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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.
| 7601. |
A : Resolving power of a microscope can be increased by choosing a medium of higher refractive index between object and objective lens. R : To increase resolving power of microscope, usually on oil having R.I colse to that of objective glass is used. |
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Answer» Both A and R are true and R is the correct EXPLANATION of A |
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| 7602. |
the radus if the rear wheel of bicycle is twice that of the front wheel. When the bicycle is moving . The angularspeed of the rear wheel comparedto thatof the front is |
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Answer» Greater ` orm_(r)r_(r)^(2) Omega_(r) = m_(f) r_(f)^(2) Omega_(f)` Wherethe subscriptsr and f denotethe rearand front. ` m(2r_(f))^(2) Omegar = mr_(f)^(2) Omega_(f) [r_(r) = 2r_(f) ` (Given)] ` therefore4Omega_(r) = Omega_(f)or Omega_(r) = (Omega_(f))/4 ` Hence, the ANGULARVELOCITY of rear wheelwill be smallercomparedto frot wheel. |
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| 7603. |
Light of wavelength 5000Å falls on a plane reflecting surface. What are the wavelength and frequency of the reflected light? For what angle of incidence is the reflected ray normal to the incident ray? |
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Answer» SOLUTION :`lamda=5xx10^(-7)m, c=3XX10^(8)ms^(-1)` `:.` Frequencey `v=c/(lamda)=(3xx10^(8))/(5xx10^(-7))=6xx10^(14)Hz` `i=r,i+r=90^(@):.2i=90^(@), i=45^(@)` |
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| 7604. |
The ratio of the numerical value of the average velocity and average speed of a body is always |
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Answer» Unity |
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| 7605. |
Which of the following is example of Gaseous Solutions |
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Answer» CHLOROFORM in Nitrogen |
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| 7606. |
The collector - base junction of a transistor offers ……….resistance to current . |
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Answer» low |
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| 7607. |
A paralle plate air capacitor has capacitance C. Now half of the space is fillel with a dielectric material with dielectric constant K as shown in figure . The new capacitance is C . Then |
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Answer» `C'=(C )/(2) [1+K]` `C_(AB)=C_(0)+C_("MED").` `=(epsilon_(0)(A)/(2))/(d)+(epsilon_(0)k(A)/(2))/(d)` `=(epsilon_(0)A)/(2D)[1+k]` `C_(AB)=(C)/(2)(1+K)` 
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| 7608. |
The force between two electrons when placed in air is equal to 0.5 times the weight of an electron. find the distance between two electrons (mass of electron= 9.1 xx 10^(-31)kg) |
| Answer» ANSWER :A | |
| 7609. |
Derive an expression for energy stored in a capacitor, |
| Answer» Solution :DERIVE an expression for the ENERGY STORED in the arrangement | |
| 7610. |
A : A series resonant circuit is also known as an acceptor circuit. R : For large value of Ohmic resistance, the quality factor of a series resonant circuit is high. |
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Answer» If both ASSERTION & Reason are TRUE and the reason is the CORRECT explanation of the assertion, then mark (1) |
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| 7611. |
If x= (a)/(b) , then the maximum percentage error in the measurement of x will be |
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Answer» `((DELTAA)/(a))/((Deltab)/b)XX100` |
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| 7612. |
A uniform disc of radius R lies in xyplanewith its centre at origin. Itsmoment of inertiaabout the axisx = 2 Rand y =0is equal to the moment of inertiaabout the axisy =d and z = 0. Whatis d equal to ? |
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Answer» ` sqrt(17)/(2) R `  The axis x = 2 R , y=0 is in Z DIRECTION as shown in figure ,Usingtheoremof parallel AXES, `I_(1)=1/2 m R^(2) + m(2 R)^(2) + m (2 R)^(2) = 9/2 m R^(2)` ` THEREFORE 1/4 m R ^(2)+ m d^(2) = 9/2 m R^(2)` `MD^(2) = ( 9/2 -1/4) m R^(2)= 17/4m R^(2) Rightarrowd= sqrt(17)/(2) R ` |
