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15401.

A particle is performing SHM. Its path length is 8 cm, amplitude is4/1 th the path length and time period 0.2s. If the particle is initially in the mean position, the time at which it will be at 1 cm will be

Answer»

Solution :Length of PATH 2A= 8 CM, A= 4 cm
Periodic time `T= 4 XX A`, Time taken to cover distance A is 1 second.
`therefore T= 4xx 1 = 4s`
Maximum VELOCITY at mean POSITION `therefore v_("max")= A omega`
`= 4xx (2pi)/(T) = 4xx (2pi)/(4)`
`= 2pi cm s^(-1)`.
Discussion
15402.

An object is executing uniform circular motion with an angular speed of (pi)/(12) radian per second. At t=0, the object starts at an angle theta=0. What is the angular displacement of the particle after 4s?

Answer»


SOLUTION :ANGULAR speed =`("Angular displacement")/("time taken")`
Angular displacement `=(PI)/(12)xx4=(pi)/(3)=60^(@)`
Discussion
15403.

Glycerine flows steadily through a horizontal tube of length 1.5m and radius 1.0 cm . If theamount of glycerine collected per second at one is 4.0xx10^(-3)kgs^(-1) , what is the pressure difference between the two ends ofthe tube? (Density of glycerine =1.3xx10^(10^(3)kgm^(-3) and viscosity of glycerine =0.83 Pa s). [ you may also like to check if the assumption of laminar flow in the tube is correct]

Answer»

SOLUTION :VOLUME of flowing glycerine in one second,
`V=(M)/(rho)`
`=(4xx10^(-3))/(1.3xx10^(3))`
`=3.1xx10^(-6)m^(3)s^(-1)`
From Poiseille.s law,
`V=(piPr^(4))/(8etal)`
`P=(8etalV)/(pie^(4))`
`=(8xx0.83xx1.5xx3.1xx10^(-6))/(3.14xx(10^(-2))^(4))`
`=9.833xx10^(2)`
`=9.8xx10^(2)P_(a)`
Value of Raynolds number for laminar flow in tube ,
`R_(e)=(rhod)/(eta)xx(m)/(rhopir^(2))`
`=(dm)/(etapir^(2))=(2rm)/(etapir^(2))`
`=(2M)/(etapir)`
`=(2xx4xx10^(-3))/(0.83xx3.14xx10^(-2))`
`=3.069xx10^(-1)`
`=0.31`
Here , `R_(e)lt2000`, the flow is steady that means laminar.
Discussion
15404.

A light clindrical vessel is kept on a horizontal surface. The base are is A. A hole of cross-sectional area a is made just at its bottom side (where alt ltA) fig. Find minimum coeffiecent of friction necessary for sliding of the vessel due to the impact of the emerging liquid.

Answer»

Solution :Let h be the height of liquid in vessel above the hole.
VELOCITY of EFFLUX, `UPSILON = sqrt(2 gh)`
Impact force of the imerging liquid on the vessel and liquid contents is
`F = upsilon (DM)/(dt) = upsilon(a rho upsilon) = a rho upsilon^(2) = a rho 2 gh`
`= 2 a rho gh `
The force of friction, `f = mu R =mu A h rho g`
For just sliding the vessel , `f =F`
so, `mu A h rho g = 2 a rho g h or mu = (2a)/(A)`.
Discussion
15405.

A body of mass 1 kg is moving with velocity 30 ms^(-1)due north. It is acted on by a force of 10 N due west for 4 seconds. Find the velocity of the body after the force ceases to act.

Answer»

`50m//s , TAN^(-1) (3/4)`
`100m//s , tan^(-1) (4/3)`
`50m//s, tan^(-1) (3/5)`
`100 m//s, tan^(-1) (4/5)`

ANSWER :A
Discussion
15406.

A vessel contains a mixture consisting of m_(1)=7g of nitrogen (M_(1)=28) and m_(2)=11g of carbon dioxide (M_(2)=44) at temperature T=300K and pressure p_(0)=1 atm. Find the density of the mixture.

Answer»


ANSWER :NA
Discussion
15407.

