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15201.

Two particles of equal masses have an elastic collision, the target particle being initially at rest. If it were not a head - on collision, the directions of their motion after the collision are

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in the same direction
in opposite directions
in MUTUALLY PERPENDICULAR directions
none of the above

Answer :C
Discussion
15202.

A saturn year is 29.5 times the earth year. How far is the saturn from the sun if the earth is 1.50 xx 10^(8) Km away from the sun ?

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SOLUTION :Given that `T_(s) = 29.5 T_(e)` and `R_(e) = 1.5 xx 10^(8) Km`
ACCORDING to Kepler.s THIRD law,
we can write `T^(2) prop R^(3)`
`rArr ((T_(s))/(T_(e)))^(2) = ((R_(s))/(R_(e)))^(3) R_(s) = R_(2)((T_(s))/(T_(e)))^(2//3)`
`rArr R_(3) = 1.5 xx 10^(8)((29.5 T_(e))/(T_(e)))^(2//3) rArr R_(3) = 1.43 xx 10^(9)Km`
Discussion
15203.

A 1 kg block is executing simple harmonic motion of amplitude 0.1 m on a smooth horizontal surface under the restoring force of a spring constant 100 N/m. Ablock of mass 3 kg is gently placed on it at the instant it passes through the mean position. Assuming that the two blocks move together, find the frequency and the amplitude of the motion.

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SOLUTION :`5/(2PI)` HZ, 5CM
Discussion
15204.

Dimension of angular velocity is :

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`MLT^(-1)`
`M^(0) L^(0)T - 1`
`MLT^(-2)`
`M^(0) L^(0) T^(-2)`

Solution :`OMEGA PROP n , "but" n prop (1)/(t)`
`omega prop (1)/(t)`
Discussion
15205.

The rms velocity of a gas at a given temperature is 300 m/s. What will be the rms velocity of a gas having twice the molecular weight and half the temperature in K.

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300 m/s
600 m/s
75 m/s
150 m/s

Answer :D
Discussion
15206.

A circular platform is mounted on a frictionless vertical axle . Its radius R = 2 m and its moment of inertia about the axle is 200 kg m^(2) . It is initially at rest . A 50 kg man stands on the edge of the platform and begins to walk along the edge at the speed of 1 ms^(-1) relative to the ground . Time taken by the man to complete one revolution is

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`pi s `
`(3pi)/(2) s`
`2pi s `
`(pi)/(2) s `

Solution :As the system is INITIALLY at REST , therefore , initial angular momentum `L_(i) =0` .
According to the principle of conservation of angular momentum FINAL angular momentum `L_(f) = 0`.
`therefore` Angular momentum of man on platform = Angular momentum of system i.e., `MVR = I omega`
or `omega = (mv R)/(I) = (50 xx 1 xx 2)/(200) = (1)/(2)` rad `s^(-1)`
Angular velocity of man relative to platform is
`omega_(r) = omega + (v)/(R) = (1)/(2) + (1)/(2) = 1 rad s^(-1)`
Time taken by the man to complete one revolution is `T = (2pi)/(omega_(r)) = (2pi)(1) = 2 pi s`
Discussion
15207.

If P represents radiation pressure, C speed of light, and Q radiation energy striking unit area per second and x,y,z are non zero integers, then P^(x) Q^(y) C^(z) is dimensionless. The values of x,y and z are respectively.

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`1,1,-1`
`1,-1,1`
`-1,1,1`
`1,1,1`

ANSWER :B
Discussion
15208.

If year is taken as the unit of time and velocity of light as the unit of velocity what is the unit of length ?

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SOLUTION :1 light YEAR = velocity of light`xx` number of SECOND in one year `=3XX10^(8)MS^(-1)xx(365xx24xx60xx60)`
`s=9.46xx10^(15)m`.
Discussion
15209.

A body is moving along a circular path . How much work is done by the entripetal force ?

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ANSWER :`W=FScos90^(@)=0`
Discussion
15210.

A block of mass m is pl aced on a smooth wedge of inclination theta. The whole system is accelerated horizontally so that the block does not slip on the wedge. Find the(i) Acceleration of the wedge(ii) Force to be applied on the wedge(iii) Force exerted by the wedge on the block.

