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 brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250 ^(@) C. if the original lengths are at 40.0 ^(@) C? Is there a "thermal stress developed at the junction ? The ends of the rod are free to expand (Co-efficient of linear expansion ofbrass = 2.0 xx 10^(-5) K^(-1) , " steel " = 1.2 xx 10^(-5) K^(-1)).

Answer»

Solution :SINCE the ends of the combined rod are not clamped, each rod expands freely.
`Delta l_("brass ") = 0.21 cm, Delta l_("steel") = 0.126 cm = 0.13 cm`
TOTAL change in length ` = 0.34` cm. No .THERMAL stress. is DEVELOPED at the junction since the rods freely EXPAND.
Discussion
2.

Whenairbubblecomesfrombottomto thetopof a lake, Itsradiusbecomesn- times. Iftemperatureremainsconstantthroughout thelake. Thedepthof lake . Is(P_0 ( n^(x+1) -1) + ( xT )/(R ) (n^(x )-1))/( rho g). Findx

Answer»


ANSWER :2
Discussion
3.

A charge q is placed at the centre of the open end of a cyclindrical vessel. Find the flux of the electric through the surfaceof the vessel.

Answer»

SOLUTION :`Q/(2in_0)`
Discussion
4.

A rocket of initial mass 6000 kg ejects mass at constant rate of 200 kg/sec. with constant relative speed of 800 m/sec. The acceleration of the rocket after 5 sec is (neglect gravity)

Answer»

`50 m//s^(2)`
`16 m//s^(2)`
`60 m//s^(2)`
`32 m//s^(2)`

Answer :D
Discussion
5.

The angle between two collinear vectors is/are,

Answer»

`0^(@)`
`90^(@)`
`180^(@)`
`0^(@)` or `180^(@)`

ANSWER :A
Discussion
6.

What is the estimated radius of the universe?

Answer»

SOLUTION :The ESTIMATED RADIUS of the UNIVERSE is `10^(25)m`.
Discussion
7.

The position of a particle is given by r=3.0thati + 2.0 t^(2)hatj+ 5.0 hatk . Where t is in seconds and the coefficients have the proper units for r to be in metres, (a) Find v(t) and aft) of the particle, (b) Find the magnitude and direction ofv(t) at t = 1.0 s.

Answer»

Solution :`v(t)(d r)/(d t) = (d)/(t) (3.0 t hat(i) + 2.0 t^(2)hat(J) + 5.0 hat(K)) = 3.0 hat(i) + 4.0t hat(j)`
`a(t) = (d v)/(d t) = +4.0 hat(j) a = 4.0 ms^(-2)` along y-direction (b) At `t = 1.0 s "" v = 3.0 hat(i) + 4.0 hat(j)`
It.s magnitude is `v = SQRT(3^(2) + 4^(2)) + 5.0 ms^(-1)`
and direction is
`theta = tan^(-1)((v_(y))/(v_(x))) = tan^(-1)((4)/(3)) = 53^(@)` with x-axis.
Discussion
8.

Find the resultant of the vectors shown in figure.

Answer»

Solution :
`VEC(R) = vec(OA) + vec(AB) + vec(BC)`
`vec(R) = (5 cos 37 hat(i) + 5 sin 37 hat(j)) + 3hat(i) + 4hat(j)`
`vec(R) + 4hat(i) + 3hat(j) + 3hat(i) + 4hat(j)`
`vec(R) = 7i + 7j, therefore R = 7sqrt(2) cm` and
`alpha = 45^(@)` with HORIZONTAL
Discussion
9.

Which technology has trigerred the computer revolution in the last three decades of twentieth century ?

Answer»

SOLUTION :The technologyof SILICON CHIP.
Discussion
10.

