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15151.

Two gases are enclosed in a container at constant temperature. One of the gases , which is monoatomic. The ratio of the rms speed of the molecules of the monoatomic gas to that of the molecules of the diatomic gas is .................... .

Answer»

8
4
`2sqrt2`
2

Solution :The RMS SPEED is independed of the ATOMICITY of the gas. `nu=sqrt((3RT)/m)`
`(nu_(mono))/nu_(DI)=sqrt(M_(di)/M_(mono))=sqrt8=2sqrt2`
Discussion
15152.

Doppler effect given an idea of a continuously expanding universe-explain .

Answer»

SOLUTION :Light coming from the distant star can be ANALYSED with the spectrometer. Experiment shows that wavelength of a spectral line for a light source situated for away from Earth is greater than that for same light source on Earth i.e., frequency is comparatively LOW .
From this apparent decrease in frequency , known as red shift in Dopplar effect , we conclude that all distant STARS are receding from Earth which indictes that the Universe is CONTINUOUSLY expanding .
Discussion
15153.

All the particles thrown with same initial velocity would strike the ground. .

Answer»

with same speed.
simultaneously
time would be least for the particle thrown with velocity `v` DOWNWARD i.e., particle `1`.
time would be maximum for the particle `2`.

Solution :(a.,c.,d.) `(KE + PE)_f = (KE + PE)_i` in all situations.
Hence, `KE_f` is also equal as `PE_f = 0`. Hence, all the particles collide with the same speed.
`-h = vt_1 -(1)/(2) "gt"_1^2` [for FIRST particle] …(i)
`-h = -vt_2 - (1)/(2) "gt"_1^2` [for second particle] ...(ii)
From Eq. (i) and Eq. (ii), `t_2 gt t_1`
`t_2` = maximum, `t_1` = MINIMUM
i.e., options ( c) and (d) are correct.
Discussion
15154.

Is dimensional analysis applicable to only SI ?

Answer»

SOLUTION :No. It is APPLICABLE to all SYSTEMS.
Discussion
15155.

A particle is placed at the point A of a frictionless track ABC as shwon in figure. It is pushed slightly towards right. Find its speed when it reaches the point B? [g=10ms^(-2)]

Answer»

`5MS^(-1)`
`sqrt(5)MS^(-1)`
`10ms^(-1)`
`sqrt(10)ms^(-1)`

Answer :D
Discussion
15156.

A certain block weighs 15N in air. But it weighs only 12N when completely immersed in water. When immersed completely in another liquid, it weighs 13N. Calculate the relative density of (i) the block and (ii) the liquid.

Answer»

Solution :(i) RELATIVE density of body = `(W_("air"))/(W_("air")-W_("WATER"))`
where `W_("air")=15N` (weight of the body in air)
and `W_("water")=12N` (weight of the body in water)
`therefore R.D_(B)=(15N)/(15N-12N)=5`
(ii) `R.D_(L)=("loss in weight in LIQUID")/("loss in weight in water")=(15-13)/(15-12)=(2)/(3)`
Discussion
15157.

Compare the equations of linear motion with rotational motion .

Answer»

SOLUTION :
Discussion
15158.

How can we reduce the random errorsin an experiment?

Answer»

Solution :Random errors can be decreased by increasing the NUMBER of times the reading is takes. Taking many readings and USING their mean in the FINAL CALCULATION can reduce the random errors in a RESULT.
Discussion
15159.

Let the angle between two nonzero vectors vecA and vecB be 120^(@) and its resultant be vecC.

Answer»

C must be EQUAL to |A-B|
C must be LESS than |A-B|
C must be greater than |A-B|
C may be equal to |A-B|

Answer :C
Discussion
15160.

A door of moment of inertia 4kgm^(2) is at rest. When a torque of 2piNm^("acts") on it find its angular acceleration. Find also its angular velocity after 1s .

Answer»

SOLUTION :`ALPHA=(tau)/(I)=(2pi)/(4)=(pi)/(2)rad//s^(2)`
`omega_(2)-omega_(1)=ALPHAT` `omega_(2)=(pi)/(2)xx1`
`omega_(2)=1.57rad//sec`
Discussion
15161.

