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14551.

A point P lies on the axis of a ring of mass M and radius a, at a distance a from its centre C. A small particle starts from P and reaches C under gravitational attraction only. Its speed at C will be

Answer»

`SQRT((2GM)/(a))`
`sqrt((2GM)/(a)(1-(1)/(sqrt(2))))`
`sqrt((2GM)/(a)(sqrt(2)-1))`
zero

Answer :B
Discussion
14552.

Four particles each of mass m are lying symmetrically on the rim of a disc of mass M and radius R. Moment of inertia of this system about an axis passing through one of the particles and I to plane of disc is

Answer»

`16 mR^2`
`(3 M + 16 m) (R^2)/(2)`
`(3 m + 12 M) (R^2)/(2)`
zero

Solution :According to the THEOREM of parallel axes , MOMENT of inertia of DISC about an axis passing through K and `bot` to plane of disc of the given figure = `(1)/(2) MR^(2) + MR^(2) = (3)/(2) MR^(2)`

Total moment of inertia of the system
`= (3)/(2) MR^(2) + m (2R)^(2) + m (sqrt2 R)^(2) +m (sqrt2 R)^(2) = (3 M + 16 m ) (R^(2))/(2)`
Discussion
14553.

When a weightlifter remains stationary with a weight lifted over his head, how much work does he do ?

Answer»

SOLUTION :To lift a weight mg above HISHEAD through a HEIGHT h, work MGH has to be done against gravity . But when the weight is held stationary above his head, no displacement occurs and no work is done by the lifter at that stage.
Discussion
14554.

Two perfectly elastic spheres of masses 2 kg and 3 kg moving in opposite directions with velocities 8 ms^(-1) and 6 ms^(-1) respectively colide with each other. Find their velocities after the impact.

Answer»

SOLUTION :Here `m_(1) = 2kg, m_(2) = 3kg`
`u_(1) 8ms^(-1), u_(2) = -6ms^(-1)`
According to the law of conservation of LINEAR momentum.
`m_(1)u_(1)+m_(2)u_(2)=m_(1)v_(1)+m_(2)v_(2)`
i.e., `(2)(8) + (3) (-6) = 2v_(1) + 3v_(2)`
`rArr2v_(1)+3v_(2)=-2 "" .....(i)`
Since the collision is elastic, `u_(1) - U _(2) = v_(2) - v_(1)`
i.e., `8 - (-6) = v_(2) - v_(1)`
`rArr v_(2) - v_(1) = 14 "" .....(ii)`
On solving the equations (i) and (ii) we get
`v_(1) = - 8.8 MS^(-1), v_(2) = 5.2ms^(-1)`
Discussion
14555.

The refractive index of glass with respect to water is 9//8. If the velocity and wavelength of light in glass are 2 times 10^8 m//s and 4000 A respectively, find the velocity and wavelength of light in water.

Answer»

SOLUTION :`_w mu_g=mu_g/mu_w=v_w/v_g implies 9/8=v_w/(2 times 10^8),`
`v_w=(9 times 2 times 10^8)/8=2.25 times 10^8 m//s`
`_m mu_g= mu_g/mu_q= lamda_w/lamda_g`
`(because mu_g= lamda/lamda_g, mu_w=lamda/lamda_w)9/8= lamda_w/4000, lamda_w=(9 times 4000)/8=4500A^c`
Discussion
14556.

Centripetal force is always in opposite direction of centrifugal force.

Answer»


ANSWER :TRUE.
Discussion
14557.

A stone is dropped into a well of 20 m deep. Another stone is thrown downward with velocity 'v' one second later. If both stones reach the water surface in the well simultan eously, v is equal to (g = 10 ms^(-2))

Answer»

`30MS^(-1)`
`15ms^(-1)`
`20MS^(-1)`
`10MS^(-1)`

Answer :B
Discussion
14558.

