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39251.

Dynamic lift due to spinning is

Answer»

MAGNUS effect
Doppler effect
Pascal effect
Torricelli's effect

Answer :A
Discussion
39252.

What is the meaning of angular diameter of the Moon ?

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Solution :From a point on the surface of the Earth, angle SUBTENDED at END point of DIAMETER of the Earth is called ANGULAR diameter of the Moon.
Discussion
39253.

A steel drill is making 180 revolutions per minute, under a constant torque of 5 N-m. If it drills a hole in 7s. in a steel block of mass 600 gm, rise in temperature of the block is (s = 0.1 cal//gml//^(@)C)

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`2.6^@C`
`1.3^@C`
`5.2^@C`
`3^@C`

ANSWER :A
Discussion
39254.

A wooden block of mass M resting on a rough horizontal surface is pulled with a force T at an angle q to the horizontal. If m is coefficient of kinetic friction between the block and the surface, the acceleration of the block is

Answer»

`(T COS THETA)/(M) - MUG`
`T/M (cos theta - sin theta) -mug `
`T/M (cos theta - mu sin theta) -mu G `
`(mu T cos theta)/(M)- mu g `

ANSWER :C
Discussion
39255.

A machine delivers power to body which is proportional to instataneous velocity V of the body. If the body starts with a velocity which is almost negligible, then the distance covered by the body is directly proportional to V. Then find the value of n.

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Kinetic energy of the bloci is `(mv^(2))/50`
POWER delivered by th external agent is `(mav)/25`
TOTAL power delivered by the string is zero
Work DONE by the string on th block is NON zero

Answer :A::B::C::D
Discussion
39256.

Weight of a body on the surfaces of two planets is the same. If their densities are d_1 and d_2, then the ratio of their radius

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`d_1/d_2`
`d_2/d_1`
`d_1^2/d_2^2`
`d_2^2/d_1^2`

ANSWER :B
Discussion
39257.

The ratio of one nanometer to one micron is……….

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`10^(-3)`
`10^(3)`
`10^(-9)`
`10^(-6)`

SOLUTION :`10^(-3)`
Discussion
39258.

The terminal velocity of a small ball falling in a viscous liquid depends upon i) its mass m ii) its radius r iii) the coefficient of viscosity of the liquid and iv) acceleration due to gravity. Which of the following relations is dimensionally true for the terminal velocity V =

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`(KMG)/(ETAR)`
`(Kmgr)/(ETA)`
`(Kmgeta)/(R )`
`(Kreta)/(mg)`

Answer :A
Discussion
39259.

The radius of a planet is R_(1) and a satellite revolve round of in a circule of radius R_(2). The time period of revolution is T. The acceleration due to the gravitation of the planet at its surface is

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Solution :Orbital velocity of the SATELLITE `V_(0) = SQRT(g^(1)R_(2))`
Where `g^(1) = g((R_(1))/(R_(2)))^(2)`
and angular velocity `omega = (V_(0))/(R_(2)) = sqrt((g^(1))/(R_(2))) = sqrt((g(R_(1))^(2))/(R_(2)^(3)))`
But time PERIOD of REVOLUTION `T = (2pi)/(omega) = 2pisqrt((R_(2)^(3))/(gR_(1)^(2)))`
(or) `T^(2) = 4pi^(2)(R_(2)^(3))/(gR_(1)^(2))` (or) `g = (4pi^(2))/(T^(2)) (R_(2)^(3))/(R_(1)^(2))`
Discussion
39260.

A sphere rolls without slipping on an incline of inclination theta. The minimum coefficient of static friction to support pure rolling is to be

Answer»

`2/3 TANTHETA`
`2/5tantheta`
`2/7tantheta`
`3/7tantheta`

ANSWER :C
Discussion
39261.

A wooden cylinder of length L is partly submerged in a liquid of specific gravity rho_() with n^(th)(n lt 1)part of it inside the liquid. Another immiscible liquid of ensity rho_(2). is poured to completely submerge the cylinder. Density of cylinder rho is the square root of the product of densities of two liquids Express the density of upper liquid in terms of density of cylinder

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`RHO//N`
`rhon`
`(n)/((n+1)rho)`
None

Answer :B
Discussion
39262.

If unit of a physical quantity is doubled, what will be its new value.