Two planets of radii r_1 and r_2 are made from the same material. The ratio of the acceleration due to gravities at the surface of the two planets is

Answer»

`r_1/r_2`
`r_2/r_1`
`(r_1/r_2)^2`
`(r_2/r_1)^2`

ANSWER :A
Discussion
15162.

Two moving particles P and Q are 10 m apart at a certain instant. The velocity of P is 8 m/s making 30º with the line joining P and Q and that of Q is 6 m/s making an angle 30º with PQ as shown in the firuge .Then angular velocity of P with respect to Q is-

Answer»

`0rad//s`
`0.1 rad//s`
`0.4rad//s`
`0.7rad//s`

ANSWER :D
Discussion
15163.

Two parallel glass plates are dipped perpendiculary in a liquid of density p. The separation between the plates is 'd' and the surface tension T. The angle of contact of glass is theta. The capillary rise of the liquid between the plates is

Answer»

` (( T cos theta) /( pd))`
` (( 2T cos theta)/( PGD) )`
` ( (2T)/( pgd - cos theta) )`
` ( ( T cos theta)/( pgd))`

ANSWER :B
Discussion
15164.

For an object moving on a straight line, draw xtotgraphs for : (i) When it is rest. (ii) When it is moving with constant velocity in positive direction. (III) When it is moving with constant velocity in · negative direction. (iv) When it performs non-uniform motion.

Answer»

Solution :Motion of an object can be represented by a position--time graph.
Such a graph is a powerful tool to represent and analyse different ASPECTS of motion of an object.
For motion along a straight line, say X-axis, only x-coordinate varies with time and we have an x-t graph.(1) Let us first consider the simple CASE in which an object is stationary, e.g. : a object STANDING still at x = 40 m. The position-time graph is a straight line parallel to the time axis.

(2) If an object moving along the straight line covers equal distances in equal intervals of time, it is said to be in uniform motion.

(3) The` XTO` graph for any object moving with constant negative velocity is as below.
(4) (a) If object covers different distances in different intervals of time, it is said to be nonuniform motion. The `x to` t graph for it is as below:

(B) Let us consider the motion of a car that starts from rest at time t = Os from the origin O.

Picks up speed till 10s and thereafter moves with uniform speed till t = 18 s. Then the brakes are applied and the car stops at t 20 s and x = 296 m. The position-time graph for this case is as above.
Discussion
15165.

In pur rolling, fraction of its total energy associatedd with rotation is alpha for a ring and beta for a solid sphere. Then

Answer»

`alpha=(1)/(2)`
`alpha=(1)/(4)`
`beta=(2)/(5)`
`beta=(2)/(7)`

ANSWER :A::D
Discussion
15166.

An asteroid of mass m is approaching earth initially at distance of 10R_(e) with speed v_(i). It hits the earth with speed vf (Re and M_(E) are radius and mass of earth then

Answer»

`v_(f)^(2) = v_(i)^(2)(2GM)/(M_(E)R)(1-(1)/(10))`
`v_(f)^(2) = v_(i)^(2)(2GM_(e))/(R_(e))(1+ (1)/(10))`
`v_(f)^(2) = v_(i)^(2)(2GM_(e))/(R_(e))(1-(1)/(10))`
`v_(f)^(2) = v_(i)^(2)(2Gm)/(R_(e))(1-(1)/(10))`

Answer :D
Discussion
15167.

A body of mass m moving with velocity u collides head on elastically with another body of mass 2m at rest. The ratio of kinetic energy of the colliding body before and after collision-is

Answer»

`4:1`
`9:1`
`2:3`
`5:4`

ANSWER :B
Discussion
15168.

The coordinates of a moving particle at any time 't' are given by x= alpha t^(3) and y= beta t^(3). The speed of the particle at time 't' is given by

Answer»

`3t SQRT(ALPHA^(2)+BETA^(2))`
`3t^(2) sqrt(alpha^(2)+beta^(2))`
`t^(2) sqrt(alpha^(2)+beta^(2))`
`sqrt(alpha^(2)+beta^(2))`

ANSWER :B
Discussion
15169.

Two bodies of masses m_(1) and m_(2) are moving with velocities 1ms^(-1) and 3ms^(-1) respectively in opposite directions. If the bodies undergo one dimensional elastic collision, the body of mass m_(1) comes to rest. Final the ratio of m_(1) and m_(2).

