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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.
| 41551. |
In the above question, the speed of each ball relative to ground just after they leave the disc is - |
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Answer» `(Romega_(0))/(sqrt3)` Let the speed of each ball just after they leave the disc be V. From conservation of energy `(1)/(2)[(1)/(2)mR^(2)]omega_(0)^(2)=(1)/(2)[(1)/(2)mR^(2)]omega^(2)+(1)/(2)[("m")/(2)]v^(2)+(1)/(2)[("m")/(2)]v^(2)` solving we get `v=(2Romega_(0))/(3)` NOTE : `v=sqrt((OMEGAR)^(2)+V_(r)^(2)) , V_(r)` = RADIAL velocity of tha ball |
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| 41552. |
Which of the following error is not systematic error ? |
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Answer» Least count ERROR |
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| 41553. |
If alpha=2N_0lambda, calculate the number of nuclei of A after one half-life of A, and also the limiting value of N as |
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Answer» `2N_0, 5/2N_0` `therefore N=1/lambda[2lambdaN_0-(2lambdaN_0-lambdaN_0)e^(-lambdat)]` or `N=(lambdaN_0)/lambda[2-e^(In(2))][Here""e^(-In(2))=2^(-1) = 1/2]` or `N=(lambdaN_0)/lambda[2-1/2] =(3N_0)/2 ` or `N=3/2N_0` When `t to oo ` and `alpha = 2lambdaN_0` `N=alpha/lambda=(2lambdaN_0)/lambda=2N_0` or `N=2N_0` |
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| 41554. |
What is the refractive index of water ? |
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Answer» Solution :` A = 60^(@) , D_(m) = 40^(@)` n ` = (SIN ((A + D_(m))/(2)) )/( sin((A)/(2))) = (sin ""((60 + 40)/(2)))/( sin((60)/(2))) = 1.53 "" n_(gw) = (n_(g))/(n_(w)) = (1.53)/(1.53) = 1.15` `n_(gw) = (sin ((A + D_(m))/(2)) )/( sin((A)/(2)))"" i.e., 1.15 = (sin"" ((60 + D_(m))/(2)))/( sin 30 )` `therefore sin ((60 + D_(m))/(2)) = 1.15 xx sin 30 = 0.573 ` ` (60 + D_(m))/(2)= 35^(@) 6.` `D_(m) = 70^(@) 12. - 60^(@) = 10^(@) 12.` |
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| 41555. |
Three point charges Q_1, Q_2and Q_3, in that order are placed equally spaced along a straight line. Q_2and Q_3 are equal in magnitude but opposite in sign. If the net force on Q_3 is zero, the value of Q1 is : |
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Answer» `Q_1 =|Q_3|` |
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| 41556. |
Current in primary coil changes from 20 A to 0 A uniformly in 0.005 sec. Calculate the emf induced in the secondary coil if coefficient of mutual inductance is 2 H. |
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Answer» |
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| 41557. |
A force P pulls on a crate of mass m that is in contact with a rough surface. The figure shows the magnitudes and directions of the forces that act on the crate in this situation. W represents the weight of the crate. Fy represents the normal force on the crate, and F represents the frictional force. What is the magnitude of F, the normal force on the crate? |
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Answer» 57 N |
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| 41558. |
A car moving with a speed of 50 km/h, can be stopped by brakes after at least 6 m. If the same car is moving at a speed of 100 km/h, the minimum stopping distance is : |
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Answer» 6 m `S=(1/2mv^2)/(F)` or`S prop v^2` `:.(S_2)/(S_1)=(v_2^2)/(v_1^2)` `:.(S_2)/(6)=((100)^2)/((50)^2)=4` `S_2=24 m` |
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| 41559. |
How much .^235U is consumed in a day in an atomic power house operating at 400 MW, provided the whole of mass .^235U is converted into energy? |
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Answer» Solution :power = 400 MW= `400xx10^6` W , time = 1 day = 86,400 s. Energy produced , E= power x time `=400xx10^6xx86,400=3.456xx10^13` J As the whole of mass is CONVERTED into energy , by Einstein.s mass-energy `E=Mc^2` `M=E/c^2=(3.456xx10^13)/(3xx10^8)^2=3.84xx10^(-4)` KG =0.384 g. |
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| 41560. |
