Explore topic-wise InterviewSolutions in Current Affairs.

This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.

1.

A non-linear triatomic gas 50% dissociates into individual atoms. Ignoring the vibrational degree of freedom and any further dissociation, the gas behaves as if the number of degrees of freedom is

Answer»

`4.67`
`4.87`
`4.54`
`4.32`

ANSWER :B
Discussion
2.

A simple pendilum is swinging in a vertical plane. The ratio of its potential energies when it is making angle 30^(@) and 60^(@) with the vertical is,

Answer»

`1:(2-sqrt(3))`
`(2-sqrt(3)):1`
`1:2`
`2:1`

ANSWER :B
Discussion
3.

Two balls A and B are thrown vertically upwards from the same location on the surface of the earth with velocities 2 sqrt((gR)/(3))and sqrt((2gR)/(3))respectively, where R is the radius of the earth and g is the acceleration due to gravity on the surface of the earth. The ratio of the maximum height attained by A to that attained by B is

Answer»

2
4
8
`4 sqrt2`

Solution :If h is the MAXIMUM HEIGHT ATTAINED , then by the lawof CONSERVATION of energy
`1/2 mv^2 - (GMm)/( R) = - (GMm)/((R + h))`
or` v^2 = (GMh)/(R(R + h)) = (2ghR)/((R+h)) ""( because g = (GM)/(R^2))`
For ball A `, v = 2 sqrt((gR)/(3)) , h = h_A`
` therefore (4gR)/(3) = (2gh_A R)/((R + h_A)) " or " h_A = 2R`
For ball B, v ` = sqrt((2gR)/(3) ) ,h = h_B`
`therefore (2gR)/(3) = (2gh_B R)/((R + h_B)) " or " h_B = R/2 therefore h_A/h_B = 4`
Discussion
4.

Find the product of 1.2, 2.54 and 3.257 with due regard to significant figures.

Answer»

Solution :Out of the three number 1.2 has GOT the least number of significant figures -TWO. We should round off the other numbers to `2 + 1= 3` significant figures and carry the multiplication. 2.54 has three significant figures and hence NEEDS no rounding off. 3.257 is to be ROUNDED off to 3.26. `1.2 xx 2.54 xx 3.26= 9.93648`. But the result should be limited to the least number of significant digits-that is two digits only. The final answer should be written as 9.9 after rounding off.
Discussion
5.

Derive an expression for the acceleration of a body of mass m moving with a uniform speed v in circular path of radius r.

Answer»

Solution :Let A = initial position of the body ,
B = position of the body after a small time INTERVAL t.
So, the angular displacement `theta= angleAOB`, is also small, and the displacement x effectively coincides with the arc AB of the CIRCULAR path. Then,
Now at A, the velocity v is horizontal , at B, the velocity makes the same angle `theta` with the horizontal direction, keeping its magnitude v to be the same, as the body moves with a uniform speed v . Again, as `theta` is small, the CHANGE of velocity u, effectively coincides with a circular arc.
So, `theta = (u)/(v)`
Then we get, `(x)/( r) = (u)/(v)`
or,u = `(v)/(r )x`
Hence, the acceleration is,
a = `("change of velocity")/("time") = (u)/(t) = (vx)/(r t)`
= `(v)/(r ) v = (v^(2))/(r )`
also SHOWS that, as `theta` is small, u is normal to the velocity v. As v is TANGENTIAL at every point on the circular path, the acceleration a is radial. This is centripetal acceleration of a uniform circular motion.
Discussion
6.

Two rods of different metals, having the same area of cross-section A, are placed end to end between two massive walls as shown in fig. If the temperature of both the rods are now raised by Delta t^@ Cthen a) Find the force with which the rods acts on each other at higher temperature. b) Find the lengths of the rods at the higher temperature. Assume that there is no change in the cross-sectional area of the rods and the rods do not bend. There is no deformation of walls.