When liquid medicine of density p is to be put in the eye, it is done with the help of dropper. As the bulb on the top of the dropper is pressed, a drop forms at the opening of the dropper. We wish to estimate the size of the drop. We first assume that the drop formed at the opening is spherical because that requires a minimum increase in its surface energy. To determine the size, we calculate the net vertical force due to the surface tension T when the radius of the drop is R. When the force becomes smaller than the weight of the drop, the drop gets detached form the dropper.If r = 5 xx 10^(-4) m,rho= 10^3 kgm^(-3) , g = 10m//s^2 , T = 0.11 Nm^(-1) , the radius of the drop when it detaches from the dropper is approximately

Answer»

`1.4 XX 10^(-3) m`
`3.3 xx 10^(-3) m`
`2.0 xx 10^(-3) m`
`4.1 xx 10^(-3) m`

ANSWER :A
Discussion
15408.

When liquid medicine of density p is to be put in the eye, it is done with the help of dropper. As the bulb on the top of the dropper is pressed, a drop forms at the opening of the dropper. We wish to estimate the size of the drop. We first assume that the drop formed at the opening is spherical because that requires a minimum increase in its surface energy. To determine the size, we calculate the net vertical force due to the surface tension T when the radius of the drop is R. When the force becomes smaller than the weight of the drop, the drop gets detached form the dropper.If the radius of the opening of the dropper is r, the vertical force due to the surface tension on the drop uf radius R (assumning

Answer»

`2 pi r T `
`2 pi r T `
` ( 2PI r ^2 T)/( R )`
` (2 piR^2 T ) /( r)`

Answer :C
Discussion
15409.

In a refrigerator one removes heat from a lower temperature and deposits to the surroundings at a higher temperature. In this process, mechanical work has to be done, which is provided by an electric motor. If the motor is of 1 kW power and heat is transferred from - 3^@ C to 27^@ C, find the heat taken out of the refrigerator per second assuming its efficiency is 50 % of a perfect engine.

Answer»

Solution :Temperature of source is `27^@C`
`therefore T_1=27+273`=300 K
Temperature of sink `T_2=-3+273`=270 K
The EFFICIENCY of a perfect ENGINE ,
`eta=1-T_2/T_1`
`therefore eta=1-270/300=30/300=0.1`
The efficiency of a REFRIGERATOR ,
`eta.`=50% efficiency of a perfect engine,
`=0.1xx50/100`
=0.05
Coefficient of performance of a refrigerator ,
`beta=Q_2/W=(1-eta.)/(eta.)`
`therefore beta=(1-0.05)/0.05 =0.95/0.05`=19
`therefore Q_2=betaW`
=19 x 1 kW
=19 kW
`=19 "KJ"/s`
Therefore , heat is taken out of the refrigerator at a rate of 19 kJ per second.
Discussion
15410.

Explain and draw the graphs of kinetic energy, potential energy and mechanical energy versus displacement for SHM.

Answer»

Solution :Kinetic energy of SHM `K(x) = (1)/(2)k(A^(2)-x^(2))`
Potentiall energy `U(x)= (1)/(2) KX^(2)` and
Mechanical energy `E= (1)/(2) kA^(2)`
The magnitude of ENERGIES at different position of SHM are shown as below :

From these values graph of energies versus displacement is obtained as below :

Following points are clear from graph :
(1) `E to x` graph is linear and is parallel to displacement axis, so we can say that mechanical energy does not depends on displacement, but remains constant.
(2) Shapes of `K(x) to x" and " U(x) to x` graphs are PARABOLIC.
(3) At fixed point x=0, potentiall energy is zero and kinetic energy is maximum and is equal to mechanical energy.
(4) As the oscillator goes towards any side increase in potentiall energy is equal to decrease in kinetic energy.
(5) At extreme point `(y= |A|)`, potential energy is maximum and kinetic energy is zero. This maximum potential is equal to mechanical energy.
(6) At any point of SHM path, the sum of potential and kinetic energy is equal to mechanical energy.
(7) Coodinates of point of intersections of graph of kinetic energy and potential energy is `(pm (A)/(sqrt(2)), (E )/(2))`.
Discussion
15411.

Three masses 0.1 kg, 0.3 kg and 0.4 kg are suspended at the end of the spring. When the 0.4 kg is removed, the system oscillates with a period of '2' sec. When 0.3 kg mass is also removed, the system will oscillate with a period

Answer»

1 sec
2 sec
3 sec
4 sec

Answer :A
Discussion
15412.

The P-V diagram of a gas undergoing a cyclic process (ABCDA) is shown in the graph, where P is in units of Nm^(-2) and V in cm^(3). Identify the incorrect statement.

