40048 + Interview Questions in Physics in Class 11 Page 760 InterviewSolution

37951.	We have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why ?
Answer» Solution :Because, for an arbitrarily SMALL INTERVAL of time, the magnitude of displacement is equal to the LENGTH of the PATH.

Discussion

37952.	Find the torque of a force 7hati + 3hatj – 5hatkabout the origin. The force acts on a particle whose position vector is hati – hatj + hatk.
Answer» Solution :Here `r=HATI-hatj+hatk` and `F=7hati+3hatj-5hatk` We SHALL use the determinant rule to FIND the torque `tau=rxxF` `tau=\|(hati,hatj,hatk),(1,-1,1),(7,3,-5)\|=(5-3)hati-(-5-7)hatj+(3-(-7))hatk` or `tau=2hati+12hatj+10hatk`

Discussion

37953.	Two identical rods of a metal are welded in series then 20 cal of heat flows through them in 4 minute. If the rods are welded in parallel then the same amount of heat will flow in
Answer» 1 minute 2 minute 4 minute 16 minute Answer :A

Discussion

37954.	Calculate rate of flow of glycerin of density 1.25xx10^(3) kg/m^(3) through the concial section of a horizontal pipe if the radii of its ends are 0.1 m and 0.04m and pressure drop across its length is 10 N//m^(2).
Answer» Solution :According to CONTINUITY equation `(v_(2))/(v_(1))=(A_(1))/(A_(2))=(r_(1)^(2))/(r_(2)^(2))=((0.1)^(2))/((0.04)^(2))=(25)/(4)` and, according to Bernouli.s equation for a horizontal tube `P_(1)+(1)/(2)rhoV_(1)^(2)=P_(2)+(1)/(2)rhoV_(2)^(2)` `V_(2)^(2)-V_(1)^(2)=2(P_(1)-P_(2))/(RHO)=2xx(10N//m^(2))/((1.25xx10^(3)kg//m^(3)))=16xx10^(-3)m^(2)//s^(2)` but `v_(2)=(25)/(4)v_(1)=6.25v_(1)` `therefore [(6.25)^(2)-t^(2)]V_(1)^(2)=16xx10^(-3)m^(2)//s^(2)` or `V_(1)~~0.0205m//s` So,the rate of VOLUME flow=`A_(1)V_(1)=PI(0.1)^(2)xx(0.02)=6.28xx10^(-4)m^(3)//s` And the rate of mass flow is `(dm)/(dt)=rhoAV=(1.25xx10^(3)kg//m^(3))xx(6.28xx10^(-4)m^(3)//s)=0.785 kg//s`

Discussion

37955.	A thin uniform rod of mass 'm' and length 'l' is kept on a smooth horizontal surface such that it can move freely. Where should a particle of mass 'm' moving horizontally strike on the rod from the centre (x) such that the point 'A' at a distance l/3 from the end of the rod is instantaneously at rest just after the elastic collision ?
Answer» l/2 l/3 l/3 l/3 Answer :A

Discussion

37956.	A satellite orbits the earth at a height of R/6, its orbital speed ?
Answer» `SQRT((2GM)/(R))` `sqrt((GM)/(R))` `sqrt((GM)/(7R))` `sqrt((6GM)/(7R))` ANSWER :D

Discussion

37957.	What is the condition to be satisfied by a mathematical relation between time and displacement to describe a periodic motion ?
Answer» SOLUTION : A perodic MOTION REPEATS after a definite TIME INTERVAL T. So, `y(t) = y(t + T)=y(t+2T)` etc.