Answer»

Solution :a) Due to heating the increases in LENGTH of the composite rod will be
`(Delta l)_1 = l_1 alpha_1 Delta t + l_2 alpha_2 Delta t = (l_1 alpha_1 + l_2 alpha_2 ) Delta t `......(1)
due to compressive force .F. from the walls, due to ELASTICITY, the decrease in length will be
`(Delta l)_D = (Fl_1)/(AY_1) + (Fl_2)/(AY_2) = F/A (l_1/Y_1 + l_2Y_2)`.....(2)
As the length of the composite rod remains unchanged the increase in length due to heating MUST be EQUAL to decrease in length due to compression.
`(l_1 alpha_1 + l_2 alpha_2) Delta t = F/A (l_1/Y_1 + l_2Y_2)`
` rArr F = (A(l_1 alpha_1 + l_2 alpha_2)Delta t)/((l_1/Y_1 + l_2/Y_2))`
b) As initially the length of one rod in `L_1`and due to heating its length increases by `(Delta l_1)_H = alpha_1 l_2 Delta t` ,while due to compression its length decreases by
`(Delta l_1)_C = (Fl_1)/(AY_1)`
so its final length` l_1^1 = l_1 + (Delta l_1)_H - (Delta l_1)_C`
` = l_1 + l_1 alpha_1 Delta t- (Fl_1)/(AY_1)`
For the other rod `l_1^1 = l_2 + (Delta l_1)_H - (Delta l_2)_C`
`l_2^1= l_2 + l_1 alpha_2 Delta t- (Fl_2)/(AY_2)`
Discussion
7.

A particle who has a constant speed of 50 m/s. it moves along a straight line from A(2,1) to B(9,25). Find its velocity vector. If at initial instant of time it is located at (2, 0), find its final location after 3 s.

Answer»

SOLUTION :`vec(AB)=vec(B)-vec(A)=7hat(i)+24hat(j)`
`vec(v)=(HAT(A)B)|vec(v)|=((7hat(i)+24hat(j)))/(25)xx50 =14hat(i)+48hat(j)`
`vec(s)_(2)-vec(s)_(1)(3), and vec(s)_(1)=2hat(i)`
`vec(s)_(2)=(42hat(i)+144hat(j))+2hat(i)=(44hat(i)+144hat(j))`
Discussion
8.

A runner jogs a along a straight road (in the +x direction) for 30 min, travelling a distance of 6 km. She then turns around and walks back towards her starting point for 20 min, travelling 2 km during this time. State true/ false: a. The final displacement of the entire trip is 0.16 km min^(-1) . b. Her average speed for the entire is 0.16 km min^(-1). c. The average velocity for the entire trip is 0.4 km min^(-1). d. The runner's average velocity while jogging is 0.4 km min^(-1). e. Her average velocity while walking is 0.1 km min^(-1).

Answer»


Solution :We first sketch `S4.37` a to DEPICT the motion and indicate our chice of locating the origin of the coordinate system where the runner starts. We also show significant positions of the runner during the motion.
,
`{:(Position""Time),(x_(0)=0""t_(0)=0), ("Jogging"""x_(1) =6 km""t_(1) =30 min),("Walking"""x_(2)=4 km""t_(2) =50 min):}`
a. True,`Delta x=(x_(2)-x_(0))=3 km-0) =4 km`
b. True , average =`(Total distance)/(Total time) =(8KM)/(50 mi n)`
c. False, the average velocity for the entire trip is
` v_(av) =(Delta x)/(Delta t) =((x_(2)-x_(0)))/((t_(2)-t_(0)))=(4-0)/(50-0) =0.080 km min^(-2)` ltbrtgt d. False, while jogging
`v_(av)=(Deltax)/(Deltat) =((x_(1)-x_(0)))/((t_(1)-t_(0))) =(6-0)/(30-0) =0.2000 km min^(-1)`
e. True, while walking
`v_(av)=(Delta x)/(Delta t) =((x_(2)-x_(1)))/((t_(2)-t_(1))) =((4 km -6 km))/((50 min-30 mi n))`
`=(-2 km)/(20 mi n) =-0.100 km min^(-1) (walki ng)`
The negative value indicates a velocity in the -`x` direction. A word of CAUTION: Please note that the runner`s average velocity for the entire trip is not SIMPLY the average of her jogging and walking velocities, because the times she spends in the these two different motions are not EQUAL.
Discussion
9.

