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15551.

An Aluminium and Copper wire of same cross sectional area but having lengths in the ratio 2: 3 are joined end to end. This composite wire is hung from a rigid support and a load is suspended from the free end. If the increase in length of the composite wire is 2.1 mm, the increase in lengths of Aluminium and Copper wires are: [Y_(Al)=20xx10^(11)N//m^(2) and Y_(CU) = 12xx10^(11)N//m^(2)]

Answer»

1.0 MM, 1.1 mm
0.6 mm, 1.5 mm
0.7 mm, 1.4 mm
0.9 mm, 1.2 mm

ANSWER :2
Discussion
15552.

A body moves from point A to B under the action of a force, varying in magnitude as shown in figure. Obtain the work done. Force is expressed in newton and displacement in metre.

Answer»

Solution :Work done = Area under F - s graph
(or) `W_(AB) = W_(AP) + W_(PQ) + W_(QR) + W_(RB)`
`= 10 xx 1 + (1)/(2) (10 + 15) xx 1 + (1)/(2) xx 1 xx 15 - (1)/(2) xx 1 xx 15`
= 22.5 J
Discussion
15553.

Which one of the following graphs represent the variation of kinetic energy (k) with time ?

Answer»




Solution :KINETIC energy VARIES with time but is NEVER NEGATIVE.
Discussion
15554.

The value of acceleration due to gravity .g_p. on the surface of a planet with radius double that of earth and same mean density as that of the earth is ( g_(e) rarracceleration due to gravity on the surface of earth)

Answer»

`g_(p)=2g_(E)`
`g_(p)=g_(e)//2`
`g_(p)=g_(e)//4`
`g_(p)=4g_(e)`

ANSWER :A
Discussion
15555.

Match the list in Column-I with the list in Column-II Column-I list shows a list of fundamental interactive forces, Column-II is list of operational area of these forces

Answer»


ANSWER :C
Discussion
15556.

Who provides the centripetal force to geostationary satellite ?

Answer»

SOLUTION : The necessary gravitational FORCE of EARTH provides the centripetal force to geostationary satellite.
Discussion
15557.

If a person goes to a height equal to radius of earth from its surface, what would be his weight relative to that on the earth ? And he goes to a same depth, what would be his weight? (i) For height h = Re

Answer»

Solution :`implies ` (i) For height `h = R_e`
`(gh)/g =(R_E^2)/((R_e+h)^2) =R_e^2/((R_e+R_e)^2) =1/2 `
Weight becomes `1/4` times
`:. Mgh =1/4 mg `
(ii) `d = R_e` for DEPTH (At the CENTRE of EARTH)
`g_d =g (1-(d)/R_e)=g (1-R_e/R_e)=0`
`:. mg_d =0`
Discussion
15558.

Water is falling from a height of 100 m at the rate of 100 kg/sec. The power delivered to the turbine is approximately equal to

Answer»

100 kW 
10 kW 
1 kW 
1000 kW

Solution :Given : Height of the water fall (h) = 100 m and quantity of water falling (Q) = 100 kg/s.
Power delivered to the TURBINE `(P) =Qgh = 100 XX 10 xx 100`
`= 100000 W = 100 kW`
Discussion
15559.

The base of an insects leg is approximately spherical in shape, with a radius of about 2.0 xx 10^(-5) m. The 0.003g mass of the insect is supported equally by the six legs. The angle thetafor an insect on the surface of water is cos^(-1) "" (0.06 xx n). Find n.(surface tension T = 0.072 N/m)

Answer»


ANSWER :9
Discussion
15560.

The apparent weight of a man in a lift is w_(1) when lift moves upwards with some acceleration and is w_(2) when it is accelerating down with same acceleration. Find the true weight of the man and acceleration of lift.

Answer»

Solution :(a) `w_(1)=w(1+(a)/(g)), w_(2)=w(1-(a)/(g))`
`w_(1)+w_(2)=2W rArr w=(w_(1)+w_(2))/(2)`
(b) `(w_(1))/(w_(2))=(cancel(w)(1+(a)/(g)))/(cancel(w)(1-(a)/(g))) "" (g+a)/(g-a)=(w_(1))/(w_(2))`
`(g)/(a)=(w_(1)+w_(2))/(w_(1)-w_(2)) "" a=g((w_(1)-w_(2))/(w_(1)+w_(2)))`
Discussion
15561.

Whatis the difference between gravitational potential and gravitational potential energy?

Answer»

SOLUTION :
Discussion
15562.

If the coefficient of performance of a refrigerator is 5 and operates at the room temperature (27^@C), find the temperature inside the refrigerator.