Answer»

Solution :`u_(1)=1ms^(-1),u_(2)=-3ms^(-1),v_(1)=0`
`v_(1)=((m_(1)-m_(2))/(m_(1)+m_(2)))u_(1)+((2m^(2))/(m_(1)+m^(2)))u_(2)`
`0=((m_(1)-m_(2))/(m_(1)+m_(2)))1+((2m^(2))/(m_(1)+m^(2)))(-3)`
`m_(1)-m_(2)=6m_(2), "" m_(1)=7m_(2),(m_(1))/(m_(2))=(7)/(1)`
Discussion
15170.

When impurities are added to a liquid it.s surface tension a. always increases b. always decreases c. If impurities are sparingly soluble surface tension decreases L d. If impurities are highly soluble surface tension increases

Answer»

a,C are correct
c,d are correct
b,d are correct
a,d are correct

Answer :B
Discussion
15171.

There is a change in length when a 33000 N tensile force is applied on a steel rod of area of cross - section 10^(-3). The change of temperature requiredto produce the sameelongation, if the steel rod is heated, is (modulus of elasticity of steel =3xx10^(11)N//m^(2), coefficient of linear expansion of steel =1.1xx10^(-5)//^(@)C........................

Answer»

`20^(@)C`
`15^(@)C`
`10^(@)C`
`0^(@)C`

Solution :`Deltal=alpha l Deltat=(FL)/(AY)`
`Deltat=(Fl)/(alpha AY)=(33000)/(1.1xx10^(-5)xx10^(-3)xx3xx10^(11))rArr Deltat=10^(@)C`
Discussion
15172.

A vessel filled with water stands on a table. In this side, the vessel has a small orifice arranged at a distance 25 cm from the bottom of the vessel and at a distance 16 cm from the level of water which is kept constant. The horizontal distance from the orifice at which the jet of water fall on the table is

Answer»

0.2 m
0.4 m
0.6 m
0.8 m

Answer :B
Discussion
15173.

A wave pulse on a string hasthe dimensions shown in Fig. 7.24 at t = 0. The wave speed is 5.0 m//s. (a) If point O is fixed end , draw the total wave on the string att = 1.0 ms , 2.0 ms , 3.0 ms , 4.0 ms , 5.0 ms , 6.0 ms and 7.0 ms. (b) Repeat part (a) for the case in which point O is a free end .

Answer»

SOLUTION :a. The wave form for the given TIMES , RESPECTIVELY , is SHOWN.

Discussion
15174.

the distance traversed by a particle moving along a stranigt line is given by x=180t+50t^(2) metre .find the initial velcoity of the particle. (ii)the accelerationof the particle .

Answer»


Answer :`(i) 180 MS^(-1) (ii) 580 ms^(-1) (iii) 100 ms^(-2)`
Discussion
15175.

Two simple harmonic motions of angular frequencies 100 and 1000 rad/s have the same displacement amplitude. The ratio of their maximum acceleration is………….

Answer»

`1:10`
`1:10^(2)`
`1:10^(3)`
`1:10^(4)`

SOLUTION :The MAGNITUDE of the maximum ACCELERATION is given by
`a_(max)=OMEGA^(2)Aimplies((a_(max))_(1))/((a_(max))_(2))=(omega_(1)^(2))/(omega_(2)^(2))=((100)/(1000))^(2)=(1)/(10^(2))`
Discussion
15176.

A man is looking at a small object placed at his near point. Without altering the position of his eye or the object. He puts a simple microscopeof magnification flying power 5X before his eyes. The angular magnification achieved is

Answer»

5
2.5
1
0.2

Answer :C
Discussion
15177.

The torque in rotational motion is analogous to _____in translational motion.

Answer»

LINEAR MOMENTUM
mass
couple
FORCE

ANSWER :D
Discussion
15178.

A loop and a disc roll without slipping with same linear velocity v. The mass of the loop and the disc is same. If the total kinetic energy of the loop 8J, find the kinetic energy of the disc (in J).

Answer»


ANSWER :6
Discussion
15179.

The far point of a shorsighted woman is 75cm from the eye. What focal length lens should the woman use in order to focus on a very distant object Also find power of the lens

Answer»

`-1.33D`
`-2.33D`
`-3.4D`
`-1.9D`

ANSWER :A
Discussion
15180.