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| 7614. |
What is radio carbon dating ? |
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Answer» Solution :Carbon dating: The interesting application of beta decay is radioactive dating or carbon dating.Using this technique, the age of an ancient object can be calculated.All living organisms absorb carbon dioxide`(CO_(2))` from air to synthesize organic molecules. In this absorbed `CO_(2)` the major PART is `""_(6)^(12)"C"` and very small fraction `(1.3 xx 10^(12))` is radioactive `""_(6)^(14)"C"` whose half - lifeis 5730 YEARS. Carbon - 14 in the atmosphere is alwaysdecaying but at the same time, COSMIC rays from outer space are continuously bombarding the atoms in the atmosphere whichproduces `""_(6)^(14)"C"`. So the continuous production and decay of `""_(6)^(14)"C"` in the atmosphere keep the ratio of `""_(6)^(14)"C"` to `""_(6)^(12)"C"` always constant. Since our human body, tree or any living organisms continuously absorb `CO_(2)` from the atmosphere, the ratio of `""_(6)^(14)"C"` to `""_(6)^(12)"C"` in the living organisms is also NEARLY constant.But when the organisms dies, it stops absorbing `CO_(2)`. Since `""_(6)^(14)"C"` starts to decay, the ratio of` ""_(6)^(14)"C"` to `""_(6)^(12)"C"` in the ancient tree pieces excavated is known, then the age of the tree pieces can be calculated. |
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| 7615. |
A galvanometer having full scale deflection current of 10 mA and resistance 8 Omegais joined in circuit as shown. When the terminals A and B are used and C remains open, the range of the ammeter is I. When the terminals A and C are used and B remains open, then the range of the ammeter is |
| Answer» ANSWER :B | |
| 7616. |
In winters ice forms on the surface of a lake. Due to abnormal expansion of water the temperature of the water at the bottom of the lake remains constant at 4^(@)C and we assume that the amount of heat required to maintain this temperamre of the bottom layer of water may come from the bed of the lake. If surrounding temperature is - 10^(@)CProve that ice formed from the surface of the lake attains a maximum thickness. Find the maximum depth from surface upto which ice is formed if the depth of lake is I m. Given that thermal conductivity of ice is 0.5 W//m^(@) C. |
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| 7617. |
The speed of light in media M_(1) and M_(2) are .5 xx 10^(8) ms^(-1) and 2 xx 10^(8)ms^(-1) respectively. A ray travels from medium M_(1) to the medium M_(2) with an angle of incidence theta. The ray suffers total internal reflection. Then the value of the angle of incidence theta is : |
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Answer» `sin^(-1)((3)/(4))` `I gt C` We have, `(1)/(sin C) = (v_(2))/(v_(1))` Given `v_(1) = M_(1) = 1.5 xx 10^(8) ms^(-1)` `v_(2) = M_(2) = 2 xx 10^(8) ms^(-1)` Here `C = sin^(-1) ((M_(1))/(M_(2)))` `THEREFORE "" i gt sin^(-1)((M_(-1))/(M_(2)))` `rArr "" i gt sin^(-1) ((1.5 xx 10^(8))/(2 xx 10^(8)))` `i gt sin^(-1) ((3)/(4))`. |
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| 7618. |
Water rises to a height of 20 xx 10^-3 m in a capillary tube. If radius of the capillary tube is made (2/3) of its previous value, the height to which the water will rise in the tube is |
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Answer» `30 XX 10^-3 m` |
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| 7619. |
Statement-I : A beaker is completely filled with water at 4^(@)C. It will overflow both when heated on cooled. Statement-II : There is expansion of water both below and above 4^(@)C. |
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Answer» STATEMENT-I is true, statement-II is true and |
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| 7620. |
The total energy of an electron revolving in the second orbit of hydrogen atom is |
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Answer» `-13.6` EV |
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| 7621. |
The fundamental radio antenna is a metal rod which has a length equal to |
