40048 + Interview Questions in Physics in Class 11

1.	A glass capillary tube sealed at both ends is 100cm long. It lies horizontally with the middle 10 cm containing mercury. The two ends of the tube which are equal in length contain air at 27°C at apressure of 76cm of Hg. Now the air column at one end of the tube is kept at 0^(@) C and the other endis maintained at 127^(@) C. Calculate the pressure of the air column at 0^(@) C. (Neglect the change in volume of Hg and glass).
Answer» 25CM of HG 35 CM of Hg 55 cm of Hg 85 cm of Hg Answer :D

Discussion

2.	A particle P moving with speed v undergoes a head - on elastic collision with another particle Q of identical mass but at rest. After the collision
Answer» <P>Both `P`and `Q` move forward with SPEED `(v)/(2)` Both `P` and `Q` move forward with speed `(v)/(sqrt(2))` `P` comes to rest and `Q` moves forward with speed `v` `P` and `Q` move in opposite DIRECTIONS with speed `(v)/(sqrt(2))` Answer :C

Discussion

3.	The average resisting force that must act on a 5kg mass to reduce its speed from 65 cms^(-1) to 15 cms^(-1) in 0.2 s is
Answer» 12.5 N 25 N 50 N 100 N ANSWER :A

Discussion

4.	Two pulses travelling on the same string are described by y_(1) = (5)/(( 3x - 4 t)^(2)+ 2) and y_(2) = ( -5) /(( 3x + 4t - 6)^(2) + 2) (a). In which direction does each pulse travel ? (b). At what instant do the two cancel everywhere ? ( c). At what point do the two pulses always cancel ?
Answer» Solution :(a). At constant phase , `phi = 3x - 4t` will be constant . Then ` x = ( phi + 4t)//3` will change : the wave moves . As `t` increases in this EQUATION , `x` increases , so the first wave moves to the right . In the same way , in the second case ` x = ( phi - 4t + 6)//3`. As `t` increases , `x` MUST decreases , so the second wave moves to the left. (b) We required that ` y_(1) + y_(2) = 0` `( 5)/(( 3x - 4t)^(2)) + ( -5)/(( 3x + 4t -6)^(2) + 2) = 0` This can be written as `( 3x - 4t)^(2) = ( 3 x + 4 t -6)^(2)` Solving for the positive root , `t = 0.750 s` ( c) The negative rootyields `( 3x - 4t) = -( 3x + 4t - 6)` The time terms cancel , leaving ` x = 1.00 m`. At this POINT , the waves always cancel. The TOTAL wave is not a standing wave , but we could call teh point at ` x = 1.00 m`a mode of the superposition.

Discussion

5.	Find the value of 2.2 + 4.08 + 3.125 + 6.3755.
Answer» Solution :Out of all the four numbers 2.2 has got the least number of decimal places-one. Hence we should retain only TWO decimal places in the REMAINING numbers. Hence 4.08 remains as it is 2.125 is to be rounded off as 3.12 (as 2 before 5 is even) 6.3755 is to be rounded off as 6.38 (as 7 before 5 is ODD). Now adding `2.2 + 4.08 + 3.12 + 6.38 = 15.78`. Finally we should have only one decimal place and hence 15.78 is to be rounded off as 15.8.

Discussion

6.	Three point masses 3kg, 2kg and 1kg are placed at A,B,C of a triangle. AB = 0.3,m AC = 0.5 m and BC = 0.4 m. The M.I of the system about an axis through A and normal to the plane of triangle is
Answer» `0.43 KGM^(2)` `0.25kgm^(2)` `0.18kgm^(2)` `0.07kgm^(2)` ANSWER :A

Discussion

7.	What is meant by 'Evaporation'?
Answer» Solution :When a liquid is heated at constant pressure to its BOILING POINT or when the pressure is reduced at a constant tempeature it will convert to a GAS. This process of a liquid CHANGING to a gas is called evaportation.