If vec(C) = vec(A) + vec(B) then

Answer»

`VEC(C)` is ALWAYS GREATER than `|vec(A)|`
C is always equal to A+B
C is never equal to A+B
It is possible to have `|vec(C)| LT |vec(A)|` and `|vec(C)| lt |vec(B)|`

Answer :D
Discussion
10.

On a 60 km straight road, a bus travels the first 30km with a uniform speed of 30kmh^(-1). How fast must the bus travel the next 30 km so as to have average speed of 40kmh^(-1) for the entire tirp?

Answer»

SOLUTION :`V_(AVG)=(S_(1)+S_(2))/(t_(1)+t_(2))=(S+S)/(S((1)/(V_(1))+(1)/(V_(2))))=(2V_(1)V_(2))/(V_(1)+V_(2))`
or `40=(2xx30xxV_(2))/(V_(1)+V_(2))impliesV_(2)=60kmh^(-1)`
Discussion
11.

Figure show the stress-strain curve for a given material. What are (a) Young's modulus and (b) approximate yield strength of the material?

Answer»

SOLUTION :a)From the GRAPH, corresponding to point P,
Stress = `150 xx 10^(6) N//m^(2), ` Strain = `0.002 = 2 xx 10^(-3)`
Young.s MODULUS,
`Y = ("stress")/("strain")= (150 xx 10^(6))/(2 xx 10^(-3))=7.5 xx 10^(10) N//m^(2)`
b) From graph, corresponding to point B (yield point), stress = `300 xx 10^(6) N//m^(2)`. THUS, approximate yield strength = `2 xx 10^(8) N//m^(2)`.
Discussion
12.

Three identical spheres each of radius .R. are placed touching each other so that their centres A, B and C lie on a straight line. The position of their centre of mass from A is

Answer»

`(2R)/(3)`
2R
`(5R)/(3)`
`(4R)/(3)`

Answer :B
Discussion
13.

An object is placed in front of a slab (mu=1.5) of thickness 6 cm at a distance 28 cm from it . Other face of the slabs silvered. Find the position final image.

Answer»

Solution :By the principleof reversibility of light, we can say if light rays are coming from the mirror and PASSING through the slab, the mirror will shift 2cm TOWARDS right for OBSERVER in fron of the slab.
The position of the object from SHIFTED mirror =32 cm
So, the position of the image formed by shifted mirror will be 32 cm BEHIND it. Hence, position of the image from surface 2 is 30 cm left to it and 36 cm left of surface 1.
Discussion
14.

A circular platform is free to rotate in a horizontal plane about a vertical axis passing through its centre. Atortoise is sitting at the edge of platform. Now the platform is given an angular velocity omega_(0). When the tortoise moves along the chord of the platform with a constant velocity (with respect to the platform), The angular velocity of platform omega(t) will vary with time.t. as

Answer»




ANSWER :B
Discussion
15.

A block of mass m is placed on the top of another block of mass M as shown in the figure. The coefficient of friction between them is mu. The maximum acceleration with which the block M may move so that m also moves along with it is

Answer»

`MU g`
`mu(M)/(m) g`
`mu(m)/(M) g`
`(g)/(mu)`

SOLUTION :(a) m will move by FRICTION,
`f_("max")=mu mg implies(a_(m))_("max")=(mu mg)/(m)=mu g`
Discussion
16.

Explain what is meant by dimensions of a physical quantity, its dimensional formula an dimensional equation.

Answer»

SOLUTION :1 (d) and 1 (d). 2
Discussion
17.

The rate of change of total momentum of a system of many particle system is proportional to the ……….on the system.

Answer»

EXTERNAL force
sum of the INTERNAL force

Answer :A
Discussion
18.

Which devices are used for measurement of length of different order?

Answer»

Solution :
Explain direct and indirect method of measurement of length. In laboratory sometimes length is to be measured. If DEVICE (INSTRUMENT) like meterscale, VERNIER CALIPERS, micrometer, screw gauge or spherometer are used to MEASURE length then it is called direct method for measurement of length. To measure very large distance like distance between celestial objects some special method are used such method are called indirect method.
Discussion
19.