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Solution :(I) For observer on the ground :

`R sin theta = ma`
`R COS theta = MG rArr a= g tan theta`
(a) Force to be applied on the WEDGE
`F=(M+m)a=(M+m)g tan theta`
(b) Force EXERTED by the wedge on the block
`= R = (mg)/(cos theta)` or `R=mg sec theta`
II. For the observer on the wedge :
Pseudo force = ma towards left.
`because` the block appears to be at rest to him
`ma cos theta = mg sin theta`
`a=g tan theta`
`R=mg cos theta + ma sin theta`
`= mg cos theta+mg tan theta sin theta`
`= mg cos theta + mg.(sin^(2)theta)/(cos theta)`

`= mg.((cos^(2)theta+sin^(2)theta))/(cos theta)`
`R=(mg)/(cos theta)=mg sec theta`
Discussion
15211.

A liquid flows through a pipe of10^(-3) m radius and 0.1m in length under a pressure of 10 Nm^(-2) Calculate the speed of the liquid coming out of the pipe. The coefficient of viscosity of the liquid of 1.25xx10^(-3) decapoise.

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`1 MS^(-1)`
`2 ms^(-1)`
`3 ms^(-1)`
` 4 ms^(-1)`

Answer :A
Discussion
15212.

A liquid is flowing with an average velocity of Vcm/s in a tube of radius rcm. Write the equation for the rate of flow

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Solution :Rate of FLOW of LIQUID = `aupsilon = PI r^2 UPSILON ."" cm^3//"sec".`
Discussion
15213.

Roofs of huts are blown up during stormy days. Why?

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Solution :During wind storm, the speed of air above the ROOF is more than that below the roof. So the pressure is low above the roof in ACCORDANCE with the Bernoulli.s equation. This pressure DIFFERENCE makes the roof to blown up.
Discussion
15214.

After perfectly inelastic collision between two identical particles moving with same speed in different directions, the spee of the combined particle becomes half the initial speed of either particle. The angle between the velocities of the two before collision is

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Solution :In perfectly inelastic collision between two PARTICLES, linear MOMENTUM is conserved. Let `theta` be the angle between the velocities of the particles before collision. Then `p^(2) = p_(1)^(2) + p_(2)^(2) costheta`
or `(2m(v)/(2))^(2)=(MV)^(2)+(mv)^(2)+2(mv)(mv)costheta`
or `1 = 1 + 1 + 2 COS theta or cos theta = - (1)/(2) , (or) thet a = 120^(@)`
Discussion
15215.

Can a body subjected to a uniform acceleration always move in a straight line.

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Solution : It will be a straight LINE in ONE dimensional MOTION but not applicable for two dimensional motion because the projectile has a parabolic path but it has a uniform ACCELERATION.
Discussion
15216.

A table is rotating along a horizontal plane about the axis passing through its centre. A thread is passed through a hole made at the centre of the table and two bodies of equal mass are tied at a distance of 20 cm thread. The body on the table is at a distance of 20 cm from the centre of the table. For what rpm of the table. will the hanging body be at rest?

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ANSWER :66.8 RPM
Discussion
15217.

(A) : Liquid molecules have greater potential energy at the melting point. (R ) : Intermolecular spacing between molecules increase at melting point.

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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 but (R ) is false
Both (A) and (R ) are false

Answer :C
Discussion
15218.

A Particle is attached to the lower end of a uniform rod which is hinged at its other end as shown in the figure. The minimum speed given to the particle so that the rod performs circular motion in a verticle plane will be [length of the rod is l, consider masses of bopth rod and particle to be same]

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`SQRT(5gl)`
`sqrt(4GL)`
`sqrt(4.5gl)`
`sqrt(2.25gl)`

ANSWER :C
Discussion
15219.

Passage - I : A uniform rod of mass m pivoted smoothly at O is resting against a vertical wall of a moving wedge. The acceleration of the wedge is a, then The reaction offered by pivot on the rod along horizontal direction is

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`m(g+a)SINTHETA`
`m/2(gcottheta+a)`
`m(gsintheta+acostheta)`
`m(gsintheta-acostheta)`

ANSWER :B
Discussion
15220.

Passage - I : A uniform rod of mass m pivoted smoothly at O is resting against a vertical wall of a moving wedge. The acceleration of the wedge is a, then The normal reaction offered by the vertical wall on the rod is

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`m(g-a)SINTHETA`
`m/2(gcottheta-a)`
`m(gsintheta-acostheta)`
`m(gsintheta+acostheta)`

ANSWER :B
Discussion
15221.

Passage - I : A uniform rod of mass m pivoted smoothly at O is resting against a vertical wall of a moving wedge. The acceleration of the wedge is a, then The minimum acceleration 'a' at which the rod looses contact with the vertical wall is

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`gcostheta`
G
`gcottheta`
`g TANTHETA`

ANSWER :C
Discussion
15222.