Discussion

37958.	A spring with 10 coils has spring constant k. It is exactly cut into two halves, then each of these new springs will have a spring constant
Answer» `k//2` `3K//2` `2K` `3k` ANSWER :B

Discussion

37959.	A circular frame made of 20 cm long thin wire is floating on the surface of water. If the surface tension of water is 70 dyne/cm. the required excess force to separate this frame from water will be :
Answer» 2800 dyne 800 dyne 280 dyne 1800 dyne Answer :A

Discussion

37960.	Two bodies of equal masses moving with momentum of first double that of the second collide completely inelastically. If the total final kinetic energy is 9 joule, the initial kinetic energy of the body of greater momentum is:
Answer» 5 JOULE 6 joule 8 joule 10 joule ANSWER :C

Discussion

37961.	A satellite is in an elliptic orbit around the earth with aphelion of 6R and perihelion of 2R where R = 6400 km is the radius of the earth . Find the velcoity of the satellite at apogee and perigee.[ G = 6.67 xx10^(-1) SI unit and M= 6 xx 10^(24)kg ]
Answer» Solution :`implies` Given , `r_p` = radius of perihelion = 2R `r_a` = radius of aphelion = 6R Hence , we can write, `r_(a) = a(1+e) = 6R "" ...(i)` `r_p = a (1-e) = 2 R ""...(ii)` SOLVING equ (i) and (ii) ,we get ECCENTRICITY , `e =1/2` Angular MOMENTUM remains unchanged `:. mv_pr_p = mv_ar_a` `:. (v_a)/(v_p) = r_p/r_a = (2R)/(6R) =1/3` Energy is same at perigee and apogee, `1/2mv_p^2-(GMm)/r_p=1/2mv_a^2-(GMm)/r_a` `:. v_p^2(1-1/9)=-2GM(1/r_a-1/r_p)=2GM(1/r_p-1/r_a)` (By putting `v_a=v_p/3`) `v_p=([2GM(1/r_p-1/r_a)]^(1/2))/([1-(v_a/v_p)^2]^(1/2))=[((2GM)/R(1/2-1/6))/((1-1/9))]^(1/2)` `=((3/2GM)/(8/9R))^(1/2)=sqrt(3/4(GM)/R) ` = 6.85 km/s `v_p = 6.85km//s` For circular orbit of radius r, `v_c` = orbita velocity = `sqrt((GM)/r)` For `r = 6R ,v_c =sqrt((GM)/(6R)) = 3.23 km//s` Hence , to increase to a circular orbit at apogee, we have to increase the velocity by (3.22-2.28) = 0.95 km/s . This can be done by suitably firing rockets from the SATELLITE .

Discussion

37962.	The magnitude of potential energy per unit mass of the object at the surface of earth is .E.. Then escape velocity of the object is
Answer» `SQRT(2E)` `4E^2` `SQRTE` `sqrt(E//2)` ANSWER :A

Discussion

37963.	The motion of a particle is given by X=Asinomegat+B cos omegat the motion of the particle is
Answer» not SIMPLE HARMONIC simple harmonicwith amplitude A+B Simple harmonic with amplitude `(A+b)/2` simple harmonic with amplitude `sqrt(A^(2)+B^(2))` ANSWER :D

Discussion

37964.	A body of mass 5kg starts from the origin with an initial velocityof barU= (30i+40j)m//s. A constant force of F= (-hati - 5 hatj)N acts on the body . Find the time in which they- component of the velocity becomes zero .
Answer» Solution :`baru= 30i + 40 j` ....... (1) ` u=u_(x) HATI + u_(y) hatj`........ (2) `F= - hati - 5hatj` ........ (3) ` F= F_(x) hati + F_yhatj` ........ (4) comparing (1) and (2), (3) and (4) we have `u_y = 40 m//s , F_y =-5N` `F_y = ma_y , a_y = -1 m//s^(2)` `v_y = u_y + a_y xx 1 , 0 = 40 -1 xx t , t = 40` sec

Discussion

37965.	The ends of a metal rod of length 40 cm are maintained at temperatures 100°C and 20°C respectively. At steady state condition, the temperature at a point at a distance 30 cm from the hot end of the rod is
Answer» `40^(@) C` `30^(@) C` `20^(@) C` `10^(@) C` ANSWER :A

Discussion

37966.	A light metal stick of square cross-section (5cmxx5cm) and length 4m mass 2.5 kg ad is shown in the figure below determine its angle of inclination when the water surface is 1m above the hinge. What minimum depth of water above high will be required to bring the metal stick in vertical position.
Answer» ANSWER :`theta=30^(@)` DEPTH of WATER `ge2m`