Answer»

SOLUTION :Coefficient of performance,
`beta=5`
`T_1`=27+273=300 K
`T_2`=?
Coefficient of performance ,
`beta=T_2/(T_1-T_2)`
`THEREFORE 5=T_2/(300-T_2)`
`therefore 1500-5T_2=T_2`
`therefore 1500=6T_2`
`therefore T_2=1500/6`=250 K
`therefore t_2`=250-273 =`-23^@`C
Discussion
15563.

(A): Bulk modulus of an incompressible liquid is infinite(R): For an incompressible liquid volume remains constant

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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 :A
Discussion
15564.

The motion of a particle in S.H.M. is described by the displacement function, x= A cos(omega t+ phi),If the initial (t=0) position of the particle is 1 cm and its initial velocity is omega cms^(-1) , what are its amplitude and initial phase angle ? The angular frequency of the particles is pi s^(-1) . If instead of the cosine function, we choose the sine function to describe the SHM , x= B sin(omega t + alpha) , what are the amplitude and initial phase of the particle with the above initial conditions.

Answer»

SOLUTION :Here , at t=0, x=1 cm and `V = omega cm s^(-1) , omega = pi s^(-1)` Given x=A `cos (omega t +PHI)`
`therefore 1 = A cos (pi xx 0 + phi) = A cos phi""….(i) ` Velocity , `V = (dx)/(dt) =- A omega sin (omega t + phi)`
`therefore =- A omega sin (pi xx 0 + phi) ` (or) `1=-Asin phi ` (or) `A sin phi =-1 ""....(ii)`
Squaring and ADDING (i) and (ii) `A^2cos^2 phi + sin^2 phi) = 1+2` (or)` A^2 =2 ` (or) `A = sqrt2 ` cm
Dividing (ii) by (i) we get `tan phi` =-1 (or) `phi = (3pi)/(4) ` (or)` (7pi)/(4) ` For , `x= B sin (omega t + alpha)"".....(iii)`
At t=0 , x=1 , so, 1= Bsin `(omega xx 0 + alpha)= B sin alpha ""....(iv)`
Differentiating (iii), w.r.t.t, we have velocity , `V= (dx)/(dt) = B omega cos (omega t+ alpha)`
Applying initial CONDITIONS i.e., at `t=0 , V= omega`
`omega = B omega cos (pi xx 0 + alpha)` or `1= B cos alpha""....(iv)`
Squaring and adding (iv) and(v) , we get `B^2 sin^2 alpha + B^2 cos^2 alpha = 1^2 + 1^2 = 2 `or `B^2 = 2 ` or `B= sqrt2 ` cm
Dividing (iv) by (v) , we have `(B sin alpha)/(B cos alpha) = 1/1 ` (or) `tan alpha = 1 `(or)` alpha = pi//4 ` or `5pi//4`.
Discussion
15565.

What is meant by order of magnitude ? Illustrate with atleast three examples ?

Answer»

SOLUTION : 1(C ).1
Discussion
15566.

The passenger section of a jet airliner has the shape of a cylindrical tube with a length of 35.0 m and an inner radius of 2.50 m. Its wall are lined with an insulating material 6.00 cm in thickness and having a thermal conductivity of 4.00xx10^(-5) cal/s cm ""^(0)C. A heater must maintain the interior temperature at 25.0^(0)C while the outside temperature is at -35.0^(0)C. What power must be supplied to the heater if this temperature difference is to be maintained ?

Answer»

9.32 kW
8.32 kW
7.52 kW
10.41 kW

Answer :A
Discussion
15567.

A person walks along a straight road from his house to a market 2.5kms away with a speed of 5 km/hr and instantly turns back and reaches his house with a speed of 7.5 kms/hr. The average speed of the person during the time interval 0 to 50 minutes is (in m/sec)

Answer»

`4""2/3`
`5/3`
`5/6`
`1/3`

ANSWER :B
Discussion
15568.

A rotor of radius (r) is rotating about its own vertical axis and a person in contact with innerwall of rotor remains in equilibrium without slipping down. If 'w' is angular velocity of rotor and mu is minimum coefficient of friction between persona and the wall of rotor. Then following is correct (A) mu alpha w^(2) (B) mu alpha (1)/(r) (C) mu alpha(1)/(w^(2)) (D) mu alpha r

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A and B are TRUE
A and D are true
C and B are true
C and D are true

Answer :C
Discussion
15569.