A force P pulls on a crate of mass m that is in contact with a rough surface. The figure shows the magnitudes and directions of the forces that act on the crate in this situation. W represents the weight of the crate. Fy represents the normal force on the crate, and F represents the frictional force. Which statement best describes the motion of the crate? |
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Answer» The crate MUST be at REST |
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| 41561. |
The magnetic flux phi (in weber) in a closed circuit of resistance 10 Ohm varies with time t according to the equation phi = 6t^2-5t+1. The magnitutde of induced current at t = 0.25 second should be: |
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Answer» -2A |
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| 41562. |
Anobject is placed at a distance of 10 cm from a coaxial combination of two lenses A and B contact. The combination forms a real image three times the size of an object. If len B is concave with a focal length of 30 cm, what is the nature and focal lenght of lens A ? |
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Answer» Convex, 6 cm v = - 3u u = - 10 cm, `therefore v = + 30 cm` `(1)/(F) = (1)/(v) - (1)/(u)` `rArr` Focal LENGTH of combination `(F) = (15)/(2) cm` Focal length of concave lens `(f_(1)) = 30 cm` `therefore "" (1)/(F) = (1)/(f_(1)) + (1)/(f_(2))` `rArr "" f_(2) = + 6 cm`. |
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| 41563. |
A satellite is revolving around the earth in a circular orbit at a distance of 10^7m from its center. The speed of the satellite is,(G=6.67xx10^-11Nm^2//kg^2,M=6xx10^24kg) |
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Answer» F |
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| 41564. |
In fig. 16-47a, string 1 has a linear density of 3.00 g/m, and string 2 has a linear density of 5.00 g/m. They are under tension due to the hanging block of mass M= 800 g. Calculate the wave speed on (a) string 1 and (b) string 2 (Hint: When a string loops halfway around a pulley, it pulls on the pulley witii a net force that is twice the tension in the string) Next the block is divided into two blocks (with M_(1)+M_(2)=M) and the apparatus in tearranged as shown in fig. 16-47b. Find (c) M_(1) and (d) M_(2) such that the wave speeds in the two strings are equal. |
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| 41565. |
The acceleration due to gravity g and mean density of the earth rho are related by which of the following relations ? (Where G is the gravitational constant and R is the radius of the earth .) |
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Answer» `rho=(3g)/(4pi GR )` `G((4//3)piR^(3)xxrho)/(R^(2))=Gxx(4)/(3)piRxxrho" or rho=(3g)/(4pi GR)` |
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| 41566. |
A copper wire and an iron wire, each having an area of cross-section A and lengths L_(1) and L_(2) are joined end to end. The copper end is maintained at a potential V_(1) and the iron end at a lower potential V_(2). If sigma_(1) and sigma_(2) are the conductivities of copper and iron respectively, then the potential of the junction will be |
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Answer» `(sigma_(1)V_(1)+sigma_(2)V_(2))/((sigma_(1)//L_(1))+(sigma_(1)//L_(2)))` |
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| 41567. |
What is the principle behind the working of a transformer ? Mention any two sources of energy loss in transformer |
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Answer» Solution :Transformer works on the principle of "mutual induction". Sources of energy losses in transformer. 1. Magnetic FLUX leakage. 2. Eddy current LOSS 3. Hysteresis loss 4. OHMIC loss due to the resistance of the windings (WIRES). |
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| 41568. |
A body falls on the ground from a height of 10 metre and rebounds to a height of 2.5 m. The ratio of the velocity of the body before collision to the velocity of the body after collision is: |
