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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your Class 11 knowledge and support exam preparation. Choose a topic below to get started.
| 36251. |
The radius vector of a particle varies with time t as vecr=vecbt(1-alpha t)where vecbis a constant vector and a is a positive factor . Find the time interval Delta ttaken by the particle to return to the initial point |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`vecr=vecbt(1-alpha t)`<br/>For <a href="https://interviewquestions.tuteehub.com/tag/initial-499269" style="font-weight:bold;" target="_blank" title="Click to know more about INITIAL">INITIAL</a> <a href="https://interviewquestions.tuteehub.com/tag/point-1157106" style="font-weight:bold;" target="_blank" title="Click to know more about POINT">POINT</a> , `vecr=<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>,vecb=t(1-alphat)=0`<br/>`t=0"and" t=1/alpha` <a href="https://interviewquestions.tuteehub.com/tag/hence-484344" style="font-weight:bold;" target="_blank" title="Click to know more about HENCE">HENCE</a>,` Deltat=1/alpha `</body></html> | |
| 36252. |
C_(P) is always greater than C_(V) for a gas. Which of the following statements provide, partly or wholly,the reason for this? a) No work is done by a gas at constant volume b) When a gas absorbs heat at constant pressure, its volume must change c) For the same change in temperature, the internal energy of a gas changes by a smallar amount at constant volume than at constant pressure d) The internal energy of an ideal gas is a function only of its temperature |
| Answer» <html><body><p>only a,<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a> are correct<br/>only b, d are correct<br/>only <a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a>, d are correct<br/>all are correct</p>Answer :D</body></html> | |
| 36253. |
A car moving on a straight path with a speed of 54 km/h, passes by a stationary listener, playing hrn with a frequency 500 Hz. If velocity of sound in air is 340 m/s, find out the differenfce in frequencies of sound of horn, heard by stationary listener when car approaches him and recedes from him. |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`<a href="https://interviewquestions.tuteehub.com/tag/f-455800" style="font-weight:bold;" target="_blank" title="Click to know more about F">F</a> _(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>) - f _(<a href="https://interviewquestions.tuteehub.com/tag/l-535906" style="font-weight:bold;" target="_blank" title="Click to know more about L">L</a>) =44.2 <a href="https://interviewquestions.tuteehub.com/tag/hz-493442" style="font-weight:bold;" target="_blank" title="Click to know more about HZ">HZ</a>`</body></html> | |
| 36254. |
Two bodies begin to fall freely from the same height. The second body begins to fall 'tau' s after the first. After what time from the beginning of first body dose the distance between the bodies equals to l? |
| Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/let-11597" style="font-weight:bold;" target="_blank" title="Click to know more about LET">LET</a> the time of fall of the 1st <a href="https://interviewquestions.tuteehub.com/tag/body-900196" style="font-weight:bold;" target="_blank" title="Click to know more about BODY">BODY</a> be t seconds Time of fall of second body `=t-tau` <br/> Distances of free fall of the bodies in the above time <a href="https://interviewquestions.tuteehub.com/tag/intervals-1049965" style="font-weight:bold;" target="_blank" title="Click to know more about INTERVALS">INTERVALS</a> respectively are <br/> `H_(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)=(gt^(2))/(2), H_(2)=(g(t-tau)^(2))/(2)` <br/> `H_(1)=(gt^(2))/(2), H_(2)=(g(t-tai)^(2))/(2)` <br/> Therefore `l=H_(i)-H_(2)=gt tau-1/2 gt^(2)""therefore t=l/(g tau)+tau /2`</body></html> | |
| 36255. |
A man weighs 75kg on the surface of the earth. His weight on a geostationary satellite is |
| Answer» <html><body><p>infinity<br/>150 <a href="https://interviewquestions.tuteehub.com/tag/kg-1063886" style="font-weight:bold;" target="_blank" title="Click to know more about KG">KG</a> <br/><a href="https://interviewquestions.tuteehub.com/tag/zero-751093" style="font-weight:bold;" target="_blank" title="Click to know more about ZERO">ZERO</a><br/>`75/2` kg </p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html> | |
| 36256. |
A solid cylinder rolls up an inclined plane of angle of inclination 30^(@) . At the bottom of the inclined plane the centre of mass of the cylinder has a speed of 5 m s^(-1) . How far will cylinder go up the plane ? (Take g = 10 m s^(-2)) |
| Answer» <html><body><p>`(<a href="https://interviewquestions.tuteehub.com/tag/15-274069" style="font-weight:bold;" target="_blank" title="Click to know more about 15">15</a>)/(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>) `m <br/>`(4)/(15)` m <br/>`(<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>)/(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)` m <br/>`(3)/(10)` m </p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :A</body></html> | |