A bodycools from 80^(@)C to 60^(@)C in 2 mintues .In how time it cools from 60^(@) to 40^(@)C ? The temperatureof the surroudingis 10^(@)C

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Solution :I CASE :
Meantemperatureof the body `= (80 + 60)/(2) = 70^(@)C`. Meanecess temperature `= 70 -10=60^(@)C`
`(d theta)/(dt) = K ( theta - theta_(0))rArr (20)/(2) = K(60)…….(1)`
II case :
Meantemperature of body`(60+40)/(2) =50^(@)C` Meanexcess temperature ` = (50-10) = 40^(@)C`.
Let.t. minutesbe thetime periodto COOL down `60^(@)C` to `40^(@)C`.
The`(20)/(t) = k (40).........(2)`
`(1)+ (2) , (t)/(2) = (60)/(40) i.e., t = 3 ` minutes.
Discussion
15223.

A shell projected from a level ground has a range R, if it did not explode. At the highest point, the shell explodes into two fragments having masses in the ratio 1:2, with each fragment moving horizontally immediately after the explosion. If the lighter fragment falls at a distance R/2 from the point of projection, behind the point of projection, what is the distance at which the other fragment falls from the point of projection ?

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ANSWER :7R/4
Discussion
15224.

Two blocks A and B of mass 10 kg and 20 kg respectively are placed as shown in figure .Coefficient of friction between all the surface is 0.2 (g=10m//s^(2)) Then,

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Tension in the STRING is 306 N
Tension in the sr=tring is 132 N
accelerration of BLOCK B is `2.6 m//s^(2)`
ASSELERATION of block B is `4.7 m//s^(2)`

Answer :a,d
Discussion
15225.

The surface energy of small liquid drop is E_s If it splits into 1000 smaller drops. Then total surface energy will be

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`10 E_s `
`E_s //2`
`E_s`
`1000E_s`

ANSWER :A
Discussion
15226.

Three bodies of black, gray and white colour are able to sustain upto 2800^(@)C temperature. If these are kept in a frame then temperature of each becomes 2000^(@)C, then which body will be unlighted maximum ?

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WHITE body
BLACK body
all three bodies
gray body

SOLUTION :Black body ABSORBS and emits maximum.
Discussion
15227.

A body is suspended from a spring a balance kept in a satellite. The reading of the balance is W_(1) when the satellite goes in an orbit of radius R and is W_(2) when it goes in an orbit of radius 2R.

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`W_(1) = W_(2)`
`W_(1) LT W_(2)`
`W_(1) gt W_(2)`
`W_(1) != W_(2)`

ANSWER :A
Discussion
15228.

Find the pressure in a bubble of radius 0.2 cm formed at the depth of 5cm from the free surface of water . The surface tension of water is 70 dyne cm^(-1) and its density is 1gcm^(-3) . Atmospheric pressure is 10^(6) dyne cm^(-2) .The gravitational acceleration is 980cms^(-2).

Answer»

<P>

SOLUTION :`{:(h=5cm,R=0.2cm),(T=70 "dyne " cm ^(-1),rho=1gcm^(-3)),(P=10^(6) "dyne" cm^(-2),g=980 cm-s^(-2)):}`
If the pressures inside and outside of the AIR bubble FORMED in water are `P_(i)andP_(o)`respectively ,
`P_(i)-P_(o)=(2T)/(R)`
(A bubble in water has one free surface only.)
`thereforeP_(i)=P_(o)+(2T)/(R)` ...(1)
But `P_(o)=` atmmospheric pressure + pressure due to watarcolumn of height h.
`thereforeP_(o)=P+hrhog` ...(2)
From equations (1)and (2),
`P_(i)=P+hrhog+(2T)/(R)`
`=10^(6)+(5xx1xx980)+(2xx70)/(0.2)`
`=10^(6)+4900+700`
`P_(i)=1.0056xx10^(6)"dyne " cm^(-2)
Discussion
15229.

A particle moves in the x-y plane with the velocity vec v = ahat i - bt hat j. At the instant t =asqrt(3)//bthe magnitude of tangential, normal and total acceleration are ….. &…….

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`(sqrt(3)B)/(2), (b)/(2), b`
`(b)/(2), (b)/(2), sqrt(2)b`
`b,b, sqrt(2)b`
`sqrt(3)b, b, 2B`

Answer :A
Discussion
15230.

