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14251.

The potential energy of a particle from a distance x from an origin, changes according to the formula U=(Asqrtx)/(x+B) where A and B are constant so the dimension of AB=……

Answer»

`M^(1)L^(5/2)T^(-2)`
`M^(1)L^(2)T^(-2)`
`M^(3/2)L^(3/2)T^(-2)`
`M^(1)L^(7/2)T^(-2)`

Solution :In `U=(Asqrtx)/(x+B)`, the DIMENSION of B
`=M^(0)L^(1)T^(0)""....(i)`
( `:.x+B & x` is distance)
The dimension of `Axx"(distance")^(1/2)`
= The dimension of U `xx` dimension of distance
`("":.Asqrtx=U(x+B))`
`:.` The dimension of A `=(M^(1)L^(2)T^(-)xxL^(1))/(L^(1/2))`
`:.[A]=M^(1)L^(5/2)T^(-2)""......(ii)`
`:.` The dimension of AB `=M^(1)L^(5/2)T^(-2)xxL^(1)`
`:.[AB]=M^(1)L^(7/2)T^(-2)`
Discussion
14252.

A tiny liquid drop is spherical but a larger drop oval shape. Why?

Answer»

Solution :In the case of tiny drop, the force of surface tension of the LIQUID dominates over gravitational force and the drop is spherical. In the case of LARGER drop, its weight PULLS the drop downwards and the drop becomes oval SHAPED.
Discussion
14253.

A block of 1.2 kg moving at 20 cm/s collides head-on with a simmilar block kept rest.The co-efficient of restitution is 3/5 .Find the loss of K.E during the collision.

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SOLUTION :`7.7*10^-3` J
Discussion
14254.

In problem 5, the CM of the plate is now in the following quadrant of xy-plane.

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I
II
III
IV

Solution :As the mass is removed from the `1^(st)` quadrant, the CM will be SHIFTED to REGION of more mass. It MEANS CM will be shifted to `3^(RD)` quadrant.
Discussion
14255.

A moter car blows a horn of frequency 100 Hz . While approaching a large reflector with a uniform velocity of50 kmh^(-1). Calculate the frequency of the echo relative to the passenger of the car . (Velocity of sound= 340 m s^(-1))

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SOLUTION :N/A
Discussion
14256.

The displacement of a particle represented by the equation y= 3 cos((pi)/(4)-2 omega t). The motion of the particle is

Answer»

SIMPLE harmonic with period `(2pi)/(omega)`
simple harmonic with period `(pi)/(omega)`
periodic but not simple harmonic
non-periodic

Solution :The displacement of motion of particle
`y= 3 cos ((pi)/(4)- 2omega t)`
`therefore v= (d)/(DT)[3 cos ((pi)/(4)- 2omega t)]`
`= 3(-2omega) [-sin((pi)/(4)- 2omega t)]`
`therefore v = 6 omega sin ((pi)/(4)-2omega t)`
Acceleration, `a= (dv)/(dt)= (d)/(dt) [6OMEGA sin ((pi)/(4)-2omega t)]`
`therefore a = 6omega xx (-2omega)cos((pi)/(4)- 2omega t)`
`= -12omega^(2) cos ((pi)/(4)- 2omega t)`
`= -4omega^(2) [3cos ((pi)/(4)- 2omega t)]`
`therefore a= -4omega^(2)y"""........"(1)`
`therefore a propto -y`
This is CONDITION of simple harmonic motion, the motion is SHM and
`omega.= 2 omega""[therefore " Comparing equation (1) with "a= -(omega.)^(2)y]`
`therefore (2pi)/(T.)= 2omega`
`therefore T.= (2pi)/(2omega)= (pi)/(omega)`
Discussion
14257.

Find the displacement of a simple harmonic oscillator at which its PE is half of the maximum energy of the oscillator.