| 36257. |
A train was moving at the rate of 54kmh^(-1)when brakes were applied. It came to rest within a distance of 225 m. Calculate the retardation produced in the train. |
| Answer» <html><body><p></p>Solution :The final <a href="https://interviewquestions.tuteehub.com/tag/velocity-1444512" style="font-weight:bold;" target="_blank" title="Click to know more about VELOCITY">VELOCITY</a> of the particle v = 0 <br/> The initial velocity of the particle <br/> `u=54xx(5)/(<a href="https://interviewquestions.tuteehub.com/tag/18-278913" style="font-weight:bold;" target="_blank" title="Click to know more about 18">18</a>)ms^(-1)=15ms^(-1)` <br/> `S=225m` <br/> Retardation is always against the velocity of the particle. <br/> `v^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)=u^(2)-<a href="https://interviewquestions.tuteehub.com/tag/2as-1836799" style="font-weight:bold;" target="_blank" title="Click to know more about 2AS">2AS</a>` <br/> `0=(15)^(2)-2a(225)` <br/> `450a=225` <br/> `a=(225)/(450)ms^(-1)=0.5ms^(-2)` <br/> Hence, retardation `=0.5ms^(-2)`</body></html> | |
| 36258. |
The period of a simple pendulum on the surface of earth is T. At an altitude of half of the radius of earth from the surface, its period will be |
| Answer» <html><body><p>`sqrt((<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)/(2))T`<br/>`(<a href="https://interviewquestions.tuteehub.com/tag/3t-311358" style="font-weight:bold;" target="_blank" title="Click to know more about 3T">3T</a>)/(2)`<br/>`(<a href="https://interviewquestions.tuteehub.com/tag/2t-301164" style="font-weight:bold;" target="_blank" title="Click to know more about 2T">2T</a>)/(3)`<br/>`sqrt((2)/(3))T`</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a></body></html> | |
| 36259. |
Power of a man who can chew 0.3 kg ice in one minute is (in Cal/s) |
| Answer» <html><body><p>400<br/>4<br/>24<br/>240</p>Answer :A</body></html> | |
| 36260. |
The measured mass and volume of a body are 53.63g and 5.8 cm^(3) respectively, with possible errors of 0.01 g and 0.1 cm^(3). The maximum percentage error in density is about |
| Answer» <html><body><p>0.002<br/>0.02<br/>0.05<br/>0.1</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a></body></html> | |
| 36261. |
One end of a long string of linear mass density10^(-2) kg m^(-1) is connected to an electrically driven tuning fork of frequency 150 Hz. The other end passes over a pulley and is tied to a pan containig a massof 90 kg. the pulley end absorbs all the incoming energy so that reflected waves from this end have neglidible amplitude , At t =0, the left end (fork end ) of the string is at x=0 has a transverse displacement of 2.5 cm and is moving along positive y-direction. the amplitude of the wave is 5 cm. write down the transverse displacement y (in cetimetres) as function of x(in metres) and t (in seconds ) that describes the wave on the string. |
| Answer» <html><body><p></p>Solution :`theta=pi//<a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a>` <br/> `<a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a>=sqrt(T//mu)=sqrt(<a href="https://interviewquestions.tuteehub.com/tag/900-341675" style="font-weight:bold;" target="_blank" title="Click to know more about 900">900</a>)/(0.01)=<a href="https://interviewquestions.tuteehub.com/tag/300-305868" style="font-weight:bold;" target="_blank" title="Click to know more about 300">300</a> m//s` <br/> <a href="https://interviewquestions.tuteehub.com/tag/frequency-465761" style="font-weight:bold;" target="_blank" title="Click to know more about FREQUENCY">FREQUENCY</a> `=150 Hz` <br/> `:.lambda=(v)/(f)=(300)/(150)=2m`<br/> Then equation will be <br/> `y=A sin{2pi((t)/(T)-(x)/(lambda))+theta}` <br/> `=5 sin{2pi(150t-(x)/(2))+(pi)/(6)}` <br/> `=5 sin{pi(300t-x)+(pi)/(6)}`</body></html> | |
| 36262. |
1 erg is equivalent to |
| Answer» <html><body><p>`10^(-<a href="https://interviewquestions.tuteehub.com/tag/7-332378" style="font-weight:bold;" target="_blank" title="Click to know more about 7">7</a>)<a href="https://interviewquestions.tuteehub.com/tag/j-520843" style="font-weight:bold;" target="_blank" title="Click to know more about J">J</a>`<br/>`1.6xx10^(-19)J`<br/>4.186J<br/>`3.6xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a>)J`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :A</body></html> | |
| 36263. |
A ladder .AB. of weight 300 N and length 5 m is lying on a horizontal surface. Its centre of gravity is at a distane of .2m. from end A.A weight of 80N is attached at end B. The work done in raising the ladder to the vertical position with end .A. in contact with the ground is, |
| Answer» <html><body><p>500J<br/>1000 J<br/>1150 J<br/>1900J</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a></body></html> | |
| 36264. |