Discussion

37967.	A horizontal uniform glass tube of 100cm length sealed at both ends contains 10cm mercury column in the middle. The temperature and pressure of air on either side of mercury column are respectively0°C and 80cm of mercury. If the air column at one end is kept at 0°C and the other end at 273°C, the pressure of air which is 0°C is (in cm of Hg)
Answer» 76 88.2 120 132 Answer :C

Discussion

37968.	A: A body is moving along a straight line such that its velocity varies with time as shown in figure. Magnitude of displacement of the body from t = 0 to t = 12s is the same as the distance travelled by it in the given time duration, R: For a unidirectional motion of a body, \|displacement\| = distance
Answer» If both A and B are TRUE and R is the correct explanation of A If both A and R are true but R is not correct EXPLANTION of A If A is true but R is false If A is false but R is true Answer :A

Discussion

37969.	For a body rolling along a level surface, without slipping the translational and rotational kinetic energies are in the ratio 2:1. The body is
Answer» HOLLOW sphere Solid cylinder Ring Solid sphere Answer :B

Discussion

37970.	A spring when compressed by 0.2 m develops a restoring force of 50 N. If a body of mass 10 kg is placed on it, calculate (i) the force constnat (I) of the spring (ii) the depression of the spring (iii) time period of oscillation when the weight is slightly displaced from equilibrium position.
Answer» Solution :(i) The FORCE constant `K=F/x=50/0.2=250Nm^(-1)` (ii) depression of the SPRING `(mg)/k=(10xx9.8)/250=0.392m` (III) Time period `T=2pisqrt(m/k)=2pisqrt(10/250)=(2xx3.14)/5=1.2568s`

Discussion

37971.	The maximum vertical height of a projectile is 10 m. If the magnitude of the initial velocity is 28 ms^(-1), what is the direction of the initial velocity. ? Take g = 9.8 ms^(-2).
Answer» ANSWER :`(30^(@))`

Discussion

37972.	If vec(P)=hat(i)+2hat(j)-4hat(k)andvec(Q)=hat(i)+2hat(j)-hat(k) then (vec(P)+vec(Q))*(vec(P)-vec(Q)) is
Answer» 5 15 25 115 Answer :B

Discussion

37973.	One parsec is .........
Answer» `3.153 XX 10^(7)` m `3.26 xx 10^(15)` m `30.84 xx 10^(15)` m `9.46 xx 10^(15)` m Solution :`30.84 xx 10^(15)` m

Discussion

37974.	What doesit mean when a height of a barometer is rising ?
Answer» Solution :A rising height of a BAROMETER indicates increasing air PRESSURE . It also indicates DRY ATMOSPHERE (less humidity).

Discussion

37975.	A machine gun has a mass of 20 kg . It fires 35 g bullets at the rate of 4bullets per second, with a speed of 400 ms^(-1). What force must be applied to the gun to keep it in position ?
Answer» SOLUTION :Mass of GUN, M=20 kg Velocity of bullet, `v=400 ms^(-1)` TIME taken to fire one bullet, `t=1/4` second If F be the average force required to hold the gun in position , then `Fxxt`= change in momentum of gun or `Fxxt = MV[`:.` initial momentum of gun is ZERO ] `:. F=(MV)/t =(20xx0.7)/(1//4)N =56N` Aliter: Number of bullet FIRED per second, n=4 momentum given to the bullets per second = nmv Same momentum per second is gained by the gun . This gives the force acting on the gun and also the force applied to keep the gun in position . So , `F=nmv=4xx0.035xx400N=56N`

Discussion

37976.	Hailstones fall from a certain height. If only 1% of the hailstones melt on reaching the ground, find the height from which they fall. (g - 10 ms^(-2)L - 80 caloric/g and I - 4.21/calorie)
Answer» 336 m 236 m 436 m 536 m ANSWER :A

