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InterviewSolution
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.
| 3701. |
A mica slab of thickness equal to the distance between the two plates of a parallel plate air capacitor is inserted in the space between the plates. Explain the changes in capacitance in the following case, When the space between the plates of the capacitor is totally filled by the mica slab. |
| Answer» SOLUTION :The CAPACITANCE will be `(in_0k ALPHA)/(d)`. | |
| 3702. |
A current canying straight wire is placed along east-west and current is passed through it eastward . The direction of the force act on it due to horizontal component of earth's magnetic field is |
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Answer» DUE west |
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| 3703. |
Where did Nanaji take Naina and Ajit to? |
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Answer» Himapet village |
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| 3704. |
Two charged particles having charge 2.0 xx 10^(-8)C. Each are joined by an isolating string of length 1m and the system is kept on a smooth horizontal task. The tension in the string. |
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Answer» `36 XX 10^(-6)N` |
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| 3705. |
What is polarisation of light ? Name any one method of producing plane polarised light. |
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Answer» Solution :The PHENOMENON of confining the vibrations of LIGHT in a single plane is CALLED POLARISATION. Reflection/scattering |
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| 3706. |
A light source of wavelength 520 nm emits 1.04 xx 10^(15) photons per second while the second source of 460 nm produces 1.38 xx 10^(15) photons per second. Then the ratio of power of second source to that of first source is……………… |
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Answer» `1.00` `P_(1) = (1.04 xx 10^(15) xx 1240)/(520) = 2.48 xx 10^(15)` `P_(2) = (1.38 xx 10^(15) xx 1240)/(460) = 3.72 xx 10^(15)` `(P_(2))/(P_(1)) = (3.72 xx 10^(15))/(2.48 xx 10^(15)) = 1.5` |
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| 3708. |
We call the semiconductor device which converts solar energy into electrical energy as ? |
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Answer» |
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| 3709. |
A short bar magnet placed with its axis inclined at 30^@to the external magnetic field of 800 G acting horizontally experiences torque of 0.016 Nm. Calculate : (i) the magnetic moment of the magnet. (ii) the work done by an external force in moving it from most stable to most unstable position.(iii) what is the work done by the force due to the external magnetic field in the process mentioned in (ii)? |
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Answer» Solution :It given that `B = 800 G = 8 xx 10^(-2) T , tau =0.016 N m and THETA =30^@` (i) Magnetic moment of the MAGNET m`=(tau)/(B sin theta) = (0.016)/(8 xx 10^(-2) xx sin 30^@) = 0.40 A m^2` (ii) In most stable position `theta_1 = 0^@`and in most unstable position `theta_2 = 180^@ `. hence , work DONE by an external force in moving the magnet from stable to unstable position `W = mB[sin theta_1 - sin theta_2] = 0.40 xx 8 xx 10^(-2) [ sin 0^@ - sin 180^@] = +0.064 J` (iii) work done by the force due to magnetic field B in moving the magnet from stable to unstable position. `W. =-W =-0.064J` [Since now force and displacement are in mutually opposite directions] |
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| 3710. |
A particle moves in x-y plane according to equations x=4t^2+5t+16 and 6y=5t. What is the acceleration of the particle? |
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Answer» Solution :`x = 4t^2+5T+16` y= 5t `v_x = dx/DT = 8t+5` v_y =5 `a_x = (dv_x)/dt = 8` (dv_y)/dt = a_y = 0 `a = SQRT(a_x^2+a_y^2) = sqrt(8^2+0)` = 8m / `s^2` |
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| 3711. |
A long cylindrical conductor of radius R has two cylindrical cavities of diameter R through its entire length, as shown in the figure. There is a current I through the conductor distributed uniformly in its entire cross section (apart from the cavity region). Find magnetic field at point P at a distance r = 2R from the axis of the conductor (see figure). |
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Answer» |
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| 3712. |
J.J. Thomson's experiment demonstrated that |
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Answer» CATHODE RAYS are streams of negatively charged ions |
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| 3713. |
An object is placed in front of a convex mirror of focal length f. Find the maximum and minimum distance of an object from the mirror such that the image formed is real magnified |
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Answer» `2F and OO` |
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| 3714. |
Whena piece of sodium or potassium are exposed to sunlight …….. |
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Answer» they will become negatively charged. |
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| 3715. |
What does indirect appeal mean in the paragraph? |
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Answer» It MEANS that one should only do things appealing to oneself |
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| 3716. |
LEAD NITRATE ON HEATING GIVES |
| Answer» Answer :D | |
| 3717. |
If the value of acceleration due to gravity on the surface of the earth is 9.8m//s^2. What is the mean value of density of earth ?(G=6.67xx10^-11nm^2//kg^2,R=6400km) |
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Answer» a)12hr. |
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| 3718. |