A large steel wheel is to be fitted on to a shaft of the same material. At 27 ^(@) C, the outer diameter of the shaft is 8.70 cm and the diameter of the central hole in the wheel is 8.69 cm. The shaft is cooled using dry Icel. At what temperature of the shaft does the wheel slip on the shaft? Assume coefficient of linear expansion of the steel to be constant over the required temperature range: alpha_("steel ") = 1.20 xx 10^(-5) K^(-1). |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :When the shaft is <a href="https://interviewquestions.tuteehub.com/tag/cooled-7329693" style="font-weight:bold;" target="_blank" title="Click to know more about COOLED">COOLED</a> to <a href="https://interviewquestions.tuteehub.com/tag/temperature-11887" style="font-weight:bold;" target="_blank" title="Click to know more about TEMPERATURE">TEMPERATURE</a> ` - 69 ^(@) C`the <a href="https://interviewquestions.tuteehub.com/tag/wheel-736315" style="font-weight:bold;" target="_blank" title="Click to know more about WHEEL">WHEEL</a> can <a href="https://interviewquestions.tuteehub.com/tag/slip-1211944" style="font-weight:bold;" target="_blank" title="Click to know more about SLIP">SLIP</a> on the shaft .</body></html> | |
| 36265. |
A,B,C and D are four points on a vertical line such that AB = BC CD. A body is allowed to fall freely from A. Prove that the respective times required by the body to cross the distances AB, BC, CD should be in the ratio 1 : (sqrt(2)-1) : (sqrt(3)-sqrt(2)). |
| Answer» <html><body><p></p>Solution :Let AB = BC = <a href="https://interviewquestions.tuteehub.com/tag/cd-407381" style="font-weight:bold;" target="_blank" title="Click to know more about CD">CD</a> = x and time taken by the body to cover these distances be`t_(1), t_(2)`and `t_(3)` respectively. <br/> Now, `x = (1)/(2)"gt"_(1)^(2) """or", t_(1) = sqrt((2x)/(<a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a>)) ""cdots(1)` <br/>`2x = (1)/(2) g(t_(1)-t_(2))^(2) """or", t_(1)+t_(2)= sqrt((<a href="https://interviewquestions.tuteehub.com/tag/4x-319255" style="font-weight:bold;" target="_blank" title="Click to know more about 4X">4X</a>)/(g)) ""cdots(2)` <br/> `3x = (1)/(2) g(t_(1)+t_(2)+t_(3))^(2)"""or" t_(1)+t_(2)+t_(3)= sqrt((6x)/(g)) "" cdots(3)` <br/> From (1) and (2) we <a href="https://interviewquestions.tuteehub.com/tag/get-11812" style="font-weight:bold;" target="_blank" title="Click to know more about GET">GET</a>, <br/> `t_(2) = sqrt((4x)/(g))-sqrt((2x)/g)=sqrt((2x)/(g))(sqrt(2)-1)` <br/> From (2) and (3) we get , <br/> `t_(3)=sqrt((6x)/(g))-sqrt((4x)/g)=sqrt((2x)/(g))(sqrt(3)-sqrt(2))` <br/> `:. "" t_(1):t_(2):t_(3)=1:(sqrt2-1):(sqrt3-sqrt(2))""`(Proved).</body></html> | |
| 36266. |
A difference of temperature of 25°C is equivalent to a difference of |
| Answer» <html><body><p>`25^(@)F`<br/>`45^(@)F`<br/>`67^(@)F`<br/>`77^(@)F`</p>Solution :`(F-32)/(9)=(<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a> )/(<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>) <a href="https://interviewquestions.tuteehub.com/tag/rarr-1175461" style="font-weight:bold;" target="_blank" title="Click to know more about RARR">RARR</a> F_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)-F_(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)=(9)/(5)(C_(2)-C_(1))=(9)/(5)xx25=45^(@)F`</body></html> | |
| 36267. |
Obtain instantaneous velocity of a particle executing SHM. |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Intantaneous velocity of SHM particle is the rate of change of displacement for a time interval. <br/> Suppose the displacement of SHM particle at time t with amplitude A and angular speed `omega` is <br/> `x(t)= A cos (omega t+ phi)"""……"(1)` <br/> where `phi` is the initial phase <br/> <a href="https://interviewquestions.tuteehub.com/tag/differentiating-953151" style="font-weight:bold;" target="_blank" title="Click to know more about DIFFERENTIATING">DIFFERENTIATING</a> <a href="https://interviewquestions.tuteehub.com/tag/equation-974081" style="font-weight:bold;" target="_blank" title="Click to know more about EQUATION">EQUATION</a> (1) respect to t gives the instantaneous velocity <br/> `therefore v(t)= (d[x(t)])/(dt) = (d)/(dt)[A cos (omega t+ phi)]` <br/> `therefore v(t)= -A omega sin (omega t+ phi)` <br/>but `sin(omega t +phi)= pm sqrt(1-cos^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>) (omega t+ phi))` <br/> `v(t)= (-A omega) (pm sqrt(1-cos^(2) (omega t+ phi)))` <br/> `= pm omega sqrt(A^(2)-A^(2) cos^(2)(omega t+ phi))` <br/> `= pm omega sqrt(A^(2)-x^(2)(t))` <br/> Generally `v= pm omega sqrt(A^(2) - x^(2))` <br/> Special Cases : <br/> (1) At mean position, x=0 <br/> Velocity `v= pm omega A` <br/> In +x-direction, v is positive and in -x-direction, v is negative. <br/> Hence, maximum velocity of SHM particle `v_("max")= A omega` <br/> (2) At extreme point, for velocity `x= |A|`, <br/> `v= pm omega sqrt(A^(2)-A^(2))""[therefore |A|^(2)= A^(2)]` <br/> `therefore v= 0` <br/> The velocity of SHM particle at extreme points is <a href="https://interviewquestions.tuteehub.com/tag/zero-751093" style="font-weight:bold;" target="_blank" title="Click to know more about ZERO">ZERO</a>.</body></html> | |
| 36268. |