Discussion

37977.	A cyclist starts from the centre O of a circular park of radius 1km , reaches the edge P of the park , then cyclesalongthe circumference and returns to the centre along QO as shown in figure . If the round trip takes 10 min. what is the (a) net displacement . (b) average velocity and (c) average speed of the cyclist ?
Answer» Solution :(a) Net displacement =zero as initial and final POSITION of the cyclist is same (b) Average VELOCITY `=("Net displacement")/("Total time taken")`=zero (c) Total distance travelled `=OP+PQ+QO` `=OP+1//4(2pir)+QO` `=(1+1+(1)/(4)xx2xx(22)/(7)xx1)`km `=(25)/(7)km` Average speed `=("Total path LENGTH")/("Total timetaken")` `=(25//7)/(10//60)km H^(-1)` `=21.43 km h^(-1)` .

Discussion

37978.	A uniform sphere of radius a and density rho is divided in two parts by a plane at a distance b from its centre. Calculate the mutual attraction between two parts.
Answer» Answer :`1/(3)PI^(2)rho^(2)G(a^(2)-b^(2))^(2)`

Discussion

37979.	A particle is executing SHM along a straight line. Its velocities at distance x_(1)" and "x_(2) form the mean position are v_(1)" and "v_(2), respectively. Its time period is………..
Answer» `2PI SQRT((x_(1)^(2)+x_(2)^(2))/(v_(1)^(2)+v_(2)^(2)))` `2pi sqrt((-x_(1)^(2)+x_(2)^(2))/(v_(1)^(2)-v_(2)^(2)))` `2pi sqrt((v_(1)^(2)+v_(2)^(2))/(x_(1)^(2)+x_(2)^(2)))` `2pi sqrt((v_(1)^(2)-v_(2)^(2))/(x_(1)^(2)-x_(2)^(2)))` Solution :For SHM particle, `V= omega sqrt(A^(2)-x^(2))` `therefore v_(1)= omega sqrt(A^(2)-x_(1)^(2))" and "v_(2)= omega sqrt((A^(2)-x_(2)^(2))` `therefore v_(1)^(2)= omega^(2)(A^(2)-x_(1)^(2))" and "v_(2)^(2)= omega^(2) (A^(2)-x_(2)^(2))` `therefore v_(1)^(2) -v_(2)^(2) = omega^(2)[A^(2)-x_(1)^(2)-A^(2)+x_(2)^(2)]` `= omega^(2) [x_(2)^(2)-x_(2)^(2)]` `therefore omega^(2)= (v_(1)^(2)-v_(2)^(2))/((x_(2)^(2)-x_(1)^(2)))` `therefore omega = sqrt((v_(1)^(2)-v_(2)^(2))/(x_(2)^(2)-x_(1)^(2)))` `therefore (2pi)/(T) = sqrt((v_(1)^(2)-v_(2)^(2))/(x_(2)^(2)-x_(1)^(2)))` `therefore T = 2pi sqrt((x_(2)^(2)-x_(1)^(2))/(v_(1)^(2)-v_(2)^(2)))`.

Discussion

37980.	Standing waves can be poduced in the air column ofa pipe. Standing waves produced in an open pipe contain._____.
Answer» SOLUTION :all HARMONICS

Discussion

37981.	Why do we slip on a rainy day ?
Answer» Solution :On a rainy DAY, the cofficient of friction between our feet and the wet ground gets much reduced. Consequently, the frictional FORCE `f_x^max=mu_sN` BECOMES very small. It MAY cause us to slip.

Discussion

37982.	A wire of length l, area of cross-section A and Young,s modulus of elasticity Y is suspended form the roof of a building. A block of mass m is attached at lower end of the wire. If the block is displaced from its mean position and released then the block starts oscillating. Time period of these oscillations will be
Answer» `2pisqrt(Al)/(mY)` `2pisqrt(AY)/(ML)` `2pisqrt(ml)/(YA)` `2pisqrt(m)/(YAL)` ANSWER :C

Discussion

37983.	When body is taken from the equator to the poles, its weight
Answer» REMAINS same increases DECREASES increases at N-pole and decreases at S-pole Answer :B

Discussion

37984.	Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0 m^(3) at a temperature of 27 ^(@)C and 1 atm pressure.
Answer» SOLUTION :`6.10 XX 10^(26)`