Show that moment of inertia of a solid body of any shape change eith temperature as I=I_0(1+2 alhatheta). Where I_0 is the moment of inertia at 0^C nad alpha is the coefficient of liear expansion of the solid.

Answer»

SOLUTION :Given, `I_0= moment of inertia at 0^@C`
` alpha= Coefficient of inertia `
expansion
` To prove I=I_0 (1+2 alpha theta)
Let the temperature changes to `theta` from `0^@C`
`Delta T= theta`
` Let R_0 be the radius of gyration `
` Now R_0=R(1+ alpha theta)`
`I_0=MR^2 where M is the MASS. `
` Now I=MR^2=MR^2 ((1+alpha theta)^2)`
` [by binominal expansion and neglecting
alpha_2 theta_2 which is a very SMALL VALUE]`
`=MR^2(1+2 alpha theta)`
` So,I=I_0(1+2 alpha theta).........(proved).`
Discussion
14559.

A refrigerator, whose coefficient of performance K is 5, extracts heat from the cooling compartment at the rate of 250 J per cycle. (a) How much work per cycle is required to operate the refrigerator cycle ? (b) How much heat per cycle is discharged to the room which acts as the high temperature reservoir?

Answer»

SOLUTION :(a) As coefficient of PERFORMANCE of a refrigerator is defined as
`K=Q_(L)//W,"So "W=(Q_(L))/(K)=(250)/(5)=50J`
(b) As `Q_(H)=Q_(L)+W" so "Q_(H)=250+50=300J`
Discussion
14560.

On a hilly region,water boils at 95^@C. The temperature expressed in Fahrenheit is

Answer»

`100^@F`
`203^@F`
`20.3^@F`
`140^@F`

ANSWER :A
Discussion
14561.

A vibrating tuning fork is held at the mouth of a cylindrical tube. The tube is dipped into water. It is found that when the level of water rises to a definite height, a sound of large intensity is heard. Explain the reason behind it.

Answer»

Solution :The empty cylindrical tube, is filled with air. The vibrating tuning fork PRODUCES forced vibration in the air column of the tube. As a result of the forced vibration of the air column, an ADDITIONAL sound is produced. The frequency of the sound produced depends on the length of the air column. By dipping the tube into water, the length of the air column can be changed. So, when the level of water RISES to particular height, the frequency of the air column in the tube and that of the tuning fork BECOME equal. Then resonance TAKES place and a sound of large intensity is heard.
Discussion
14562.

If a constant torque of 500 Nm turns a wheel of moment of inertia 100 kg-m^2 about an axis through its centre, find the angular velocity and the kinetic energy gained in 2 seconds,

Answer»

SOLUTION :`prop = tau//l = 500//100 = 5 "rads"^(-2), OMEGA = at = 10 "rads"^(-1), K.E =(1//2)Iomega^(2) = 500 J`
Discussion
14563.

Mean solar day is the time interval between two successive noon when sun passes through zenith point (meridian). Sidereal day is the time interval between two successive transit of a distant star through the zenith point (meridian). By drawing appropriate diagram showing the earth's spin and orbital motion, show that mean solar day is 4 min longer than the sidereal day. In other words, distant stars would rise 4 min early every successive day. As per diagram, the earth moves from the point P to Q in one solar day.

Answer»

Solution :`implies` As per diagram , the earth moves from the point P to Q in one SOLAR DAY

Every day the earth advances in the orbit by approximately `1^@` . Then, it will have to ROTATE by `361^@`(which we define as 1 day) to have the sun at zenith point again.
`:. ` To COVER `361^@` , time taken = 24 h
`:. ` To cover `1^@` , time taken = t,
`t = (24)/(361) xx 1 = 0.066 h `
`= 3.99 min`
`~~ 4 min`
Hence, distant stars would rise 4 min early every SUCCESSIVE day .
Discussion
14564.

Kinetic energy, with any reference, must be

Answer»

negative
zero
POSITIVE
both (2) and (3)

ANSWER :C
Discussion
14565.