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Solution :If the amplitude of oscillator is A and DISPLACEMENT x at time t, then TOTAL energy
`E= (1)/(2) kA^(2)` and
Potential energy `U= (1)/(2) kx^(2)`
but potential energy `PE= (1)/(2) E` is given,
`THEREFORE (1)/(2) kx^(2)= (1)/(2)xx (1)/(2) kA^(2)`
`therefore x^(2)= (A^2)/(2)`
`therefore x= pm (A)/(sqrt(2))`.
Sign `pm` indicates that oscillator will be in one side from mean position.
Discussion
14258.

(A): Two rods of same material but of different lengths are fixed between two rigid points and subjected to same temperature rise. The thermal stress will be more for longer rod.(R): Thermal stress is independent of the length and area of cross-section of the rod.

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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 :B
Discussion
14259.

A body of mass 6kg is under a force which causes displacement in it which is given by s= (tau^(2))/(4)m, where 't' is time. The work done by the force in 2s is

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12J
9J
6J
3J

Answer :D
Discussion
14260.

Unit of moment of inertia

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kgm
`mkg^(-2)`
`kgm^(2)`
`kgm^(-1)`

ANSWER :C
Discussion
14261.

Which of the following is the case of an adiabatic process, where gamma = C_(p)//C_(v)?

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<P>`P^(1-GAMMA)T^(gamma)=`CONSTANT
`P^(gamma)T^(1-gamma)=` constant
`PT^(gamma)`= constant
`P^(gamma)T=`constant

Answer :A
Discussion
14262.

If mass of He atom is 4 times that of hydrogen atom then mean velocity of He is

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2 TIMES of H-mean value
1/2 times of H-mean value
4 times of H-mean value
Same as H- mean value

Answer :B
Discussion
14263.

Chemical reactions are invariably associated with the transfter of energy either in the form of heat or light. In the laboratory, heat changes in physical and chemical processes are measured with an instrument called calorimeter. Heat change in the process is calculated as {:(q = ms DeltaT,,s ="Specific heat"),(=cDeltaT,,c ="Heat capacity"):} Heat of reaction at constant volume is measured using bomb calorimeter. q_(V) = DeltaU = Internal energy change Heat of reaction at constant pressure is measured using simple or water calorimeter. q_(p)= DeltaH q_(p) = q_(V) +P DeltaV DeltaH = DeltaU +DeltanRT The enthalpy of fusion of ice is 6.02 kJ mol^(-1). The heat capacity of water is 4.18 J g^(-1)C^(-1). What is the smallest number of ice cubes at 0^(@)C, each containing one molw of water, the are needed to cool 500g of liquid water from 20^(@)C to 0^(@)C?

Answer»

Solution :`C _(P) =r + C _(V) = (R//M) + C _(V) = (8.3//4xx 10 ^(-3)) + 3129 = 5204 Jkg ^(-1) K ^(-1)`
Discussion
14264.

How does the velocity of a particle in SHM vary? Where will it be maximum ?

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Solution :In SHM velocity of a particle at MEAN POSITION is MAXIMUM. When the particle moves from mean position to extreme position, velocity gradually decreases. At extreme position velocity of the particle is zero.
Discussion
14265.

A body of mass 100 kg falls on the earth from infinity. What will be its velocity on reaching the earth? What will be its K.E.. ? Radius of the earth is 6400 km and g=9.8 ms^(-2). Air friction is negligible.

Answer»


ANSWER :`11.2 KMS^(-1),6.27xx10^(9)`J
Discussion
14266.

Choose the correct option (s)

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with the rise in temperature viscosity decreases
viscosity of liquid increases with increase in pressure but in case of water, viscosity decreases with rise in pressure
viscosity of liquid is about 100 times GREATER than that of gases
viscosity of gases is INDEPENDENT of pressure

Answer :B::C::D
Discussion
14267.