Discussion

37985.	A disc is at rest at the top of a rough incined plane. It rolls without slipping. At the bottom of inclined palne there is a vertical groove of radius .R.. In order to loop the groove, the minimum height of incline required is:
Answer» `(15R)/4` `(9R)/4` `(5R)/2` `(7R)/5` Answer :A

Discussion

37986.	A uniform rectangular plate is heated from 0^(@) to 100^(@)C. Initial area of plate is 10 cm^(2). The linear coefficient of the material of the plate is 18xx10^(-6)K. What is the shift of the centre of mass ?
Answer» ZERO 1 cm 2cm 3cm Answer :A

Discussion

37987.	Show that a nonhomogeneous cylinder on a horizontal plane has a position of stable equilibrium when its centre of gravity is vertically below the axis O of the cylinder.
Answer» ANSWER :N//A

Discussion

37988.	in the given progressive wave y = 5 sin (100 pi t -0.4 pi x) where yand x are in metre, t is in second. What is the (a) amplitude (b) wave length (c) frequency (d) wave velocity (e) particle velocity amplitude,
Answer» Solution :Comparing give equation, `y = 5 SIN (100 pi t 0.4 pix)` with `y = a sin (omega t - kx + phi)` we get `a= 5m, k = 0.4pi =- (2pi)/(lamda ) implies lamda = 5m` `omega = 100 pi = 2pi f implies f = 50 Hz` `V = (omega )/(k) = (100 pi)/(0.4pi) = 250 (m)/(s)` Maximum velocity of particle performing S.H.M. and taking part in the propagation of wave, `v _(max) = a omega` `= 5 xx 100 pi` `= 500 pi m//s`

Discussion

37989.	Derive an expression for the gravitational potential energy of a body of mass 'm' raised to a height 'h' above the earth's surface.
Answer» SOLUTION :The gravitational potential energy (U) at some height h is equal to the aount of WORK required to take the object from the GROUND to that height h. `U=intvecF_(a),dvecr=underset(0)overset(h)int\|vecF_(a)\|\|dvecr\|costheta` If the DISPLACEMENT and the applied force are in the same upward direction then, the angle betweeen them is `theta=0^(@)`. Since, `cos0^(@)=1and\|vecF_(a)\|=mgand\|dvecr\|=dr` `U=mgunderset(0)oversethintdrimpliesmg[r]_(0)^(h)=MGH`

Discussion

37990.	Assume that if the earth were made of lead of relative density 11.3, then what would be the value of acceleration due to gravity on the earth surface ?
Answer» Solution :Since, DENSITY of earth can be given as `RHO=` Relative density `xx` density of water `=11.3 xx 10^(3) "kgm"^(-3)` `:.` Acceleration due to GRAVITY on the earth's surface, `g=(GM)/(R^(2))=(G)/(R^(2)).(4)/(3)piR^(3).rho` `=(4)/(3)xx(22)/(7)xx6.67xx10^(-11)xx6.4 xx10^(6)xx11.2xx10^(3)` `=22.21 MS^(-2)`

Discussion

37991.	What is capillarity? Derive an expression for the ascent of liquid in a capillary.
Answer» Solution :The RISE or fall of a liquid in a narrow tube is called capillarity. Let us consider a capillary tube which is held vertically in a beaker containing water, the water rises in the capillary tube to a height h due to surface tension. The surface tension force `F_(T)` acts along the tangent at the point of contact downwards and its reaction force upwards. Surface tension T, is resolved into two components. (i) Horizontal component T `sintheta` and (ii) Vertical component T `costheta` acting upwards, all along the whole circumference of the meniscus. Tota upward force `=(Tcostheta)(2pir)=2pirTcostheta` Where `THETA` is the angle of contact, R is the RADIUS of the tube. Let `rho` be the density of water and h be the height to which the liquid rises inside the tube. Then, `{:(("the volume of"),("liquid column"),("in the tube,V")):}={:(("volume of the"),("liquid column of"),("radius r height h")):}+{:(("volume of the liquid of radius"),("r and height h-volume of"),("the hemisphere of radius of r")):}` `V=pir^(2)h+(pir^(2)xxr-(2)/(3)pir^(3))` `rArrV=pir^(2)h+(1)/(3)pir^(3)` The upward force supports the weight of the liquid column above the free surface, therefore, `2pirTcostheta=pir^(2)(h+(1)/(3)r)rhog` `rArr""T=((h+(1)/(3)r)rhog)/(2costheta)` If the capillary is a very FINE tube ofradius (i.e., radius is very small) then `(r)/(3)` can be neglected when it is compared to the height h. Therefore, `T=(rrhogh)/(2costheta)`