Discussion

8.	A projectile is fired vertically upward from the surface of earth with a velocity Kupsilon_e where upsilon_e is the escape velocity and K lt 1. Neglecting air resistance, show that the maximum height to which it will rise measured from the centre of earth is R/(1-K^2) where R is the radius of the earth.
Answer» Solution :If a body is projected from the surface of earth with a velocity `UPSILON` and REACHES a height H, by conservation of ENERGY `-(GMm)/R+1/2mv^2 =-(GMm)/(R+h) therefore h=R/(1-K^2)`

Discussion

9.	A liquid takes 30s to cool from 95^(0)C to 90^(0)C and 70s to cool from 55^(0)C. If time taken by the liquid to cool from 50^(0)C to 45^(0)C is 21k (in seconds), then find the value of k.
Answer» ANSWER :4

Discussion

10.	The temperature of the sun is measured with
Answer» PLATINUM thermometer gas thermometer pyrometer vapour PRESSURE thermometer Answer :C

Discussion

11.	Find the work done in lifting a stone of mass 10kg and specific gravity 3 from the bed of a lake to a height of 6m in water.
Answer» Solution :Mass of STONE= 10kg, Specific GRAVITY `=(d_(s))/(d_(w))= 3` `W= mgh (1- (d_(w))/(d_(s)))= 10 xx 9.8 xx 6 xx (1- (10/(3))` `W= 98 xx 6 xx (20/(3) = 392J`

Discussion

12.	State Kepler's law of period in planetory motion.
Answer» Solution :The SQUARE of the TIME period of revolution of a PLANET around the SUN in its elliptical orbit is directly proportional to the cube of the semi-major AXIS of the ellips. `T^(2)propa^(3)` `(T^(2))/(a^(3))=` constant

Discussion

13.	Two particles are projected with same velocity but at angle of projection 35^(0) and 55^(0). Then their horizontal ranges are in the ratio of
Answer» `1 : 2` `2 : 1` `1 : 1` `4 : 1` ANSWER :C

Discussion

14.	A ball hits the ground and rebounds after collision. In this process, momentum of which bodies is conserved ?
Answer» Solution :TOTAL MOMENTUM of the earth and the BALL system is CONSERVED.

Discussion

15.	A block of mass 2kg is moving with speed v_(0) towards a massles sunstretched spring (K=10N/m). It is found that for maximum speed v_(0)=sqrt(6.4)m/s, the block compresses the spring, stops at that position and does not return back. Friction coefficient at surface is mu=1/x. Value of x is
Answer» ANSWER :5

Discussion

16.	A car is moving with a speed of 4-0km//h^(-1). As the car reaches a circular turn on the road of radius 100m , the driver applies brakes and reduces his speed at the rate of 1ms^(2). Calculate the magnitude of net acceleration of the car.
Answer» ANSWER :`1m//s^(2)`

Discussion

17.	Two masses m_(1) and m_(2) connected by a light spring of natural length l_(0) iscompressed completely and tied by a string. This system while moving with a velocity v_(0) along +ve x-axis pass through the origin at t=0. At this position the string snaps. Postion of mass m_(1) at time t is given by the equation. x_(1) (t)=v_(0)t-A(1-cosomegat) calculate : (a) Position of the particle m_(2) as a function of time. (b) l_(0) in terms of A.
Answer» ANSWER :A::B::C

Discussion

18.	Name the situation where the speed of an object is constant while the velocity is not?
Answer» SOLUTION :In UNIFORM CIRCULAR MOTION .

Discussion

19.	Two blocks of masses 10 kg and 20 kg are connected by a massless spring and are placed on a smooth horizontal surface. A force of 200 N is applied on 20 kg mass as shown in the diagram. At the instant, the acceleration of 10 kg mass is 12 ms^(-2), the acceleration of 20 kg mass is,
Answer» `4 MS^(-2)` `12 ms^(-2)` `20 ms^(-2)` `8 ms^(-2)` Answer :A

Discussion

20.	In Q.No 6.2, if the planet rotates in counter clockwise direction, then areal velocity has a direaction :
Answer» GIVEN by "RIGHT HAND Thumb Rule" Given by "Left Hand Thumb Rule" NORMAL to the plane of orbit upwards Normal to the plane of orbit downwards Answer :A::C

Discussion

21.	Give names of elementary physical quantities.
Answer» Solution :The seven elementary physical quantities are (i) mass (ii) length (III)time (IV)electric current (V) TEMPERATURE (vi) luminous INTENSITY (vii) amount of substance.