A physical quantity is given by h=(Fv^2)/L is the force, v is the velocity and Lis the angular momentum. Find the dimensions of h.

Answer»

SOLUTION :`H=(Fv^2)/L[h]=([MLT^-2][LT^-1]^2)/[[ML^2T^-1]]=[M^0LT^-3]`
Discussion
15181.

Arrange the following materials in the increasing order of elasticity a) steel b) lead c) rubber d) glass

Answer»

C,b,d,a
a,b,c,d
a,d,b,c
b,d,a,c

Answer :1
Discussion
15182.

A particle moving with a constant acceleration describes in the last second of its motion 9/25th of the whole distance. If it starts from rest, how long is the particle in motion and through what distance does it move if it describes 6 cm in the first sec.?

Answer»

5S , 150 CM
10s , 150 cm
15 s , 100 cm
None

Answer :A
Discussion
15183.

A body 'A' with a momentum 'P' collides with another identical stationery body 'B' one dimensionally. During the collision, 'B' given an impulse 'J' to the body 'A'. Then the coefficient of restitution is

Answer»

Solution :ACCRODING to law of conservation of linear momentum, `m_(1)u_(1) + m_(2)u_(2)= m_(1)v_(1) + m_(2)v_(2)`
i.e., `m u + m(0)= mv_(1) + mv_(2)`
`rArr P- P_(1) = P_(2) " where " P_(2)=J` given
`THEREFORE` Coefficient of restitution,
`e= (v_(2)-v_(1))/(u)= (mv_(2)-mv_(1))/(m u)= (P_(2)-P_(1))/(P)`
`=(P_(2) -(P- P_(2)))/(P)= (2p_(2)-P_(1))/(P) therefore e= (2J-P)/(P)= (2J)/(P)-1`
Discussion
15184.

A piece of metal weighs 46 g in air. When it is immersed in a liquid of specific gravity 1.24 at 27^(@)C, it weight 30 g. When the temperature of the liquid is raised to 42^(@)C, the metal piece weighs 30.5 g. Specific gravity of liquid at 42^(@)C is 1.2. Calculate the coefficient of linear expansion of the metal.

Answer»

Solution :Loss of weight of a BODY at `27^(@)C=46-30=16g`
Loss of weight of body at `42^(@)C=46-30.5=15.5g`
If V is the VOLUME of body at `27^(@)C`,
volume at `42^(@)C(V^(1))=V(1+gamma_(s)t)`
`V^(1)=V(1+gamma_(s)(15))` where `gamma_(s)` is the volume coefficient of expansion of SOLID.
Since loss of weight of body = weight of the liquid
displaced = `Vd_(1)G`
We have 16 = Vd {d = density at `27^(@)C`}
15.5 = `V^(1)d^(1)" "{d^(1)="density at "42^(@)C}`
`16/15.5=V/V^(1)*d/d^(1)or32/31=V/V^(1)*d/d^(1)`
or `32/31=1/(1+15gamma_(s))xx31/30`
`rArr1+15gamma_(s)=31/30xx31/32=961/960`
`15gamma_(s)=961/960-1=1/960`
`gamma_(s)=1/960xx1/15=3alpha_(s)`, where `alpha_(s)` is the linear coefficient of expansion of solid.
`alpha_(s)=1/(960xx45)=0.000024=2.4xx10^(-5)//K`
Discussion
15185.

Newton's first law of motion describes the

Answer»

energy
work
inertia
momentum

Answer :C
Discussion
15186.

At what temperatures (in 0^(@)C) will the speed of sound in air be 3 times its value at 0^(@)C?

Answer»

Solution :We have `v prop sqrtT implies (v _(2))/( v _(1)) = SQRT (( T _(2))/( T _(1)))`
` therefore (v _(1))/( 3 v _(1)) = sqrt (( 0 + 273)/( T _(2)))`
` therefore (1)/(9) = sqrt ((0 + 273)/(T _(2)))`
`therefore 1/9 = (273)/(T _(2))`
`therefore T _(2) = 2457 K = (2457 - 273)^(@)C`
`therefore T_(2) = 2184^(@)C`
Discussion
15187.

A person with dark skin when compared to a person with white skin will experience

Answer»

Less BEAT, less cold
Less HEAT, more cold
More heat, more cold
more heat, less cold

Answer :C
Discussion
15188.