Discussion

37992.	For one wire tied at both the ends with a tension of 450 N, linear density of mass is 0.05 g/cm. It sounds resonantly for two consecutive frequencies 420 Hz and 490 Hz. Find length of this wire.
Answer» Solution :In the PRESENT CASE, FREQUENCY of `N^(th)` harmonic `f _(n) = n f _(1) = 420 Hz` `f _(n +1) = (n +1) f _(1)= 490 Hz ` Taking subtraction, `n f _(1) + f _(1) -nf _(1) = 490 -420` `therefore f _(1) = 70 Hz` `therefore (1)/(2L) sqrt ((T)/( mu ))=70` `therefore L = (300)/(140) =2.1428 m ~~2.143 m`

Discussion

37993.	Three identical vessels are filled to the same height with three different liquids A, B and C(rho_(A) gt rho_(B) gt rho_(C)). The pressure at the base will be
Answer» EQUAL in all vessels Maximum in VESSEL A Maximum in vessel B Maximum in vessel Answer :B

Discussion

37994.	Give the location of the centre of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv) cube, each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body ?
Answer» Solution :The GEOMETRICAL centre of each. No, the CM may lie OUTSIDE the body, as in case of a ring, a HOLLOW sphere, ahollow cylinder, a hollow cube ETC.

Discussion

37995.	A person of mass 60 kg wants to lose 5 kg by going up and down a 10 m high stairs. Assume he burns twice as much fat while going up than coming down. If 1 kg of fat is burnt on expending 7000 kilo calories, how many times must he go up and down to reduce his weight by 5 kg ?
Answer» Solution :Height of the stairs H = 10 m Work done to burn 1 kg of fat = 7000 K cal `therefore` Work done to burn 5 kg of fat =5 x 7000 = 35000 k cal Work done towards burning of fat in up and down the stairs, `=mgh+1/2mgh` `=3/2mgh` `=3/2xx60xx10xx10` =9000 J `=9000xx1/4.2` cal =2142.85 cal `therefore` For energy of 2142.85 cal one trip (up and down) of stairs needed Hence , for energy of `35xx10^6` cal N trip (up and down) of stairs needed. `therefore n=(35xx10^6)/(2142.85)=0.01633xx10^6` =16330 TIMES

Discussion

37996.	If A=B+ (C)/(D+E)the dimensions of B and Care [M^(0)LT^(-1)] and [M^(0)LT^(0)], respectively. Find the dimensions of A, D and E.
Answer» `[A]= [M^(0)L^(0)T^(-1)], [D]= [T], [E]= [LT]` `[A]= [MLT^(0)], [D]= [T^(2)], [E]= [T^(2)]` `[A]= [M^(0)LT^(-1)], [D]= [MT], [E]= [MT]` `[A]= [M^(0)LT^(-1)], [D]= [T], [E]= [T]` Solution :As `A= B+ (C)/(D+E)` `:. [D]= [E]` `:. [A]= [B]= [(C)/(D+E)]= [(C)/(D)]= [(C)/(E)] :. [A]= [B]= [M^(0)LT^(-1)]` `[(C)/(D)]= [A]= [M^(0)LT^(-1) ]` `[D]= [E]= [(C)/(LT^(-1)] = ( [M^(0)LT^(0) ])/( [M^(0)LT^(-1) ])= [T]`