Two glass bulbs of volumes 500 c.c and 200 c.c are connected by a narrow tube. These bulbs are filled with air at 17^(@)C and 750 mm of Hg pressure. If the small bulb and big bulb are placed in melting ice and boiling water respectively, find the pressure of the air in bulbs.

Answer»

Solution :If the two bulbs are at the same and temperature
Total volume `V_(0)=V_(1)+V_(2)=500+200=700 cm^(3)`, Total mass of the gas = m
Pressure `P_(0)=750mm` of Hg , Temperature `T_(0)=273+17=290K`
From the gas equation, `P_(0)V_(0)=m/M""RT_(0)" (or) "(P_(0)V_(0))/T_(0)=m/MR"....(1)"`
If the two bulbs are different temperatures let P be the temperature in the bulbs.
For big bulb, `PV_(1)=m_(1)/M""RT_(1)`
Here, `V_(1)=200CM^(3)" and "T_(1)=273+100=373K`
For small bulb, `P_(2)=m_(2)/M""RT_(2)`
Here, `V_(2)=200 cm^(3)" and "T_(2)=273K`
`(PV_(1))/T_(1)+(PV_(2))/T_(2)=m_(1)/M""R+m_(2)/M""R=m""R/M`
For gas equation, `(P_(0)V_(0))/T_(0)=(PV_(1))/T_(1)+(PV_(2))/T_(2)=P(V_(1)/T_(1)+V_(2)/T_(2))`
`P=(P_(0)V_(0))/T_(0) times 1/(V_(1)/T_(1)+V_(2)/T_(2))=(750 times 700)/(290) times 1/((500/373+200/273))`
`THEREFORE P=873mm` of Hg.
Discussion
20.

Explain with illustration meaning of work , energy and power in everyday life and state that this meaning in physics

Answer»

Solution :A farmer FARMING his farm , a worker carrying bricks , a PAINTER a picture ETC . Is said that they all are doing work but there is no work DONE in the physics .
If a person work works 14 to 16 hours in a day then we say that his energy is more for work but in physics work and ability to do work has different meaning .
In a quarrel between a father and his son , another person gets involves and we have often heard the saying that " being a son on whom you are showing your power" .Then meaning of power here is different than power of in physics .
Discussion
21.

If the kinetic energy of the body is increased by 300 % then , theincrease in momentum will be ……

Answer»

1
1.5
2
1.75

Solution :INITIAL KINETIC ENERGY `E_(1)=E`
When kinetic energy increases300 % then
Now kinetic energy
`E_(2)=E+ 2E = 4E`
Now `E = (p^(2))/(2m)`
` :. P prop sqrt€ ` (m = constant)
`:. (p_(2))/(p_(1))=sqrt((E_(2))/(E_(1)))=sqrt((4E)/E)`
` :. p_(2) =2p_(1) =p_(1)+p_(1)`
` :. ` Momentum increased by 100 %
Discussion
22.

What will be the shape of speed of sound in air versus pressure at constant temperature ?

Answer»

SOLUTION :STRAIGHT line, parallel to pressure axis (because SPEED of sound does not DEPEND on pressure at constant temperature).
Discussion
23.

An object is thrown from Earth is such a way that it reaches a point at infinity with non-zero kinetic energy [K.E(r=infty)=(1)/(2)MV_(infty)^(0)], with what velocity should the object be thrown from Earth?

Answer»

SOLUTION :`V_(E) = sqrt(2g(R_(E)+H)) = sqrt(2g(R_(E) + oo))`
`:. H = oo`
`V_(e) = sqrt(V_(oo) + 2ghR_(E))`.
Discussion
24.

A train is moving with constant acceleration. When the ends of the train pass by a signal their speeds are 30 km/h and 40 km/h respectively. Calculate the speed of the midpoint of train while passing the signal.