A rigid bar of mass M is supported symmetrically by three wires each of length l. Those at each end are of copper and the middle one is of iron. The ratio of their diameters, if each is to have the same tension, is equal to |
| Answer» <html><body><p>`("Ycopper")/("Yiron")`<br/>`sqrt(("Yiron")/("Ycopper"))`<br/>`(Y^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>) "iron")/(Y^(2)"copper") `<br/>`("Yiron")/("Ycopper")`</p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`Y=(<a href="https://interviewquestions.tuteehub.com/tag/f-455800" style="font-weight:bold;" target="_blank" title="Click to know more about F">F</a>)/(<a href="https://interviewquestions.tuteehub.com/tag/pi-600185" style="font-weight:bold;" target="_blank" title="Click to know more about PI">PI</a>((D)/(2))^(2))xx(1)/(((Deltal)/(l)))=(4Fl)/(piD^(2)Deltal)` <br/> `impliesD=sqrt((4Fl)/(piDeltalY))or D prop sqrt((1)/(Y))` <br/> Hence `(D copper)/(D iron) = sqrt((Y iron)/(Ycopper))`</body></html> | |
| 36269. |
A particle is executing simple harmonic motion with an amplitude of 2m . The difference in the magnitudes of its maximum acceleration and maximum velocity is 4. The time period of its oscilation and its velocity when it is 1m away from the mean position are respectively. |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/2s-301125" style="font-weight:bold;" target="_blank" title="Click to know more about 2S">2S</a>, 2sqrt(3)ms^(-1)`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/7-332378" style="font-weight:bold;" target="_blank" title="Click to know more about 7">7</a>/(<a href="https://interviewquestions.tuteehub.com/tag/22-294057" style="font-weight:bold;" target="_blank" title="Click to know more about 22">22</a>)s,4sqrt(3)ms^(-1)`<br/>`(22)/7s,2sqrt(3)ms^(-1)`<br/>`(44)/7s,4sqrt(3)ms^(-1)`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :A</body></html> | |
| 36270. |
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 Fig. 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» <html><body><p><br/></p>Solution :Mass of ladder `=M-m`, because net mass on both sides of string is same. Let in the given process, block `M` rises by height `y`, then ladder will <a href="https://interviewquestions.tuteehub.com/tag/come-409636" style="font-weight:bold;" target="_blank" title="Click to know more about COME">COME</a> down by <a href="https://interviewquestions.tuteehub.com/tag/distance-116" style="font-weight:bold;" target="_blank" title="Click to know more about DISTANCE">DISTANCE</a> `y` and when man comes in the <a href="https://interviewquestions.tuteehub.com/tag/level-1072714" style="font-weight:bold;" target="_blank" title="Click to know more about LEVEL">LEVEL</a> of `M`, then height raised by man w.r.t ground will be `h+y`. <br/> Now cenre of mass (man`+`ladder) system will be raised by `y` , then <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/BMS_VOL2_C01_E01_069_S01.png" width="80%"/> <br/> `y=(m(h+y)+(M-m)(-y))/(2(M-m))` <br/> Now work done by man <br/> `W=`total change in potential energy of system <br/>`implies W=Mgy+mg(h+y)-(M-m)<a href="https://interviewquestions.tuteehub.com/tag/gy-469735" style="font-weight:bold;" target="_blank" title="Click to know more about GY">GY</a>` <br/> Put the values of `y` and solve to get `W=(Mmgh)/(M-m)`</body></html> | |
| 36271. |
For the arrangment shown in the figure, tension in the string is |
| Answer» <html><body><p>`("<a href="https://interviewquestions.tuteehub.com/tag/mg-1095425" style="font-weight:bold;" target="_blank" title="Click to know more about MG">MG</a>")/(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)` <br/>`("<a href="https://interviewquestions.tuteehub.com/tag/2mg-1835961" style="font-weight:bold;" target="_blank" title="Click to know more about 2MG">2MG</a>")/(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)`<br/>`("<a href="https://interviewquestions.tuteehub.com/tag/3mg-1862197" style="font-weight:bold;" target="_blank" title="Click to know more about 3MG">3MG</a>")/(3)`<br/>`("Mg")/(3)`</p>Answer :D</body></html> | |
| 36272. |
A vessel contains two non reactive gases : neon (monatomic) and oxygen (diatomic). The ratio of their partial pressures is 3:2. Estimate the ratio of (i) number of molecules and (ii) mass density of neon and oxygen in the vessel. Atomic mass of Ne=20.2 u, molecular mass of O_(2) = 32.0 u. |