An astronaut of mass m is working on a satellite orbiting the earth at a distance h from the earth.s surface. The radius of the earth is · R, while its mass is M. The gravitational pull F_G on the astronaut is :

Answer»

Zero since astronaut FEELS WEIGHTLESS
`(GMm)/((R+h)^2)lt F_(G) lt (GMm)/R^2`
`F_(G) = (GMm)/((R+h)^2)`
`0 lt F_(G) lt (GMm)/R^2`

Answer :C
Discussion
14268.

A guided missile is fired to strike an object at the same level 38 km away. It may be assumed that it rises vertically 1.5 km and then for the remainder of the flight it follows a parabolic path at an elevation of 45^(@) . Calculate its velocity at the begining of its parabolic path.

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ANSWER :[2177 km/h]
Discussion
14269.

A4 kg blocks A is plced on the top of 8 kg block B which rests on a smoth table .A just slips on B when a force of 12 N is applied on B to make Both A and B move together is (4x) N Find the value of x.

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ANSWER :6
Discussion
14270.

In case of resonance, the characteristic property which is the same for free vibrationof the bodyandthe external periodic force is

Answer»

amplitude
phase
velocity
frequency

Answer :D
Discussion
14271.

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

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`((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
14272.

Can the moment of inertia of a body be different about different axes?

Answer»

Solution :MOMENT of inertia of a BODY DEPENDS on the mass of the body position of the axis of its rotation and the distribution of mass of the body about its axis of rotation. So the moment of inertia of a particular body MAY be different about different AXES.
Discussion
14273.

Applications of Bernoulli's theorem can be see in

Answer»

dynamic lift of aeroplane
hydraullic press
helicopter
none of the above

Solution :Ac cording to Bernoulli's theorem, when aeroplane MOVES, the AIR streams are created on upper and lower face of the wings of aeroplane. Then VELOCITY of air streams on the upper face is more than on lower face. Ac cording to Bernoulli's theorem, the pressure is more on lower face than on upper face of the WINS of aeroplane. This PROVIDES a lift on the aeroplane for its flight.
Discussion
14274.

A uniform rod of length l is held vertically on a horizontal floor fixing its lower end, the rod is allowed to fall onto the ground. Find (i) its angular velocity at that instant of reaching the ground (ii) The linear velocity with which the tip of rod falls to floor.

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SOLUTION :(i) The rod rotates about an axis through one END. From the principle of conservation of mechanical energy. Loss of P.E. of the rod is EQUAL to its gain in rotational K.E.
`therefore mg.(1)/(2)=(1)/(2) I omega^(2) RARR mg.(1)/(2).=(1)/(2).(ML^(2))/(3)omega^(2)`
on solving `omega = sqrt((3g)/(l))`
(ii) `V = r omega` or `V = l omega = l sqrt(3g//l)=sqrt(3gl)`
Discussion
14275.

(a)Pressure decreases as oneascends the atmosphere . If the density of air is rho ,what is the change in pressure dp over differential height dh ? (b)Considering the pressure P to be proportional to the density find the pressure P at a height h if the pressure on the surface of the earth is P_(o).(c ) If P_(o)=1.03xx10^(5)N//m^(-2),rho_(o)=1.29kg//m^(3)andg=9.8m//s^(2) what height will the pressure frop to (1)/(10) the value at the surface of earth ? (d) This model of the atmosphere works for relatively small distance .Identify the underlying assumption that limits the model.

Answer»

SOLUTION :Since the air is less dense at higher up and hence pressure is also less.
(a) Consider a HORIZONTAL portion of air with cross section A and height dh.