Answer»

35 km/hr
`(25)/(sqrt2) km//hr`
`25 sqrt2 km//hr`
`25 km//hr`

Solution :Velocity of midpoint is
`V. = SQRT (( u ^(2) + v ^(2))/( 2 )) = sqrt (((30 ) ^(2) + (40 ^(2))^(2))/( 2))`
`therefore v. = sqrt (( 900+ 1600)/(2)) = sqrt ((2500)/(2))`
`therefore v. = (50)/(sqrt2) = 25 sqrt2 (m)/(s)`
Discussion
25.

A galvanometer has resistance1 Omega. Maximum deflection current through it is 0.1 A

Answer»

To measure a current of 1A a resistance of `(1)/(10)Omega` is put in PARALLEL with galvanometer
To measure a current of 1A a REISTANCE of `(1)/(9) Omega` is put in parallel with galvanometer
To measure a potential difference of 10 V a resistance of `99 Omega` is put in series with galvanometer
To measure a potential difference of 10 V a resistance of `100 Omega` is put in series with galvanometer

Answer :B::C
Discussion
26.

The resistance R of a wire is found by determining its length 1 and radius r. The percentage errors in measurement of 1 and r are respectively 1% and 2%. The percentage error in measurement of R is

Answer»

0.04
0.05
0.03
0.015

Answer :B
Discussion
27.

A satellite which is geostationary in a particular orbit is taken to another orbit. Its distance from the center of earth in new orbit is two times of the earlier orbit. The time period in second orbit is ......... hours.

Answer»

4.8
`48sqrt(2)`
24
`24sqrt(2)`

Solution :`T prop r^(3/2)` if r becomes DOUBLE then TIME PERIOD will become `(2)^(3/2)` times so new time period will be `24xx2sqrt(2)` HOURS , i.e ., `48sqrt(2)` hours.
Discussion
28.

The average resisting force that must act on a 5 kg mass to reduce its speed from 65cms^(-1) to 15cms^(-1) in 0.2 s is

Answer»

12.5 N
25 N
50 N
100 N

Answer :A
Discussion
29.

The force between two objects decreases by 36% when the distance between them is increasedby 4 m. The original distance between them in meter is

Answer»

4
16
10
12

Answer :B
Discussion
30.

The position vector for a particle is represented be vecr=3t^(2)hati+5thatj+6hatk, find the velocity and speed of the particle at t=3 sec.

Answer»

Solution :`vecv=(BAR(dr))/(dt)=6thati+5hatj`
The velocity at any time .t. is GIVEN by `vecv=6thati+5hatj`. The MAGNITUDE of velocity is speed. The speed at any time .t. is then given by
speed `=sqrt((6t)^(2)+5^(2))=sqrt(36t^(2)+25)`
Now the velocity at t=3 sec is given by
`vecv=6(3)hati+5hatj=18hati+5hatj`
and speed at t=3 sec, is given by
speed `sqrt(349)m//s`.
Discussion
31.

Which of the following statement is/are correct about location of centre of mass ?

Answer»

The CENTRE of mass of three particles of equal weight placed at the vertices of a triangle ABC COINCIDES with the CENTROID of triangle ABC.
A uniform wire is bent to form the sides of a triangle ABC. The centre of mass of wire coincides with centroid of triangle ABC.
The centre of mass of four PARTICLE of equal weight placed at the vertices of a quadrilateral coincides with centroid of quardilateral.
The centre of mass of a body can exist outside the material of the body.

ANSWER :A::D
Discussion
32.

A train of mass 4000 metric tons is moving with a velocity of 18 KMPH. With the application of a force, its velocities changes to V. in 2 minutes. IF the power of the engine is 10 MW, the value of V. is (in KMPH)

Answer»

72
90
36
54

Answer :B
Discussion
33.

The unit and dimensions of impedance in terms of charge Q are

Answer»

`ohm, ML^(2) T^(-2) Q^(-2) `
`ohm, ML^(2) T^(-1) Q^(-2)`
`ohm, ML^(2) T^(-2) Q^(-1) `
`ohm, MLT^(-1) Q^(-1)`

Answer :B
Discussion
34.

The average kinetic energy of nitrogen molecules is E, at room temperature. The corresponding value for helium molecules will be: (Relative molecular weights of N_2 and He are 28 and 4 )

Answer»

`7E`
`E//7`
`E`
`(3E)/( 5)`

ANSWER :D
Discussion
35.

