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14051.

An object falls from a bridge that is 45 m above the water. It falls directly into a small row - boat moving with constant velocity that was 12m from the point of impact when the object was released. What was the speed of the boat ?

Answer»

2 m/s
3 m/s
5 m/s
4 m/s

Answer :D
Discussion
14052.

The force acting on a body is mesured as 4.25 N. Round it off with two significant figure............

Answer»

4.3
4.2
both (a) or (B)
4.25

Answer :B
Discussion
14053.

One mole of an ideal gas at a temperature T_(1) expands according to the law (p)/(V)= a constant.Findthe work donw when the final temperature is T_(2)

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Solution :`(p)/(V)= a CONSTATN = k`
initial temperature - `T_(1) `
Final temperature - `T_(2)`
`W = intPdv`
`W=overset(V_2)underset(V_i)intkvdv`
`W = (k(V_(2)^(2)-V_(1)^(2)))/(2)`
`P_(2) = kV_(2),P_(1) =kV_(1)`
`W = (P_(2)V_(2)-P_(1)V_(1))/(2) =(R(T_(2)-T_(1)))/(2)`
Discussion
14054.

A flywheel having a radius of gyration of 2m and mass 10 kg rotates at an angular speed of 5 rad s^(-1) about an axis perpendicular to it through its centre. The kinetic energy of rotation is

Answer»

500 J
2000 J
1000 J
250 J

Solution :`I=mK^(2)=10(2)^(2)=40 "kg-m"^(2)`
`K_(R)=(1)/(2)I omega^(2)=(1)/(2)(40)(5)^(2)=500J`
Discussion
14055.

A mass m is vertically suspended from a spring of negligible mass, the system oscillates with a frequency n. What will be the frequency of the system, if a mass 4m is suspended from the same spring?

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`(n)/(2)`
2n
`(n)/(4)`
4n

Solution :`n=(1)/(2pi)SQRT((k)/(m)),n.=(1)/(2pi)sqrt((k)/(4m))=(n)/(2)`
Discussion
14056.

A block sliding along a horizontal frictionless surface with a velocity of 2 ms^(-1) comes to the bottom of a frictionless inclined plane making an angle of 30° with the horizontal and comes to a stop after ascending the inclined plane as indicated. If g = 10 ms^(-2), the value of .h. is

Answer»

5 m
10 m
0.2 m
2 m

ANSWER :C
Discussion
14057.

If the linear momentum is increased by 50%, then kinetic energy will be increased by

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50% 
100% 
125% 
25% 

Solution :Kinetic energy of a BODY, `K = (p^2)/(2m)`
where p is the momentum of the body and m is the mass of the body.
`p. = p + (50)/(100) p = 3/2 p`
Assuming m remains constant
`:. (K^.)/(K) = ((P.)/(P))^(2) = 9/4`
% INCREASE in kinetic energy = `(K. - K)/(K) xx 100%`
`= ((K.)/(K) - 1) xx 100% = (9/4 - 1) xx 100% = 125%`.
Discussion
14058.

A ball A moving with a certain velocity collides, with another ball B of the same mass at rest. If the coefficient of restitution is e, the ratio of the velocities of A and B just after the collision is

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`(1 + E)/(1-e)`
`(1+e)/(2)`
`(1-e)/(2)`
`(1-e)/(1+e)`

ANSWER :D
Discussion
14059.

Calculate the change in internal energy of a block of copper of mass 200 g when it is heated from 25^@ C to 75^@ C. Specific heat of copper = 0.1 cal /g/ ""^(@)C and assume change in volume is negligible.

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SOLUTION :`DQ = cm DeltaT = 0.1 xx 200 ( 75-25) = 100` calorie
` DW = PDV =0`
`dU = dQ - dW= 100-0 =100 ` calorie `= 4200 J`
Discussion
14060.

Two boys simultaneoulsy aim their guns at a bird sitting on a tower. The first boy releases his shot with a speed of 100 m/s at an angle of projection of 30^(@). The second boy is ahead of the first by a distance of 50 m and releaaes hisshot with a speed of 80 m/s. How must he aim his gun so that both the shots hit the bird simultaneously? What is the distance of the foot of the tower from the first boy and the height of the tower?