Discussion

22.	Rate of cooling (d theta)/(dt)depends upon
Answer» NATURE of the SURFACE and surface. AREA of the body Temperature of body and surroudings Mass and Spheat of the body All the above ANSWER :D

Discussion

23.	The air column in a pipe which is closed at oneend will be in resonance with a vibrating tuning fork at a frequency 260Hz, if the length of the air column is (speed of sound in air =330ms^(-1))
Answer» `31.73`CM `62.5`cm `35.75`cm `12.5`cm Solution :`n=(V)/(4L)impliesl=(V)/(4n)=(330)/(4xx260)=3.17cm`

Discussion

24.	Who decoupled the motion and force?
Answer» GALILEO Aristotle NEWTON JOULE SOLUTION :Galileo

Discussion

25.	An enclosed ideal gas is taken through a cycle as shown in the figure. Then
Answer» ALONG AB, temperature decreases while along BC temperature INCREASES Along AB, temperature increases while along BC the temperature decreases. Along CA work is done by the gasand the INTERNAL energy reamins constant. Along CA WORKD is done on the gas and internal energy of the gas increases Solution :AB `rarr` isobaric BC `rarr` isocharic CA `rarr` isothermal

Discussion

26.	Statement I: When a person walks on a rough surface, the net force exerted by surface on the person in the direction of his motion. Satement II: It is the force exerted by the road on the person that causes the motion.
Answer» Statement I is true, statement II is true, statement II is a CORRECT EXPLANATION for statement I. Statement I is true, statement II is true, statement II is not a correct explanation for statement I. Statement I is true, statement II is FALSE. Statement I is false, statement II is true. Answer :D

Discussion

27.	Explain the beat phenomenon.
Answer» SOLUTION :When two or more waves superimpose each other with slightly different frequencies, then a sound of periodically varying amplitude at a POINT is observed. This PHENOMENON is KNOWN as beats. The number of amplitude maxima per second is CALLED beat frequency . If we have two sources, then their difference in frequency gives the beat frequency. Number of beats per second n `= \|f_(1) - f_(2)\|` per second .

Discussion

28.	If velocity of light c, Planck's constant h and gravitational constant G are taken as fundamental quantities, then express mass, length and time in terms of dimensions of these quantities.
Answer» Solution :We know that, dimensions of`(h)=[M^(1)L^(2)T^(-1)]` (From E=hf) dimensions of `(c) =[L^(1)T^(-1)](c)` (c is velocity) Dimensions of gravitational constant (G) `=[M^(-1)L^(3)T^(-2)] ""("From" F=(Gm_(1)m_(2))/(r^(2)))` (i) Let `m prop c^(a)h^(b)G^(c)` `rArr m=kc^(a)h^(b)G^(c)` where, k is a dimensionless constant of proportionality. Substituting dimensions of each term in Eq. (i), we get `[ML^(0)T^(0)]=[LT^(-1)]^(x)xx[M^(2)L^(-1)]^(y)[M^(-1)K^(-3)T^(-2)]^(z)` `=[M^(b-c)L^(a+2b+3c)T^(-a-b-2c)]` COMPARING powers of same TERMS on both sides, we get `b-c=1 ""...(ii)` `a+2b+3c=0 ""...(iii)` `-a-b-2c=0""...(IV)` Adding EQS. (i), (iii) and (iv), we get `2b=1 rArr b=(1)/(2)` Substituting value of b in eq. (i), we get `c=-(1)/(2)` From eq. (iv) `a=-b-2c` Substituting values of b and we get `a=-(1)/(2)-2(-(1)/(2))=(1)/(2)` Putting values of a, b and c in equ. (i), we get `m=kc^((1)/(2))h^((1)/(2))G^(-(1)/(2))` `:.m=ksqrt((ch)/(G))` (ii) Let `L prop c^(a)h^(b)G^(c)` `:.L=kc^(a)h^(b)G^(c) ""..(iv)` Where, k is a dimensionless constant. Substituting dimensions of each term in equ. (v), we get `[M^(0)LT^(0)]=[LT^(-1)]^(a)xx[ML^(2)T^(-1)]^(b)xx[M^(-1)L^(3)T^(-2)]^(c)` `=[M^(b-c)L^(a+2b+3c)T^(-a-b-2c)]` On comparing powers of same terms, we get `b-c=0 ""...(vi)` `a+2b+3c= ""...(vii)` `-a-b-2c=0""...(viii)` Adding equ. (vi), (vii) and (viii), we get `2b=1` `:.b=(1)/(2)` Substituting value of b in equ. (vi), we get `c=(1)/(2)` From equ. (viii), `a=-b-2c` Substituting values of b and c, we get `a=-(1)/(2)-2((1)/(2))=-(3)/(2)` Putting values of a, b and c in equ. (v), we get `L=kc^((3)/(2)h^((1)/(2))G^((1)/(2))` `L=ksqrt((hG)/(c^(3)))` (iii) Let `T prop c^(a)h^(b)G^(c)` `:.T=kc^(a)h^(b)G^(c)` where, k is a dimensionless constant, Substituting dimensions of each term in equ. (ix), we get `[M^(0)K^(0)T]=[LT^(-1)]^(a)xx[ML^(2)T^(-1)]^(b)xx[M^(-1)L^(3)T^(-2)]^(c)` `=[M^(b-c)L^(a+2b+3c)T^(-a-b-2c)]` On comparing powers of same terms, we get `b-c=0 ""...(x)` `a+2b+3c=0 ""...(xi)` `-a-b-2c=1 ""...(xii)` Adding equ. (x). (xi) and (xii), we get `2y=1` `:.y=(1)/(2)` Substituting value of b in equ. (x), we get `c=b=(1)(2)` From eq. (xii) `a=-b-2c-1` Substituting values of b and c we get `x=-(1)/(2)-2((1)/(2))-1=-(5)/(2)` Putting values of a,b and c in equ. (ix) we get `T=kc^((5)/(2))h^((1)/(2))G^((1)/(2))` `T=ksqrt((hG)/(c^(5)))`