Figure shows a point charge +Q, located at a distance R/2 from the centre of a spherical metal shell. Draw the electric field lines for the given system. |
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Answer» SOLUTION :INSIDE OUTSIDE 
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| 3719. |
A plane longitudinal wave having angular frequency co = 500rad/sec is travelling in positive x-direction in a medium of density rho =1 kg//m^(3)and bulk modulus 4 xx 10^4 N//m^2. The loudness at a point in the medium is observed to be 20 dB. Assuming at x = 0 initial phase of the medium particles to be zero, find the equation of the wave |
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Answer» `y = 2 xx 10^(-9) sin(500 t - (5x)/2)` |
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| 3720. |
No current flows betweentwo charged particleswhen connected if they have same ______ . |
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Answer» If they have the same capacitance |
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| 3721. |
Photon is a quanta of light. Who introduced the concept of photon? |
| Answer» SOLUTION :MAX PLANK | |
| 3722. |
The circular boundary of the concave mirror subtends a cone of half angle theta at its centre of curvature.The minimum value of thetafor which ray incident on this mirror parallel to the principle axis suffers reflection more than one is |
| Answer» ANSWER :B | |
| 3723. |
Statement - I : The angular dispersion produced by a prism increases if the mean refractive index of the matetial of prism increases. Statement - II : A prism prodouces deviation as well as dispersion when polychromatic light is incident on it. |
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Answer» Statement I is TRUE, statement II is FALSE. |
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| 3724. |
Two sound waves having wave numbers 0.98 m^(-1)" and "1 m^(-1) when superposed produce 7 beats per second. The speed of sound in air is |
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Answer» `330 m//s` |
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| 3725. |
When sound is produced in an aeroplane with a velocity of 200 m/s horizontally its echo is heard after 10sqrt(5)seconds. If velocity of sound in air is 300 ms^(-1)the elevation of aircraft is |
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Answer» 250 m Velocity of the plane = 200 m/s OC `=200 xx 5sqrt(5) = 2236`m BC = velocity of sound ` xx 5sqrt(5)` `rArr` BC `=300 xx 5sqrt(5) = 3354`m `therefore OB = sqrt(BC^(2) - OC^(2))` `rArr OB ~= 2500 m` The plane is 2500 m abvoe the ground. 
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| 3726. |
The total number of tubes of force passing normally through unit area is called as : |
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Answer» ELECTRIC flux |
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| 3727. |
Statement -1 : The magnetic field at the centre of a circular loop carrying current is zero. Statement -2 : Due to symmetry the field prpduced by the diametrically opposite points add to each other. |
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Answer» Statement -1 is True Statement -2 is True , Statement -2 is a CORRECT explanation for Statement -1 |
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| 3728. |
Mention any two importance of speed of light. |
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Answer» Solution :(i) To calculate the frequency, wavelength of electromagnetic wave using `c = flambda`. (ii) To calculate the REFRACTIVE index of the medium `n=(c)/(v)` (III) In Einstein.s mass-energy RELATION `E = mc^(2)` (iv) To determine the RELATIVISTIC mass in theory of relativity. |
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| 3729. |
If no heat flows between the system and surrounding in a thermodynamic process, then the process is |
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Answer» isothermal |
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| 3730. |
Show the trajectory of a particle of different impact parameter and using it how did Rutherford determine the upper limit of the nuclear size ? |
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Answer» Solution :IMPACT parameter is the perpendicular distance of the initial velocity vector of the a-particle from the centre of the nucleus and the trajectory of the `alpha`-particle depend on the impact parameter of collision b with the nucleus. A GIVEN beam of a particles has a DISTRIBUTION of impact parameter b, so that the beam is scattered in various directions with different probabilities. In a beam all particles have nearly same kinetic energy.  The smaller the impact parameter b of the C-particle the closer it is to the nucleus and its scattering is larger means the angle of scattering is larger. In case of head on collision, the impact parameter is minimum (b = 0) and the C-particle rebounds back `(theta=pi)`. As the impact parameter of the `alpha`-particle becomes larger, the scattering angle decreases and a particle for a larger impact parameter keeps the motion on its original trajectory without scattering that is scattering angle `theta=0^(@)`. Hence, only a small FRACTION of the number of incident particles rebound back indicates that the number of `alpha`-particles undergoing head on collision is small. This means that the mass of the atom is concentrated in a small volume. Therefore, Rutherford scattering is a powerful way to determine an upper limit to the SIZE of the nucleus. From this experiment, Rutherford suggested the dimension of nucleus is `10^(-15)m` to `10^(-14)m` |
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| 3731. |
Four point charges are placed at the four corners of a square in the two ways (i) and (ii) as shown below: Will the i. electric field ii. electric potential, at the centre of the square, be the same or different in the two configurations and why? |
Answer» SOLUTION :i. 