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SOLUTION :We KNOW that,
`n_(1)u_(1)=n_(2)u_(2)`
`:. n_(2)=(n_(1)u_(1))/(u_(2))`
`=(n_(1)u_(1))/(2u_(1)) "" [ :.u_(1)=2u_(1)]`
`=(n_(1))/(2)`
Discussion
39263.

A wooden cylinder of length L is partly submerged in a liquid of specific gravity rho_() with n^(th)(n lt 1)part of it inside the liquid. Another immiscible liquid of ensity rho_(2). is poured to completely submerge the cylinder. Density of cylinder rho is the square root of the product of densities of two liquids Calculate the fraction of the cylinder (less than one) submerged in the lower liquid after the upper liquid is poured in the vessel

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`(N)/((n+1))`
`((n-1))/((n+1))`
`(n(n-1))/((n+1))`
`(n+1)/((n-1))`

ANSWER :A
Discussion
39264.

A geo-stationary satellite orbits around the earth in a circular orbit which is at a height of 36000 km from the surface of earth. Then the period of spy satellite orbiting a few hundred km above the earth.s surface (R_("earth") = 6400 km ) will aproximately be

Answer»

`(1)/(2)` HR
`1(1)/(2)`hr
2 hr
4 hr

ANSWER :B
Discussion
39265.

A box of mass 50kg at rest is pulled up on an inclined plane 12m long and 2m high by a constant force of 100N. When it reaches the top of the inclined plane if its velocity is 2ms^(-1) , the work done against friction in Joules is (g = 10ms^(-2))

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50
100
150
200

Answer :B
Discussion
39266.

A man is waiting at a bus stop by holding a box on his head. Work done by him is

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ZERO
POSITIVE
NEGATIVE
INFINITY

ANSWER :A
Discussion
39267.

Force means change in momentum.

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ANSWER :FORCE
Discussion
39268.

A rod of original length L_(1) changes its temperature from T_(2) to T_(2). If alpha is temperature coefficient of linear expansion and L_(2) is length of temperature T_(2). Choose the correct options(s).

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`L_(2) = L_(1) [1+ALPHA (T_(2) - T_(1) ) ]`, where `T_(2) - T_(1) and alpha` are not large
`L_(2) = L_(1) e^( alpha (T_(2) -T_(1) ) )`, where `alpha` is CONSTANT
`L_(2) = L_(1) e^( int_(T_1)^(T_2)alpha d T) `, where `alpha` is a function of temperature
Formula in option (a) is valid in all conditions

Answer :A::B::C
Discussion
39269.

A body of mass 5 xx 10^(-3) kg is launched upon a rough inclined plane making an angle of 30° with the horizontal. Find the coefficient of friction between the body and the plane if the time of ascent is half of the time of descent.

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0.346
0.436
0.463
0.364

Solution :0.346
Discussion
39270.

A pendulum is released from point A as shown in figure. At some instant net force one the bob is making an angle theta with the string. Then Match the following {:(,"Table-1",,"Table-2"),("(A)","For "theta=30^(@),"(P)","Particle may be moving along BA"),("(B)","For "theta=120^(@),"(Q)","Particle may be moving along CB"),("(C)","For "theta=90^(@),"(R)","Particle may be at A"),("(D)","For "theta=0^(@),"(S)","Particle is at B"),(,,"(T)","None"):}

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ANSWER :(A)QP,(B)T,(C)R,(D)S
Discussion
39271.

Select the correct statement(s) for a given vernier callipers, when the two jaws of the apparatus are in contact.

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If the ZERO of the vernier CALLIPERS, does not coincide with the zero of the MAIN scale, then the vernier callipers is said to possess zero error.
Zero error and zero correction are same numerically, but opposite in SIGN.
Zero error is positive, when zero of the vernier scale lies to the left of the zero of the main scale.
Zero error is negative, when zero of the vernier scale lies to the left of the zero of the main scale.

ANSWER :A::B::D
Discussion
39272.

A body is pushed up on a rough inclined plane making an angle 30^@to the horizontal. If its time of ascent on the plane is half the time of its descent, find coefficient of friction between the body and the inclined plane.

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`(SQRT3)/(5)`
`SQRT(3//5)`
`(3)/(SQRT5)`
`3//5`

Answer :A
Discussion
39273.