The frequency of oscillation of a mass m suspended from a spring having force constant k is given by f = Cm^(x)k^(g), where C is a dimensionless quantity. The value of x and y

Answer»

`x=(1)/(2),y=(1)/(2)`
`x=-(1)/(2),y=-(1)/(2)`
`x=(1)/(2),y=-(1)/(2)`
`x=-(1)/(2),y=(1)/(2)`

Solution :`f=C m^(x)k^(y)[k]=M^(1)L^(0)T^(-2)`
Comparing dimensional FORMULA on both sides,
`M^(0)L^(0)T^(-1)=M^(x)(M^(1)L^(0)T^(-2))^(y)`
`=M^(x)xxM^(y)M^(y)L^(0)T^(-2y)`
`M^(0)L^(0)T^(-1)=M^(x+y)L^(0)T^(-2y)`
Comparing dimensional formula on both SIDE
`0x+y and -1=-2y`
`:.y=(1)/(2)So, 0=x+(1)/(2)`
`:.x=-(1)/(2)`
Discussion
11.

Two particles having position vectors vecr_1=(3hati+5hatj)m and r_2 =(-5hati + 3hatj) m are moving with velocities V_1 = (4hati +3hatj)m/s and V_2 = (-ahati + 4hatj) m/s . If they collide after 2 s, the value of a is

Answer»

2
4
6
8

Answer :D
Discussion
12.

Fundamental frequency of one closed pipe is 300 Hz. What will be the frequency of its second overtone ?

Answer»

Solution :Frequency of second overtone = frequency of third hermonic
`= (2n -1) f _(1) `(here n=3)
`= 5 f _(1) = 5 XX 300 = 1500 HZ`
Discussion
13.

Which of the following is expressed correctly

Answer»

MAGNETIC PERMEABILITY `to "FM"^(-1)`
Permittivity `to Hm^(-1)`
Relative permittivity `to Vm^(-1`
Intensity of MAGNETIZING field `to Am^(-1)`

Answer :D
Discussion
14.

Calculate the change in intensity level when the intensity of sound increases by 10^(6) times its original intensity.

Answer»

SOLUTION :Here,
`(I)/(I_(0))=("Final INTENSITY")/("Original intensity")=10^(6)`
INCREASE in intensity level,
`:.L=10log_(10)((I)/(I_(0)))`
`=10log_(10)(10^(6))=10xlog_(10)(10^(6))`
`=10xx6xx1`
`impliesL=60` DECIBEL
Discussion
15.

From what height must a piece of lead be dropped if the heat generated on impact with the ground is just sufficient to melt it. Assume that all the heat is confined to lead. The initial temperature of lead is 27^(@)C and its final temperature is 327^(@)C. Specific heat capacity of lead 0.125 xx 10 ^(3) J//Kg//K. Latent heat of fusion of lead =31 xx 10 ^(3) J //kg.

Answer»

SOLUTION :`mcd T + ML = mgh, 0.125 XX 10 ^(3) xx 307 + 31 xx 10 ^(3) = 9.8 xx h , h = 7. 08 xx 10 ^(3) m`
Discussion
16.

(A): Hollow shaft is found to be stronger than a solid shaft made of same material and same quantity. (R) : Rigidity modulus is the ratio of shearing stress to shearing strain.

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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 :A
Discussion
17.

One end of a horizontal string is attached to a vibrating blade , and the other end passes over a pulley as in Fig. 7.49 (a). A sphere of mass 2.00 kg hangs at the end of the string . The string is vibrating in its second harmonic . A container of water is raised under the sphere so that the sphere is completely submerged . In this configuration , the string vibrates in its fifth harmonic as shown in Fig . 7.49 (b). What is the radius of the sphere ?

Answer»