Let the pressure on the top surface and bottom surface be P and `P_(1)+dP`. If the portion is in equilibrium then the net upward force must be balanced by the weight.
`(P+dP)A-PA=-mg"" (because"mass"="volume"xx"density")`
`therefore(dP)A=-rho(Adh)g`
`(rho=` density of air)
`thereforedp=-pgdh`...(1)
Negative sign indicates that pressure decreases at height increases .
(b)Let the density of air on the EARTH.s surface be `rho_(o)`, then
Pressure `prop` density
`thereforePproprhoandP_(o)proprho_(o)`
`therefore(P)/(P_(o))=(rho)/(rho_(o))`
`thereforerho=((P)/(P_(o)))rho_(o)`....(2)
PUTTING value of equation (2) in (1),
`thereforedp=-((P)/(10))rho_(o)dgh`
`therefore(dP)/(P)=-(rho_(o)g)/(P_(o))dh`

`thereforeh=(P_(o))/(rho_(o)g)ln(10)`
`thereforeh=(P_(o))/(rho_(o)g)[2.363log_(10)^(10)]`
`=((1.093xx10^(5)))/(1.22xx9.8)xx2.363(1)=0.16xx10^(5)m`
`thereforeh=16xx10^(3)m`
(d)Pressure `Pproprho` (For isothermal prosess)temperature remains constant only NEAR the surface of earth at greater height relation `Pproprho` does not obay.Hence ,for small distances near earth `Pproprho`is obay, this is not for greater distances . This is the limitation of this model .
Discussion
14276.

A block of mass 2 kg rests on a rough horizontal plank , the coefficient of friction between the plank and block is 0.3. If the plank is pulled horizontally with a constant acceleration of 4 m//s^(2) , the distance moved by the block on the plank in 5secondstarting from rest (in m) is : (Take g = 10 m//s^(2))

Answer»

5
10
25
50

Solution :`s = (1)/(2) at^(2)`
Discussion
14277.

An artificial satellite is revolving in a circular orbit at a height of 1200km above the surface of the earth. If the radius of the earth is 6400km, mass is 6xx10^24 kg find the orbital velocity (G=6.67xx10^(-11) Nm^2//kg^2)

Answer»

SOLUTION :`V_0=SQRT((GM)/(R+h))=sqrt(((6.67xx10^(-11))(6xx10^24))/((6400xx10^3)+(1200xx10^3)))`
`=sqrt((6.67xx10^(-11)xx6xx10^24)/(7600xx10^3))`=7.26 km/s
Discussion
14278.

A heavy body and a light body have same momentum. Which one of them has more kinetic energy and why?

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Solution :Lighter body has more KE as `KE = (p^2)/(2M) ` and for constant p , `KEprop l/m`
Lighter body have more KINETIC ENERGY .
Discussion
14279.

When a satellite is lifted from a lower orbit to a heigher orbit (a) Gravitational potential energy increases KE decreases (c) Gravitational PE decreases (d) KE increases

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a is only correct
a & d are correct
a and C are correct
a & B are correct

Answer :D
Discussion
14280.

Why a given sound is louder in a hall than in the open?

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Solution :In a hall, repeated REFLECTIONS of sound take place from the walls and the ceiling. These REFLECTED sounds mix with original sound which RESULTS in increase the INTENSITY of sound. Butin OPEN, no such a repeated reflection is possible. sound will not be louder as in hall.
Discussion
14281.

A body is thrown with the velocity u=12.0ms^(-1) at an angle of alpha=45^(@) to the horizon. It touches the ground at the distance s from the point where it was thrown. From what height h should stone be thrown in a horizontal direction, with the same initial velocity u so that it falls at the same spot ?

Answer»

17 m/s
1.7 m/s
3.4 m/s
7.3 m/s

Answer :D
Discussion
14282.

In the figure a and b AC, DG and GF ar fixed inclined planes BC=EF=x and AB=DE=y. A small block of mass m is released from the point A. It slides down AC and reaches C with speed V_(C). The same block is released from rest from the point D, It slides down DGF and reaches the point F with speed V_(F), Coefficient of kinetic frictions between the block and bot the surfaces AC and DGF are mu. The values of V_(C) and V_(F) is sqrt(kgy-mugx) then find the value of k.

Answer»


ANSWER :2
Discussion
14283.