The upper and lower fixed points of a faulty mercury thermometer are 210^@F and 34^@Frespectively . Find the correct temperature read by this thermometer .

Answer»


ANSWER :`122^@F`
Discussion
15189.

A body of mas 1 kg is taken from infinilty to a point P. When the body reaches that point, it has a speed of 2m^(-1). The work done by coservative force is -5J. Which of the following is true (assuming non conservation and pseudo-forces to be absent).

Answer»

<P>Work done by applied force is +7J.
The TOTAL energy possessed by the body at P is+7J
The potential energy possessed by the body at P is +5J
Work done by all forces TOGETHER is EQUAL to the change in kinetic energy.

Answer :A::B::C::D
Discussion
15190.

In which of the following processes, heat is neither absorbed nor released by a system ?

Answer»

isochoric
isothermal
adiabatic
isobaric

Solution :Adiabatic PROCESS is the process in which no exchange of HEAT energy TAKES place between the gas and the surroundings, i.e., `DELTA Q = 0`.
Discussion
15191.

By increasing the volumeof water coming outper secondthe water pumpto n times , what is the powerof motorhas to beincreases ?

Answer»

Solution :Power `P = W/t =(1/2mv^(2))/t`
Where m = mass of WATER coming out from pump .
` :. P =(mv^(2))/(2t)`
` =(Vrhov^(2))/(2t) "" [ :. m = rhoV] `
` :. P = (Alrhot^(2))/(2t) "" [ :. V = Al] `
V = volumeof water coming from the pumpin time t
A = CROSS sectional area of pipe ( is constant )
l = length of FLOW of water
`v = l/t` velocity of water thrown by pump
t = density of water
` :. P = 1/2 A rhov^(2) xx1/t`
` =1/2 Arhov^(3) "" [ :. l/t =v] `
` :. P = 1/2 Arho v^(3)` , here A and `rho `are constant
` :. P prop v^(3) ""....(1)`
` rArr`VOLUME of water coming from cross SECTION of pipe of length l per second
`V = V/t =(Al)/t`
` :. V = Av - to [l/t = v] `
If by increasing the value of V to n times then V = Av (here A is constant )
` :. ` The value of v has to done to n times so from equation (1) ,
` :. ` The value of v has to done t n times so from equation (1) , New power of pump ,
`P prop (nv)^(3)`
` :. P prop n^(3) v^(3)`
` :. P prop n^(3) P `
Discussion
15192.

The angle between vecF =(1,-3, 1)and vecd =(2,-3,-11) will b e ...... rad.

Answer»

0
`(pi)/(4)`
`(pi)/(2)`
`pi`

SOLUTION :`cos theta = (VEC(F).vec(D ))/(Fd ) = ((1,-3,-1).(2,-3,-11))/(sqrt(11)sqrt(134))`
`cos theta= (2+ 9 - 11)/(sqrt(11) squr(134))`
`=(0)/(sqrt(1474))`
`=0`
`theta= (pi)/(2)`
Discussion
15193.

In the arrangement shown in figure match the following {:("column1","column2"),("A Velocities of center of mass","P 2SI unit"),("B Velocity of combined mass when compression in the spring is maximum","1 SI unit"),("C Maximum compression in the spring","4 SIunit"),("D Maximum potential energy stored in the spring","0.5SI unit"):}

Answer»


Solution :`v_(CM)=(m_(1)v_(1) + m_(2)v_(2)) = 1ms^(-1)`
During maximum COMPRESSION ALSO, velocity of combined MASS is `ms^(-1)`.
Now, `U_(max) = K_(1) - K_(2)`
`1/2KX_(max)^(2) = 2J`
We have, `X_(max) = 1m`
Discussion
15194.

Water is filled into a container with hexagonal cross section of side 6 cm. If the surface tension of water is 0.075Nm^(-1)then the surface energy of water will be

Answer»

`8.5 xx 10^(-4) J`
`2.8 xx 10^(-4 )J`
`6.4 xx 10^(-4) J`
`7 xx 10^(-4) J`

ANSWER :D
Discussion
15195.