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Answer» `2:1` and height of rebound `h_2=25m` LET the velocity of strike be `v_1` and that of rebound be `v_2` then `(v_1)/(v_2)=sqrt((2gh)/(2gh_2))=sqrt((h_1)/(h_2))` =`sqrt((10)/(2.5))=2//1` `v_1:v_2: :2:1` |
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| 41569. |
The length l, breadth b and thickness t of a block of wood were measured with the help of a metre scale. The results after calculating the errors are given as l=15*12+-0*01cm, b=10*15+-0*01cm t=5*28+-0*01cm. The percentage error in volume upto proper significant figure is : |
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Answer» Solution :`V=lbt` `(DeltaV)/(V)XX100=(Deltal)/(l)xx100+(Deltab)/(b)xx100+(Deltat)/(t)xx100` `=(0*01)/(15*12)xx100+(0*01)/(10*15)xx100+(0*01)/(5*28)xx100` `0*36%` So the correct choice is `(a)`. |
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| 41570. |
Image formation by a system of thin prism and lens A beam parallel to the principal axis is incident on a thin prism (Fig. 34-60). Find the location of final image. |
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Answer» Solution :(1) Here , two events occur before the formation of IMAGE, the first is refraction through the thin prism which will only deviate the parallel beam of rays. This parallel beam will be incident on the lens which would form the image (2) When the rays pass through the prism all the rays undergo the same deviation. A parallel beam incident on the lens will form an image somewhere on the focal plane. CALCULATION : The deviation of the rays while passing through the prism is given by `delta=(n-1)A=0.5xx2=1^(@)` The parallel beam of light will pass through the lens. So, the image will be formed in focal plane at `20cm` fromthe lens . The image will be located at an angle `theta` to the principal axis. `theta=(pi)/(180)xx1^(@)` The distance of the image from the focal plane is given by `(h)/(20)=(pi)/(180)` `h=(pi)/(9)cm` Learn : If an object is viewed through a thin prism , all the rays coming from the object are deviated by the same angle. So, the image is formed at some point above the object as shown in Fig. 34-61. Note that this is a VIRTUAL image. Shift of image due to prism `=xdelta=x(n-1)A` 
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| 41571. |
Give an equation representing the decay of a free electron. |
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Answer» where `bar(v)` is anti neutrino zero MASS. HALF life period of neutron is 10.8 minute. |
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| 41572. |
From the top of a tower, a stone is thrown up and reaches the ground in time t_(1)=9s. a second stone is thrown down with the same speed and reaches the ground in time t_(2)=4s. A third stone is released from rest and reaches the ground in time t_(3), which is equal to |
| Answer» Answer :B | |
| 41573. |
Obtain the equaiton for resultant intensity due to interference of light. |
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Answer» Solution :Let us cosider two light waves from the sources `S_(1) and S_(2)` meeting at a point P. The wave from `S_(1)` at an INSTANT t at P is, `y_(1) = a_(1) sin omegat` The WAVR from `S_(2)` at an instant t at P is, `y_(2) = a_(2) sin (omegat + phi)` The two waves have different amplitudes `a_(1) and a_(2),` same angular frequency `omega` and a phase difference of `phi` between them. The resultant displacement will be given by, `y = y_(1) + y_(2) = a_(1) sin omega t + a_(1) (omega t + phi)` The simplification of the above equation by using trigonometric indentites gives the equation, `y = A sin (omega t + theta)` `"Where", A = sqrt(a_(1)^(2)+a_(2)^(2)+2a_(1)a_(2)cosphi)` `theta=tan^(-1)""(a_(2)sinphi)/(a_(1)+a_(2)cosphi)` The resltant amplitude is maximum, `A_(min)=sqrt((a_(1)+a_(2))^(2)),"when"phi=o,pm2pi,pm4pi...,` The resultant amplitude is minimum, `A_(min)=sqrt((a_(1)-a_(2))^(2)):"when"phi=pmpi,pm3pi,pm5pi...,` The intensity of light is prportional to square of amplitude, `1 PROP A^(2)` Now, equation (5) becomes, `1 prop I_(1) + I-(2) + 2 sqrt(I_(1)I_(2))cosphi` In equaiton (10) if the phase difference, `phi = 0, pm 2pi, pm4pi` .... it CORRESPONDS to the condition for maximum intensity of light called as constructive interference. The resultant maxiumum intensity is, `IpropI_(1)(a_(1)+a_(2))propI_(1)+I_(2)+2sqrt(I_(1)I_(2))` In equation (10) if the phase difference, `phi = pm pi, pm 3 pi, pm 5 pi`... it corresponds to the condition for minimum intensity of light called destructive interference. The resultant minimum intensity is, `I_(min), prop (a_(1) - a_(2))^(2) prop I_(1) + I_(2) - 2 sqrt(I_(1)I_(2))` 