Solution :Imagine what happens when the sphere is immersed in the water . The buoyant force acts upward on the sphere , reducing the tension in the string . The change in tension causes a change in the SPEED of waves on the string , which in turn causes a change in the wavelength . This altered wavelength results in the string vibrating in its normal mode rather than the second .
The hanging sphere is modelled as a particle in equilibrium . One of the forces acting on it is the buoyant force from the water . we also apply the waves under boundary conditions model to the string .
Apply the particle in equilibrium model to the sphere in Fig . 7.49 (a) , identifying ` T_(1)` as the tension in the string as the sphere hangs in air .
`T_(1) = MG`(i)
Apply the particle in equilibrium model to the sphere in Fig . 7.49 (B) , where `T_(2)` is the tension in the string as the sphere is immersed in water . LTBR. `B + T_(2) = mg rArr B = mg - T_(2)`(ii)
The desired quantity , the radius of the sphere , will appear in the expression for the buoyant force `B` . Before proceeding in this direction , however , we must evalute `T_(2)` from the information about the standing wave .
WRITE the equation for the frequency of a standing wave on a string twice , once before the sphere is immersed and once after . Notice that the frequency `f` is same in both the cases becauseit is determined by the vibrating blade . In addition , the linear mass density `mu` and the length `L` of the vibrating portion of the string are same in both cases . Divide the two equations.
` f = (n_(1))/( 2L) sqrt((T_(1))/( mu))`(iii)
` f = (n_(2))/( 2L) sqrt ((T_(2))/(mu))`(iv)
` 1 = (n_(1))/( n_(2)) sqrt ((T_(1))/( T_(2)))`(v)
solve for `T_(2)`:
`T_(2) = ((n_(1))/(n_(2)))^(2) T_(1) = ((2)/(5))^(2) ( 19.6 N) = 3.14 N`
Substituting this result into Eq. (ii)
` B = mg - T_(2) = 19.6 N - 3.14 N = 16.5 N`
Now`B = rho _( water) g V _(sphere) = rho _(water) g ((4)/(3) pi r^(3))`
Solve for the radius of the sphere :
` r = (( 3B)/( 4 pi rho_(water) g ))^(1//3)`
` = ( 3( 16.5 N ))/( 4 pi ( 1000 kg //m^(3)) ( 9.80 m//s^(2)))^(1//3)`
` = 7.38 xx 10^(-2) m = 7.38 cm`
Notice that only certain radii of the sphere will result inthe string vibrating in a normal mode , the speed of waves on the string must be changed to a value such that the length of the string is integer multiple of half wavelengths . The radii of the sphere that cause the string to vibrate in a normal mode are quantized.
Discussion
18.

A lift is movingup with a constant acceleration a.A man standingon the lift, throws a ball vertically upwards with a velocity v, which returns to the thrower after a time t. show that v=(a+g) t/2 where g is the acceleration due to gravity.

Answer»

Solution :The downward acceleration of theball with respect to the LIFT =g(-a)=g+a.
The initialvelocity of the ball is v upward. As the ball returns to the thrower, its relative DISPLACEMENT is ZERO.
Hence, from the equation H=`ut-1/2"gt"^2,` we get ,
`0=vt-1/2(g+a)t^2 or, 0=r{v-1/2(g+a)t}`
Since `tne0`,
`therefore v-1/2(g+a)t=0 or ,v=(a+g)t/2.`
Discussion
19.

You are given a thread and a metre scale. How will you estimate the diameter of the thread?

Answer»

SOLUTION : The diameter of a THREAD is so small. THEREFORE we cannot measure it using metre SCALE. We wind a number of turns of the thread on the metre scale so that the turns are closely touching one another.
Measure the length (I) of the windings on the scale which contains number of turns.
`therefore ` Diameter of thread `=(l)/(N)`
Discussion
20.

The refractive index of glass is 1.520 for red light and 1.525 for blue light. Let D_1 and D_2 are the angles of minimum deviation for red and blue respectively in a prism of the glass. Then

Answer»

`D_1 lt D_2`
`D_1 gt D_2`
`D_1=D_2`
`D_1` can LESS than or greater than DEPENDENT upon the ANGLE of prism.

Answer :A
Discussion
21.

A player caught a cricket ball of mass 150 g moving at a rate of 20 m/s. The cathcing process is completed in 0.1 second. What is the force exerted by the ball on the hand of the player?

Answer»

15 N
20 N
30 N
40N

Solution :Impulse = Change in MOMENTUM of `F xx t= m(v-0)`
`Fxx0.1= 150xx10^(-3)[20-0]`
`THEREFORE F= (150xx10^(-3)xx20)/(0.1)= 30N`
Discussion
22.