{:("Column-I","Column-II"),("(A) Solid sphere rolling with slipping","(P) Total kinetic energy is conserved on a rough horizontal surface"),("(B) Solid sphere in pure rolling on a rough horizontal ","(Q) Angular momentum about CM is conserved"),("(C) Solid sphere in pure rolling on a smooth horizontal surface","(R) Angular momentum about a point on contact surface is conserved"), ("(D) Solid sphere in pure rolling on a rough incline","(S) Momentum is conserved"):}

Answer»


ANSWER :A-R;B-QRS;C-PQRS;D-S
Discussion
15570.

For an adiabatic process, the relation between V and T is given by

Answer»

`TV^(GAMMA)="a CONSTANT"`
`T^(gamma)V="a constant"`
`TV^(1-gamma)="a constant"`
`TV^(gamma-1)="a constant"`

ANSWER :D
Discussion
15571.

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 direction of their motion after collision are

Answer»

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

Answer :C
Discussion
15572.

Pick out the two scalar quantities in the following list : Force, angular momentum, work, current linear momentum, electric field , average velocity, magnetic moment , relativevelocity .

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SOLUTION :WORK and the only VECTOR quantity in the FOLLOWING list :
Discussion
15573.

Which of the following is the dimensions of coefficient of friction ?

Answer»

`[MLT^(-2)]`
`[M^@L^@T^@]`
`[M^2LT^(-2)]`
`[M^2LT]`

ANSWER :B
Discussion
15574.

Three samples of the same gas A, B and C (gamma = 3//2) have initially equal volume. Now the volume of each sample is doubled. The process is adiabatic for A, isobaric for B and isothermal for C. If the final pressures are equal for all three samples, find the ratio of their initial pressures

Answer»

<P>

Solution :Let the initial PRESSURE of the THREE SAMPLES be `P_(A), P_(B) and P_(C ) ` then `P(A) (V)^((3)/(2)) =(2V)^((3)/(2)) P`
`P_(B) =P"" P_(C ) (V) = P(2V)`
`:.P_(A):P_(B):P_(C )=(2)^(3//2) : 1:2 = 2sqrt(2) :1:2`
Discussion
15575.

A mass of 0.1kg is rotated in a vertical circle using a string of length 1m. When the string makes an angle 30^(@) with the vertical, the speed of the mass is 2ms^(-1). The radial acceleration of the mass at that instant is

Answer»

`4MS^(-2)`
`0.1 MS^(-2)`
`4ms^(-2)`
`4ms^(-2)`

ANSWER :D
Discussion
15576.

An ideal gas is heated from 20^(@)C to 40^(@)C underconstant pressure. The change in internal energy is

Answer»

zero under constant pressure
Double the ORIGINAL value
PROPORTIONAL to CHANGE in VOLUME
Proportional to change in temperature

Answer :D
Discussion
15577.

A stone is thrown down the slope as shown. Determine the magnitudeu and direction of its initial velocity so that the stone will rise 12 m and still have a range of 50 m down the slope.

Answer»


ANSWER :17.74m/s, `theta=59.8^(@)`
Discussion
15578.

The density of lead at 0^(@)C is 11.34 g/cm^(3). The density of lead at 100^(@) (if the coefficient of linear expansion of lead = 28 xx 10^(-6) //^(@) C) is =28 xx 10^(-6)//^(@) Cis

Answer»

13.25 GM/`CM^(3)`
17.25 gm /`cm^(3)`
18.25 gm /`cm^(3)`
11.25 gm /`cm^(3)`

ANSWER :D
Discussion
15579.

What is the acceleration of train moving with speed of 50 ms^(-1) on circular path of radius 250 m ?

Answer»

Solution :`a_(V )(v^(2))/(R ) = ((50)^(2))/ (250) = (2500)/(250)`
`=10 ms^(-2)`
Discussion
15580.

hati and hatj are unit vector along x-axis and y-axis respectively. It changes with time.

Answer»


ANSWER :FALSE. REMAINS CONSTANT.
Discussion
15581.

Show that the total momentum of system of particles is equal to the product of total mass of system and velocity of centre of mass.

Answer»

Solution :According to equation (2) as in QUESTION 19.
`MvecV=m_(1)vec(v_(1))+m_(2)vec(v_(2))+...m_(n)vec(v_(n))""...(1)`
According to chapter 4, `vecp=mvecvand(dvecp)/(dt)=VECF` Now, the linear momentum of system of n particle is equal to the vector sum of individual linear momentum of all particles of the system.
`THEREFORE vecP=vec(p_(1))+vec(p_(2))+...vec(p_(n))`
`vecP=m_(1)vec(v_(1))+m_(2)vec(v_(2))+...m_(n)vec(v_(n))""...(2) [because vecp=mvecv]`
COMPARING equation (1) and (2),
`vecP=MvecV""...(3)`
Hence, the linear momentum of system of particle is equal to the product of total mass of system and velocity of centre of mass.
Discussion
15582.