| Answer» <html><body><p></p>Solution :Partial pressure of a gas in a mixture is the pressure it would have for the same volume and temperature if it alone occupied the vessel. (The total pressure of a mixture of non-reactive gases is the sum of partial pressures due to its constituent gases.) Each gas (assumed <a href="https://interviewquestions.tuteehub.com/tag/ideal-1035490" style="font-weight:bold;" target="_blank" title="Click to know more about IDEAL">IDEAL</a>) obeys the gas law. Since V and T are common to the two gases, we have `P_(1) V = mu_(1), <a href="https://interviewquestions.tuteehub.com/tag/rt-615359" style="font-weight:bold;" target="_blank" title="Click to know more about RT">RT</a>` and `P_(2)V= mu_(2)RT`, i.e. `(P_(1)//P_(2)) = (mu_(1) // mu_(2))`. Here 1 and 2 referto neon and oxygen respectively. Since`(P_(1)//P_(2)) = (<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>//2)`(<a href="https://interviewquestions.tuteehub.com/tag/given-473447" style="font-weight:bold;" target="_blank" title="Click to know more about GIVEN">GIVEN</a>),`(mu_(1) // mu_(2))= 3//2`. <br/> (i) By definition `mu_(1) = (N_(1)//N_(A))` and `u_(2) = (N_(2)//N_(A))`where `N_(1)` and `N_(2)` are the number of molecules of 1 and 2, and `N_(A)` is the Avogadro.s number. Therefore, `(N_(1)//N_(2)) = (mu_(1) // mu_(2)) = 3//2`.<br/>(ii) We can also write `mu_(1) = (m_(1)//M_(1))` and `mu_(2) = (m_(2)//M_(2))` where `m_(1)` and `m_(2)` are the masses of 1 and 2, and `M_(1)` and `M_(2)` are their molecular masses. (Both `m_(1)` and `M_(1)`, as well as `m_(2)` and `M_(2)` should be expressed in the same units). If `rho_(1)` and `rho_(2)` are the mass <a href="https://interviewquestions.tuteehub.com/tag/densities-948213" style="font-weight:bold;" target="_blank" title="Click to know more about DENSITIES">DENSITIES</a> of 1 and 2 respectively, we have <br/> `(rho_(1))/(rho_(2))= (m_(1)//V)/(m_(2)//V)= (m_(1))/(m_(2))= (mu_(1) )/ (mu_(2)) xx (M_(1)/(M_(2)))` <br/> `= (3)/(2) xx (20.2)/(32.0)= 0.947`</body></html> | |
| 36273. |
The correctness of equation can be checked using the principle of homogeneity in dimensions. The velocity V of a particle depends on time V as V=At^2+BtFind the dimensions and units of A and B. |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`[A]=[[<a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a>]]/[[t^2]]=[[M^0LT^-1]]/[[M^0L^0T^2]]=[M^0LT^-3]` <a href="https://interviewquestions.tuteehub.com/tag/unit-1438166" style="font-weight:bold;" target="_blank" title="Click to know more about UNIT">UNIT</a> of`A=ms^-3[<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>]=[[V]]/[[t]]=[[M^0LT^-1]]/[[M^0L^0T]]=[M^0LT^-2]` Unit `B=ms^-2`</body></html> | |
| 36274. |
A capillary tube of diameter 0.4mm is dipped in a beaker containing mercury. If density of mercury is 13.6 xx 10^3 kg//m^3, surface tension 0.49Nm^(-1), angle of contact 135^@, the depression of the meniscus in the capillary tube will be |
| Answer» <html><body><p>`0.13 m`<br/>`0.013 m`<br/>`0.026m`<br/>`0.0026 m`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html> | |
| 36275. |
Find the weight of water supported by surface tension in a capillary tube with a radius of 0.2mm. Surface tension of water is 0.072Nm^(-1) and angle of contact of water is 0^(@) |
| Answer» <html><body><p></p>Solution :Assume the weight of <a href="https://interviewquestions.tuteehub.com/tag/water-1449333" style="font-weight:bold;" target="_blank" title="Click to know more about WATER">WATER</a> to be .F. <br/> weight of water in capillary tube = upward force due to surface tension <br/> i.e., `F=2pir(Scostheta)` <br/> Surface tension of water `S=0.072Nm^(-1)` <br/> <a href="https://interviewquestions.tuteehub.com/tag/angle-875388" style="font-weight:bold;" target="_blank" title="Click to know more about ANGLE">ANGLE</a> of contact`theta=0^(@)` <br/> <a href="https://interviewquestions.tuteehub.com/tag/radius-1176229" style="font-weight:bold;" target="_blank" title="Click to know more about RADIUS">RADIUS</a> of capillary tube `(r)=(0.2)/(1000)m=0.2xx10^(-3)m` <br/> `F=2pir(Scostheta)=2xx(22)/(7)xx0.2xx10^(-3)xx0.072xx1` <br/> `=90.51xx10^(-<a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a>)NimpliesF=90.51xx10^(-6)N`</body></html> | |
| 36276. |
Derive an expression for the potential energy of a body near the surface of the Earth. (OR) Calculate the potential energy of the object of mass m at a height h. |
| Answer» <html><body><p></p>Solution :(i) The gravitational potential energy (U) at some <a href="https://interviewquestions.tuteehub.com/tag/height-1017806" style="font-weight:bold;" target="_blank" title="Click to know more about HEIGHT">HEIGHT</a> h is equal to the amount of work required to take the object from ground to that height h with constant velocity. <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/SUR_PHY_XI_V01_C04_E02_105_S01.png" width="80%"/><br/>(ii)Consider a body of mass m being moved from ground to the height h against the gravitational force as shown in Figure. <br/> (<a href="https://interviewquestions.tuteehub.com/tag/iii-497983" style="font-weight:bold;" target="_blank" title="Click to know more about III">III</a>)The gravitational force `vec(F)_(g)` acting on the body is, `vec(F)_(g)=-mg hat(j)` (as teh force is in y direction, unit vector is used). Here,negative sign implies that the force is acting vertically downwards. In order to move the body without acceleration (or with constant velocity), an external applied force `vec(F)_(a)` equal in magnitude but opposite to that of gravitational force `vec(F)_(g)` has to be applied on the body i.e., `vec(F)_(a)-vec(F)_(g)`. This implies that `vec(F)_(a)=-mg hat(j)` <br/>(iv)The positive sign implies that the applied force is in