Discussion

29.	Match Column-I with Column-II : {:("Column-I","Column-II"),("(1) Static friction","(a) Boundary friction"),("(2) Rolling friction","(b) Ball-bearing"),(,"(c) Object moving on road"):}
Answer» ANSWER :(1-a), (2-b)

Discussion

30.	A string of length L, fixed at its both ends is vibrating in its 1^(st) overtone mode. Consider two elements of the string of the same small length at positions l_(1)=0.2L"and"l_(2)=0.45L from one end. If K_(1)"and" K_(2) are their respective maximum kinetic energies,then
Answer» `K_(1)=K_(2)` `K_(1)gtK_(2)` `K_(1)ltK_(2)` it is not possible to DECIDE the relation Answer :B

Discussion

31.	A string of length L, fixed at its both ends is vibrating in its 1^(st) overtone mode. Consider two elements of the string of the same small length at positions l_(1)=0.2L"and"l_(2)=0.45L from one end. Follow situation of part (a) If K_(1) "and" K_(2) are respective force constants of their simple hrmonic motions, then
Answer» `k_(1)=K_(2)` `K_(1)gtK_(2)` `K_(1)ltK_(2)` it is not possible to decide the relation Answer :A

Discussion

32.	Select the correct statement from the following statements.
Answer» Number of DEGREE of FREEDOM for diatomic MOLECULE is 8 Number of degree of freedom for diatomic molecule is 3. Number of degree of freedom for diatomic molecule is 6 Number of degree of freedom for diatomic molecule is 7 Answer :B

Discussion

33.	When does a solid dropped in a liquid attain terminal velocity?
Answer» SOLUTION :When the NET FORCE on the SOLID is ZERO.

Discussion

34.	A car sounding its horn at 480 Hz moves towards a high wall at a speed of 20 m/s, the frequency of the reflected sound heard by the man sitting in the car will be nearest to :
Answer» 510 HZ 570 Hz 480 Hz 540 Hz Solution :After reflection from the wall, the SOUND moves towards OBSERVER in the car. `v'=(v+v_(0))/(v-v_(s))xx v=(340+20)/(340-20) xx480` `=540 Hz`

Discussion

35.	If, in Q.19, coefficient of friction between m_(1) and the wall is mu,and f_(1) and f_(2) are respectively, the force of friction on m_(1) and m_(2), then :
Answer» `f_(1)` and `f_(2)` both are downward. `f_(1)` and `f_(2)` both are downward. `f_(1)` is UPWARD and `f_(2)` upward. `f_(1)` is downward and `f_(2)` upward. Solution :In upward DIRECTION .