Answer»

0.4 J of work is done by the gas from A to B
0.2 J of work is done on the gas from C to D.
No work is done by the gas from B to C.
Work is done by the gas from B to C and on the gas from D to A.

Solution :In going from B to C and from D to A, V = CONSTANT.
So, `DELTA V = 0. :. W = P Delta V = 0`
Discussion
15413.

For a satellite to orbit around the earth, which of the following must be true ?

Answer»

It cannot pass over the poles at any time
It must above the equator at SOMETIME
Its PERIOD of rotation must be greater than `2pisqrt((R )/(G))
Its height above the surface of earth cannot exceed 36000 km

ANSWER :B::C
Discussion
15414.

A bob of mass m is attached to one end of the rod of negligible mass and length r, the other end of which is pivoted freely at a fixed centre O as shown in the figure. What initial speed must be given to the object to reach the top of the circle?

Answer»

Solution :
ACCORDING to law of conservation of energy
`(1)/(2)mv_(0)^(2)+0=0+mg(2r)`
`v_(0)=sqrt(4gr)ms^(-1)`
This value is LESS than
`v_(1)=sqrt(5gr)` and greater than
`v_(2)=sqrt(gr)`
Discussion
15415.

vec(A) and vec(B) are two vectors. Then correct options are

Answer»

If `vec(A) + vec(B)`magnitude is equal to magnitude of `vec(A) - vec(B)` magnitude angle between `vec(A) and vec(B)` is `90^(@)`
If angle between `vec(A) + vec(B) and vec(A) - vec(B) " is " 60^(@)` their resultant is `SQRT(3A^(2) + B^(2))`
If `vec(A) + vec(B) and vec(A) - vec(B)` vectors are PERPENDICULAR `vec(A) - vec(B)` is a null vector
`(vec(A) + vec(B)).(vec(A)- vec(B)) = 0` if `vec(A) and vec(B)` have equal magnitude

Answer :A::B
Discussion
15416.

If Y=a-b, then maximum percentage error in the measurement of Y will be

Answer»

`((DELTA a)/(a)+(Delta B)/(b))XX100`
`((Delta a)/(a-b)+(Delta b)/(a-b))xx100`
`((Delta a)/(a)-(Delta b)/(b))xx100`
`((Delta a)/(a-b)-(Delta b)/(a-b))xx100`

ANSWER :B
Discussion
15417.

Give the value of coefficient of volume expansion at 0^(@)C for ideal gas.

Answer»

Solution :`alpha_(V)=3.7xx10^(-3)K^(-1)`
Discussion
15418.

Potential energy U (for a pair of atoms) versus interatomic separation r for two substance A and B are as shown. For both substance, graphs of U v/s t are on same scale : Where E_(1) , E_(2) and E_(3) are total energy line (for a pair of atoms) and T_(1) , T_(2) and T_(3) are the corresponding temperatures. Also T_(1)gt T_(2) gt T_(3). {:("Column-I","Column-II"),("A) Expands of heating","P) Substance A"),("B) Contracts on heating","Q) Substance B"),("C) Liquifies above temperature" T_1 ,),("D) Liquifies above temperature" T_2,):}

Answer»


ANSWER :A-P; B-Q; C-P,Q; D-P
Discussion
15419.

When two bodies A and B collide, A exerts a force F on B. What is the force applied by B on A ?

Answer»

SOLUTION :F. Since, by Newton.s third law of motion, action and REACTION are EQUAL and OPPOSITE.
Discussion
15420.

On the surface of a solid cylinder of mass 2 kg and radius 25 cm a string is wound several times and the end of the string is held stationary while the cylinder is released from rest. What is the acceleration of the cylinder and the tension in the string ?

Answer»

Solution : Let a be the acceleration of the centre of mass. Then mg -T = ma……(i)
where Tis the tension in the string. (SEE Fig.) For the ROTATIONAL motion about the centre of mass,
`tau = l alpha`
`Tr = 1/2 mr^(2) alpha`………(ii)
But `a = r alpha`
`T =1/2 mr a/r =1/2 ma`.......(iii)
From EQNS (i) mg-T = ma
`mg -1/2 ma = ma`
`3/2 ma = mg`
`a =(2g)/3 = (2 xx 9.8)/3 = 6.53 m//s^(2)`
`T = 1/2 ma = 1/2 xx 2 xx (19.6)/3 = 6.53 N`
Discussion
15421.