vertically upward direction. Hence, when the body is lifted up its velocity remains <a href="https://interviewquestions.tuteehub.com/tag/unchanged-2316761" style="font-weight:bold;" target="_blank" title="Click to know more about UNCHANGED">UNCHANGED</a> and thus its <a href="https://interviewquestions.tuteehub.com/tag/kinetic-533291" style="font-weight:bold;" target="_blank" title="Click to know more about KINETIC">KINETIC</a> energy also remains constant. <br/> (v) The gravitational potential energy (U) at some height h is equal to the amount of work required to take the object from the ground to that height h. <br/> `U = int vec(F)_(a).d vec(r )=int_(0)^(h)|vec(F)_(a)||d vec(r )|cos theta` <br/> (vi)Since the displacement and the applied force are in the same upward direction, the angle between then, `theta = 0^(@)`.<br/>Hence, `cos theta^(@)=1` and `|vec(F)_(a)|=mg` and <br/> `= |d vec(r )|dr`. <br/> `U = ms int_(0)^(h)dr` <br/> `U=mg [r]_(0)^(h)=mgh`.</body></html> | |
| 36277. |
A body moving ina curved path possessesa velicty 3m//stowards north at any instant of its motion . A fter 10s, the velocity of the body was found to be 4 m//s towards west. Calculate the average acceleration during this interval . |
| Answer» <html><body><p></p>Solution :To <a href="https://interviewquestions.tuteehub.com/tag/solve-647535" style="font-weight:bold;" target="_blank" title="Click to know more about SOLVE">SOLVE</a> this problem the vector nature of velocity must be <a href="https://interviewquestions.tuteehub.com/tag/taken-659096" style="font-weight:bold;" target="_blank" title="Click to know more about TAKEN">TAKEN</a> into account. <br/> In the figure, the initial velocity`v_(<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>)`and the final velocity v arfe <a href="https://interviewquestions.tuteehub.com/tag/drawn-2062388" style="font-weight:bold;" target="_blank" title="Click to know more about DRAWN">DRAWN</a> from a common origin . The vecto difference of them is found by the parallelogram method. <br/> The magnitude of difference is <br/> `|v-v_(0)|=OC=sqrt(OA^(2)+AC^(2))`<br/> `=sqrt(4^(2)+3^(2))=5m//s`<br/> The direction is given by <br/> `tan theta =3/4=0.75, theta =37^(0)`<br/> Average acceleration `=|vecv-vecv_(0)|/t=5/(10)=0.5m//s^(2)`at`37^(@)`south of <a href="https://interviewquestions.tuteehub.com/tag/west-1451899" style="font-weight:bold;" target="_blank" title="Click to know more about WEST">WEST</a><br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_ELT_AI_PHY_XI_V01_A_C04_SLV_010_S01.png" width="80%"/></body></html> | |
| 36278. |
Same heat is given to Helium and Oxygen gasof equal masses under constant volume. Then the temperature of the gas which increases more is |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/oxygen-1144542" style="font-weight:bold;" target="_blank" title="Click to know more about OXYGEN">OXYGEN</a> <br/>Helium<br/>Rise in <a href="https://interviewquestions.tuteehub.com/tag/temperatures-1241229" style="font-weight:bold;" target="_blank" title="Click to know more about TEMPERATURES">TEMPERATURES</a> are <a href="https://interviewquestions.tuteehub.com/tag/equal-446400" style="font-weight:bold;" target="_blank" title="Click to know more about EQUAL">EQUAL</a> <br/><a href="https://interviewquestions.tuteehub.com/tag/none-580659" style="font-weight:bold;" target="_blank" title="Click to know more about NONE">NONE</a> </p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :A</body></html> | |
| 36279. |
In the case of a body freely falling from small height |
| Answer» <html><body><p>the <a href="https://interviewquestions.tuteehub.com/tag/changes-913881" style="font-weight:bold;" target="_blank" title="Click to know more about CHANGES">CHANGES</a> of position are equal in equal intervals of time <br/>the changes of velocity are equal in <a href="https://interviewquestions.tuteehub.com/tag/unequal-1437090" style="font-weight:bold;" target="_blank" title="Click to know more about UNEQUAL">UNEQUAL</a> intervals of time<br/>the changes of <a href="https://interviewquestions.tuteehub.com/tag/acceleration-13745" style="font-weight:bold;" target="_blank" title="Click to know more about ACCELERATION">ACCELERATION</a> is <a href="https://interviewquestions.tuteehub.com/tag/zero-751093" style="font-weight:bold;" target="_blank" title="Click to know more about ZERO">ZERO</a> in equal or unequal intervals of time<br/>none</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html> | |
| 36280. |
If a room has dimensions 3m x 4m x 5m, what is the mass of air in the room if density of air at NTP is 1.3kg/m^(3) |