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Answer» `lambda` in free SPACE at the frequency of OPERATION |
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| 7622. |
The work function of caesium metal is 2.14 eV. When light of frequency 6xx10^(14)Hz is incident on the metal surface, photoemission of electrons occurs. What is the a.maximum kinetic energy of the emitted electrons, b. stopping potential, and c. maximum speed of the emitted photoelectrons ? |
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Answer» SOLUTION :`phi=2.14eV=2.14xx1.6xx10^(-19)J, upsilon=6xx10^(14)Hz` a. `(1)/(2)mv_("max")^(2)=h upsilon-phi_(0)=6.62xx10^(-34)xx6xx10^(14)-2.14xx1.6xx10^(-19)` `=39.72xx10^(-20)-34.24xx10^(-20)=5.48 xx 10^(-20)=0.548xx10^(-19)J` b. `eV_(0)=(1)/(2)mv_("max")^(2)` `V_(0)=(1)/(2)(m)/(e )v_("max")^(2)=(0.548xx10^(-19))/(1.6xx10^(-19))=0.3425V` C. `(1)/(2)mv_("max")^(2)=0.548xx10^(-19), v_("max")^(2)=(2xx0.548xx10^(-19))/(9.1xx10^(-31))=0.1204xx10^(12)` `v_("max")=344xx10^(3)MS^(-1)` |
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| 7623. |
A person can see objects clearly when they are at a distance lesser that 2 m from him. What is the power of the lens which he must use to see a star clearly? |
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| 7624. |
You given two planoconvex lenses of focal lengths 0.03m and 0.05m and asked to design an eye-piece free from chromatic aberration. Explain how you will achieve it . Calculate the equivalent focal length of the eye-piece formed and its magnifying power if it is used by a person for whom least distance of distinct vision is 0.25m. Will your eye-piece be free from spherical aberration? |
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| 7625. |
What is the value of Rydberg constant for hydrogen ? |
| Answer» Solution :VALUE of Rydberg.s constant for HYDROGEN is `R = 1.097 x 10 m^(-1)` | |
| 7626. |
Binding energy of an atomic electron. |
| Answer» Solution :The binding energy of an atomic electron is defined as the MINIMUM energy that should be provided to an orbital electron to REMOVE it from the ATOM such that its total energy is zero. | |
| 7627. |
Twelve straight uniform wires of length a and resistance R are joined to form the edges of a cube of side a. Current enters the system at one corner andd leaves from a point on one of the edges meeting at the opposite corner at a distance Ka(ltKlt1) from it. Q. The maximum value of the equivalent resistance is |
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Answer» <P>`(9R)/(5)`  `Ry+R(y-q)-RP-RZ=0` (or) `2y-Z=p+q` ..(i) `Rx+Rr-R(z-p)-RZ=0` (or) `2z-x=r+P`.(II) `Ry+Rq-R(x-R)-Rx=0` (or) `2x-y=q+r` ..(iii) THEREFORE `2p=3y+z-3x`..(iv) `2q=3x+y-3z`..(v) `2r=3z+x-3y`..(vi) Again `(x-r)+(x-r+q)=r+(z-p+r)` (or) `2x-z=4r-p-q` ..(vii) substituting values of p,q,r, we FIND `z=y` ..(viii) again `=(z-p)+(z-p+r)+K(z+x+q-p)` `=p+(1-K)(y-q+p)` this gives `K(2y+x)+8(x-y)=0` (or) `V=R[x+(x-r)+(x-r+q)+K(z+x+q-p)]` `V=R[(7x-2y)/(2)+2K(2x-y)]` ..(10) `I=x+2y` ..(11) `(V)/(i)=Req=R[((7x-2y)+4K(2x-y))/(2(x+2y))]` `=(R)/(12)[10+4K-5K^(2)]` R is maximum when `K=(2)/(5)` and `R_(max)=(9R)/(10)` |
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| 7628. |
At resonance, V_(L) and V_(C ) both very much greater than the applied potential, V itself. The quantity factor for an LCR circuit in resonance is given by Q = (X_(L))/(R ). In pratice, Q = 200 has been achieved. At resonance, the capacitor has been adjusted for (1). 200 xx 10^(-6) mu F (2) 0.00013 mu F (3). 0.0012 mu F (4). 0.0013 F At resonance, the potential difference across the inductance is (1) 1.3 V (2) 13 V (3). 0.3 V (d) none of these The potential across the capacitor at resosnance is (1) 1.3 V (2) 13 V (3) lt 13 V (4) none of these The Q factor is (1) (V_(L))/(V_(C )) (2) (V_(C ))/(V_(L)) (3) (V_(C ))/(V) (d) (V_(L))/(V) (e) choose the right statement. (1) V_(L)+V_(C) can be greater than V_("applied") (2) V_(L)+V_(C)=V_("applied") (3) V_(L)+V_(C) lt V_("applied") (4) none of these |
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Answer» |
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| 7629. |