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| 41574. |
In 1908-1910 Perrin determined the Avogadro number. He did it by observing the distribution of tiny gumboge gum balls in water with the aid of a short-focus microscope (Fig.). By adjusting the focus of the microscope to observe a definite layer he was able to count the number of particles in each layer. In one of the experiments the following data were obtained: {:("Height of the layer above the",,,,),("tray's bottom," mu,5,35,65,95),("Number of particles in the layer",100, 47, 23, 12):} Knowing the ball's radius to be 0.212 um, the density of gumboge gum to be 1.252 xx 10^3kg//m^3 the density of water at 27^@C to be 0.997 xx 10^3 kg//m^3 find the Avogadro number. |
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Answer» `(N_1)/(N_2) = e^((mg(h_2 -h_1))/(kt))”log” (N_1)/(N_2) = (0.434mg(h_2 - h_1))/(kT)` ONE should take into account that in addition to the force of gravity, the particles are acted upon by the Archimedean force. Expressing the BOLTZMANN constant in terms of the gas constant and the Avogadro number, we obtain for the latter `N_A = (3RT “log” (N_1//N_2))/(0.434 xx 4 pi r^3 g(rho - rho_0) (h_2 - h_1))` Substituting the known data, we obtain the WORKING formula `N = 5.79 xx 10^(22) (“log” (N_1//N_2))/(h_2 - h_1)` The values obtained for the Avogadro number are: `6.32 xx 10^26, 5.98 xx 10^26` and `5.45 xx 10^26` The average value is 5.92 X 1026, the maximum error is `0.3 xx 10^26`. Hence from data obtained in this experiment `N_A = (5.9 pm 0.3) xx 10^(26) “k mol”^(-1)`. |
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| 41575. |
Fig. shows a small plane strip suspended from a fixed support through a string of length 1. A continuous beam of monochromatic light is incident horizontally on the strip and is completely absorbed. The energy falling on the strip per unit area per sec is P. The deflection of the string from the vertical, if the strip stays in equilibrium. |
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Answer» `theta=sin^(-1)""(P)/(mgC)` `T cos theta= mg` ALSO `T sin theta=(P)/(c)` Dividing we get, `tan theta=(P)/(mgc)` `:. theta= "tan"^(-1)(P)/(mgc)` |
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| 41576. |
A capacitor of 250 mu F is connected parallel with a inductor of 0.16mH. IF the effective resistanceis 20 Omega then resonant frequency……. Hz. |
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Answer» `9 times 10^4` `THEREFORE f_0=1/(2 times 3.14 times sqrt(0.16 times 10^-3 times 250 times 10^-6))` `therefore f_0=1/(6.28 times sqrt(16 times 10^-5 times 25 times 10^-5))` `therefore f_0=1/(6.28 times sqrt(400 times 10^-10))` `therefore f_0=1/(6.28 times 20 times 10^-5)` `therefore f_0=1/(6.28 times 20) times 10^5` `therefore f_0=7.96 times 10^-3 times 10^5` `therefore f_0=7.96 times 10^2 Hz` |
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| 41577. |
Assertion: Diode lasers are used as optical sources in optical communication. Reason: Diode layers consume less energy. |
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Answer» If both the assertion and reason are true and reason is a true explantion of the assertion. |
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| 41578. |
If barE and barB represent electric and magnetic field vectors of the electromagnetic waves , the direction of propagation of the electromagnetic wave is that of |
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Answer» `BARE` |
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| 41579. |