On a smooth inclined plane a body of mass M is attached between two springs. The other ends of the springs are fixed to firm supports. If each spring has a force constant k, find the period of oscillation of the body

Answer»

Solution :`k_("eff")= 2K`
`:. T= 2pisqrt((M)/(2k))`
PERIOD (or `k_("eff")`) is INDEPENDENT of `theta`
Discussion
14284.

A vessel contains 6 xx 10^(26) molecules/m^(3). Mass of each molecule is 6 xx 10^(-27) kg. Assume that, on an average, one-sixth of the molecules move with a velocity 10^(3) ms^(-1) perpendicularly towards each wall. If the collisions with the walls are perfectly elastic, then which of the following statements are correct?

Answer»

CHANGE in momentum of each molecule is `12 xx 10^(-24)` kg `ms^(-1)` in each collision
The number of MOLECULES hitting 1 `m^(2)` of the WALL per second is `10^(29)`
Total change in momentum of all molecules is `10^(31)` SI units
The number of molecules hitting ONE square metre of the surface is `6 xx 10^(29)`

Answer :A::B
Discussion
14285.

x=a+bt+ct^(2)=xin metre and tis in sec then write unit of a, b, c.

Answer»

SOLUTION :Unit of a is METRE, unit of B is m/s and unit of cis `m//s^(2)`
Discussion
14286.

Change in momentum is given by

Answer»

FORCE `XX` mass
force `xx` DISTANCE`
force `xx` velocity
force `xx` time

Answer :D
Discussion
14287.

In Column-I number and in Column-II their significant digit are given. Match them properly : {:("Column-I","Column-II"),((1) 2.85xx10^26kg,(a)1),((2)0.009m^2,(b)2),((3)0.060s,(c)3):}

Answer»


ANSWER :(1-c), (2-a), (3-b)
Discussion
14288.

The mirror which is used as a rear view mirror is

Answer»

CONVEX MIRROR
CONCAVE mirror
plane mirror
parabolaidal mirror

Answer :A
Discussion
14289.

A system of four masses m_(1),m_(2),m_(3),m_(4),two springs and a fixed pulley is shown in figure. The system is kept at rest by attaching the lower thread to a rigid peg . Determine the acceleration of all the masses after the lower thread has been cut . The springs are massless, pulley is frictionless and massless.

Answer»

Solution :Clearly `(m_(1)+m_(2))gt(m_(3)+m_(4))`
Let `T_(1)` be the tension in left spring and `T_(2)` that in right spring and T is tension in the string passing over the pulley.
Then for equilibrium of system
`T_(1)=m_(2)G,T=(m_(1)+m_(2))g`….(2)
and `m_(3) g+T_(2)=T`
From lastequation
`T_(2)=T - m_(3)g=(m_(1)+m_(2)-m_(3))g`....(3)
After the lower thread is cut off
EQUATIONOF MOTION of mass `m_(1) ` is
`m_(1)g+T_(1)-T=m_(2)a_(3)`....(4)
Equation of motion of mass `m_(2)` is
`m_(2)g-T_(1)=m_(2)a_(2)`....(5)
Equation of motion mass `m_(3)` is
`m_(3)g+T_(2)-T=m_(1)a_(1)`....(6)
Equation of motion of mass `m_(4)` is
`m_(4)g-T_(2)=m_(4)a_(4)`....(7)
USING (1),(2) and (3) we GET
`a_(1)=a_(2)=a_(3)=0`
and `a_(4)=(m_(4)g-T_(2))/m_(4)=(m_(4)g-(m_(1)+m_(2)-m_(3)g)/m_(4)=((m_(3)+m_(4)-m_(1)-m_(2))g)/m_(4)`
Discussion
14290.

For a body travelling with uniform acceleration, its final velocity is v=sqrt(180-7x) , where x is the distance travelled by the body. Then the acceleration is

Answer»

`-8m//s^(2)`
`-3.5m//s^(2)`
`-7m//s^(2)`
`180m//s^(2)`

ANSWER :B
Discussion
14291.