Twelve straight uniform wires of length a and resistance R are joined to form the edges of a cube of side a. Current enters the system at one corner andd leaves from a point on one of the edges meeting at the opposite corner at a distance Ka(ltKlt1) from it. Q. The equivalent resistance is maximum, when |
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Answer» `K=(4)/(5)`  `Ry+R(y-q)-RP-RZ=0` (or) `2y-Z=p+q` ..(i) `Rx+Rr-R(z-p)-RZ=0` (or) `2z-x=r+P`.(ii) `Ry+Rq-R(x-R)-Rx=0` (or) `2X-y=q+r` ..(iii) THEREFORE `2p=3y+z-3x`..(iv) `2q=3x+y-3z`..(v) `2r=3z+x-3y`..(vi) Again `(x-r)+(x-r+q)=r+(z-p+r)` (or) `2x-z=4r-p-q` ..(vii) substituting values of p,q,r, we find `z=y` ..(viii) again `=(z-p)+(z-p+r)+K(z+x+q-p)` `=p+(1-K)(y-q+p)` this gives `K(2y+x)+8(x-y)=0` (or) `V=R[x+(x-r)+(x-r+q)+K(z+x+q-p)]` `V=R[(7x-2y)/(2)+2K(2x-)]` ..(10) `I=x+2y` ..(11) `(V)/(i)=Req=R[((7x-2y)+4K(2x-y))/(2(x+2y))]` `=(R)/(12)[10+4K-5K^(2)]` R is maximum when `K=(2)/(5)` and `R_(max)=(9R)/(10)` |
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| 7630. |
The velocity of a body of mass 1 kg is 1m/s. Calculate de Broglie wavelength? h = 6.62 xx 10^(-34) Js. |
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Answer» SOLUTION :`m=1kg, v-1 m/s` `lambda=h/(MV)=(6.62xx10^-34)/(1xx1)=6.62xx10^-34m` |
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| 7631. |
In a region , an electroc field E = 10NC^(-1) making an angle of theta = 30^@ with the horizontal plane is present. Coefficient of restitution between ball and surface is e = 0.2. Ball has charge q = -2C over it and mass m= 2 kg . Ball is projected at an angle of theta = 30^@ with the horizontal with speed u = 10ms^(-1) Horizontal distance traveled up to first rebound is |
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Answer» `20 sqrt3 m ` `5t-(1)/(2)15t^(2)=0` or `t=(2)/(3)s` At time `t=2//3` s the displacement in x-direction `x=5sqrt(3)((2)/(3))-(1)/(5)5sqrt(3)((2)/(3))^(2)=(20)/(3sqrt(3))m` After IMPACT with the surface, VELOCITY along vertical direction `u'^(y)=eu_(y)=0.2xx5=1ms^(-1)` velocity along the horizontal direction after time `T_(1)` `u'_(x)=eu_(x)+a_(x)T_(1)=5sqrt(3)-5sqrt(3)xx(2)/(3)=(5)/(sqrt(3))ms^(-1)` The ball rebounds in the vertical direction with velocity1 `ms^(-1)` |
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| 7632. |
An object is 5 m to the left of a flat screen. A converging lens for which the focal length is 0.8m is placed between object and screen.Show that for two positions of lens form images on the screen and determine how far these positions are from the object? |
Answer» SOLUTION :Usingthe lensformula`(1)/(V) -(1)/(u) =(1)/(F)`  we have`(1)/(5-u) +(1)/(u )=(1 )/(0.8) or (1)/(5-u) +(1)/(u) =1.25` ` thereforeu+5 -u =1.25u(5-u) or 1.25u^2 -6.25u + 5=0` u= 4mand 1m Both the values are real, which MEANS there exist two POSITIONS of lens that form images of object on the SCREEN. |
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| 7633. |
In a region , an electroc field E = 10NC^(-1) making an angle of theta = 30^@ with the horizontal plane is present. Coefficient of restitution between ball and surface is e = 0.2. Ball has charge q = -2C over it and mass m= 2 kg . Ball is projected at an angle of theta = 30^@ with the horizontal with speed u = 10ms^(-1) After first impact with the ground, |
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Answer» the ball REBOUNDS at an angle `30^@` with the HORIZONTAL `5t-(1)/(2)15t^(2)=0` or `t=(2)/(3)s` At time `t=2//3` s the displacement in x-direction `x=5sqrt(3)((2)/(3))-(1)/(5)5sqrt(3)((2)/(3))^(2)=(20)/(3sqrt(3))m` After impact with the surface, velocity along vertical direction `u'^(y)=eu_(y)=0.2xx5=1ms^(-1)` velocity along the horizontal direction after time `T_(1)` `u'_(x)=eu_(x)+a_(x)T_(1)=5sqrt(3)-5sqrt(3)xx(2)/(3)=(5)/(sqrt(3))ms^(-1)` The ball rebounds in the vertical direction with velocity1 `ms^(-1)` |
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| 7634. |
An object is 5 m to the left of a flat screen. A converging lens for which the focal length is 0.8m is placed between object and screen.How do the two images differ from each other? |