A gas under constant pressure of 4.5 xx 10^(5) Pa when subjected to 800 kJ of heat, changes the volume from 0.5 m^(3) to 2.0 m^(3). The change in internal energy of the gas is

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`6.75 xx 10^(5) J`
`5.25 xx 10^(5) J`
`3.25 xx 10^(5) j`
`1.25 xx 10^(5) J`

SOLUTION :Here, `P = 4.5 xx 10^(5) Pa, DELTA Q = 800 kJ, V_(1) = 0.5 m^(3), V_(2) = 2m^(3)`
The work done by the gas is
`Delta W = P Delta V = P(V_(2) - V_(1)) = 4.5 xx 10^(5) (2-0.5) = 6.75 xx 10^(5) J`
From the first law of thermodynamics, the change in internal energy is
`Delta U = Delta Q - Delta W = 800 xx 10^(3) - 6.75 xx 10^(5) = 1.25 xx 10^(5)J`
Discussion
39274.

A lawn roller has a mass of 80 kg and radius of gyration of k_(G)=0.175 m. The coefficient of friction between the roller the ground is mu_(A)=0.12. If it is pushed forward with a force of 200 N when the handle makes an angle of 45^(@), determine the acceleration of the roller. The radius of the roller is 200 mm.

Answer»

SOLUTION :1.18 `m//s^(2)`
Discussion
39275.

A disc rolls on ground slipping Velocity of centre of mass is v. If the point P lies on the circumference of disc at angle theta as shown in the figure then the speed v_(p) at the point P, when theta=90^(@),120^(@),180^(@)

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`sqrt(2)V,sqrt(3)v,2V`
`v,2v,sqrt(3)v`
`2v,sqrt(3)v,sqrt(2)v`
`sqrt(3)v,v,sqrt(5)v`

ANSWER :A
Discussion
39276.

S.I. unit of Electric intensity is

Answer»

`Hm^(-1)`
`NM^(-1)`
`Vm^(-1)`
`NC`

Answer :C
Discussion
39277.

A dimensionally wrong equation in which principle of homogenity of dimensions is violated must be wrong. Can you say that, a dimensionally correct question should always be right?

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SOLUTION :NEED not be.
Discussion
39278.

A force of 20 N acts on a body for 2 micro second. Calculate the impulse. If the mass of the body is 1 g, calculate the change of velocity.

Answer»


Answer :`4 xx 10^(-5) Ns; 4 xx 10^(-1) ms^(-2)`
Discussion
39279.

An electric heater supplies heat to a system at a rate of 100J/s. If system performs work at the rate of 75 Joule per second, the rate of increase of internal energy is

Answer»

`50 "JS"^(-1)`
`25 "JS"^(-1)`
`12.5 "JS"^(-1)`
`175 "JS"^(-1)`

ANSWER :B
Discussion
39280.

The surface tension of a liquid is 0.04 Nm^(-1) its value in C.G.S system is

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`4"DYNE"cm^(-1)`
`40 "dyne"cm^(-1)`
`400"dyne"cm^(-1)`
`0.044"dyne"cm^(-1)`

Answer :B
Discussion
39281.

A sphere of mass 3 kg moving with a velocity of 2 ms- collides head-on with another sphere of mass 4 kg moving in opposite direction with a velocity of 1 ms. After collision if they stick together and move with a velocity v, find (i) the total initial momentum and (ii) the value of velocity v.

Answer»


ANSWER :`2 KGMS ^(-1);2//7 MS ^(-1)`
Discussion
39282.

Particles of masses m, 2m , 3m …..nm grams are placed on the same line at distances l,2l,3l,…nl cm from a fixed point. The distance of centre of mass of the particles from the fixed point in centimetre is

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`((2n+1)l)/(3)`
`(l)/(n+1)`
`(n(n^(2)+1)l)/(2)`
`(2L)/(n(n^(2)+1))`

ANSWER :A
Discussion
39283.

Time for 20 oscillations of a pendulum is measured as t_(1)= 39.6 s, t_(2) = 39.9 s and t_(3) = 39.5 s. What is the precision in the measurements ? What is the accuracy of the measurement ?