Two rods identical in length and cross-sectional area, but one of brass and the other of lead are coated with wax and then fitted into the holes in th sides of a bathl. Boiling water is poured into the bath. The wax coating of which rod wil melt first and to what extent? Thermal conductivity of brass =110 SI units and theat of lead =32 SI units. Sp. heat capacity of brass =370Jkg^(-1)K^(-1) and that of lead =126J kg^(-1)K^(-1). Densitiy of brass =8500kgm^(-3) and that of lead =11340 kg ,m^(-3)

Answer»


ANSWER :NA
Discussion
36.

A bus starts from rest with a constant acceleration of 5 "m/s"^2. At the same time a car travelling with a constant velocity 50 m/s overtakes and passes the bus. How fast is the bus travelling when they are side by side?

Answer»

10m/s
50m/s
100m/s
None

Answer :C
Discussion
37.

Stirred liquid comes to rest after some time. Why?

Answer»

Solution :DIFFERENT layers of a stirred liquid destroy the relative MOTION AMONG themselves DUE to intenal VISCOUS force.
Discussion
38.

A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 m s^(-1) . How long does the body take to stop ?

Answer»

Solution :RETARDING forceF= - 50 N
Mass m= 20 kg
Finalvelocityv= 0
ByNewtonsecondlaw of motion
`:.A = (F )/( m ) =(-50)/(20)-2.5m//s^(2)`
Accordingto equationof motion
`V= v_(0)+ at `
`0=15+ (-2.5)t`
`t= (15)/( 2.5)=6s`
Discussion
39.

An organ pipe open at one end is vibrating in first overtone and is in resonance with another pipe open at both ends and vibrating in thrid harmonic. The ratio of length of 2 pipes is

Answer»

`1:2`
`4:1`
`8:3`
`3:8`

ANSWER :A
Discussion
40.

A rubber ball is thrown downwards from the top of a tower with a velocity of14 m/s. A second ball is dropped from the same place 1 s later. The firstball reaches the ground in 2 s and rebounds with the same magnitude of velocity . How much later wouldthe two balls collide with each other ?

Answer»

Solution :HEIGHT of thetower,
h=Distance travelley bythe FIRST ball in 2 s(t=0 to t=2s )
`=14xx2+1/2xx9.8xx2^2=47.6 m`
The second ball startsat t=1 s. From t=1 s tot=2 s, this ball comes down through a height,
`h. =1/2xx9.8xx(2-1)^2=4.9m`
`therefore`Att=2 s, the distance between the two BALLS,
`H=h-h.=47.6-4.9=42.7 m `
For the first ball, velocity at t=2s is,
`v_1=14+9.8xx2=33.6 m//s`
As it rebounds with thesame instant , i.e ,at t=2s is
`v_(1)^(.) =-33.6 m//s.`
For the second ball , att=1 s, the velocity u=0.
So, at t=2s , the velocity,
`v_(2)=0+9.8xx(2-1)=9.8m//s`
`therefore` RELATIVE velocity of the second ball with respect to the first at t=2 s is,
`V=v_(2)-v_(1)^(.)=9.8 -(-33.6)=43.4 m//s`
Relative acceleration =g-g=0. So, for relative motion , the velocity V is uniform . Thus TIME required totravel the height of H =42.7m is ,
`t=H/V=(42.7)/(43.4)=0.98 s`
Therefore , theballs will collide at `T=2+t=2.98s.`
Discussion
41.

The dimensions of which of the quantities of the following pairs are not the same ?

Answer»

light YEAR and WAVELENGTH
TORQUE and work
energy and work
angular momemtum and work

ANSWER :D
Discussion
42.

What is the relation between the average kinetic energy and pressure?