Two identically sized rooms A and B are connected by an open door. If the room A is air conditioned such that its temperature is 4^(@) lesser than room B, which room has more air in it?

Answer»

ROOM A
Room B
Both room has same AIR
cannot be DETERMINED

ANSWER :A
Discussion
14566.

At what temperature, a perfect black body will radiate energy at the rate of 6.48 W//cm^2 . Take, sigma= 5.67 xx 10^(-8) Wm^(-2)K^(-4)

Answer»

Solution :Given `E = 6.48 W cm^(-2) = 6.48 XX 10^4 WM^(-2)`
` sigma = 5.67 xx 10^(-8) Wm^(-2) K^(-4)`
Using stefan.s law`E= sigma T^4`
` therefore T= (E/sigma)^(1/4) = ( (6.48 xx 10^4)/(5.67 xx 10^(-8)))^(1/4)`
`= 1.033 xx 10^3 = 1033 K`
Discussion
14567.

Calculate the height upto which an insect can crawl up a fixed bowl in the from of a hemisphere of radius r Given coefficient of friction = 1 sqrt(3) .

Answer»

Solution :B is botton of the bowl of RADIUSR = OB . The insect can crawl up the bowl From B to P through a HEIGHT
` BA = h `
As is clear from ` R = W cos alpha `
and`F = W sin alpha ` where W is weight of insect and F is force of friction
`mu = (F)/(R) = (Wsin alpha)/(W cos alpha) = TAN alpha (1)/(sqrt3) `
In ` triangleOPA , tan alpha = (PA)/(OA) `
`:. tan alpha = sqrt(r^(2) - Y^(2))/(y)= (1)/sqrt3 `
` (r^(2) - y^(y))/(y^(2)) = (1)/(3) `
` y^(2) = (3 r^(2))/(4) , y = (sqrt3 r)/(2) `
` h = BA = OB = OB - OA = r - y `
` h = r - (sqrt3 r )/(2) = 0 .134 r `
`h = 13 .4 %` of r.
Discussion
14568.

Twoidentical wires of equal lengths are stretched in such a way that their simultaneous vibrations produce 6 beats per second. The tension in one of the wires is changed slightly and it is observed that the beat frequency remains the same. How is it possible ?

Answer»


Answer :Thefrequency of a stretchedstring is proportionalto the SQUARE root of the tension. If the FUNDAMENTAL frequency in the 1st wire is more than that in the 2ND wire, tension in the 1st wire `(T_(1) ) gt`the tension in the 2nd wire`(T_(2))` .
Then , `n_(1) = n_(2) + 6 ` ,where ` n_(1) and n_(2)` arethe fundamentalfrequencies of thetwo wires respectively . Now, the tension `T_(2)` is increased gradually until the fundamental frequency of the 2nd wire CHANGES to `n_(2)^(') = n_(2) + 12` . Under these circumstanes, thefrequencydifference becomes,
`n_(2)^(') - n_(1) = (n_(2) + 12) - (n_(2) + 6) = 6 `
This means that 6 beats he heard again PER second .
Discussion
14569.

In the aboveproblem the momentof inertia of four bodiesabout an axis perpendicularto the planeof frameand passingthrougha corneris

Answer»

`ML^(2)`
`2ML^(2)`
`2sqrt(2)ML^(2)`
`4ML^(2)`

Solution :`I = 2 [ML^(2) ] = M [ L sqrt(2)]^(2) "" = 4ML^(2)`
Discussion
14570.

In the aboveproblem the momentof inertiaof four bodiesabout an axisperpendicularto the plane of frameand passing through a corneris

Answer»

`ML^(2)`
`2ML^(2)`
`2sqrt(2)ML^(2)`
`4ML^(2)`

SOLUTION :`I = 2 [ M (l/(SQRT(2)))^(2)] = ML^(2)`
Discussion
14571.