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| 3732. |
Find equivalent resistance between points A and B in the circuit shown. |
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Answer» |
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| 3733. |
Two cells of emf 1.25V, 0.75V and each of internal resistance 1 ohms are connected in parallel. The effective voltage will be |
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Answer» 1 V |
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| 3734. |
White light is passed through a double slit aand interference pattern is observed on a screen 2.5 away. The separation between the slits is 0.5 mm. The first violet and red fringes are formed 2.0 mm and 3.5 mm away from the central white fringe. Calculate the wavelengths of the violet and the red light. |
| Answer» Answer :B | |
| 3735. |
Equivalent resistance across A and B in the given circuit is |
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Answer» `7R ` |
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| 3736. |
A ray of light passing through a glass prism of refracting angle of 60^@ undergoes a minimum deviation of 30^@. Calculate the RI of glass prism? |
| Answer» SOLUTION :`MU=(SIN 1/2((A+D_m)))/(sinA/2)=(sin(60+30))/2/sin60/2=(SIN45^@/sin 30^@)=1/sqrt2xx2=sqrt2=1.414` | |
| 3737. |
A short bar magnet has a magnetic moment of 0.48 JT^(-1) . Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on (a) the axis, (b) the equatorial lines (normal bisector) of the magnet. |
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Answer» Solution :`m = 0.48 JT^(-1),r = 10 cm` a . Field on axial LINE= `(mu_0)/(4pi)(2M)/r^3=(10^(-7)xx2xx0.48)/((10xx10^(-2))^3)=(10^(-7)xx2xx0xx0.48)/(10^(-3))` `=0.96xx10^(-4)T,` in the DIRECTION of `barm` b. Field on equatorial line = `=(mu_0)/(4pi)xxm/r^3=0.48xx10^(-4)T, ` OPPOSITE to `barm` |
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| 3738. |
In the given figure, a long platform of mass m is placed on frictionless surface. Two blocks of masses 4 m and m (where m=10 kg) are placed on the platform. For both blocks, the coefficient of static friction with platform equal to 0.16 and the coefficient of kinetic friction equal to 0.10 the blocks are connected by a light ideal string through a light pulley (mounted at a movable massless stand). Which is acted upon by an unknown horizontal force F. if the acceleration of the platform is 2m//s^(2). Find the value of unknown force F and acceleration of blocks 4 m and m on a_(1) and a_(2) respectively. |
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Answer» `{:(F(N),,a_(1)(m//s^(2)),,a_(2)(m//s^(2))),(180N,,2m//s^(2),,8m//s^(2)):}`  Both block 4 m and m has tendency of motion towards rightwards so friction `(F_(1)` and `F_(2))` will act on both blocks LEFTWARD, hence on the platform rightward.So `a=(F_(1)+F_(2))/(m)=2m//s` `F_(1)+F_(2)=20N` `F_(1)le64N_(1)" "F_(2)le16N` Case I: Suppose blok of mass 4 m has a RELATIVE motion with respect to platform So, `F_(1k)=40N` `F_(2)=-20N` (not possible) Case II: Suppose block of mass m has a relative motion with respect to platform, so, `F_(2k)=10NimpliesF_(1s)=10N` (possible) `(F)/(2)-F_(1s)=4mxx2implies(F)/(2)=1+4xx10xx2=90` `impliesF=180Nimplies(F)/(2)-F_(2k)=maimplies90-10=10aimpliesa=8m//s^(2)` |
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| 3739. |
A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects whena) the telescope is in normal adjustment (i.e., when the final image is at infinity) ? |
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Answer» Solution :Here, `f_(0)=140cm, f_(E)=5.0cm` Magnifying power = ? a) In NORMAL adjustment, Magnifying power `=(f_(0))/(-f_(e))=(140)/(-5)=-28` |
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| 3740. |
A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects whenb) the final image is formed at the least distance of distinct vision (25 cm) ? |
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Answer» Solution :Here, `f_(0)=140cm, f_(e)=5.0cm` Magnifying POWER = ? B) When final IMAGE is at the least distance of distinct vision, Magnifying power `=(-f_(0))/(f_(e))(1+(f_(e))/(d))=(-140)/(5)(1+(5)/(25))=-33.6` |
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| 3741. |
A thin glass (refractive index 1.5) lens has optical power of - 5D in air. Its optical power in a liquid medium with refractive index 1.6 will be ..... |