Figureshows an aluminium wire of length 60 cm joined to a steel wire of length 80 cm and stretched between two fixed supports. The tension produced is 40 N. The cross-sectional area of the steel wire is .1.0 mm^2and that of the aluminium wire is 3.0 mm 2. What could be the minimum frequency of a tuning fork which can produce standing waves in the system with the joint as a node ? The density of aluminium is 2.6 g cm^-3and that of steel is7.8gcm^-3

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Solution :`P_s=7.8(gm)/cm^3`
`P_A=2.6(gm)/cm^3`
`m_s=P_sA_s`
`=7.8xx10^-2(gm)/cm`
(m=mass per unit length)
`7.8xx10^-3kg/m`
`m_A=P_A`
`=21.6xx10^-2xx3(gm)/cm`
`=7.8xx10^-2(gm)/cm`
`=7.8xx10^-3(kg)/m`
A node is always placed in the joint. Since ALUMINIUM and steel rod has same mass per unit lenght, velocity of wave in both of them is same
`rarr v=sqrt((T/m))`
`=sqrt({40/((7.8xx10^-3))})`
`=sqrt((((4xx10^4)))/7.8)`
`=71.6m/s`
For minimum frequency there would be MAXIMUM WAVELENGTH. for maximu wavelength minimum no. of loops are to be produced.
`:.` Maximum distance of a LOOP =20cm
`rarr` wavelength =`lamda=2xx20`
`=40cm=0.4m`=180Hz`
Discussion
15231.

The position vector of a particle isvecr = (a cos omega t)hati + (a sin omegat)hatj.the velocity of the particle is

Answer»

directed TOWARDS the origin
direction away from the origin
parallel to the position vector
PERPENDICULAR to the position vector

ANSWER :D
Discussion
15232.

A bullet of mass 20 g strikes a pendulum of mass 5kg. The centre of mass of pendulum rises a vertical distance of 10 cm. If the bullet gets embedded into the pendulum, calculate its initial speed.

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Solution : Given DATA: `m_1= 20 g = 20xx10^(-3) kg, m_2 = 5 kg, s = 10 xx 10 ^(-2)` m.
Let the SPEED of the bullet be v. The common velocity of bullet and pendulum bob is V. According to law of conservation of linear momentum.
`v = ( m_1 v ) /((m_1 +m_2))=( 20 xx 10^(-3)v )/( 5 +20 xx 10^(-3)) = ( 0.02)/( 5.02 ) v= 0.004v `
The bob with bullet go up with a deceleration of `g=9.8.ms^(-2)`. Bob and bullet come to me at a height of `10 xx10^(-2) m.`
from III rd equation of MOTION .
` V^2= u^2 + 2as` ,
here ` v=u + at `
` v^2- 2gs =0 ""s=((u+v) /(2))t `
` ( 0.004 v)^2= 2 xx 9.8xx 10 xx 10^(-2) ""v^2= u^2+ 2as `
`v^2 = ( 2xx 9.8xx 10xx 10^(-2))/(( 0.004 )^2)"" s = ut+ 1/2at^2 `
` v= 350ms^(-1)`
Discussion
15233.

An ideal gas expands in such a way that its pressure and volume satisfy with the condition PV^(2) = constant. During this process, the gas is

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HEATED
Cooled
First heated then cooled
Neither '1' nor '2'

Answer :B
Discussion
15234.

What is needed to produce pure rotation ?

Answer»


ANSWER :TORQUE
Discussion
15235.

The work done by a force vecF =(-6x^(2)hati) Nin displacing a particle from x = 4mtox = -2mis

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`-240 J`
`360 J`
`420 J`
`100 J`

Answer :B
Discussion
15236.

A body is projected vertically up and h is the maximum height raised bythe body compare column-I with column -II in which the height from the ground are given

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ANSWER :A::B::C::D
Discussion
15237.

On observing light from three different stars P, Q and R, it was found that intensity of voilet colour is maximum in the spectrum of P, the intensity of green colour is maximum in the Then spectrum of R and the intensity of red colour is maximum in the spectrum of Q. If T_(P),T_(Q)andT_(R) are the respective absolute temperatures of P, Q and R, then it can be concluded from the above observations that:

Answer»

`T_(P)ltT_(R)ltT_(Q)`
`T_(P)ltT_(Q)ltT_(R)`
`T_(P)gtT_(Q)gtT_(R)`
`T_(P)gtT_(R)gtT_(Q)`

ANSWER :D
Discussion
15238.