The loss of weight of a solid when immersed in a liquid at 0^(0) C is W_(0) . If alpha and beta are the volume coefficients of expansion of the solid and the liquid respectively, then the loss of weight at t^(0C is approximately

Answer»

SOLUTION :Loss of wt at `0^(0) C = V_(0) rho_(0) g = w_(0)`
`V_(0)` = volume of body at `0^(0) `C
`V_(1) ` = volume of body at `t_(1)^(0) ` C
loss of weigth at `t_(1)^(0) C = V_(1) rho_(1) g = w_(1)`
`(w_(1))/(w_(0)) = (V_(1)rho_(1))/(V_(0)rho_(0)) = (d_(0) )/(d_(1)) xx (rho_(1))/(rho_(0))` where `d_(0)` is DENSITY of solid at `0^(0)` C and `rho_(0)` is density of LIQUID at `0^(0)` C.
`(w_(1))/(w_(0)) = (d_(0) (1 + alpha t))/(d_(0))(rho_(0))/((1 + beta t)rho_(0)) = (1 + alphat t)/(1 + beta t)`
`w_(1) = w_(0) (1 + alpha t ) (1 - beta t ) = w_(0) [ 1 + (alpha - beta ) t] `
Discussion
14572.

Four identical particles each of mass m are arranged at the corners of a square of side .a.. If mass of one of the particle is doubled, the shift in the centre of mass of the system is

Answer»

`(a)/(2sqrt(2))`
`(a)/(3sqrt(2))`
`(a)(4sqrt(4))`
`(a)/(SQRT(3))`

ANSWER :D
Discussion
14573.

A massless string passes over a frictionless pulley and carries mass m_(1) hanging at one end and mass m_(2) connected by another massless string to mass m_(3) at other end as shown in figure . Calculate the tension in string joining masses m_(2)"and " m_(3).

Answer»

Solution :Let `T_(1)` be the tension in the string joining `m_(1)"and" m_(2)` , while` T_(2)`the tension is string joining `m_(2) "and " m_(3)`
Let a be acceleration of masses. The resultant force on MASS `m_(3)` is `(m_(3)G-T)` downward, therefore , we have
`m_(3g-T_(2)= m_(3)a`....(1)
Resultant force on mass `m_(2)` is
`(m_(2) g +T_(2) - T_(1)) ` downward , therefore , we have
` m_(2) g + T_(2)-T_(1)=m_(2) a ` ....(2)
Resultant force on mass `m_(1) ` is `(T_(1)-m_(1) g) `UPWARD , therefore , we have
`T_(1)-m_(1) g= m_(1) a`....(3)
Adding (1),(2) and (3) , we get , `m_(3)g+m_(2)g-m_(1)g=(m_(3)+m_(2)+m_(1))a`
`:. "acceleration " a =((m_(2)+m_(3)-m_(1)))/(m_(1)+m_(2)+m_(3)) g
substituting this VALUE of a in (1) , we get
`m_(3)g-T_(2)=m_(3).((m_(2)+m_(3)-m_(1))g)/(m_(1)+m_(2)+m_(3))`
This gives `T_(2)=m_(3)g-(m_(3)(m_(2)+m_(3)-m_(1))g)/(m_(1)+m_(2)+m_(3))`
`:. "Required Tension , " T_(2)=(2m_(1)m_(3)g)/(m_(1)+m_(2)+m_(3))`
Discussion
14574.

Two bodies of masses 5kg and 3kg are moving towards each other with 2ms^(-1) and 4ms^(-1) respectively. Then velocity of centre of mass is

Answer»

`0.25ms^(-1)` towards `3KG`
`0.5ms^(-1)` towards `5kg`
`0.25ms^(-1)` towards `5kg`
`0.5ms^(-1)` towards `3kg`

ANSWER :C
Discussion
14575.

When one end of a glass rod is heated its shape becomes spherical. Its reason is

Answer»

Friction
Viscosity
SURFACE TENSION
Gravitation

Answer :C
Discussion
14576.

A body hanging from a spring stretches it by 1 cm at the earth's surface. How much will the same body stretch the spring at a place 1600km above the earth's surface?