Answer»

Solution :GIVEN `t_(1)=39.6s t_(2)=39.9 s and t_(3)=39.5 s`
Least count of measuring instrument = 0.1 s (As measurements have only one significant FIGURE after decimal POINT)
Precision in the measurement = Least count of the instrument = 0.1 s

Average time for 20 oscillations is given by
`t=(t_(1)+t_(2)+t_(3))/(3)`
`=(329.6+39.9+39.5)/(3)=39.7 s`
Absolute errors in the measurements
`Deltat_(1)=t-t_(2)=39.7-39.6-39.6=0.1s `
`Deltat_(2)=t-t_(2)=39.7-39.39.5 = 0.2 s`
Average absolute error `=(|Deltat_(1)|+|Deltat_(2)|+|Deltat_(3)|)/(3)`
`=(0.1+0.2+0.2)/(3)`
`=(0.5)/(3)=0.17~~0.2`
(rounding off upto one decimal place)
`:.` Accuracy of measurement `=+-0.2s`
Discussion
39284.

The casting of a rocket in flight burns up due to friction. At whose expense the heat energy required for burning obtained ? The rocket or the atmosphere ,or both /

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SOLUTION :The only sourceof ENERGY is the kinetic energy of the ROCKET.The atmosphere does not contribute any energy . Its role is only to provide friction to convert the kinetic energyof the rocket into heat energy .
Discussion
39285.

A rocket accelerates straight up by ejecting gas downwards . In small time interval Deltat , it ejects a gas of mass Deltamat a relative speed u . Calculate kE of the entire system at t+Deltat and tand show that the device that ejects gas does work = (1/2) Delta"mu"^(2) in this time interval (negative gravity ).

Answer»

Solution :M =mass of rocket at t
v = velocity of rocket at t
`Deltam ` = mass of ejected GAS in `Deltat`
u = relative SPEED of ejected gas
CONSIDER at time t + `Deltat`
`(KE)_(t) +(KE)_(Deltat) = `KE of rocket + Ke of `= 1/2 (M-Deltam) (v+Deltav)^(2) +1/2Deltam (v-u)^(2)`
`1/2 Mv^(2) +Mv Deltav -Deltamvu +1/2 Delta"mu"^(2)`
`(KE)_(t) =` KE of the rocket at time t = `1/2 Mv^(2)`
`DeltaK - (KE)_(t) +(KE)_(Deltat) -(KE)_(t)`
`(MDeltav - Delta"mu") v +1/2 Delta" mu"^(2)`
Hene , `M (dv)/(dt) =(dc)/(dt) |u|`
` :. M Deltav = Delta"mu"`
`:. DeltaK = 1/2 Delta"mu"^(2)`
Now , by work - ENERGY theorem ,
`DeltaK = Delta W `
` :. Delta W = 1/2 Delta"mu"^(2)`
Discussion
39286.

40V A constant force vecF is acting on a body of mass m makes it move with constant velocity vecv as shown in the M figure. The power P exerted is 

Answer»

`F cos theta V`
`(F cos theta)/(MG)`
`(F mg cos theta)/(v)`
`(mg sin theta)/F`

Solution :
POWER = (COMPONENT of FORCE in the direction of velocity) `|vecv|`
`= F cos theta v`.
Discussion
39287.

A tuning fork A, marked 512 Hz, produces 5 beats per second, where sounded with another unmarked tuning fork B. If B is loaded with wax the number of beats is again 5 per second. What is the frequency of the tuning fork B when not loaded ?

Answer»

Solution :Here `F _(A) = 512 Hz implies f _(B) = 517 Hz or 507 Hz`
`(because ` Initial beat FREQUENCY between A and B is 5 Hz)
Now, when some wax is applied on B, its frequency will decrease and will assume that value `f._(B)` such that,
`(i) f ._(B) lt f _(B.) (ii) f _(A)~f._(B) = 5 Hz`
`(because` Final beat frequency beetween A and B is also 5 Hz)
`implies f_(B)=517 Hz and f._(B) = 507 Hz`
because `507 lt 517 impliesf._(B) lt f _(B)`
`and f _(B) -f _(A) =517 -512 =5 Hz`
and `f _(A) -f _(B) = 512 -507 = 5 Hz`
Note: Here `f _(B) ne 507Hz` because otherwise after applying wax on `B, f ._(B) lt 507Hz` and then `f _(A) -f._(B) gt 5 Hz` which would be wrong as per the data given in the statement.
Discussion
39288.

(A) : The order of accuracy of a measurement depends on the least count of the instrument. (R) : The smaller the least count, more number of significant figures are in the measured value.