| Answer» <html><body><p></p>Solution :`m=rhoxxv=(1.3kg//m^(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>))(3xx4xx5m^(3))=78 kg`<br/>i.e. ,mass of <a href="https://interviewquestions.tuteehub.com/tag/air-852233" style="font-weight:bold;" target="_blank" title="Click to know more about AIR">AIR</a> in the <a href="https://interviewquestions.tuteehub.com/tag/room-25779" style="font-weight:bold;" target="_blank" title="Click to know more about ROOM">ROOM</a> is 78 kg which is unbelievable (as air is assumed to be <a href="https://interviewquestions.tuteehub.com/tag/weightless-3272644" style="font-weight:bold;" target="_blank" title="Click to know more about WEIGHTLESS">WEIGHTLESS</a>) but is <a href="https://interviewquestions.tuteehub.com/tag/true-713260" style="font-weight:bold;" target="_blank" title="Click to know more about TRUE">TRUE</a>.</body></html> | |
| 36281. |
A small hole in the wall of an enclosure behaves as |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/good-1009017" style="font-weight:bold;" target="_blank" title="Click to know more about GOOD">GOOD</a> absorber & <a href="https://interviewquestions.tuteehub.com/tag/poor-602678" style="font-weight:bold;" target="_blank" title="Click to know more about POOR">POOR</a> <a href="https://interviewquestions.tuteehub.com/tag/emitter-970223" style="font-weight:bold;" target="_blank" title="Click to know more about EMITTER">EMITTER</a> <br/>Good absorber & good emitter <br/>Poor absorber & poor emitter <br/>Poor absorber & good emitter </p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a></body></html> | |
| 36282. |
The frequency of the whlstle of a train Is 512 Hz . Thetrain crosses a station at a speed of 72 km . h^(-1) Calculate the frequency of the sound heard by a Iistener, standing on the platform , before and after the train crosses the station. Neglect the effect of wind . Velocity of sound is 336 , m s^(-1) . |
| Answer» <html><body><p></p>Solution :Velocity of the <a href="https://interviewquestions.tuteehub.com/tag/train-1425202" style="font-weight:bold;" target="_blank" title="Click to know more about TRAIN">TRAIN</a> ` = 72 km . `h^(-1)`<br/> ` = (72 xx1000)/(60xx60) m. s^(-1)` <br/> `= 20 m. s^(-1)`<br/> When the train approaches the station , the <a href="https://interviewquestions.tuteehub.com/tag/distance-116" style="font-weight:bold;" target="_blank" title="Click to know more about DISTANCE">DISTANCE</a> covered hy <a href="https://interviewquestions.tuteehub.com/tag/512-1895092" style="font-weight:bold;" target="_blank" title="Click to know more about 512">512</a> sound waves = 336 - 20 = <a href="https://interviewquestions.tuteehub.com/tag/316-307105" style="font-weight:bold;" target="_blank" title="Click to know more about 316">316</a>` m<br/> ` therefore ` Apparent wavelength , ` lambda . = (316)/(512) m ` <br/> ` therefore ` Apparent frequencydue to Doppler effect<br/> ` = (" velocity of sound (V))/(lambda .)`<br/> `(336)/((316)/(512)) = (336xx512)/(316) = 544.4 Hz ` <br/> ` Again , when the train recedes from the station , the distance covered y 512 sound waves= 366 + 20 = <a href="https://interviewquestions.tuteehub.com/tag/356-1858306" style="font-weight:bold;" target="_blank" title="Click to know more about 356">356</a> m <br/>` therefore ` Apparent wavelength , ` lambda . = (356)/(512) m`<br/> ` therefore ` Apparent freaquency= `(V) /(lambda .) = (336)/((356)/(512)) = (336xx 512)/(356) = 483.2 Hz .`</body></html> | |
| 36283. |
A body constrained to move along the z-axis of a coordinate system is subject to a constant force F give by F = hat(i)+2hat(j)+3hat(k). What is the work done by this force in moving the body a distance of 4 m along the z- axis ? |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/12-269062" style="font-weight:bold;" target="_blank" title="Click to know more about 12">12</a> <a href="https://interviewquestions.tuteehub.com/tag/j-520843" style="font-weight:bold;" target="_blank" title="Click to know more about J">J</a> <br/><a href="https://interviewquestions.tuteehub.com/tag/24-295400" style="font-weight:bold;" target="_blank" title="Click to know more about 24">24</a> J <br/><a href="https://interviewquestions.tuteehub.com/tag/48-318016" style="font-weight:bold;" target="_blank" title="Click to know more about 48">48</a> J <br/><a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a> J</p>Answer :A</body></html> | |
| 36284. |
Water is taken in one beaker and glycerine in another. Both are stirred well and kept on the table. Which will come to rest first? |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :The <a href="https://interviewquestions.tuteehub.com/tag/liquid-1075124" style="font-weight:bold;" target="_blank" title="Click to know more about LIQUID">LIQUID</a> with <a href="https://interviewquestions.tuteehub.com/tag/larger-1067345" style="font-weight:bold;" target="_blank" title="Click to know more about LARGER">LARGER</a> <a href="https://interviewquestions.tuteehub.com/tag/viscosity-13767" style="font-weight:bold;" target="_blank" title="Click to know more about VISCOSITY">VISCOSITY</a> i.e. <a href="https://interviewquestions.tuteehub.com/tag/glycerine-474196" style="font-weight:bold;" target="_blank" title="Click to know more about GLYCERINE">GLYCERINE</a></body></html> | |
| 36285. |
Define angle of friction |