An inclined plane is inclined at an angle theta when the block placed on it is just at the point of moving down the plane. What can be minimum acceleration with which the block can be moved up the inclined plane? |
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Answer» `g sin THETA` `=g(sin theta+mu cos theta)=g(sin theta+tan theta cos theta)` `=g(sin theta+sin theta)=2g sin theta` (b) is the choice |
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| 41580. |
The mass defect in a particular nuclear reaction is 0.3 grams. The amount of energy liberated in kilowatt hour is [Velocity of light =3xx10^8 m/s] |
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Answer» `1.5xx10^6` |
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| 41581. |
In a sinusoidal wave, the time required for a particular point to move from maximum displacement to zero displacement is 0.170 s. the frequency of the wave is : |
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Answer» 1.47 Hz `therefore` FREQUENCY= `(1)/(0.68)= 1.47` Hz. CORRECT choice is (a). |
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| 41582. |
A particle is executing S.H.M., with the length of its path as as 8 cm. At what displacement from the mean position half the energy is kinetic and half is potential ? |
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Answer» At 2 cm At `y=( r )/(sqrt(2))`,P.E. = K.E. `=(1)/(2)` T.E. `:.""y=(4)/(sqrt(2))=2 sqrt(2)` cm. Correctchoice is (b). |
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| 41583. |
Pick out the statement which is not true? |
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Answer» While taking reading with a tangent galvanometer, the readings are repeated by reversing the current to TAKE care of the fact that the plane of the coil may not be exactly ALONG the earth's magnetic meridian. |
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| 41584. |
In a crystal, atomic separation is around 2A to 3A. At this separation due to interatomic interaction, energies of |
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Answer» outermost ELECTRONS is changed |
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| 41585. |
When a current of 0.5 A is passed through two resistances in series, the potential difference between the ends of the series arrangement is 12.5 V. On connecting them in parallel and passing a current of 1.5 A, the potential difference between their ends is 6 V. Calculate the two resistances. |
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| 41586. |
The ratio of electric force and gravitational force between a proton and an electron at a certain distance is…… |
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Answer» `10^(41)` |
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| 41587. |
100pi phase difference = ...... path difference. |
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Answer» `50 lambda` `:.`Phase differencer `=(lambda)/(2PI)xx` path difference `=(lambdaxx100pi)/(2pi)=50 lambda` |
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| 41588. |
A : Transverse wave nature of light is proved by polarisation. R : According to Maxwell, light is an electromagnetic wave but not mechanical wave. |
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Answer» Both A and R are TRUE and R is the correct explanation of A |
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| 41589. |
A resistance R draws power P when connected to an AC source. If an inductance is now placed in series with the resistance, such that the imped-ance of the circuit becomes Z, the power drawn will be |
| Answer» Answer :A | |
| 41590. |
A convex lens of focal length 15 cm and concave mirror of focal length 30 cm are kept their optical axes PQ and RS parallel but separated in vertical direction by 0.6 m, as shown. The distance between the lens and mirror is 30 cm. An upright object AB of height 1.2 m is placed on the optic axic PQ of the lens at a distance of 20 cm from the lens. Find the linear magnification of the second image after reflection from the mirror. |
| Answer» ANSWER :C | |
| 41591. |