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

Answer :B
Discussion
39289.

Calculate the specific heat capacities at constant volume and constant pressure for CO (2),Given gamma = 1.30, R =8.33 J mol ^(-1) K ^(-1).

Answer»

<P>

Solution :`C _(p) = C _(v) =R, gamma = ( C _(p))/( C _(v))`
`gamma C _(gamma) - C _(v) = R`
`C _(v) =(R )/(gamma - 1) = (8.33)/(1.3 -1) = (8.3)/(3) = 27.77 J mol ^(-1) K ^(-1)`
`C _(p) = C _(v) + R = 27.77 + 8. 33 = 36.1 J mol ^(-1) K ^(-1)`
Discussion
39290.

What is the condition for the vectors 2i + 3j - 4k and 3i - aj + bk to be parallel ?

Answer»

`a= -9//2, B = - 6`
`a= -6, b= - 9//2`
`a=4, b= 5`
`a= 8, b= 2`

ANSWER :A
Discussion
39291.

A ball is thrown up with a certain given velocity at angle theta to the horizontal. The kinetic energy, KE of the ball varies with horizontal displacement x as:

Answer»




ANSWER :C
Discussion
39292.

The most interesting and exciting part of applied physics with great ingenuity and persistence of effort in

Answer»

Admiring the SCIENTIST who established the PHYSICAL laws
Application of physical laws to make devices USEFUL
Exploitation of physical laws to make useful devices
Practicing theories established

ANSWER :C
Discussion
39293.

A ball is thrown downwards with a speed of 20m//s from top of a building 150m high and simultaneously another ball is thrown vertically upwards with a speed of 30m//s from the foot of the building . Find the time when both the balls will meet . (g=10m//s^(2))

Answer»

Solution :(i)`S_(1)=20t+5t^(2)+S_(2)=30t-5t^(2)`
`150=50t`
`t=3s`.

(II)RELATIVE acceleration of both is zero since both have acceleration in DOWNWARD direction.
`veca_(AB)=veca_(A)-veca_(B)=g-g=0`
`vecV_(BA)=30-(-20)=50`
`S_(BA)=V_(BA)xxt`
`t=(S_(BA))/(V_(BA))=(150)/(50)=3s`
Discussion
39294.

When a soapwater bubble is given a positive charge it expands. If it is given a negative charge, then it

Answer»

expands
contracts
remains same
does not HOLD NEGATIVE charge

Answer :A
Discussion
39295.

A particle moves along x-axis from x=0 "to"x=8 under the influence of a force given by F=3x^(2)-4x+5. Find thework done in the process.

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Solution :WORK donein the moving a PARTICLE from x=0 to x=8 wil be
`W=underset(0)overset(8)INT(3x^(2)-4x+5)dx=[(3x^(3))/3-(4x_(2))/2+5x]_(0)^(8)`
`W=[3""(8)^(3)/3-4".(8^(2)/2)+40]=[512-128+40]=424J`
Discussion
39296.

Find the gravitational attraction between two H-atoms of a hydrogen molecule. Given that mass of atom = 1.67xx10^(-27)kg and distance between two H-atoms = 1A^@

Answer»

zero
100N
600N
3600N

Answer :B
Discussion
39297.

Explain when small pieces of ice pressed together, form a single block.

Answer»

Solution :This occurs due to the regelations. On increasing the pressure,the melting POINT of ice is LOWERED,and water is formed at the surface of contact. On removal of the pressure water freezes and THUS the pieces are JOINED TOGETHER.
Discussion
39298.

A 50 kg man is standing at one end of a 25 m long boat. He starts runing towards the other end. On reaching the other end his velocity is 2ms^(-1). If the mass of the boat is 200 kg, final velocity of the boat is (in ms^(-1))

Answer»

`(2)/(5)`
`(2)/(3)`
`(8)/(5)`
`(8)/(3)`

ANSWER :A
Discussion
39299.

(A) Optical instruments are assembled using lenses with very wide apertures. ( R) Wide aperture of lenses enables them to collect more light and produce bright images.

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
Both A and R is false

Answer :A
Discussion
39300.

Calculate the change in weight of a body of mass 5 kg when it is taken from equator to pole of earth. The time period of rotation of earth around its own axis is 24 hours.

Answer»


ANSWER :0.17 N
Discussion