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Answer» Solution :`m=(v)/(U ) ` ` therefore m_1= ((5-4))/((-4)) = -0.25 and m_2 =((5-1))/((-1)) =-4.00` Hence, both the images are real and inverted, the FIRST has magnification -0.25 and the SECOND -4.00. |
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| 7635. |
Describe expriments to demonstrate electromagnetic induction. |
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Answer» Solution :Correct figure Brief explanation: Detailed Answer: (i) The arrangement CONSISTED of a coil with a GALVANOMETER and a bar magnet.c (ii) When NORTH pole of the bar magnetic moved towards the coil, galvanometer should deflect in one direction indicating flow of current in the coil (iii) When North pole of the bar magnet moved away from the coil galvanometer again showed the deflection but now in the opposite direction indicating flow of current in the coil but in the opposite direction. (iv) The deflection of the galvanometer was large when the bar magnet was moved faster towards or away from the coil. (v) When south pole of the magnet was brought near the coil, or was moved away from the coil the galvanometer showed deflection in the opposite direction as compared to the available with the similar movement of the north pole of the north. (vi) Conclusion: (i) Whenever THREE is a relative motion between a closed and a magnetic, INDUCED e.m.f. is set up across the coil or induced flows throught the coil. 
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| 7636. |
What is the resolving power of a telescope whose objective lens has a diameter 1.22 m for a wavelength 4000 A ? |
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Answer» `2 XX 10^6` |
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| 7637. |
The intensity of solar radiation reaching the illuminated side of the Earth each second is J = 1.36 kW//m^2. Find the docrease in the internal energy and the muss of tho Sun per second. How long will it take for the Sun's mass to decrease due to radiation by 10% ? The volume of the Sun is to be assumed to remain constant. |
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Answer» `TAU=(0.1M_(o.))/mu=(M_(o.)c^2)/(40piJR^2)` |
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| 7638. |
A satellite is revolving around earth in a circular orbit of radius 3 R. Which of the following is incorrect? ( M is mass of earth, R is radius of earth m is mass of satellite) |
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Answer» Its orbital velocity is `sqrt((GM)/(3R)` |
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| 7639. |
An electron with kinetic energy T~~4eV is confirned within a region whose linear dimension is l=1 mu m. Using the uncertianty principle, evaluate the relative uncertainty of its velocity. |
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Answer» Solution :The momentum the electron is `Deltap_(X)=sqrt(2mT)` Uncertainty in its momentum is `DELTA p_(x)geħ//Deltax=ħ//l` Hence relative nucertainty `(Delta P_(x))/(p_(x))=(ħ)/(lsqrt(2mT))=sqrt((ħ^(2))/(2ml^(2)))//T=(Deltav)/(v)` SUBSTITUTION GIVES `(Delta v)/(v)=(Delta p)/(p)= 9.75xx10^(-5)~~10^(-4)` |
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| 7640. |
What is nuclear fall out? How can it be reduced? |
| Answer» Solution :Nuclear EXPLOSION INVOLVES emission of penetrating radiations like X-rays, U.V. rays, alpha, beta, gamma rays and number of radioisotopes. All these radiations and particles constitute nuclear fall out. It is very harmful for the LIVING beings. The adverse EFFECT of nuclear fall out can be reduced by carrying out explossion in deserts or deep inside the earth. | |
| 7641. |
A disc of mass 0.5kg is kept floating horizontally in mid air by firing bullets of mass 5 g each, vertically at it at the rate of 10 per second. If the bullets drop dead. The speed of the bullet striking the disc is (g=10 ms^(-2)) |
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Answer» `100 MS^(-1)` |
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| 7642. |
What is meant by mobile communication ? |