A body of mass 0.5 kg is whirled with a velocity of 2 ms^(-1)using 0.5 m length string which can be withstand a tension of 15 N, Neglecting the force of gravity on the body predict whether or not the string will break.

Answer»


SOLUTION :`F=mv^(2)//R = 0.5 xx 4//0.5 = 4 N`. String will not break.
Discussion
23.

A mass is suspended from the end of a spring. When the system is oscillating the amplitude of oscillation is 4 cm and maximum kinetic energy of oscillation of the system is 1J. Then the force constant of the spring is

Answer»

`2500 N"/"m`
`1250 N"/"m`
`500 N"/"m`
`250 N"/"m`

Answer :B
Discussion
24.

From the top of a tower of height 100m a 10 gm block is dropped freely and a 6 gm bullet is fired vertically upwards from the foot of the tower with velocity 100 ms^(-1) simultaneously. They collide and stick together. The common velocity after collision is (g= 10 ms^(-2))

Answer»

`27.5 MS^(-1)`
`150 ms^(-1)`
`40 ms^(-1)`
`100 ms^(-1)`

ANSWER :A
Discussion
25.

A wire is stertched with negligible tension at 30^(@)Cbetween two rigid supports. Find the velocity of the transverse wave in the wire at 20^(@)C . Given, the coefficient of linear expansion, the Young's modulus and the density of the material of the wire ofalpha = 18 xx 10^(-6@) C^(-1), Y = 12 xx 10^(11) dyn * cm^(-2)andrho = 6 g* cm^(-3)respectively.

Answer»

Solution :Decrease in temperature, `theta = 30 - 20 = 10^(@) C ` ,due to this decrement, the wire will TEND to contract. As aresult, a themal stress will be developed.
This thermal stress ` Y alpha theta` ,
So the tension in the wire, `Y = Y alpha theta A ` , where a = area of CROSS section of the wire.
Again, mass per unit LENGTH of the wire,
` m = A * 1* RHO = Arho`
`:.`The velocity of transverse wave in the wire is
` V = sqrt((T)/(m)) sqrt((Y alpha theta A)/(A rho))`
` sqrt((Yalpha theta)/(rho)) = sqrt(((12 xx 10^(11)) xx (18 xx 10^(-6)) xx 10)/(6))`
` = 6000 cm* s^(-1)`
Discussion
26.

The heat dissipated in a resistor can be obtained by the measurement of resistance, current and time. If the maximum error in the measurement of these quantities is 1%, 2% and 1% respectively, the maximum error in the determination of the dissipated heat is

Answer»

`4%`
`6%`
`(4)/(3)%`
`2%`

SOLUTION :Heat dissipated in a RESISTOR is given by `H= I^(2)Rt`
`:. (DeltaH)/(H)= 2(DeltaI)/(I)+ (DeltaR)/(R)+ (Deltat)/(t)`
For MAXIMUM percentage error,
`(DeltaH)/(H)xx100= 2(DeltaI)/(I)xx100+ (DeltaR)/(R) xx 100+(Deltat)/(t)xx100= 2 xx 2%+ 1%+ 1%= 6%`
Discussion
27.

In a harbour , wind is blowing at the speed of 72 km/h and the flag on the mast of a boat anchored in the harbour flutters along the N- E direction . Ifthe boat starts moving at a speed of 51 km/h to the north , whatis the direction of the flag on the mast of the boat?

Answer»


ANSWER :EAST (approcimately)
Discussion
28.

A thin disc rotates about an axis passing through its centre and perpendicular to its plane a constant angular velocity 'omega' I is the moment of inertia of that disc and'L' is its angular momentum about the given axis. Then rotational kinetic energy of the disc 'E' is (A) EalphaL^(2)"" (B) EalphaL^(-2) (C)EalphaI"" (D) EalphaI^(-1)("if L constant")

Answer»

A and C are CORRECT
B and C are correct
B and D are correct
A and D are correct

ANSWER :D
Discussion
29.