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ANSWER :`theta=sin^(-1)(5//8),179.26, 82.5 m`
Discussion
14061.

What is the density of lead under a pressure of 2 xx 10^8Nm^(-2) , if the bulk modulus of lead is 8 xx 10^9 Nm^(-2)and initially density of lead is 11.4 g cm^(-3)?

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`12.01 G CM^(-3)`
`11.89 g cm^(-3)`
`11.69 g cm^(-3)`
`12.42 g cm^(-3)`

ANSWER :C
Discussion
14062.

A rectangular tank contains water to a height h. A metal rod is hinged to the bottom of the tank so that it can rotate freely in the vertical plane. The length of the rod is L and it remains at rest with a part of it lying above the water surface. In this position the rod makes an angle theta with the vertical. Assume that y=costheta and find fractional change in value of y when temperature of the system increases by a small value DeltaT.Coefficient of linear expansion of material of rod and the tank arealpha_(1) and alpha_(2) respectively. Coefficient of volume expansion of water is lambda. What is necessary condition for theta to increase?

Answer»


Answer :`1/2 (alpha_(1)+gamma-4alpha_(2))DELTA T; 4alpha_(2) gt alpha_(1)+gamma`
Discussion
14063.

All harmonics are overtones but all overtone are not harmonics. How?

Answer»

SOLUTION :The OVERTONES with frequncies which are overtones. But overtones which are non-integral MULTIPLES of FUNDAMENTAL are not HARMONICS.
Discussion
14064.

Twelve forces each of magnitude 10N acting on a body at an angle of 30^(@) with other forces then their resultant is

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(a) 10 N
(B) 120N
(c) `(10)/(SQRT3)`
(d) ZERO

Discussion
14065.

The graph of potential energy and kinetic energy of a particle in SHM with respect to position is a parabola. Potential energy and kitnetic energy do not vary linearly with position .

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Both 'A' and 'R' are ture 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
A' is false and 'R' is true

Answer :B
Discussion
14066.

A block of metal is heated to a temperature much higher than the room temperature and allowed to cool in a room free from air currents, Which of the following curves correctly represents the rate of cooling ?

Answer»




ANSWER :B
Discussion
14067.

On heating a liquid having coefficient of volume expension alpha in a container having coefficient of linear expansion alpha//2, the level of the liquid in the container would

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rise
fall
remains ALMOST stationary
Cannot be bredicted

SOLUTION : Coefficient of volume EXPANSION of container will become `3((alpha)/2)'` which is grater than coefficient of volume expansion of liquid. Hence, container EXPANDS more.
Discussion
14068.

Viscous force exerted by the liquid floqing between two plates in a streamline flow depends upon the

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AREA of the plates
Pressure of the LIQUID
Temperature of the liquid
Level of the liquid surface

Answer :A
Discussion
14069.

The orbitalperiod of a satellite in a circular orbit of radius r about a spherical planet of mass M and density rho for a low altitude orbit r will be..

Answer»

`SQRT((3PI)/(Grho))`
`sqrt(3piGrho)`
`sqrt((PI)/(Grho))`
`sqrt(22Grho)`

ANSWER :A
Discussion
14070.

The equation of motion of a particle is given by (dp)/(dt)+mw^(2)x=0 when 'p' is the momentum and 'x' is the position. Then the particle is

Answer»

Moves ALONG a STRAIGHT line
Moves along a parabola
Executes SHM
FALLS FREELY under gravity

Answer :C
Discussion
14071.

Assertion: For a mixture of non reactive ideal gases, the total pressure gets contribution from each gas in the mixture. Reason: In equilibrium, the average kinetic energy of the molecules of different gases will be equal.

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If both ASSERTION and reason are true and reason is the correct EXPLANATION os assertion.
If both assertion and reason are true but reason is not be correct explanation of assertion.
If assertion is true but reason is FALSE.
If both assertion and reason are false.

ANSWER :B
Discussion
14072.