Discussion

36.	The young's modulus for a perfectly rigid body is
Answer» infinity zero 1 -1 Answer :A

Discussion

37.	The force 7 hati + 3hatj - 5hatk acts on a particle whose position vector is hati - hatj + hatk . What is the torque of a given force about the origin ?
Answer» `2 HATI + 12 HATJ + 10 hatk` `2 hati + 10 hatj + 12 hatk` `2 hati + 10 hatj + 10 hatk` `10 hati + 2hatj + hatk` ANSWER :A

Discussion

38.	For which of the two pairs, the angle of contact is same?
Answer» WATER and GLASS, glass and MERCURY PURE water and glass, glass and alcobol silver and water, mercury and glass silver and chromium, water and chromium Answer :B

Discussion

39.	During an experiment an ideal gas is found to obey an additional law VP^2= constant. The gas is initially at a temperature 'T' and volume 'V'. When it expands to a volume 2V, the temperature becomes
Answer» T 2T `SQRT2` T `(T)/( sqrt2)` ANSWER :C

Discussion

40.	What will be the difference in volume of water when it is heated from 0^(@)C to 10^(@)C ?
Answer» Solution :INITIALLY (from `0^(@)C` to `4^(@)C`) VOLUME decreases and the (from `4^(@)C` to `01^(@)C`) volume increase.

Discussion

41.	By using the expression you derived above, show that motion of satellite obeys Kepler's law of periods.
Answer» SOLUTION :`T^2 PROP r^3 or Kepler.s `3^rd` LAW.

Discussion

42.	Let I_(A) and I_(B) be moments of inertia of a body about two axes A and B respectively. The axis A passes through the centre of mass of the body but B does not
Answer» `I_(A)ltI_(B)` If `I_(A)ltI_(B)` the axes are PARALLEL If the axes are PARALEL `I_(A)ltI_(B)` if the axes are not parallel the `I_(A)geI_(B)` SOLUTION :`I_(B)=I_(A)+Md^(2)`

Discussion

43.	Show that in the absence of any external force, the velocity of CM remains constant.
Answer» SOLUTION :`vecF_("ext")=M.veca_(CM)` SINCE `vecF_(ext)=0` `Mveca_(CM)=0` i.e.`veca_(CM)=0` `veca_(CM)=d/(dt)(vecV_(CM))` `:.vecV_(CM)=` constant

Discussion

44.	A calorie is a unit of heat or energy and it equals 4.2 J where 1J = 1 kg m^(2)s^(-2) . Suppose we employ a system of units in which unit of mass is alpha kg. unit of length is beta m, unit of time is gamma . What will be magnitude of calorie in terms of this new system?
Answer» Solution :`n_(2) = n_(1)[(M_(1))/(M_(2))]^(a) [(L_(1))/(L_(2))]^(B) [(T_(1))/(T_(2))]^(c) = 4.2 ((KG)/(alpha kg))^(1) ((m)/(betam))^(2) ((s)/(gammas))^(-2)` `n_(2) = 4.2 alpha^(-1) beta^(-2) gamma^(2)`

Discussion

45.	………….. Errors arise on account of shear……………… of the ………….. .
Answer» ANSWER :GROSS ; CARELESSNESS ; OBSERVER.

Discussion

46.	Explain the reflection of wave at rigid support.
Answer» Solution :A pulse (WAVE) moving in +x-direction and reflecting wave from fixed support are shown in figure. ? If we suppose that the energy is not ABSORBED at end, then the shape of the reflected pulse will be same as incident but the phase will be changed by `180^(@)(pi).` The reason behind it is the end is fixed. So that the displacement of pulse should be zero. Supppose, the incident progressive wave displacement at .t. time is `y _(1) (x,t)= a sin (kx-omegat).` Suppose, the displacement of reflected wave is `y _(i)` According to superposition principle, `y (x,t) =y _(i) (x,t) + y _(r)(x,t)` But `y (x,t) =0` `(because` DISPLACEMTN of fixed end is zero) `therefore 0= y _(i) (x,t) + y _(r) (x,t)` `therefore y _(r) (x,t) =-y _(i) (x,t)` `=- a sin (kx - omega t )` `therefore y _(r) (x,t) = a sin (kx - omega t + pi)` Thus, when the pulse (wave) reaches to end the string applies FORCE on fixed end. According to third law of Netton the end will also apply the same force in opposite direction. Hence, there will be difference in phase of `pi` in reflected wave. It will be progressive in -x-direction. Thus, the plane of reflected wave will be increased by `pi.` Hence, the shape of wave will be reversed. It means the crest will be through and through will be crest.