Discussion

37997.	At what angle two vectors of magnitudes vecA +vecBand A - B act, so that their resultant is sqrt(3A^2 + B^2)?
Answer» `30^@` `45^@` `60^@` `90^@` ANSWER :C

Discussion

37998.	The moment of inertia of a wheel of radiu 20 cm is 40 kg m^(2). If a tangential force of 80 N applied on the wheel, its Rotational K.E. after 4s is
Answer» 16.2J 51.2J 25.6J 24.8J Answer :B

Discussion

37999.	0.1m^3 of water at 80^@C is mixed with 0.3m^3 of water at 60^@C. The final temperature of the mixture is
Answer» `65^@C` `70^@C` `60^@C` `75^@C` ANSWER :A

Discussion

38000.	A particle has a time period of 1s under the action of a certain force and 2 s under the action of another force. Find the time period when the forces are acting in the same direction simultaneously.
Answer» Solution :LET the period of oscillation be `T_1`under tre action of the force `F_1 and T_2`under the action of force `F_2`when acting SEPARATELY. Lot F be the RESULTANT force and the period of oscillation under the action of the resultant force `F=F_1+F_2` `momega^2y= momega_1^2y+momega_2^2y` `((2pi)/T)^2=((2pi)/T_1)^2+((2pi)/T_2)^2` `1/T^2=1/T_1^2+1/T_2^2` `T^2=(T_1^2T_2^2)/(T_1^2+T_2^2)` `T=(T_1T_2)/(SQRT(T_1^2+T_2^2))` `T_1=1 s, T_2 = 2S` `T = (1xx2)/(sqrt(1^2+2^2))` = 0.894 s

Discussion

Explore topic-wise InterviewSolutions in .

We have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why ?

Find the torque of a force 7hati + 3hatj – 5hatkabout the origin. The force acts on a particle whose position vector is hati – hatj + hatk.

Two identical rods of a metal are welded in series then 20 cal of heat flows through them in 4 minute. If the rods are welded in parallel then the same amount of heat will flow in

Calculate rate of flow of glycerin of density 1.25xx10^(3) kg/m^(3) through the concial section of a horizontal pipe if the radii of its ends are 0.1 m and 0.04m and pressure drop across its length is 10 N//m^(2).

A satellite orbits the earth at a height of R/6, its orbital speed ?

What is the condition to be satisfied by a mathematical relation between time and displacement to describe a periodic motion ?

A spring with 10 coils has spring constant k. It is exactly cut into two halves, then each of these new springs will have a spring constant

A circular frame made of 20 cm long thin wire is floating on the surface of water. If the surface tension of water is 70 dyne/cm. the required excess force to separate this frame from water will be :

Two bodies of equal masses moving with momentum of first double that of the second collide completely inelastically. If the total final kinetic energy is 9 joule, the initial kinetic energy of the body of greater momentum is:

A satellite is in an elliptic orbit around the earth with aphelion of 6R and perihelion of 2R where R = 6400 km is the radius of the earth . Find the velcoity of the satellite at apogee and perigee.[ G = 6.67 xx10^(-1) SI unit and M= 6 xx 10^(24)kg ]

The magnitude of potential energy per unit mass of the object at the surface of earth is .E.. Then escape velocity of the object is

The motion of a particle is given by X=Asinomegat+B cos omegat the motion of the particle is

A body of mass 5kg starts from the origin with an initial velocityof barU= (30i+40j)m//s. A constant force of F= (-hati - 5 hatj)N acts on the body . Find the time in which they- component of the velocity becomes zero .

The ends of a metal rod of length 40 cm are maintained at temperatures 100°C and 20°C respectively. At steady state condition, the temperature at a point at a distance 30 cm from the hot end of the rod is

A light metal stick of square cross-section (5cmxx5cm) and length 4m mass 2.5 kg ad is shown in the figure below determine its angle of inclination when the water surface is 1m above the hinge. What minimum depth of water above high will be required to bring the metal stick in vertical position.

For a body rolling along a level surface, without slipping the translational and rotational kinetic energies are in the ratio 2:1. The body is

A spring when compressed by 0.2 m develops a restoring force of 50 N. If a body of mass 10 kg is placed on it, calculate (i) the force constnat (I) of the spring (ii) the depression of the spring (iii) time period of oscillation when the weight is slightly displaced from equilibrium position.