A barometer ·with brass scale, which is correct at 0°C, reads 75.000 cm on a day when the temperature is 20°C. Calculate correct reading at 0°C. (Coefficient of real expansion of mercury = 0.00018/°C and coefficient of linear expansion of brass = 0.0000189/0^(@)C.)

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`74.758` CM
`75.748` cm
`74.825` cm
`75.847` cm

Answer :A
Discussion
14073.

A car moving with a velocity of 72 KMPH stops its engine just at the foot of a hill. Ifthe car ascends a height of 12m, the energy lost in overcoming friction is (g = 10 m//s^(-2))

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0.4
0.25
0.6
0.5

Answer :A
Discussion
14074.

The instantaneous angular position of a point on a rotating wheel is given by the equationtheta (t) = 2t^3 - 6t^2 .The torque on the wheel becomes zero at : .......... .

Answer»

`t = 1s`
`t = 0.5 s`
`t= 0.25 s `
`t =2s`

Solution :According to equation , TORQUE ` TAU = 0` it means that ` PROP = (d^2)/( dt^2)`
here ,` THETA = 2t^3 - 6t^2`
` alpha = (d^2)/(dt^2) (2t^3 - 6t^2) = 12T = 12`
` [ prop= (d^2 t)/(dt^2) = 0 ] = 12t - 12 = 0`
` therefore t = 1s `
Discussion
14075.

Passage-II: Coefficient apparent expansion of a liquid is gamma_(H) = ( "mass excelled")/( " (remaining mass) ( change in temperature)") A glass bulb of volume 250 cc is completely filled with mercury at 20°C. Temperature of the system is raised to l 10°C. If the coefficient of linear expansion of glass is 9 xx 10^(-6)//^(@) C and the coefficient of absolute expansion of mercury is 1.8 xx 10^(-4)//""^(@) C. Volume of mercury that overflows is

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`3.44 CC`
`4.34 cc`
`7.88 cc`
`6.88 cc`

ANSWER :A
Discussion
14076.

Passage-II: Coefficient apparent expansion of a liquid is gamma_(H) = ( "mass excelled")/( " (remaining mass) ( change in temperature)") In the above question coefficient of apparent expansion of the liquid is

Answer»

`10^(-4)//""^(@) C`
`4 XX 10^(-4)//""^(@) C`
`6 xx 10^(-4)//""^(@) C`
`8 xx 10^(-4)//""^(@) C`

ANSWER :B
Discussion
14077.

Passage-II: Coefficient apparent expansion of a liquid is gamma_(H) = ( "mass excelled")/( " (remaining mass) ( change in temperature)") If 52 gm of a liquid is heated in a vessel from 0^(@) C to 100^(@) C. 2 gm of the liquid is expelled. If 104 gm of the same liquid taken in the same vessel and heated from 0^(@) C to 50^(@)C. Then the mass of the liquid expelled is

Answer»

4 gm
3 gm
2 gm
1 gm

Answer :C
Discussion
14078.

Statements :(a) Allgebric sum of momentum of mass about centre of mass is equal to zero(b) x-co-ordinate of centre of mass of system of particles in a plane is represented byx_(cm)=(1)/(M)sum m_(i)x_(i)(C ) x-co-ordinate of a rigidbody of continuous mass distribution represented byx_(cm)=(1)/(M)int x.dm

Answer»

a and B are TRUE
b and C are true
a and c are true
all a, b, c are true

ANSWER :D
Discussion
14079.

Assume that the acceleration due to gravity on the surface of the moon is 0.2 times the acceleration due to gravity on the surface of the earth. If Re is the maximum range of a porjectile on the earth.s surface then what is the maximum range on the surface of the moon for the same velocity of projection:

Answer»

`0.2 R_(E)`
`2R_(e)`
`0.5 R_(e)`
`5 R_(e)`

ANSWER :D
Discussion
14080.

A monochromatic beam of electromagnetic radiation has an intensity of 1W//m^2. Then the average number of photons per m^3 for a 10MeV gamma rays is

Answer»

4166
3000
5000
2083

Answer :D
Discussion
14081.

Assertion:Soft steel can be made red hot by continued hammering on it, but hard steel cannot.Reason: Energy transfer in case of soft iron is large as in hard steel.