Answer»

Solution :The internal energyof the gas is GIVEN by
` U = 3/2NkT`
The aboveequation can alsobe writtenas
` U = 3/2 PV`
sincePV = NkT
`P= 2/3 U/V= 2/3 u`
From the equation (1),the pressure of the gas is equal to twothirds of internal ENERGYPER unitvolume or internal energy per unit volume or internal energydensity ( u = UV)
Writingpressure in TERMS of meankineticenergy densityusing equation
` P = 1/3 nmv^(bar2) = 1/3 pv^(bar2)`
Where p= nm = massdensity( Note n is number f density)
Multiplyand divideR.H.S of equation (2) by 2,we get
` P = 2/3 ( P/2 bar(v^(2)))`
` P = 2/3 K E `
From the equation (3) , pressureis equalto 2/3of mean kineticenergyper unitvolume.
Discussion
43.

A stone is projected from the point of a ground in such a direction so as to hit a bird on the top oftelegraph post of height h and then attain the maximum height 2h above the ground. If at the instant of projection, the bird were to fly away horizontally with a unifrom speed. Find the ratio between the horizontal velocities of the bird and stone, if the stone still his the bird while ascending.

Answer»

Solution :Let the stone is projected with a velocity u at an angle `theta`with the horizontal, we have
`(2h)=((usin theta)^(2))/(2g)` or `u SIN theta=2sqrt(gh)`
Suppose t is the time taken by the stone to reach the height h above the ground. Then
`h=u sin theta t-1/2"GT"^(2)` (or) `("gt"^(2))/2- u sin theta+h=0`
As we have `u sin theta =2sqrt(gh)"" :.("gt"^(2))/2-2sqrt(gh)t+h=0`

Solving above equation for t we get
`t=(2sqrt(TH)+-sqrt((2sqrt(gh))^(2)-4xxg/2h))/(2xxg/2)=(2sqrt(gh)+-sqrt(2gh))/g`
Let `t_(1)=sqrt((2h)/g)(sqrt(2)-1)` and `t_(2)=sqrt((2h)/g)(sqrt(2)+1)`
Where `t_(1)` and `t_(2)` correspond to P and Q in the figure. Suppose V is the horizontal velocity of the bird. Then `PQ=vt_(2)`
For the motion of stone `PQ=u cos theta (t_(2)-t_(1))"":.vt_(2)=ucos theta(t_(2)-t_(1))`
which gives `v/(u cos theta)=((t_(2)-t_(1))/(t_(2)))=(sqrt((2gh)/g)(sqrt(2)+1)-(sqrt(2)-1))/(sqrt((2h)/g)(sqrt(2)+1))=2/(sqrt(2)+1)`
Discussion
44.

A constant power is supplied to a rotating disc. The relationship of angular velocity (w) of disc and number of rotations (n) made by the disc is governed by:

Answer»

`W PROP N^(1//3)`
`w prop n^(2//3)`
`w prop n^(3//2)`
`w prop n^(2)`

ANSWER :A
Discussion
45.

Explain the role of lakes in an ecosystem.

Answer»

ice is a good conductor of heat
ice REFLECTS heat and light
of ANOMALOUS expansion of water between `4^@C` to `0^@C`
NOTHING can be said

Solution :of anomalous expansion of water between `4^@C` to `0^@C`
Discussion
46.

A man is sitting in a boat, which is floating on a pond. If the man drinks some water from the pond, the level of water in the pond:

Answer»

Solution :No, When the man drinks WATER, say m KG, he DISPLAYS m kg of water and HENCE the level TENDS to increase. But m kg of water has already gone inside his stomach. So the level remains the same.
Discussion
47.

The value of 44.8- 21.235 is

Answer»

23.565
23.56
23.57
23.6

Answer :D
Discussion
48.

If the gas in an open vessel is heated what will be the relation between its temperature and mass of the gas in the container?

Answer»

SOLUTION :When temperature increases some of the GAS flows out of the container which is to be kept OPEN to maintain PRESSURE constant. Mass of the gas in the container is inversely proportional to its temperature on Kelvin Scale.
Discussion
49.

Calculate the percentage increase in the moment of inertia about the axis of symmetry of a flywheel when the diameter of the flywheel is increased by 2%.

Answer»


ANSWER :`4%`
Discussion
50.

State Pascal's law in fluids.

Answer»

Solution :Pascal's law STATE that, if the pressure in a liquid is changed ata PARTICULAR POINT, the change is transmitted to the entire liquid without being diminished in magnitude.
Discussion