Two particle A and B (of masses m and 4m) are released from rest in the two tunnels as shown in the figure-6.93. Which particle will cross the equatorial plane first?

Answer»

A
B
Both SIMULTANEOUSLY
DATA INSUFFICIENT

ANSWER :C
Discussion
14292.

When two spheres of equal masses undergo glancing elastic collision with one of them at rest after collision they will move

Answer»

opposite to ONE another
in the same direction
together
at right angle to each other

Solution :When two spheres of EQUAL masses UNDERGO a glancing ELASTIC COLLISION with one of them at rest, after the collision they will move at right angle to each other
Discussion
14293.

Which of the following quantities is always negative is SHM ? Here, is displacement , from mean position.

Answer»

F.a
v.s
a.s
F.v

Solution :(C) a and s are ALWAYS in OPPOSITE DIRECTIONS.
Discussion
14294.

Give examples of nearly perfect classic & plastic bodies ?

Answer»

Solution :There is not perfectly ELASTIC body. But behaviour of QUARTZ fiber is very nearer to perfectly elastic body. Real bodies are not perfectly PLASTIC, but behaviour of wet CLAY, butter etc can be taken as exmples for perfectly plastic bodies.
Discussion
14295.

If vec(A)=4N, vec(B)=3N the value of |vec(A)xx vec(B)|^(2)+|vec(A).vec(B)|^(2) then

Answer»

5 N
25 N
144 N
169 N

Answer :C
Discussion
14296.

A 75kg man stands in a lift. What force does the floor exert on him when the elevator starts moving upward with an acceleration of 2ms^(-2) Given : g=10 ms^(-2).

Answer»

SOLUTION :`R-mg = ma`,
`R=mg+ma=m(g+a)`
`=75(10+2)N=900N`
`=(900)/(10)` kg wt. = 90 kg. wt
Discussion
14297.

The length and breadth of a plate are (6 + -0.1)cm and (4 +- 0.2)cm respectively. The area of the plate is

Answer»

`(24 +- 1.6)CM^(2)`
`24.4 cm^(2)`
`23.6 cm^(2)`
`(24 +- 0.02)cm^(2)`

ANSWER :A
Discussion
14298.

If the unit of force is 100N, unit of length is 10m and unit of time is 100s. What is the unit of Mass in this system of units?

Answer»

SOLUTION :`[F] = [MLT^(-2)]`
`[M] = ([F])/([L][T^(-2)]) = ([100N])/ ([10m][100s]^(-2)) =10^(5) KG`
Discussion
14299.

IF your eyeglasses have f=60cm what is your near point?

Answer»

43cm
34cm
46cm
36cm

Answer :A
Discussion
14300.

A railway engine weighing 40 metic ton is travelling along a level track at aspeed of 54 km H^(-1) What additional power is required to maintain the same speed up an incline of1 in 49 given mu = 0.1g= 9.8ms^(-2) .

Answer»

Solution :Here ` m =40` metric ton `= 40 xx 10^(3) kg`
`UPSILON = 54 km h^(-1) = (54 xx 1000)/(60 xx 60) = 15 ms^(-1)`
Now, `cos theta = sqrt(1 - sin^(2) theta) = sqrt(1 - (1//49)^(2)) = 1`
Power required on level TRACK ` P_(1) = F xx upsilon = mu mg xx upsilon`
Power required up an incline `P_(2) = (mg sin theta + mu mg cos theta) upsilon`
Additional power required
`P =P_(2) P_(1) = [ mg sin theta + mu mg cos theta - mu mg] upsilon = [ mg sin theta + mu mg xx 1 - mu mg] upsilon = mg sin theta xx upsilon`
`=40 xx 10^(3) xx 9.8 xx (1)/(49) xx 15 = 120 xx 10^(3) "watt" = 120 kW`.
Discussion