Answer»

If both ASSERTION and reason are true and the reason is the correct explanation of the assertion
If both assertion and reason are true but reason is not the correct explanation of the assertion
If asserti on is true but reason is false
If the assertion and reason both are false

Answer :A
Discussion
14082.

If the vectors vec(P)=a hat(i)+a hat(j)+3hat(k) and vec(Q)=a hat(i)-2hat(j)-hat(k) are perpendicular to each other then the positive value of .a. is

Answer»

ZERO
1
2
3

Answer :D
Discussion
14083.

A rotating wheel changes angular speed from 1800 rpm to 3000 rpm in 20 s. What is the angular acceleration assuming to be uniform?

Answer»

`2pi rad s^(-2)`
`60 pi rad s^(-2)`
`40 pi rad s^(-2)`
`90 pi rad s^(-2)`

Solution :Here , Initial angular speed of the wheel , `omega_(0) = 1800 xx (2pi)/(60) "rad" s^(-1) = 60 pi rad s^-1`
Final angular speed of the wheel ,
`omega= 3000 xx (2pi)/(60) rad s^(-1) = 100 pi rad s^(-1)`
Time during which this CHANGE of speed takes PLACE , t = 20 s
Let `alpha` be angular acceleration of the wheel .
`therefore alpha = (omega - omega_(0)) /(t) = (100 pi - 60 pi)/(20) rad s^(-2) = 2pi rad s^(-2)`
Discussion
14084.

A particle is thrown with a velocity u at an angle theta from the horizontal. Another particle is thrown wih the same velocity at an angle a from the vertical. The ratio of times of flight of the two particles will be

Answer»

`Tan 2 theta :1`
`Cot 2 theta :1`
`Tan theta : 1`
`Cot theta : 1`

Answer :C
Discussion
14085.

If the span of a day on the earth would have been 6 days, what would be the percentage decrease in the acceleration due to gravity at the equator compared to that at the poles ? (Radius of the earth =6400 km, value of g at the poles =981 cm*s^(-2)

Answer»


ANSWER :0.055
Discussion
14086.

A box of mass 50 kg at rest is pulled up on an inclined plane 12 m long and 2 m 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))

Answer»

50
100
150
200

Answer :B
Discussion
14087.

A 3 m long organ pipe both at both ends is driven to third harmonic standing wave . If the amplitude of pressure oscillation is 0.1% of the mean atmospheric pressure (P_(0) = 10^(5) N//m^(2)). Find the amplitude of i. particle oscillation and ii. density oscillation. Speed of soundv = 330 m//s , density of air rho_(0) = 1.0 kg//m^(3)

Answer»

<P>

Solution :` 3 = 3(LAMBDA)/(2)`
`lambda = 2 m`
`P_(m) = 100 N//m^(2) , V = 330 m//s , rho_(0) = 1 kg//m^(3)`
i. `P_(m) = Bs_(0) k = rho_(0) v^(2) s_(0) ( 2pi)/( lambda)`
`s_(0)= (lambda P_(m))/( rho_(0) v^(2) 2 pi) = ( 2 xx 100)/( 1 xx 330 xx 330 xx 2 pi)`
ii.`B = -( dp)/( dV//V) = (dp)/( d rho//rho)`
`[ m = rho v rArr 0 = (d rho)/(rho)+ ( d v)/( v)]`
`rArr d rho = (rho d p)/(B)`
`( d rho)_(max) = ( rho p_(m))/( rho v ^(2)) = (100)/(108900) kg//m^(3) = (1)/( 1089) kg//m^(3)`
Discussion
14088.

Length, breadth and thickness of a block is measured using vernier calipers. The percentage errors in the measurements are 2%,1% and 3% respectively. Estimate the percentage error in its volume

Answer»

SOLUTION :`(DELTAV)/Vxx100=(DELTAL)/lxx100+100+(DELTAT)/txx100=2+1+3==6%`
Discussion
14089.

A thin lens is made with a material having refractive index mu=1.5 Both the sides are convex, It is dipped in water, It will behave like

Answer»

a CONVERGENT lens
a DIVERGENT lens
a RECTANGULAR slab
a prism

Answer :A
Discussion
14090.

