICULAR | |
| 40. |
A hollow metal sphere of radius 5 cm is charged such that the potential on its surface is 10 V. What is the potential at the centre of the sphere? |
| Answer» SOLUTION :Potential at the centre of SPHERE = Potential on its surface = 10 V. | |
| 41. |
A charge +2 muC is placed at a point A (2,1,0). Another charge -3muC is placed at a point B (4,2,-1). Calculate net force on a charge +5 muC placed ata point C (-1,3,2). |
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Answer» Solution :If `vecr` represents a vector joining the two charges along the direction of force. Then the force in vector form can be written as follows : `vecF=(QQ)/(4piepsilon_0)[vecr/r^3]` Notice that is the given figure the points are not shown on appropriate locations as per their coordinates . But the fact is that you do not require this , as you can WORK just on the basis of given coordinates . Here , you should note that the direction of `vecr_1` and `vecr_2`are selected as per the direction of forces `vecF_1` and `vecF_2` , respectively.  `vecr_1=(-1-2)hati+(3-1)hatj+(2-0)hatk=-3hati+2hatj+2hatk` `vecr_2=(4+1)hati+(2-3)hatj+(-1-2)hatk=5hati-hatj-3hatk` `r_1=sqrt((-3)^2+(2)^2+(2)^2)=sqrt17` `r_2=sqrt((5)^2+(-1)^2+(-3)^2)=sqrt35` `vecF_1=(9xx10^9)(2xx10^(-6))(5xx10^(-6))vecr_1/r_1^3 =0.09 vecr_1/r_1^3` `vecF_2=(9xx10^9)(3XX10^(-6))(5xx10^(-6))vecr_2/r_2^3=0.135 vecr_2/r_2^3` Now substituting the values `vecF_1=0.09vecr_1/r_1^3=0.09/17^(3//2) (-3hati+2hatj+2hatk)` `=(0.00130)(-3hati+2hatj+2hatk)` `vecF_2=0.135 vecr_2/r_2^3 =0.135/35^(3//2)(5hati-hatj-3hatk)` `=(0.00065)(5hati-hatj-3hatk)` On solving we get `vecF=vecF_1+vecF_2=0.00065(-hati+3hatj+hatk)` MAGNITUDE of net force F=0.00065 `(sqrt11)` =0.0022 newton |
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| 42. |
What kind of blueprint does the natural structure provide ? |
| Answer» Solution :The NATURAL structure provides a blueprint for creating ultra-durable synthetic MATERIALS that COULD be useful for mechanical components in electronics, and in other DEVIES that undergo repetitive movement, abrasion and contact stress. | |
| 43. |
A bady increases in weight from 5.58 kg to 6.16 kg . How much weight has the bady gained ? |
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Answer» `triangleQ=triangleQ_(1)+triangleQ_(2)+triangleQ_(9)=(i_(1 av)^(2)+i_(2av)^(2)+i_(9av)^(2))R trianglet= 2580 xx 40 x 1 =1.03 xx 10^(5)J` (b) Integration. The current according to the low i=5+2t. The quantity of heat is `Q=underset(0)overset(10)int i^2 R dt=40 underset(0)overset(10)int (5+2t)^(2) dt-20 underset(0)overset(10)int (5+2t)^(2) d (5+2t)=`  `=(20)/3 (5 2t)^(3) underset(0)overset(10)int =(20)/3 (25^(3)-5^(3))=(20 xx 775 xx 20)/(3)=1.03 xx 10^(5)J` We SEE that the result of the numerical calculation was accurate. |
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| 44. |
The meaning of 'observe'... |
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Answer» In a SCIENTIFIC kind of way |
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| 45. |
(A): Shake and light year, both measure time. (R): Both velocity and velocity of light have same dimensions. |
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Answer» Both (A) and (R) are true and (R) is the CORRECT EXPLANATION of (A) |
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| 46. |
A sonometer wire under tension of 64N vibrating in its fundamental mode is in resonance with a vibrating tuning fork. The vibrating portion of that sonometer wire has a length of 10cm and a mass of 1 gm. The vibrating tuning fork is now moved away from the vibrating wire with a constant speed and an observer standing near sonometer hears one beat per second. The speed with which the tuning fork is moved is 752 xx 10^y (speed of sound in air is 300 m/s). Then y |
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Answer» |
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| 47. |
A 6-volt battery of negligible internal resistance is connected across a uniform wire AB of length 100cm. The positive terminal of another battery of emf 4V and internal resistance 1(Omega)is joined to the point A as shown in figure.Take the potential at B to be zero. (a) What are the potentials at the poits A and C ? (b) At which point D of the wire AB, the potential is equal to the potential at C?(c ) If the point C and D are connected by a wire, what will be the current through it ? (d) If the 4V battery is replaced by 7.5V battery,wht would be the answer of parts (a) and (b) ? |
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Answer» SOLUTION :Potential difference between AB is 6V. B is at 0 potential. Thus potential of a point is 6V. The potential difference between AC is 4V. `V_A- V_C = 4 ` ` V_C = V_A -4 = 6-4 = 2V. ` (b) The potential difference at D =` 2V = V_AD = 4V`. ` V_BD = 0V ` Current through the resisters R_1 & R_2 are equal. Thus , ` 4/R_1 = 2/R_2 ` ` rArr R_1/R_2 = 2 ` ` rArr l_1/l_2 = 2 ` (Acc.to the law of potentiometer) ` rArr l_1 + l_2 = 100 cm ` ` rArr l_1 + (l_1/2) = 100 cm ` ` rArr 3l_1/2 = 100cm ` ` rArr = l_1 = 200/3 cm = 66.67 cm .` ` AD = 66.67 cm` (c)When the points C and D are connected by a wire, current FLOWING through it is 0, since the points are equipotential . (d) Potential at A = 6V. ` Potential at C = 6 - 7.5 =- 1.5 V. ` The potential at B= 0 and towards A potential INCREASES. Thus negative potential point does not come within the wire. |
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| 48. |
For whom is life hell according to Padma's mother? |
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Answer» Widow |
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| 49. |
According to the corpuscular theory which of the following is not the property of light ? |
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Answer» LIGHT TRAVELS in straight of lines |
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| 50. |
The magnetic field at the centre of a circular coil of radius r is pi time that due to a long straight wire at a distance r from it for equal currents. The following figure show three cases. In all cases the circular part has radius r and straight ones are infinitely long. For same current, the B field at the centre P in cases 1, 2, 3, has the ratio |
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Answer» `(-(PI)/(2)):((pi)/(2)):((3pi)/(4)-(1)/(2))` |
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