A narrow tube of radius R and length L is connected in series with another tube of radius R/2 and length L/4. If the pressure across the two tubes taken together is the pressure across the two tubes separately are (Assume steady and stream line flow)

Answer»

<P>`(4)/(5) P and (P)/(5)`
`(P)/(3) and (2)/(3)P`
`(P)/(4) and (3)/(4)P`
`(P)/(2) and (P)/(2)`

ANSWER :A
Discussion
30.

Sometimes a cycle with well inflated tyre left in the sun has its tube burst open because:

Answer»

The volume of AIR inside the TUBE increases
The PRESSURE of air inside the tube increases
The tube containing air melts
The volume and pressure both increase

Answer :B
Discussion
31.

A steel plate of face area2 cm^(2) and the thickness 1.0cm is fixed at the lower face. Atangential force of 10N is appliedon the upper surface. Find thelateral displacement of the upper surface withrespect to the lowersurface. Rigidity modulus of steel =8.4xx10^(10)Nm^(-2)

Answer»

Solution :`A = 2cm^2 = 2 xx 10^(-4) m^2 , L = 1.0cm = 1 xx 10^(-2)m, F = 10N, G = 8.4 xx 10^10 Nm^(-2) , l= ? , G = (FL)/(A DELTA x)`
The lateral displacementof the upper face with respect to the lower face is `Delta x = (FL)/(AG)`
` = (10 xx 1 xx 10^(-2))/(2 xx 10^(-4) xx 8.4 xx 10^10)`
` (1)/(16.8) xx 10^(-7) m = 5.952 xx 10^(-9) m`
Discussion
32.

A steel wire, 2mm in diameter, is just stretchedbetween two fixedpointsat a temperatureof 30^(0)C. Determineits tensionwhenthe temperaturefalls to 20^(0)C. (Coefficient of linearexpansionof steel =0.000011//""^(0)C, Young.s modulusfor steel=2.0xx10^(11)Nm^(-2))

Answer»

Solution :Theremalstress`=(F)/(A) =Y alpha Delta t`
Tensionin the wire`=Y alpha A(t_(2)-t_(1))`.
Here,`Y=2.0xx10^(11)Nm^(-2) , alpha =11xx10^(-6) //""^(0)C`,
`t_(2) =30^(0)C, t_(1) =20^(0)C, `radius = 1mm `=1xx10^(-3)m`,
`A=pi r^(2)=pi(1xx10^(-3))^(2) = pi xx 10^(-6) m^(2) `
The tensionin the wire
`=2.0xx10^(11) xx11 xx 10^(-6) xx pi xx 10^(-6) xx (30-20)=69.14 N`
Discussion
33.

A steel rod of length 5m is fixed rigidly between two supports. a of steel = 12xx10^(-6)//""^(@)C,Y=2xx10^(11)Nm^(-2). With the increase in its temperature by 40^(@)C, the stress developed in the rod is

Answer»

`9.6 xx 10^(5) "Nm"^(-2)`
`9.6 xx 10^(6) "Nm"^(-2)`
`9.6 xx 10^(7)"Nm"^(-2)`
`9.6 xx 10^(8) "Nm"^(-2)`

Answer :C
Discussion
34.

Imagine a light planet is revolving round a very massive star in a circular orbit of radius R with a time period of revolution T. If the gravitational force of attraction between the star and planet is proportional to R^(-n), then T^2 is proportional to

Answer»

`R^(n+1)`
`R^(n+2)`
`R^(n-1)`
`R^(n-2)`

ANSWER :A
Discussion
35.

A small body of mass 10 gram is making simple harmonic oscillations along a straight line with a time period of pi/4 and the maximum displacement is 10cm. Find the energy of oscillations. What is the average kinetic energy during one oscillation ?

Answer»

SOLUTION :`32xx10^(-2) J , 16xx10^(-2) J`
Discussion
36.

A paricle is moving east wards with a velocity of 5 ms^(-1). In 10S, the velocity changes to 5 ms^(-1) north wards. The average acceleration in this time is

Answer»

`(1)/(SQRT(2)) MS^(-2)` TOWARDS north east
`(1)/(2) ms^(-2)` towards north
ZERO
`(1)/(sqrt(2)) ms^(-2)` towards northwest

Answer :D
Discussion
37.