To measure the temperature of a furnace, a platinum ball of mass 80 g is kept inside it. When the ball attains the temperature of the furnace, it is dropped into water at 15^(@)C in a container. The amount of water in it and its water equivalent together is 400 g. Find the temperature of the furnace. Specific heat of platinum ="0.0365 cal . g"^(-1).^(@)C.

Answer»

SOLUTION :`704.9^(@)C`
Discussion
15239.

For a body to escape from earth, angle from horizontal at which it should be fired is

Answer»

`45^(@)` only
`0^(@)` only
`90^(@)` only
any ANGLE

ANSWER :D
Discussion
15240.

Consider the system shown in figure. The pulley and the string are light and all the surface are frictionless. The tension in the string is ("Take g" = 10 m//s^(2))

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0N
1N
2N
5N

Solution :For the BLOCK on the SURFACE `T=m_(1)a`
For the HANGING block `m_(2)g-T=m_(2)a`
Discussion
15241.

A car moving over a straight path, covers a distance d with constant speed of 40 km/hr and then the same distance with speed 60km/hr. The average speed of car is .....

Answer»

50 km/hr
48 km/hr
52 km/hr
None of the given

Solution :To cover DISTANCE d with constant speed 40 km /hr,
Time `t_(1) = (d)/(40) hr.`
And to cover distance d with constant speed 60 km/hr required time `t _(2) = (d)/(60) hr.`
`therefore ` TOTAL time required to cover distance 2D is
`t = t _(1)+t _(2)`
`therefore t = (60 d )/(40) + (d)/(60)`
`therefore t = (60 d + 40 d )/(2400)`
`therefore t = (d)/(24)`
Now average velocity `bar V = (2d)/(t) = 2d xx (24)/(d) = 48 Km//hr`
Discussion
15242.

The work done against friction in moving a body from A to B is 200J. If the body is again brought back to A along the same path, the total work done for the trip against friction is

Answer»

0J
400J
200J
800J

Answer :B
Discussion
15243.

A uniform of mass m and length L is tied to a vertical shaft. It rotates in horizontal plane about the vertical axis at angular velocity omega. How much horizontal force does the shaft exert on the rod?

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Solution :Let `T` be the force applied horizontally on the rod. The acceleration of the centre of MASS is `omega^(2)(L/2)` as it moving in a circle of radius `L/2`.

Using `vecF_(ext)=Mveca_(CM)impliesT=Momega^(2)(L/2)`
THUS, the horizontal force exerted is `1/2Momega^(2)L`.
Discussion
15244.

A room has dimensions 3m xx 4m xx 5m. A fly starting at one corner ends up at the diametrically opposite corner. The magnitude of the displacement of the fly is

Answer»

12m
60 m
`2sqrt(5) m`
`5sqrt(2)m`

ANSWER :D
Discussion
15245.

The radius of gyration of a body about an axis at a distance of 4 cm from its centre of mass is 5 cm. The radius of gyration about a parallel axis through centre of mass is. . . . cm

Answer»


ANSWER :3
Discussion
15246.

The mass of a uniform ladder of length 5 m is 20 kg. A person of mass 60 kg stands on the ladder at a height of 2 m from the bottom.The position of centre of mass of the ladder and man from the bottom nearly is

Answer»

1m
2.5 m
3.5 m
2.125 m

ANSWER :D
Discussion
15247.

A person of mass 50 kg carrying a load of 20 kg walks up a staircase. If width and height of each step are 0.25m and 0.2 m respectively, the work done by the man in walking up 20 steps is, (g=10 ms^(-2))

Answer»

1400 J
800 J
2000 J
2800 J

Answer :D
Discussion
15248.

In the previous problem, if theta is the angle sbtended by the Sun at the planet, then temperature at which planet radiates energy is proportional to

Answer»

`THETA`
`theta^(2)`
`theta^(1//2)`
`theta^(-1//2)`

ANSWER :C
Discussion
15249.

Beats occur because of

Answer»

INTERFERENCE
REFLECTION
REFRACTION
Doppler effect

SOLUTION :interference
Discussion
15250.

A wheel of radius 'r' rolls without slipping with a speed 'v' on a horizontal road. When it is at point 'A' on the road , a small lump of mud separtes from wheen at its heighest point B and deops at point 'C' on the road. The distance AC will be

Answer»

`vsqrt(r//g)`
`2vsqrt(r//g)`
`4vsqrt(r//g)`
`vsqrt(3r//g)`

ANSWER :C
Discussion