Answer»

poles
equator
centre of earth
same everywhere

Answer :A
Discussion
14577.

What is satellite ? Gives its types and uses.

Answer»

SOLUTION :`implies` A body revolving around a planet in the direction of revolution of planet is CALLED SATELLITE.
`implies` Motion of satellite is very similar to the motion of planets around the sun hence Kepler.s law of PLANETARY motion are equally applicable to them. Their orbits around the earth are circular or elliptic.
`implies`There are two types of satellite :
(i) Natural satellite :
Moon is the natural satellite of the earth with a time period of approximately 27.3 days. It is EQUAL to the rotational period of the moon about its own axis. Planet Jupiter has many satellites.
(ii) Artificial satellite :
The first artificial satellite made by the man kind was Sputnik put into orbit around the earth by Russian scientist in 1957. India (19, April 1975) also successfully launched "Aryabhatta. and .INSAT" series of satellite.
`implies`Satellite are used for scientific, engineering, communication, whether forecast, spying andmilitary geophysics and meteorology.
Discussion
14578.

Which of the following pairs represent units of the same physical quantity?

Answer»

KELVIN and Joule
Kelvin and CALORIES
newton and calories
joule and CALORIE

ANSWER :D
Discussion
14579.

Which of the following statement is false regarding the vectors ?

Answer»

The magnitude of a vector is ALWAYS a scalar
Each COMPONENT of a vector is always a scalar.
Two VECTORS having different magnitudes cannot have their RESULTANT zero
Vectors obey triangle law of addition

Answer :B
Discussion
14580.

A block ofice is sliding down the sloping roof of a house and the angle of inclination of the roof with the horizontal is 30^@ . The maximum and minimum heights of the roof from the ground are 8.1 m and 5.6 m. How far from the starting point, measured horizontally , does the block land ?[ignore friction]

Answer»

Solution :LET the highest point of the roofbe A and the lowest point be P as shown in Fig.2.69.
`therefore AC=8.1m, PD =5.6m`
`therefore AB=AC-BC=AC-PD=8.1-5.6=2.5m`
`AP=(AB)/(sin 30^@)=(2.5)/(1/2)=5m`

Let the velocity of the block at P be v .
Considering the motion of the block from A to P,
`v^2=2g sin 30^2xxAP =2xx9.8xx1/2xx5 =49 or, v=7m*s^(-1)`.
The horizontal and VERTICAL components of the velocity at P are `v cos 30^@ =(7sqrt(3))/(2) m*s^(-1) and v sin 30^@ =7/2 m*s^(-1)` respectively.
Let the total taken by the block tocome from P to E be t. Considering the vertical motion of the block ,
`5.6=7/2t+1/2xx9.8xxt^2 or, 7t^2 +5t-8=0`
`therefore t=(-5pmsqrt(25+224))/(14)=0.77s`[takingthe POSITIVE value of t]
Now, `DE=v cos30^@ xxt=(7sqrt(3))/(2)xx0.77=2.7sqrt(3) m`
`therefore CE=CD+DE=2.5sqrt(3)+2.7sqrt(3)`
[`tan 30^@ =(AB)/(PB) or, 2.5 SQRT(3)m and , DC =PB =2.5 sqrt(3)m]`
=`5.2 sqrt(3)=9m`.
Discussion
14581.

In an experiment to measure density of object mass and volume are measured as 22.42 gram and 4.7cm^(3) respectively. Error in measurement of mass and volume is 0.1 gram and 0.1cm^(3) respectively. Find maximum error in measurement of density.

Answer»

0.22
0.02
`2%`
`0.02%`

SOLUTION :Density `rho=(M)/(V)`
`:.` Percentage error,
`(DELTADELTA)/(delta)xx100%=(DELTAM)/(M)xx100%+(DELTAV)/(V)xx100%`
`=(0.1)/(22.42)xx100%+(0.1)/(4.7)xx100%`
`=0.446%+2.04%`
`=2.45+2.04%`
`=2.49%`
`~~2%`
Discussion
14582.