Distance between the centres of two stars is 10a. The masses of these star are M and 16 M and their radii a and 2a, respectively. A body of mass m is fired straight from the surface of the largr star towards the smaller star. What should be its minimum initial speed to reach the surface of the smaller star ? Obtain the expression in terms of G, M and a.

Answer»

Solution :The distance x (from the smaller planet) where the gravitational pull of two planet.s balance each other will be given by
`(-GMm)/(x^(2)) = (-G(16M))/((10a-x)^(2))`
i.e., So the body will REACH the smaller planet due to planet.s gravitational field if it has SUFFICIENT energy to CROSS the point B(x=2a), i.e.,
`(1)/(2) mv^(2) gt m(V_(B)-V_(S))`
But `V_(S) = -[(16 GM)/(2a) + (GM)/((10-2a))] = (65 GM)/(8a)`

and `V_(B) = -[(16 GM)/(8a) + (GM)/(2a)] = -(20 GM)/(8a)`
so `(1)/(2) mv^(2) gt m [(65 GM)/(8a) - (20 GM)/(8a)]`
i.e., `v_("min") = (3)/(2)sqrt((5GM)/(a))`
Discussion
15422.

If the focal length of a magnifier is 5cm calculate (a) the power of the lens (b) the magnifying power of the lens for relaxed and strained eye.

Answer»

Solution :(a) As POWER of a LENS is RECIPROCAL of FOCAL length in m.
`P=1/(( 5 TIMES 10^-2 m))=1/0.0` diopter =20D
(b) For relaxed eye, MP is minimum and will be
`MP=D/f=25/5=5`
While for strained eye, MP is maximum and will be
`MP=1+D/f=1+5=6`
Discussion
15423.

A cyclic process ABCD is shown in the p-Vdiagram. Which of the following curves represent the same process?

Answer»




ANSWER :A
Discussion
15424.

If mu is the permeability and in is the permittivity then (1)/(sqrt(mu in)) is equal to

Answer»

Speed of SOUND
Speed of light in VACCUM
Speed of sound in MEDIUM
Speed of light in medium

ANSWER :D
Discussion
15425.

A geostationary satellite orbits the earth at a height of nearly 36,000 km from the surface of earth. What is the potential due to earth.s gravity at the site of this satellite? (Take the potential energy at infinity to be zero). Mass of the earth= 6.0 xx 10^24 kg, radius = 6400 km, G = 6.67 xx 10^(-11)Nm^2//kg^2

Answer»

Solution :GRAVITATIONAL potential at height h from the surface of earth is
`V = -(GM)/((R+h)) = (-6.67 xx 10^(-11) xx (6 xx 10^24))/((6.4 xx 10^6 + 36 xx 10^6) ) = -9.4 xx 10^6 J//kg`
Discussion
15426.

A uniform lamina ABCDE is made from a square ABDE and anequilaterial triangle BCD. Find the centre of mass of the lamina.

Answer»

Solution :Here, `A_(1)`= area of square ABDE =(a)(a) =`a^(2)`
`A_(2)` = area of triangle `=1/2(a)((sqrt(3)a)/2) = sqrt(3)/4 a^(2)`
`(x_(1),y_(1))` = co-ordiantes of centre of MASS of square `=(a/2,0)`
`(x_(2),y_(2))` =co-ordinates of centre of mass of the triangle = `(a + a/(2sqrt(3)),0)`
Now, `x_(CM) =(A_(1)x_(1) + A_(2)x_(2))/(A_(1) + A_(2)) =((a^(2)).(a/2) + (sqrt(3)a^(2))/4 (a+a/(2sqrt(3))))/(a^(2) + sqrt(3)/4 a^(2)) = 0.74 a`
`Y_(CM) =0` as `y_(1)` and `y_(2)` both are zero.
Therefore co-ordinates of CM of the given lamina are (0.74a, 0).
Discussion
15427.

A string is wrapped on a uniform disc and the other end of the string connected to a wall. The system is placed on a smooth plane, inclined at an angle theta, with the string parallel to the plane, as shown in the figure. The acceleration of the disc is

Answer»

`(1)/(3)gsintheta`
`(1)/(2)gsintheta`
`(2)/(3)gsintheta`
`(1)/(3)gsintheta`

ANSWER :C
Discussion
15428.

A mass M is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes SHM of time period T. If the mass is increased by m, the time period becomes 5T/3. What is the ratio m/M?