Find the concentration of holes and electrons in a P-type semiconductor at room temperature if conductivity is 100 mhos.Given that hole mobility mu_(p)=1800 and intrinsic concentration is 2.5xx10^(13)//m^(3) |
| Answer» SOLUTION :`1.8xx10^(9)CM^(-3)` | |
| 41592. |
The position of a particle moving along an x axis is given by x=12t^(2)-2t^(3), where x is in meters and t is in seconds. Determine (a) the position, (b) the velocity, and (c) the acceleration of the particle at t=3.5s. (d) What is the maximum positive coordinate reached by the particle and (e) at what time is it reached ? (f) What is the maximum positive velocity reached by the particle and (g) at what time is it reached ? (h) What is the acceleration of the particle at the instant the particle is not moving ( other than at t=0 ) ? (i) Determine the average velocity of the particle between t=0 and t=3s . |
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| 41593. |
A convex lens of focal length 15 cm and concave mirror of focal length 30 cm are kept their optical axes PQ and RS parallel but separated in vertical direction by 0.6 m, as shown. The distance between the lens and mirror is 30 cm. An upright object AB of height 1.2 m is placed on the optic axic PQ of the lens at a distance of 20 cm from the lens. Find the linear magnification of the first image after refraction from the lens. |
| Answer» ANSWER :B | |
| 41594. |
What is meant by geographic north pole ? |
| Answer» Solution :The northern Most part of earth . ( The north end POINT of a DIAMETRICAL line, perpendicular to EQUATOR PLANE) | |
| 41595. |
The wavelenths involved in the spectrum of deuterium (""_(1)^(2)H) are slightly different from that of hydrogen spectrum because |
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Answer» sizes of the two NUCLEI are DIFFERENT |
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| 41596. |
A body is projected with a velocityof 30m/s to haveto horizontal range of45m. Findthe angleof projection . |
| Answer» SOLUTION :`15^(@)` & `75^(@)` | |
| 41597. |
A rope AB of linear mass density lamda is placed on a quarter vertical fixed disc of radius R as shown in the figure. The surface between the disc and rope is rough such that the rope is just is equilibrium. Gravitational acceleration is g. Choose the correct option (s). |
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Answer» Coefficient of static friction between rope and disc is `mu=1` `lamda Rg int_(0)^((pi)/2) cos theta d theta =mu lamda Rg int_(0)^((pi)/2) SIN theta d theta` `:.mu=1` At the position of maximum tension in the rope `lamda R d THETAG cos theta=mu(lamda R d theta g sin theta)` `:. theta=45^(@)` At any `theta` `dT=lamda R d theta g cos theta -mu lamda d theta g sin theta` `int_(0)^(T_("max")) dT =lamda Rg int_(0)^((pi)/4) (cos theta-sintheta)d theta` `T_("max")=lamda Rg[sintheta+costheta]_(0)^((pi)/4)=lamda Rg[1/(sqrt(2))+1/(sqrt(2))-1]=lamdaRg(sqrt(2)-1)` 
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| 41598. |
The radii of curvature of the faces of a double convex lens are 10 cm and 15 cm. Its focal length is 12 cm. What is the refractive index of glass ? |
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Answer» 1 `(1)/(12) = (n_(2) - 1) ((1)/(10) - (1)/(=15))` or `(1)/(12) = (n_(2) - 1) ((3+2))/(30) = ((n_(2) - 1))/(6)` THUS, refractive index of glass, `n_(2) = 15` |
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| 41599. |
Luminosity of sun is 3.9xx10^(26) watts. Mean distance of earth from the sun is 1.496xx10^(11)m. What is the value of solar constant ? |
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Answer» `1.388xx10^(5)W` `:.` Solar const. `=(E)/(4piR^(2))=(3.9xx10^(26))/(4xx(22)/(7)xx(1.496xx10^(11))^(2))` `=1.388xx10^(3)"watt/m"^(2)`. Thus correct CHOICE is (b). |
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| 41600. |
A gun of mass 10 kg fires 4 bullets per second. The mass of each bullet is 20 g and the vlocity of the bullet when it leaves the gun is 300 ms^(-1) The force required to hold the gun while firing is : |
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Answer» 6N (C) is the choice, |
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