Two wheels of M.I. 3kg m^(2) and 5 kg m^(2) are rotaing at a rate of 600 rpm and 800 rpm respectively in opposite directions. If they are coupled so as to rotate with the same axis of rotation, the resultent speed of rotation (in rpm)

Answer»

275
350
210
420

Answer :A
Discussion
14583.

If the spring is cut in two equal piece the spring constant of every piece decreases.

Answer»

SOLUTION :FALSE, INCREASES.
Discussion
14584.

Three uniform spheres each having mass 'm' and radius 'R' kept in such a way that each two touches the other. The magnitude of the gravitational force on any sphere, due to the other two is

Answer»

`sqrt(3)(Gm^(2))/(4R^(2))`
`(Gm^(2))/(4R^(2))`
`(sqrt(3)Gm)/(2R^(2))`
`sqrt(3)(Gm)/(R^(2))`

Answer :A
Discussion
14585.

The volume of 1 kg of hydrogen gas at N.T.Pis 11.2 m^(3).Specific heat of hydrogen at constant volume is 10046J kg^(-1) K^(-1). Find the specific heat at constant pressure in Jkg^(-1)k^(-1).

Answer»

12020
14220
16340
18230

Answer :B
Discussion
14586.

Considering the sun as a perfect sphere of radius 6.8xx10^(8) m, calculate the energy radiated by it one minute. Take the temperature of sun as 5800 k and sigma = 5.7xx10^(-8) S.I. unit.

Answer»


Solution :`Q=sigmaT^(4)At= 5.7xx10^(-8)XX(5800)^(4)xx4pixx(6.8xx10^(8))^(2)xx60=2.248xx10^(28)` J
Discussion
14587.

(A): Even when orbit of a satellite is elliptical, its plane of rotation passes through the centre of earth. (R) : According to law of conservation of angular momentum plane of rotation of satellite always remain same.

Answer»

Both ‘A’ and .R. are TRUE and .R. is the CORRECT EXPLANATION of ‘A’
Both ‘A’ and .R. are true and .R. is not the correct explanation of .A.
‘A’ is true and .R. is false
.A. is false and .R. is true

ANSWER :A
Discussion
14588.

A hot body will radiate heat most rapidly ,if its surface is

Answer»

WHITE and polished
white and rough
black and polished
black and rough

Answer :D
Discussion
14589.

A particle has initial velocity (2hati+3hatj)ms^(-1) when it was at origin and has constant acceleration (3hati+2hatk)ms^(-2). Find angle made by displacemnt after 2 sec with XY plane {sin^(-1)sqrt((k)/(21))}. Find the value of k.

Answer»


Solution :`vecu=(2hati+3hatj) vecd=vecut+(1)/(2)vecat^(2)`
`VECA=(3hati+2hatk)`
Now, if ANGLE is `THETA`, then
`theta=sin^(-1)((2)/(21))`
Discussion
14590.

In case of mechanical wave a partical oscillates and during sacillation its kinetic energy and potentialenergy changes.

Answer»

<P>

ANSWER :(a)P,Q (B)R,S (c )P,S,(d)Q,R
Discussion
14591.

The normal density of a metal is d and its bulk modulus is K. Find the increase in density of that metal piece. When a pressuer P is applied uniformly from all sides.

Answer»


ANSWER :`(PD)/(k-d)`
Discussion
14592.

A 500 kg horse pulls a cart of mass 1500 kg along a level road with an acceleration of 1 m//s^(2). If coefficient of sliding friction is 0.2, then force exerted by the earth on horse is

Answer»

3000N
4000N
5000 N
6000N

Solution :Force of friction `F_(1)=MU(m_(1)+m_(2))g`
force producing ACCELERATION
`F_(2)=(m_(1)+m_(2))a`
Total foce APPLIED by the horse `=F_(1)+F_(2)`
Discussion
14593.