(A): The melting point of ice decreases with increase of pressure.(R ): Ice contracts on melting. (2004)

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Both (A) and(R ) are TRUE and (R ) is the correct explanation of (A)
Both (A) and (R ) are (R ) true and is not the correct explanation of (A)
(A) is true but (R ) is FALSE
Both (A) and (R ) are false

Answer :A
Discussion
15583.

In process Tprop(1)/(V), pressure of the gas increases from p_(0) to 4p_(0). Match the following.

Answer»

<P>

ANSWER :(A) R, (B) T, (C) Q, (D) P
Discussion
15584.

Three elastic wires PQ, PR and PS support a body P of mass M, as shown in figure. The wires are of the some material and cross sectional area, the middle one being vertical. Find the loads by each wire.

Answer»

`Mg//1+ 2 cos^(2)THETA`
`Mg//1+ 2 cos^(3)theta`
`Mg cos^(2)theta//1+ 2cos^(3)theta`
`Mg costheta//1+ 2cos^(3)theta`

Solution :Figure shows the situationwhentheload (boby P) descends dueto thestretchinload wires. The new positionsof wires PQ andPsare shown by thedotted lines. Let `T_(1), T_(2)` and`T_(3)`be the loads(tensions) carried by the three wires PQ, PR and PS respectivelywith thewires PQ and PS makingangle `theta` with the vertical .

Consideringthe horizontal equilibrium ofpointP,
`T_(1)sin theta= T_(3) sin rArr T_(1) = T_(3) = T("SAY")`
Considering the veritcalequilibrium of point P, `T_(2) +2T cos theta = Mg""......(i)`
PR =a
`THEREFORE ` PQ = PS = a sec`theta`
I f`deltal_(1)` and `deltal_(2)` bethe elongations in thewires PR and PQ (or Ps ) respectively then
`deltal_(2) = deltal_(1) cos theta` (from geometry)
If A nd Y be the cross- sectional area and Young'smodaulus of reachof the each of the three wires, then
`(T)/(AY)(a sectheta) = (T_(2))/(AY ) a cos theta rArr T =T_(2) cos ^(2) theta""....(ii)`
Sloving eqns. (i) and (ii) we get `T_(2) = (Mg)/(1+2 cos^(3) theta)`
Discussion
15585.

According to Hooke's law within the elastic limit (stress)/(strai n)= constant. The constant depends on the type of strain or the type of force acting. Tensile stress might result In compressional or elongative strain, however, a tangential stress can only cause a shearing strain. After crossing the elastic limit. the material undergoes elongation and beyond a stage beaks. All modulus of elastically are basically constants for the materials under stress Two wires of same material have length and radius l,r and 2l, r/2 respectively. The ratio of their Young's modulus is

Answer»

`1:2`
`2:3`
`2:1`
`1:1`

ANSWER :D
Discussion
15586.

According to Hooke's law within the elastic limit (stress)/(strai n)= constant. The constant depends on the type of strain or the type of force acting. Tensile stress might result In compressional or elongative strain, however, a tangential stress can only cause a shearing strain. After crossing the elastic limit. the material undergoes elongation and beyond a stage beaks. All modulus of elastically are basically constants for the materials under stress After crossing the yield region the material will have

Answer»

REDUCED stress
increased stress
breaking stress
constant stress

Answer :A::C
Discussion
15587.

The coffecients of thermal expansion in solids are mainly coffecient of volume expansion (gama) . Invar is used for making pendulum of clocks . Why ?

Answer»

SOLUTION :Due to its SMALL COFFECIENT of linear expansion.
Discussion
15588.

According to Hooke's law within the elastic limit (stress)/(strai n)= constant. The constant depends on the type of strain or the type of force acting. Tensile stress might result In compressional or elongative strain, however, a tangential stress can only cause a shearing strain. After crossing the elastic limit. the material undergoes elongation and beyond a stage beaks. All modulus of elastically are basically constants for the materials under stress If (stress)/(strai n) is x in elastic region and y in yield region, then

Answer»

X=y
`x GT y`
`x LT y`
`x=2y`

ANSWER :B
Discussion
15589.

If charge distribution within a Gaussian surface changes inside it, will electric field strenght change inside and outside the Gaussian surface ?