Answer»

Solution :With mass M, the TIME period of the SPRING is `T=2pisqrt((M)/(k))`
With mass `M+m`, the time period becomes
`(5T)/(3)=2pisqrt((M+m)/(k))or(5)/(3)xx2pisqrt((M)/(k))=2pisqrt((M+m)/(k))`
`(25)/(9)M=M+m,(16)/(9)M=mor(m)/(M)=(16)/(9)`
Discussion
15429.

A body is projected with an initial speed of 100 sqrt(3) ms^(-1) at an angle of 60^(@) above the horizontal. If g = 10ms^(-2) then velocity of the projectile (a) Is perpendicular to it's acceleration at the instant t = 15 sec. (b) Is perpendicular to initial velocity of projection at t = 20 sec. (c) Is minimum at the highest point (d) Changes both in magnitude and direction, during its flight. Mark the answer as.

Answer»

If a, B, C and d are correct
If a, c and d are correct
If b, c and d are correct
If a, b and d are correct

Answer :A
Discussion
15430.

The vectors vecA and vecB are such that |vecA+vecB| =|vecA-vecB| . The angle between the two vector is .......... .

Answer»

`45^(@)`
`60^(@)`
`75^(@)`
`90^(@)`

Solution :The ANGLES between two vector is ALWAYS `90^(@)` .
Discussion
15431.

Match Column -I with column -II : ltBRgt

Answer»


SOLUTION :(a) `GMeMs=Fxxr^(2)`
(b) `(3RT)/(M)=v^(2) ("metre")^(2)("second")^(2)`
(c) `F q vB""` Hence, `v^(2)=F^(2)/(q^(2)B^(2))`
(d) Since `F=(GM_(E)m)/(R_(e)^(2))""`Hence`=(FxxR_(e))/m=(Gme)/R_(e)`
Discussion
15432.

A rectangle block rests on a horizontal table. A horizontal force is applied on the block at a height h above the table to move the block. Does the line of action of the normalforce N exerted bythe table on block depend on h?

Answer»

Solution :(i) Yes. The LINE of action of the NORMAL force N exerted by the table on the block depend on h.

(II) When height of the APPLIED force 'h' INCREASES, a torque is produced by the applied force and frictional force such that block start to tilt.
(iii) To balance this effect, line of action of normal force shift away from applied force and make a opposite torque, joining with gravitational force 'W'.
Discussion
15433.

For which of the following colour, the magnilying power of a microscope will be maximum

Answer»

WHITE colour
Red colour
Violet colour
Yellow colour

Answer :C
Discussion
15434.

A person travelling on a straight line moves with a uniform velocity v_1 for a distance x and with a uniform velocity v_2 for the next equal distance. The average velocity v is given by

Answer»

`V=(v_(1)+v_2)/(2)`
`v=sqrt(v_(1)v_(2))`
`2/v=1/v_(1)+1/v_(2)`
`1/v=1/v_(1)+1/v_(2)`

ANSWER :C
Discussion
15435.

Can a liquid offer resistance to bulk stress?

Answer»

Solution :YES. It can offer very GREAT resistance to forces TENDING to DECREASE its volume.
Discussion
15436.

Obtain relation between coefficient of volume expansion (alpha_(V)) and coefficient of linear expansion (alpha_(l)).

Answer»

Solution :Suppose there is a cube of side of length .l.. When its TEMPERATURE is increased by `DeltaT`, it expands EQUALLY in all DIMENSIONS.
Hence from `V=l^(3)`
`DeltaV=(l+Deltal)^(3)-l^(3)`
`=l^(3)+3l^(2)Deltal+3l(Deltal)^(2)+(Deltal)^(3)-l^(3)`
But `(Deltal)^(2)` and `(Deltal)^(3)` are much more less than l, hence by neglecting them
`DeltaV=3l^(2)Deltal`. . .(1)
But, from LINEAR expansion.
`Deltal=alpha_(l)lDeltaT`. . .(2)
`:.` By using value of equation (1) in equation (2),
`DeltaV=3l^(2)(alpha_(l)lDeltaT)`
`=3l^(3)alpha_(l)DeltaT`
`DeltaV=3Valpha_(l)DeltaT""`. . .(3)
`""(becausel^(3)=" VOLUME of cube")`
By comparing equation (3) with general equation of volume expansion `DeltaV=alpha_(V)VDeltaT`.
`alpha_(V)=3alpha_(l)` which is relation between coefficient of volume and linear expansion.
Discussion
15437.