Anobjectcontainsmoreheat-Is it a right statement ? Ifnot why ?

Answer»

Solution :When heated, an OBJECT receives HEAT from the agency. Now object has more INTERNAL energy than before. Heat is the energy in transit and which flows from an object at HIGHER temperature to an object lower temperature. Heat is not a quantity. So the statement I would prefer "an object contains more thermal energy".
Discussion
14594.

Maximum static friction force depend on area of surfaces in contact.

Answer»


ANSWER :False. Maximum static FRICTION force do not depend on AREA of surfaces in contact.
Discussion
14595.

Two small and heavy sheres, each of mass M, are placed at a distance r apart on a horizontal surface. The gravitational potential at the midpoint on the line joining the centre of the spheres is:

Answer»

zero
`-(GM)/(r)`
`-(2GM)/(r)`
`-(4GM)/(r)`

SOLUTION :Gravitational potential at O DUE to MASS M at A = `-(GM)/(r//2)=-(2GM)/(r)`

Gravitational potential at O due to mass M at `B=-(GM)/(r//2)=-(2GM)/(r)`
`therefore` TOTAL gravitational potential at `O=-(4GM)/(r)`
Discussion
14596.

A child is playing on a sliding board. If he is sliding down: (a) Mention the forces acting on the child. (b) Draw FBD (Free Body Diagram). (c) Write the force equation.

Answer»

Solution :(a) `ma= MG SINTHETA- f_(K)`
(B)
(c) F= mg sin THETA
Discussion
14597.

The force of attraction between two points 1 kg masses I_m apart proposed as the unit of force and call it neodyne, the first Bohr orbit (0.5xx10^(-10) m) as the unit of length and call it neometre and the mass of electron (9xx10^(-31) kg) as unit of mass and call it neogram.Find the value of 'neosecond' in this system.

Answer»


ANSWER :`8.2xx10^(-16)s`
Discussion
14598.

Define orbital velocity and establish an expression for it.

Answer»

Solution :(i) ORBITAL velocity is the velocity given to artificial satellite so that it may startrevolingaround the earth.
Expression for orbital velocity
Consider a satellite of mass .m. is revolving around the earth in a circularorbit of radius .r. at aheight .h. from the SURFACE of the earth.
Let .M. be the mass of the earth and .R. be the radius of the earth
Therefore , r = R + h
The centripetal force that is required to revolve the satellite ` = (mv_0^2)/(r )`
where `V_0` is orbital velocity
Orbital velocity is produced by the gravitational force b/w the earth and the satellite ` = (GMm)/(r^2)`

` therefore (mv_2^2)/(r ) = (GMm)/(r^2)`
` v_0^2 = (GM)/( r) = (GM)/( R + h)`
` v_0 = sqrt((GM)/(r )) = sqrt((GM)/(R +h))`
This is the expression for the orbital velocity.
(ii)Mass of the earth `M = 6 xx10^24 kg `
Radius of the earth R = 64000 km
The height of the artificial satellite from the earth (h) = 1000 km
Gravitational constant ` (G) = 6.67 xx 10^(-11) Nm^(-2) kg^(-2)`
Orbitalvelocity ` V_0 = sqrt((GM)/(R+h))`
` = sqrt( (6.7 xx 10^(-11) xx 6 xx 10^24)/(64000+1000)) = sqrt( (40.02 xx 10^13)/(65 xx 10^3)) = sqrt( (400.2)/(65)) = sqrt(6.1569)`
` V_0 = 2.48 ms^(-1)`
Discussion
14599.

Statement I: A large force is required to draw apart normally two glass plates enclosing a thin water film. Statement II: Water works as glue and sticks two glass plates.

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 :B
Discussion
14600.

A particle is acted by a force F = kx, where 'k' is a positive constant. Its P.E at x = 0 is zero. Which curve correctly represents the variation of potential energy of the block with respect to 'x'?

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ANSWER :A
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