Discussion

47.	A particle of mass 4g is in a gravitational potential field given by V= (800x^(2)+150)erg"/"g. Its frequency of oscillations is
Answer» `100 PI HZ` `(20)/(pi)Hz` `50 pi Hz` `(50)/(pi)Hz` Answer :B

Discussion

48.	An elevator can carry a maximum load of 1800 kg (elevator + pasenger) is moving up with a constant speed of 2ms^(-1). The frictional force opposing the motion is 4000 N. Determine theminimum power delivered by the motor to the elevator in watts as well as in horse power..
Answer» <P> Solution :The downword force on the ELEVATOR is `F=mg+F_(f)=(1800xx10)+4000=22000N` The MOTOR must SUPPLY enough POWER to balance this force. Hence, `P = Fv = 22000xx2` `= 44000 W = 59 hp`.

Discussion

49.	If 10^(20) oxygen molecules per second strike 4cm^(2) of wall at an angle of 30^(@) with the normal when moving at a speed of 2xx10^(3)ms^(-1), find the pressure exerted on the wall. (mass of 1O_(2) atom = 2.67xx10^(-26)kg)
Answer» Solution :Mass of `1O_(2)` atom `=2.67xx10^(-26)kg` Mass of `10^(20)O_(2)` atom = `2.67xx10^(-26)kg` `=26.72xx10^(-27)xx10^(20)` Momentum P = mv `=26.72xx10^(-7)xx8xx2xx10^(3)` `=427.5xx10^(-4)kgms^(-1)` COMPONENT of momentum normal to wall is `30^(@)` `=427.5xx10^(-4)xxcos30^(@)` `=427.5xx10^(-4)xx(SQRT(3))/(2)` `=370.2xx10^(-4)kgms^(-1)` PRESSURE = `(F)/(A) = ("Force (change in momentum)")/("area")` `=(370.2xx10^(-4))/(4xx10^(-4))` `P=92.55Nm^(-2)`

Discussion

50.	Magnitude of force experienced by an object moving with speed v is given by F = kv^(2) . Find dimensions of k
Answer» Solution :`[K] = ([F])/([V^(2)]) = ([M^(1)L^(1)T^(-1)])/([LT^(-1)]^(2)) = ([M^(1)L^(1)T^(-2)])/([M^(0)L^(2)T^(-2)]) = [M^(1)L^(-1)]`

Discussion

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For which of the two pairs, the angle of contact is same?

During an experiment an ideal gas is found to obey an additional law VP^2= constant. The gas is initially at a temperature 'T' and volume 'V'. When it expands to a volume 2V, the temperature becomes

What will be the difference in volume of water when it is heated from 0^(@)C to 10^(@)C ?

By using the expression you derived above, show that motion of satellite obeys Kepler's law of periods.

Let I_(A) and I_(B) be moments of inertia of a body about two axes A and B respectively. The axis A passes through the centre of mass of the body but B does not

Show that in the absence of any external force, the velocity of CM remains constant.

A calorie is a unit of heat or energy and it equals 4.2 J where 1J = 1 kg m^(2)s^(-2) . Suppose we employ a system of units in which unit of mass is alpha kg. unit of length is beta m, unit of time is gamma . What will be magnitude of calorie in terms of this new system?

………….. Errors arise on account of shear……………… of the ………….. .

Explain the reflection of wave at rigid support.

A particle of mass 4g is in a gravitational potential field given by V= (800x^(2)+150)erg"/"g. Its frequency of oscillations is

An elevator can carry a maximum load of 1800 kg (elevator + pasenger) is moving up with a constant speed of 2ms^(-1). The frictional force opposing the motion is 4000 N. Determine theminimum power delivered by the motor to the elevator in watts as well as in horse power..

If 10^(20) oxygen molecules per second strike 4cm^(2) of wall at an angle of 30^(@) with the normal when moving at a speed of 2xx10^(3)ms^(-1), find the pressure exerted on the wall. (mass of 1O_(2) atom = 2.67xx10^(-26)kg)

Magnitude of force experienced by an object moving with speed v is given by F = kv^(2) . Find dimensions of k