The phase difference between the instantaneous velocity & acceleration of a particle executing simple harmonic motion is

Answer»

`0.5pi`
`PI`
`0.707pi`
zero

Answer :A
Discussion
38.

A uniform spherical shell rolls down a fixed inclined pla ne without sliping. Find the ratio of | rotational kinetic energy to translational kinetic energy as it reaches lowest point of the incline.

Answer»

Solution :As there is no slip,
`v=Romega` is valid and rotational KINETIC energy
`K_(R)=(1)/(2)I_(CM)omega^(2)=(1)/(2)mk^(2)omega^(2)`
and translational kinetic energy
`K_(T)=(1)/(2)MV^(2)=(1)/(2)mR^(2)omega^(2)` therefore `(K_(R))/(K_(T))=(K^(2))/(R^(2))=(2)/(3)`
Discussion
39.

The temperature of source and sink of a carnot engine are 400K and 300k respectively its efficiecy is

Answer»

1
0.333
0.75
0.25

Answer :D
Discussion
40.

Find the value 15 joule in a system which has 10 cm, 100 gm and 20sec as a fundamental units?

Answer»


Solution :In S.I. `n_(1)=15, M_(1)=1KG, L_(1)=1m, T_(1)=1`s Energy `=ML^(2)T^(-2),x=1,y=2,z=-2`
In the NEW SYSTEM `n_(2)=?,M_(2)=100g, L_(2)=10cm,T_(2)=20s`
`n_(2)=n_(1)(M_(1)//M_(2))^(x)(L_(1)//L_(2))^(y)(T_(1)//T_(2))^(z)=15(10^(3)g//100g)^(1)(100cm//10cm)^(2)(15//20s)^(-2)`
`1J=6xx10^(6)` new UNITS.
Discussion
41.

Force is given by the expression, F= A cos (Bx) + C cos (Dt) where x is displacement and t is time. The dimension of ((D)/(B)) is same as that of

Answer»

velocity
velocity gradient
angular velocity
Angular momentum

Answer :A
Discussion
42.

If we imagine two ideally smooth surfaces & if they are kept in contact, the frictional force acting between them is,

Answer»

zero
a small FINITE value but not zero
very HIGH
We can't PREDICT

Answer :C
Discussion
43.

A body is moving unidirectionally under the influence of a source of constant power supplying energy. Which of the diagrams shown below correctly shows the displacement time curve for its motion ?

Answer»




ANSWER :B
Discussion
44.

A cylindrical vessel has three identical horizontal tubes of length 40 cm, each coming out at heights 0, 5 and 9 cm, respectively. What will be the length of a single overflow tube of the same radius as that of identical tubes which can replace the three tubes if placed at the bottom of the vessel if the level of liquid in the cylindrical vessel is kept constant at 20 cm.

Answer»

Solution :Let the three tubes be P, Q and R.
Height of tube P=0 cm from bottom
Pressure head for tube, `h_p= 20 cm` of water column
Height of tube Q = 5 cm from bottom
Pressure head for tube Q. `h_Q= 20 -5= 15` cm of water column
Height of tube R =9 cm from bottom
Pressure head for tube R, `h_r=` 20 - 9 =11 cm of water column
Let r cm be the radius of each tube.
Given, length of each tube, l= 40 cm
Rate of flow of liquid through DIFFERENT tubes will be `Q_(p)=(pi p_(1) r_(1)^(4))/(8 eta n)=(pi x h_(p) pg xx r^(4))/(8 eta l)`
`=(pi xx 20 xx pg xx r^(4))/(8 eta xx 40)`
Similarlry, `Q_(Q)=(pi x h_(Q) pg xx r^(4))/(8 eta l)=(pi xx 15 xx pg xx r^(4))/(8 eta xx 40)`
`Q_(R)=(pi xx h_(R) pg xx r^(4))/(8 eta l)=(pi xx 1 xx pg xx r^(4))/(8 eta xx 40)`
Total VOLUME of liquid FLOWING through the three tubes P,Q and R is `Q=Q_(P)+Q_(Q)+Q_(R)`
`=(pi xx pg r^(4))/(8 eta xx 40) xx 46`
If l. is the length of the required single tubem then `Q.=(pi h pg r^(4))/(8 eta l.)=(pi xx 20 xx pg r^(4))/(8 eta l.)`
Since Q.=Q
`RARR (pi xx 20 xx pg r^(4))/(8 eta l.)=(pi xx pg r^(4) xx 46)/(8 xx 40 eta)`
`rArr (20)/(l.)=(46)/(40)`
`rArr l.=(20 xx 40)/(46) =17.39cm`
Discussion
45.