Derive an expression for excess of pressure in a liquid drop.

Answer»

Solution :Let us consider a water sample of cross sectional area in the form of a cylinder. Let `h_(1)andh_(2)` be the depths from the air-water interface to level 1 and level 2 of the cylinder, respectively as shown in Figure(a). Let `F_(1)` be the force acting downwards on level 1 and `F_(2)` be the force acting upwards on level 2, such that,`F_(1)=P_(1)AandF_(2)=P_(2)A` Let us mass of the sample to be m and under equilibrium condition, the total upward force `(F_2)` is balanced by the total downward force `(F_(1)+MG)`, otherwise, the gravitational force will act downward which is being exactly balanced by the difference between the force `F_(2)=F_(1)`
`F_(2)-F_(1)=mg=F_(G)""...(1)`
Where m is the mass of the water available in the sample element. Let `rho` be the density of the water then, the mass of water available in the sample element is
`m=rhoV=rhoA(h_(2)-h_(1))`
`V=A(h_(2)-h_(1))`

Hence, gravitational force,
`F_(G)=rhoA(h_(2)-h_(1))g`
On substituting the value of W in equation (1)
`F_(2)=F_(1)+mg`
`rArr""P_(2)A=P_(1)A+rhoA(h_(2)-h_(1))g`
Cancelling out A on both sides,
`P_(2)=P_(1)+rho(h_(2)-h_(1))g""...(2)`

If we choose the level 1 at the surface of the liquid (i.e, air-water interface) and the level 2 at a depth 'H' below the surface (as shown in Figure), then the value if `h_(1)` becomes zero `(h_(1)=0)` when `P_(1)` assumes the value of atmospheric pressure (say `P_(a)`). In addition, the pressure `(P_2)` at a depth becomes P. Substituting these values in equation,
`P_(2)=P_(1)+rho(h_(2)-h_(1))g`
we get`P=P_(a)+rhogh`
Which means, the pressure at a depth h is greater than the pressure on the surface of the liquid, where `P_(a)` is the atmospheric pressure `=1.013xx10^(5)P_(a)`. If the atmospheric pressure is neglected then
`P=rhogh`
For a given liquid, `rho` is fixed and g is also CONSTANT, then the pressure due to the fluid column is DIRECTLY proportional to vertical distance or height of the fluid column.
Discussion
15438.

In a cubical vessel (1 m xx 1 m xx 1m) the gas molecules of diameter 1.7 xx 10^(-8) cm are at a temperature of 300 K and a pressure of 10^(-4) mm of mercury. The mean free path of the gas molecule is

Answer»

1 metre
4 metre
2.42 metre
1 cm

Answer :C
Discussion
15439.

States and explain the inverse square Law of radiation.

Answer»

SOLUTION :The intensity of HEAT radiation at a point varies inversely as the square of that point from the source.
i.e.,`I prop (1)/(r^(2))`: If `I_(1)`and `I_(2)` be the intensities of the thermal radiation at distance `r_(1)` and `r_(2)`from the source then `I_(1) prop (1)/(r_(1)^(2)) and I_(2) prop (I)/(r_(2)^(2)) therefore(I_(1))/(I_(2)) prop (r_(2)^(2))/(r_(1)^(2))`
Discussion
15440.

What is Oscillatory motion?

Answer»

Solution :When an object or a PARTICLE moves back and forth REPEATEDLY for some duration of TIME, its motion is said to be oscillatory (or VIBRATORY).
Discussion
15441.

A vehicle was moving at 32 km*h^(-1) towards north. A passenger noticed that rain was strikinghis body from the east. When the velocity of the vehicle was increasedto 64 km*h^(-1) , rain appeared to come from the north-east direction. Find the actual velocity and direction of rain.

Answer»


ANSWER :`32 sqrt(2) KM*h^(-1)` ; from south-east
Discussion
15442.

If the energy input to a carnot engine is thrice the work it performs then, the fraction of energy rejected to the sink is

Answer»

`(1)/(3)`
`(1)/(4)`
`(2)/(3)`
`(3)/(4)`

Solution :If `Q_(1)` is the energy input and `Q_(2)` is the energy REJECTED to the sink, then work done `W = Q_(1) - Q_(2)`
Dividing by `Q_(1)` on both SIDES, we get
`(W)/(Q_(1)) = 1 - (Q_(2))/(Q_(1))` or `(Q_(2))/(Q_(1)) = 1 - (W)/(Q_(1))`
As `Q_(1) = 3W` (given) `:. (Q_(2))/(Q_(1)) = 1 - (W)/(3W) = 1 - (1)/(3) = (2)/(3)`
Thus the fraction of energy rejected to the sink is `(2)/(3)`.
Discussion
15443.