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Answer» `-1D` `P_2=(1)/(f_2)=((n_2-n_1)/(n_1))((1)/(R_1)-(1)/(R_2))` for LIQUID `therefore(P_1)/(P_2)=((n_2-n_1)/(n_1))((n_1)/(n_2-n_1))` `=((1.5-1.0)/(1))((1.6)/(1.5-1.6))` `=0.5xx(-16)=-8` `therefore(-5D)/(P_2)=-8` `therefore P_2=5/8=0.625` D |
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| 3742. |
Two conductors are made of the same material and have the same length. Conductor A is a solid wire of diameter 1 mm. Conductor B is a hollow tube of outer diameter 2 mm and inner diameter 1 mm. Find the ratio of resistances R_A to R_B |
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Answer» Solution :Here ` rho_1 = rho_2 =rho ` and ` l_1 = l_2 = l` Conductor A is a solid wire of DIAMETER D=1 mm, its cross-section area`A_1 = (PI D^2)/(4)` Conductor B is a hollow tube of OUTER diameter `D_1` = 2 mm and INNER diameter `D_2` = 1 mm, henceits cross-section area `A_2 = pi/4 (D_1^2 - D_2^2)` As `R= (rho l)/(A) `, hence `(R_A)/(R_B) = A_2/A_1 =(pi/3 (D_1^2 - D_2^2) )/(pi/4D^2) = (D_1^2 - D_2^2)/(D^2) = ( (2mm)^2 - (1mm)^2)/((1mm)^2) = 3/1` ` rArr R_A = 3R_B` |
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| 3743. |
A particle moving along straight line has velocity v = mu s^2 where s is in the displacement . If s = s_0 then which of the following graph best represent s versus t |
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Answer»
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| 3744. |
Draw a simple circuit of a CE transistor amplifier. Explain its working. Show that the voltage gain, A_V of the amplifier is given by A_v = - ( beta_(ac) R_L )/(I_i)where beta_(Ac) is the current gain, R_L is the load resistance and r_1 is the input resistance of the transistor. What is the significance of the negative sign in the expression for the voltage gain? |
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Answer» Solution :When an a.c. input SIGNAL `V_i` is superimposed on the bias `v_(B E)` the OUTPUT, which is measured between collector and ground, increases `V_(C C) = V_(CE)+I_C R_L` ` V_(B B) = V_(B E) +I_B R_B` when`v_i`is notzerowe have ` V_(BE )+V_i = V_(BE)+I_BR_B+DeltaI_B(R_B +R_i)` `impliesV_i =Delta I_B(R_B +R_I)` ` V_i= r DeltaI_B` changein `I_B ` causesa changein ` I_C` HENCE `beta_(a.c) = (DeltaI_C)/( DeltaI_B ) = (I_C)/(I_B)` As ` DeltaV_(C C)= DeltaV_(CE )+R_LDelta I_C = 0` ` impliesDelta V_(CE ) =- R_LDelta L_C` `impliesV_0=- R_L Delta I_C` ` = beta _(ac )DeltaI_B R_L` ` implies `Voltagegainof theamplifer ` A_0= (V_0)/( V_i )= (Delta V_(CE ))/( rDelta I_B) = ( - beta_(ac) Delta I_bR_l)/( rDelta I_E)` ` = beta_(ac )(R_L )/(r)` negative signin tehexperessionshowsthat ouputvoltageand inputvoltagehaephasedifferenceof `pi` |
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| 3745. |
At a temperature of 20^(@)C, the volume of a certain glass flask, up to a reference mark on the cylindrical stem of the flask is exactly is 100 cm^(3). The flask is filled to this point with a liquid of coefficient of cubical expansion 120xx10^(-5).^(@)C^(-1)., when both the flask and the liquid are at 20^(@)C. The corss section ofthe stem is 1mm^(@) and the coeffiocient of linear expansion of glass is 8xx10^(-5).^(@)C^(-1). Fidn the rise of fall of the liquid level in the stem, when the temperature is raised to 40^(@)C. |
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| 3746. |
Ratiram was the son of ............... |
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Answer» SOHANLAL Ratiram |
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| 3747. |
A maximum current of 1.5 mA can be passed through a galvanometer of resistance 30 Omega. Calculate the resistance to be connected in series to convert it into a voltmeter of range 0-9V. |
| Answer» SOLUTION :5070 `OMEGA` | |
| 3748. |
Aclinic alines are of |
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Answer» ZERO DECLINATION |
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| 3749. |
What did the grandmother do in her last minutes? |
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Answer» She was TELLING the BEADS of her rosary |
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| 3750. |
A condenser of capacity 2mF charges to a potential 200V is connected in parallel with a condenser of same capacity but charged to a potential 100V. The percentage loss of energy of system is |
| Answer» ANSWER :D | |