Answer»

Solution :As the total charge INSIDE the GAUSSIAN surface remains UNCHANGED, the same electiec FLUX will PASS through the Gaussian surface. However, due to the change in charge distribution, the value of E will change inside as well as outside the Gaussian surface.
Discussion
15590.

A satellite of mass m is orbiting in an orbit of radius r_1. If it is given an extra impulse in the direction of motion, it goes to the orbit of radius r_2. Find the extra velocity given to the satellite.

Answer»

SOLUTION :`SQRT((GM)/( r_1)) [ sqrt((2r_2)/(r_1+r_2)) - 1 ]`
Discussion
15591.

Which one of the following processes depends on gravity?

Answer»

conduction
convection
radiation
none of these

Answer :B
Discussion
15592.

The period of oscillation of a simple pendulum of length 'L' suspended from the ceiling of a vehicle which moves with out friction down a fixed inclined plane of inclination is given by

Answer»

`2PI SQRT((L)/(g cos ALPHA))`
`2pi sqrt((L)/(g SIN alpha))`
`2pi sqrt((L/g))`
`2pi sqrt((L)/(g tan alpha))`

Answer :A
Discussion
15593.

If heat is supplied to an ideal gas in anisothermal process

Answer»

The internal energy of the GAS increases
The gas performs POSITIVE work
The gas performs will NEGATIVE work
The SAID process is not possible

Answer :B
Discussion
15594.

During an isobaric heating process the work done by oxygen gas is 4 J. Calculate the amount of heat transferred to the gas. [gamma = 1.4]

Answer»

Solution :HEAT transferred `= Q = ?`
`Q = DELTA U + Delta W , Delta W = 4 J`
`Delta U = n C _(V )Delta T = (n R Delta T )/( gamma -1) = (P Delta V )/( gamma -1) = (Delta W)/( gamma -1)`
`Q = (Delta W)/(gamma -1)+ Delta W = Delta W [ (1+ gamma -1 )/( gamma -1) ] = (4 xx 1.4)/(1. 4 -1) =1.4 J`
Discussion
15595.

In force oscillation of a particle, the amplitude is maximum for a frequency omega_1, while the energy maximum for a frequency omega_2, of the force. Then

Answer»

`omega_1 lt omega_2`
`omega_1 gt omega_2`
`omega_1 lt omega_2` when DAMPING is small and `omega_1 gt omega_2` when damping is LARGE
`omega_1 = omega_2`

Answer :B
Discussion
15596.

A wooden block of cross-sectional area 10cm^(2) is floating vertically on water. The volume of the immersed portion of the block is 200cm^(3). The block is depressed slightly inside water and then released. Calculate the time period of vibration of the block.

Answer»

SOLUTION :Volume of displaced water = `200CM^(3)`
`therefore" "` Mass of displaced water = 200 g, and mass of the wooden BLOCK = 200 g
Let the block be depressed inside water through x cm and then RELEASED.
`therefore` Upward restoring force on the block,
`F=10x xx1xxg=10xg`
`therefore` Acceleration of the block,
`a=F/m=(10xg)/200=(xg)/20`
`therefore` Time period of vibration of the block,
`T=2pisqrt(x/a)=2pisqrt(20/g)=2pisqrt(20/980)=0.897s`.
Discussion
15597.

A liquid of mass m and specific heat c is heated to a temperature 2 T . Another liquid of mass m/2 and specific heat 2c is heated to a temperatureT. If these two liquid are mixed , the resulting temperature of the mixture is

Answer»

(2/3)T
(8/5)T
(3/5)T
(3/2)T

Answer :D
Discussion
15598.

The limbs of a manometer consist of uniform capillary tubes of radii 1.4 xx 10^(-3) m and 7.2 xx 10^(-4) m. The correct pressure when the level of the liquid in the narrower tube stands 0.2m above that in the broader tube is X(621Pa). Find the value of X. (density = 10^3 kg//m^3, surface tension = 72 xx 10^(-3) N/m)

Answer»


ANSWER :3
Discussion
15599.

Inthe above problem what is the time taken to cover that distance .

Answer»

`(2v_(0))/(k)`
`(2 sqrt(v_(0)))/(k)`
`(2v_(0))/(3K)`
`(2sqrt(v_(0)))/(3k)`

Answer :B
Discussion
15600.

A pump holding water is designed as a horizontal cylinder with a piston having an area of A and an outlet orifice having an area of .a. arranged near the cylinder axis. Determine the velocity of outflow of water in ms from the pump if the piston moves with a constant velocity under the action of a force F(A=2m^(2) and F=4KN. Assume A gt gt a).

Answer»


ANSWER :2
Discussion