By kinetic theory, we can measure

Answer»

viscosity
conductivity
diffusion
All of these

Answer :D
Discussion
46.

A vernier callipers has 1 mm marks on the main scale. It has 20 equal divisions on the Vernier scale which match with 16 main scale divisions. For this vernier callipers, the least count is

Answer»

`0.02` MM
`0.05` mm
`0.1` mm
`0.2` mm

Answer :D
Discussion
47.

In the above example the angular velocity of S_(2) as actually observed by an astronaut in S_(1) is -

Answer»

`3xx10^(-4)`
`3xx10^(23)`
`3xx10^(-6)`
`3xx10^(-7)`

Solution :(a) Angular SPEED of `S_(2)` as observed by astronaut in `S_(1)`.
`omega_(r )=(|v_(1)-v_(2)|)/(|r_(2)-r_(1)|)`
`|v_(1)-v_(2)|=(2pixx10^(4)-pixx10^(4))km//h`
`or |v_(1)-v_(2)|=(pixx10^(4)xx1000)/(60xx60)m//s`
`or |v_(1)-v_(2)|=(3.14xx10^(5))/(36)=0.0872xx10^(5)=8.72xx10^(3) m//s`
`|r_(2)-r_(1)|=3xx10^(7) m`
`:. omega_(r )=(8.72xx10^(3))/(3xx10^(7))=2.9xx10^(-4)~~3xx10^(-4) rad//sec`
`:. omega_(r )=3xx10^(-4) rad//sec.`
Discussion
48.

If the inertial mass and gravitational mass of the simple pendulum of length l are not equal, then the time period of the simple pendulum is :

Answer»

`T= 2PI sqrt((m_(i)l)/(m_(g)g))`
`T= 2pi sqrt((m_(g)l)/(m_(i)g))`
`T=2pi (m_(g))/(m_(i)) sqrt((l)/(g))`
`T=2pi (m_(i))/(m_(g)) sqrt((l)/(g))`

SOLUTION :`T= 2pi sqrt((M)/(K))`
`:. T_(1)=2pi sqrt((M_(1))/(k)): T_(2)= 2pi sqrt((M_(2))/(k))`
INERTIAL mass `M_(1) prop m_(i) l`
Gravitational mass `M_(2) prop m_(g)g`
`:. (T_(1))/(T_(2))= 2pi sqrt((M_(2)l)/(k))xx(1)/(2pi) sqrt((k)/(m_(g)g))`
`:. T= 2pi sqrt((m_(i)l)/(m_(g)g)`
Discussion
49.

A combination of two resisters R_(1) and R_(4) are placed in an electrical circuit. In an experimental arrangement, the resistance are measured as R_(1)=(100 pm 3)Omega , R_(2)=(200 pm 4)OmegaThe equivalent resistance, when they are connected in parallel, (1)/(R_(P))=(1)/(R_(1))+(1)/(R_(2)) can be expressed as

Answer»

`(66.7 PM 12)Omega`
`(66.7 pm 7)Omega`
`(66.7 pm 4.9)Omega`
`(6.7 pm 1.8)Omega`

ANSWER :C
Discussion
50.

Two billiard balls each of mass 0.05 kg moving in opposite directions with speed 6 m s^(-1)collide and rebound with the same speed.What is the impulse imparted to each ball due to the other ?

Answer»

Solution :Impulse on each BALL `= 0.05 xx 12 = 0.6 KG m s^(-1)`in magnitude.The two impulses are OPPOSITE in direction
Discussion