Expalin the motion of blocks connected by a string (i) verticla motion (ii) Horizontal motion

Answer»

Solution : Verticalmotion : Let us considertwoblocksof masses `m_(1)` and `m_(2) (m_(1) gtm_(2))`connectedby alightandinextensiblestringthatpassesovera pulley

Let thetensionin thestringbe Tand ACCELERATION aWhenthe systemisreleasedboth theblocksboth thestartmoving`m_(2)`verticallyupwardwithsameaccelerationa. The gravitationalforce `m_(1)`gon MASS `m_(1)` is usedto lift the mass `m_(2)`
theupwarddirection is selectedas YDIRECTION. the free bodydiagrams ofbothmasses.

Horizontal motion : in this case mass `m_(2)` is kept on a horizontal table and mass `m_(1)`is hanging through a smalll pulley it is assumed that there is no friction on the surface

As both the blocks are connectedto the unstrechable stringif `m_(1)` moves with an acclereation a downward athen `m_(2)`also moves iwth the same acceleration a horizontally
The forces acting on mass `m_(2)`are
(i) horizontal tension (T) exerted by the string
(ii) upward normal force (N) exerted by the surface
(III) Downard normal gravitational force `(m_(2)g)`
The forces acting on mass `m_(1)` are
(i)Tension (T) acting upwards
(ii) Downward gravitational force `(m_(1)g)` the free body diagrams for both the masses is
Discussion
15444.

Three particles of masses 1kg,2kg and 3kg are placed at the vertices A,B and C of an equilateral triangle ABC. If A and B lie at (0,0) and (1,0)m, the co-ordinates of their centre of mass are

Answer»

`((SQRT(3))/(2)m"and"(7)/(6)m)`
`((7)/(6)m"and"(sqrt(3))/(4)m)`
`((7)/(12)m"and"(sqrt(3))/(4)m)`
`((7)/(12)m"and"(7)/(12)m)`

ANSWER :C
Discussion
15445.

Match list - I with list - II in uniform circular motion {:("List-I","List-II"),("(a) changes only in direction ","(e) external torque"),("(b) zero magnitude","(f) angular momentum"),("(c) acts along axis of rotation","(g) kinetic energy"),("(d) non zero and constant (having no direction)","(h) linear momentum"):}

Answer»

a-e, b0h, c-f, d-g
a-h, b-e, c-g, d-f
a-h, b-f, c-e, d-g
a-h, b-e, c-f, d-g

ANSWER :D
Discussion
15446.

In the previous question

Answer»

`W_(1)+W_(2)=W_(3)+W_(4)`
`W_(1)+W_(2)=W_(2)+W_(4)`
`W_(1)+W_(4)=W_(2)+W_(4)`
`W_(1)+W_(4)=W_(2)+W_(3)`

ANSWER :A
Discussion
15447.

Three masses 700 gm, 500 gm and 400 gm are suspended at the end of the spring and they are in equilibrium. When the 700 gm mass is removed, the system oscillates with a period of 3sec. when the 500 gm mass is also removed, it will oscillate with a period of

Answer»

1 SEC
2 sec
3 sec
`SQRT((12)/(5))` sec

Answer :B
Discussion
15448.

A block weighing 10kg is at rest on a horizontal table. The coefficient of static friction between the block and the table is 0.5. If a force acts downward at 60^(@) with the horizontal, how large can it be without causing the block to move ? (g=10ms^(-2))

Answer»

346 N
446 N
746 N
846 N

Solution :`F COS THETA=f, f=mu_(s)N, N=mg+F sin theta`
Discussion
15449.

Is it possible for any moving object that speed is constant but velocity is changing ?

Answer»

Solution :YES, if MOVING object is changing the DIRECTION, then it is POSSIBLE.
Discussion
15450.

A bodyweight 7000gms on the surface of the earth. How much will it weight on the surface of a planet whose mass is 1/7 and radius half of the earth

Answer»

200 GM wt
400 gm wt
500 gm wt
300 gm wt

Answer :B
Discussion