A student performs experiment with a simple pendulum and measure time period for 20 vibrations. If he measures time for 100 vibrations the error in the measurement of time period will be reduced by a factor of

Answer»

10
`20`
`5`
`80`

ANSWER :C
Discussion
14091.

The moment of inertia of a circular disc of mass m and radius r about an perpendicular axis passing through its centre is

Answer»

`MR^(2)`
`(mr^(2))/(4)`
`(mr^(2))/(2)`
`(5)/(4) mr^(2)`

Answer :C
Discussion
14092.

Calculate the power of an engine which can pull a mass of 500 metric ton up an incline rising I in 100 with a velocity of 10ms^(-1).

Answer»

Solution :`SINTHETA=(1)/(100).Vm//s,m=500xx10^(3)KG`
`F=mg sin theta=500xx10^(3)xx9.8xx(1)/(100)=49xx10^(3)N`
power `P=FV=49 xx 10^(3)xx10=49xx10^(4)` walt = 490 KW
Discussion
14093.

A Stationary light, smooth pulley can rotate without friction about a fixed horizontal axis. A light rope passes over the pulley. One end of the rope supports a ladder with man and the other end supports a counterweight of mass M. Mass of the man is m. Initially, the centre of mass of the counterweight is at a height h from that of man as shown in figure. If the man starts to climb up the ladder slowly, calculate work done by him to reach his centre of mass in level with that of the counterweight.

Answer»


ANSWER :`(M m G H)/((M - m))`
Discussion
14094.

State in the following cases, whether the motion is one, two or three dimensional: An insect crawling on a globe.

Answer»

SOLUTION :TWO DIMENSIONAL
Discussion
14095.

Guess the possible position of the centre of mass of the body shown here ?

Answer»

Solution :Here .B.is HEAVIER than .A. and .C. So the CENTRE of MASS is NEARER to .B..
Discussion
14096.

What is the change in the potential energy going away from earth?

Answer»


ANSWER :INCREASES
Discussion
14097.

Two projectiles of same mass and with same velocity are thrown at an angle of 60^(@) and 30^(@) with the horizontal then which of the following will remain same?

Answer»

TIME of flight
range of projectile
maximum HEIGHT reached
all the above

Answer :A::C
Discussion
14098.

Find the workrequired to be done to increase the volume of a bubble of soap solution having a radius1mm to 8 times . (surface tension of soap solution is 30 dyne cm^(-1))

Answer»

SOLUTION :INITIAL volume `V_(1)` and radius `R_(1)`
New volume will be `V_(2)`
`R_(1)=1CM,T=30" dyne"//cm`
`V_(2)=8V_(1)W=?`
Now `V_(2)=8V_(1)`
`therefore(4)/(3)piR_(2)^(3)=8((4)/(3)piR_(1)^(3))`
`thereforeR_(2)^(3)=(2R_(1))^(3)`
`thereforeR_(2)=2R_(1)`
Now `W=DeltaA*T=8piT[R_(2)^(2)-R_(1)^(2)]`
`=8piT[4R_(1)^(2)-R_(1)^(2)]=24piTR_(1)^(2)`
`=24xx3.14xx30xx1`
`thereforeW=2261erg`
Discussion
14099.

In the horizontal projection, the range of the projectile is

Answer»

`sqrt((2H)/(G))`
`ROOT(u)((h)/(g))`
`root(u)((g)/(2h))`
`root(u)((g)/(2h))`

Answer :B
Discussion
14100.

Attempt to foumulate your 'moral' views on the parctice of science. Imagine yourself stumbling upon a discovery, which has great academic interest but is certain to have nothing but dangerous consequences for the human society. How if at all, will you resolve your dilemma ?

Answer»

Solution :Science is search for truth. If a discovery is of great academic INTEREST but is sure to have dangerous
consequences for the human society, it must be made public. To reveal the truth and the means to prevent
its misuse, both are the responsibilities of the DEVELOPMENT of an atom BOMB, a WEAPON of mass destruction.
The humanity at large has to be EDUCATED to use nuclear energy for peacefulpurposes.
Discussion