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Answer» SOLUTION :Mobile communication is used to communicate with others in DIFFERENT locations without the use of any physical connection like wires or cables. It allows the transmission over a wide RANGE of area without the use of the physical link. It enables the people to communicate with each other regardless of a PARTICULAR location like office, house, etc. It also provides communication access to remote areas. It provides the facility of roaming that is, the user may move from one place to another without the need of compromising on the communication. The maintenance and cost of installation of this communication network are also cheap. Applications (i) It is used for personal communication and cellular phones OFFER voice and data connectivity with high speed. (ii) Transmission of news across the globe is done within a few seconds. (iii) Using Internet of Things (IoT), it is made possible to control various devices from a single device. (iv) It enables smart classrooms, online availability of notes, monitoring student activities etc. in the field of education . |
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| 7643. |
A planet goes round the sun three times as fast as the earth. If r_(p) and r_(e) are the radii of orbit of the planet and the earth respectively then |
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Answer» <P>`r_(E)^(3) = 8 r_(p)^(3)` `((T_(p))/(T_(e)))^(2) = ((r_(p))/(r_(e)))^(3) rArr r_(e)^(3) = 9r_(p)^(3)` |
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| 7644. |
In the circuit shown in figure I_(1),I_(2) and I_(D_(2)) are respectively- |
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Answer» `0.212 MA, 3.32 mA, 3.108 mA` By KVL, `- V_(T_(1)) - V_(T_(2)) - V_(2)+E = 0`  `:. V_(2) = 20 - 0.7 - 0.7 = 18.6 V` `I_(2) = (V_(2))/(R_(2)) = 3.32 mA` |
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| 7645. |
A carrierwave of frequency 1.5 MHz and amplitude 50 V is modulated by a sinusoidal wave of frequency 10 kHZ producing 50% modulation . Calculate the amplitude of AM wave and frequencies of the side bands produced . |
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Answer» SOLUTION :`f_c= 1.5 MHz` ` E=50V ` `f_m= 10 kHz` `m_a= 50% ` `m_a= ( E_("max")-E_("min"))/(E_("max") + E_("min"))` ` (50)/(100)= (50-E_("min"))/(50 + E_("min"))` `50 + E_("min") = 100 - 2E_("min")` ` 3E_("min") = 50` ` E_("min")= 16.66V` Side BAND `= f_(C) PM f_(m)` `= 1500 pm 10 kHz` `=1510 & 1490 kHz`. |
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| 7646. |
If I_(1) is the moment of inertia of a thin rod about an axis perpendicular to its length and passing through its centre of mass, and I_(2) is the moment of inertia (about central axis) of the ring formed by bending the rod, then |
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Answer» `I_(1):I_(2)=1:1` and `I_(2)=m[(L)/(2pi)]^(2)""(because 2pir=l)` `=(ml^(2))/(4pi^(2))""...(ii) [r=l//2pi]` `therefore (I_(1))/(I_(2))=(ml^(2))/(12)xx(4pi^(2))/(ml^(2))=(pi^(2))/(3)` |
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| 7647. |
An electrons is revolving in n=3 orbit. What will be the magnetic field at the centre of hydrogen atom. |
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Answer» a) 0.1 T `=(mu_(0))/(2r)XX(eV)/(2pir)=(mu_(0)eV)/(4pir^(2))` `V=2.18xx10^(6)xx(1)/(3)=0.73xx10^(6)=7.3xx10^(5)m//s` `r=0.529`Å`xx(3^(2))/(1)=7.461`Å `B=(10^(7)xx1.6xx10^(-19)xx7.3xx10^(5))/(4.761xx4.761xx10^(-20))=0.515T` |
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| 7648. |
A carnot engine working between 300K and 400K has 800 j of useful work . The amount of heat enegy supplied to the engine from the source is |
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Answer» 2400 J `(800)/(Q_(1))=1-(300)/(400)` `Q_(1)=3200J` |
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| 7649. |
If electrons are accelerated through a potential of V volt, the minimum wavelength of the continuous X-ray radiation is |
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Answer» `(HC)/(Ve)` |
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| 7650. |
A charged particle enters a magnetic field B with its initial velocity v making an angle of pi/4 and B. The path of the particle will be |
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Answer» a STRAIGHT line |
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