A refrigerator works between 4^(@)C and 30^(@)C. It is required to remove 600 calories of heat every second in order to keep the temperature of the refrigerator space constant. The power required is (Takae 1 cal = 4.2 Joules)

Answer»

236.5 W
2365 W
2.365 W
23.65 W

Solution :Given, `T_(2) = 4^(@)C = 277 K, T_(1) = 30^(@)C = 303 K`
`Q_(2) = 600` cal per second
Coefficient of performance.
`alpha = (T_(2))/(T_(1) - T_(2)) = (277)/(303 - 277) = (277)/(26)` Also, `alpha = (Q_(2))/(W)`
`:.` Work to be done second = power REQUIRED
`= W = (Q_(2))/(alpha) = (26)/(277) xx 600` cal per second
`= (26)/(277) xx 600 xx 4.2 J` per second = 236.5 W
Discussion
15196.

Discuss about the formation of waves on stretched string?

Answer»

Solution :Take a long string and tie ONE end of the string to the wall as shown in Figrue. 11.4 (a). A QUICK jerk, is 92 a bump (LIKE pulse) is produced in the strig as shown in Figure 11.4(b). Such a disturbance is sudden and it LASTS for a short duration, hence it is known as a wave pulse. If jerks are given continuously then the waves produced are standing waves. Similar waves are produced by a PLUCKED string in a guitar.
Discussion
15197.

An ornament weighting 36 gin air, weighs ony 34 g in water. Assuming that some copper is mixed wilth godl to prepare the ornament, find the amount of coper in it. Specific gravity of godl is 19.3 and that of copper is 8.9.

Answer»


SOLUTION :Given that `m_1a=36gm`
`m_2 water =34GM`
Density of GOLD `rho_(An)=19.3gm/cc`
` Density of copper, rho_(Cu)=8.9gm/cc`
we know that loss of `wt=wt` of displaced water
`=36-34=2gm
Here, `m_c=m_(Au)+m_(Cu)`
`=36gm`…….i
Let v be the volume of the orN/Ament in cm.
S, `Vxxrho_wxx=2xxg`
`rarr (v_(Au)+V_(Cu))xxrho_wxxg=2xxg`
`rarr (m/(rho_(Au))+m/(rho_(Cu)))rho_wxxg=2xxg`
`rarr ((m_(Au))/19.3+(m_(Cu))/8.9)x1=2`
`rarr 8.9m_(Au)+19.3m_(Cu)=2xx19.3x8.9`
`=343.54`..........ii
`Form eq i and ii ltbr. `8.9(m_(Au)+m_(Cu))=8.9xx36`
i. `8.9m_(Au)+8.9m_(Cu)=320.40`.........iii ltbr. `from ii and iii we get ,
`10.4m_(Cu)=23.14`
`=m_(Cu)=2.225gm`
so the AMOUNT of copper in the oerN/Ament is 2.2g.
Discussion
15198.

The equation of motion of particle is described in column-I . At t=0 particle is at origin and at rest {:("Column-I","Column-II"),("A) "x=3t^(2)+2t,"P) Velocity of particle at "t=1sec " is " 8m//s),("B) " V=8t,"Q) Particle moves with uniform acceleration"),("C) "a=16t,"R) Particle moves with variable acceleration"),("D) "V=6t-3t^(2),"S) Acceleration of the particle at " t=1 " is " 2 m//s),(,"T) Particle will change its direction of motion"):}

Answer»


ANSWER :A-PQ; B-PQ; C-PR; D-RT
Discussion
15199.

An electric pump on the ground floor of a building taken 15 minutes to fill a tank of volume 30 m^(3) with water. If the tank is 40m above the ground and the efficiency of the pump is 30%, find the electric power consumed by the pump in filling the tank. [density of water = 10^(3) kg//m^(3)]

Answer»

Solution :mass of the water to be filled= (VOLUME of the TANK) `xx` density of water
`m= 30 xx 10^(3)kg `
H= 40m, t `=15 xx 60 = 99sec`
Efficiency `eta= 30%`
Inputpower `p_(i) = (100)/(eta) xx (mgh)/(t)`
`=(100 xx 30 xx 10^(3) xx 9.8 xx 40)/(30 xx 900)= 43.56KW`
Discussion
15200.

A needle floats in clean water, but sinks in soap solution. Why?

Answer»

Solution :The SURFACE TENSION of water being large as compared to SOAP solution. When the surface tension is high, the upper surface of water acts as stretched membrane. It can support the weight of the needle. So, a needle floats in clean water.
Discussion