The maximum vertical height of a projectile is 10 m. If the magnitude of the initial velocity is 28 ms^(-1), what is the direction of the initial velocity. ? Take g = 9.8 ms^(-2).

If vec(P)=hat(i)+2hat(j)-4hat(k)andvec(Q)=hat(i)+2hat(j)-hat(k) then (vec(P)+vec(Q))*(vec(P)-vec(Q)) is

One parsec is .........

What doesit mean when a height of a barometer is rising ?

A machine gun has a mass of 20 kg . It fires 35 g bullets at the rate of 4bullets per second, with a speed of 400 ms^(-1). What force must be applied to the gun to keep it in position ?

Hailstones fall from a certain height. If only 1% of the hailstones melt on reaching the ground, find the height from which they fall. (g - 10 ms^(-2)L - 80 caloric/g and I - 4.21/calorie)

A uniform sphere of radius a and density rho is divided in two parts by a plane at a distance b from its centre. Calculate the mutual attraction between two parts.

A particle is executing SHM along a straight line. Its velocities at distance x_(1)" and "x_(2) form the mean position are v_(1)" and "v_(2), respectively. Its time period is………..

Standing waves can be poduced in the air column ofa pipe. Standing waves produced in an open pipe contain._____.

Why do we slip on a rainy day ?

When body is taken from the equator to the poles, its weight

Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0 m^(3) at a temperature of 27 ^(@)C and 1 atm pressure.

A disc is at rest at the top of a rough incined plane. It rolls without slipping. At the bottom of inclined palne there is a vertical groove of radius .R.. In order to loop the groove, the minimum height of incline required is:

A uniform rectangular plate is heated from 0^(@) to 100^(@)C. Initial area of plate is 10 cm^(2). The linear coefficient of the material of the plate is 18xx10^(-6)K. What is the shift of the centre of mass ?

Show that a nonhomogeneous cylinder on a horizontal plane has a position of stable equilibrium when its centre of gravity is vertically below the axis O of the cylinder.

in the given progressive wave y = 5 sin (100 pi t -0.4 pi x) where yand x are in metre, t is in second. What is the (a) amplitude (b) wave length (c) frequency (d) wave velocity (e) particle velocity amplitude,

Derive an expression for the gravitational potential energy of a body of mass 'm' raised to a height 'h' above the earth's surface.

Assume that if the earth were made of lead of relative density 11.3, then what would be the value of acceleration due to gravity on the earth surface ?

What is capillarity? Derive an expression for the ascent of liquid in a capillary.

For one wire tied at both the ends with a tension of 450 N, linear density of mass is 0.05 g/cm. It sounds resonantly for two consecutive frequencies 420 Hz and 490 Hz. Find length of this wire.

Three identical vessels are filled to the same height with three different liquids A, B and C(rho_(A) gt rho_(B) gt rho_(C)). The pressure at the base will be

Give the location of the centre of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv) cube, each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body ?

A person of mass 60 kg wants to lose 5 kg by going up and down a 10 m high stairs. Assume he burns twice as much fat while going up than coming down. If 1 kg of fat is burnt on expending 7000 kilo calories, how many times must he go up and down to reduce his weight by 5 kg ?

If A=B+ (C)/(D+E)the dimensions of B and Care [M^(0)LT^(-1)] and [M^(0)LT^(0)], respectively. Find the dimensions of A, D and E.

At what angle two vectors of magnitudes vecA +vecBand A - B act, so that their resultant is sqrt(3A^2 + B^2)?

The moment of inertia of a wheel of radiu 20 cm is 40 kg m^(2). If a tangential force of 80 N applied on the wheel, its Rotational K.E. after 4s is

0.1m^3 of water at 80^@C is mixed with 0.3m^3 of water at 60^@C. The final temperature of the mixture is

A particle has a time period of 1s under the action of a certain force and 2 s under the action of another force. Find the time period when the forces are acting in the same direction simultaneously.