| Answer» <html><body><p></p>Solution :The anglebetweenthe normalforce (N) <a href="https://interviewquestions.tuteehub.com/tag/andthe-875214" style="font-weight:bold;" target="_blank" title="Click to know more about ANDTHE">ANDTHE</a> resultantforce (<a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a> )of normalforce andmaximum <a href="https://interviewquestions.tuteehub.com/tag/frictionforce-2660117" style="font-weight:bold;" target="_blank" title="Click to know more about FRICTIONFORCE">FRICTIONFORCE</a> `(f_(<a href="https://interviewquestions.tuteehub.com/tag/x-746616" style="font-weight:bold;" target="_blank" title="Click to know more about X">X</a>)^(max))`</body></html> | |
| 36286. |
State and prove theorem of parallel axis. |
| Answer» <html><body><p></p>Solution :The moment of <a href="https://interviewquestions.tuteehub.com/tag/inertia-1043176" style="font-weight:bold;" target="_blank" title="Click to know more about INERTIA">INERTIA</a> of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its centre of <a href="https://interviewquestions.tuteehub.com/tag/mass-1088425" style="font-weight:bold;" target="_blank" title="Click to know more about MASS">MASS</a> and the product of its mass and the square of the distance between the two parallel axes. <br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/KPK_AIO_PHY_XI_P1_C07_E01_051_S01.png" width="80%"/> <br/> As shown in figure <a href="https://interviewquestions.tuteehub.com/tag/z-750254" style="font-weight:bold;" target="_blank" title="Click to know more about Z">Z</a> and Z. are two parallel axes separated by a distance a. `therefore OO.=a` <br/> The Z-axis passes through the centre of mass O of the rigid body. <br/> From theorem of parallel axes <br/> `I_(Z)=I_(Z)+Ma^(2)orI_(Z)=I_(C)+M(OO.)^(2)` <br/> where `I_(Z)andI_(Z)` are the moments of inertia of the body about the Z and Z. axes respectively. <br/> M = total mass of the body <br/> a = <a href="https://interviewquestions.tuteehub.com/tag/perpendicular-598789" style="font-weight:bold;" target="_blank" title="Click to know more about PERPENDICULAR">PERPENDICULAR</a> distance between the two parallel axes. <br/> If moment of inertia of object of any <a href="https://interviewquestions.tuteehub.com/tag/shape-1204673" style="font-weight:bold;" target="_blank" title="Click to know more about SHAPE">SHAPE</a> about the axis passing through its centre is given, then moment of inertia about the axis parallel to the axis passing through its centre can be determined.</body></html> | |
| 36287. |
A particle is executing SIIM with an amplitude of 0.2 m. At what distance from the mean position the potential energy of the particle will be equal to its kinetic energy ? |
| Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :P.E. at the displacement x <br/> `P.E.= (1)/(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)m^(2)omega^(2)x^(2)`…….(1) <br/> K.E. at the displacement x is given by <br/> `K.E.= (1)/(2)m^(2)omega^(2)(A^(2)-x^(2))` …………(2) <br/> But PE=KE i.e., `(1)/(2)m^(2)omega^(2)x^(2)= (1)/(2)momega^(2)(A^(2)-x^(2))` <br/> `<a href="https://interviewquestions.tuteehub.com/tag/implies-1037962" style="font-weight:bold;" target="_blank" title="Click to know more about IMPLIES">IMPLIES</a> x^(2)= A^(2)-x^(2)` or `implies <a href="https://interviewquestions.tuteehub.com/tag/2x-301182" style="font-weight:bold;" target="_blank" title="Click to know more about 2X">2X</a>^(2)=A^(2) x^(2)= (A^(2))/(2)` <br/> `:. x= (A)/(sqrt(2)).` But A=0.2m <br/> `:. x= (0.2)/(sqrt(2))= 0.1 xx sqrt(2)= 0.1414m`</body></html> | |
| 36288. |
(A) : The velocity of a body at the bottom of an inclined plane of given height , is more when it slides down the plane, compared to when it rolling down the same plane. (R) : In rolling down, a body acquires both, kinetic energy of translation and rotation |
| Answer» <html><body><p>Both 'A' and 'R' are <a href="https://interviewquestions.tuteehub.com/tag/true-713260" style="font-weight:bold;" target="_blank" title="Click to know more about TRUE">TRUE</a> 'R' is the correct explanation of 'A' <br/>Both 'A' and 'R' are true 'R' is not the correct explanation of 'A' <br/>A' is true and 'R' is <a href="https://interviewquestions.tuteehub.com/tag/false-459184" style="font-weight:bold;" target="_blank" title="Click to know more about FALSE">FALSE</a> <br/>A' is false and 'R' are true </p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :A</body></html> | |
| 36289. |
Describe the construction and working of venturimeter and obtain an equation for the volume of liquid flowing per second through a wider entry of the tube. |
| Answer» <html><body><p></p>Solution : <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/FM_PHY_XI_SP_01_E01_042_S01.png" width="80%"/> <br/>Venturimeter : This device is used to measure the rate of flow ( or say flow speed) of the incompressible fluid flowing through a pipe. It works on the principle of Bernoulli.s theorem. It consists of two wider tubes A and A. (with cross sectional area A) connected by a narrow tube B ( with cross sectional area a). A <a href="https://interviewquestions.tuteehub.com/tag/manometer-15863" style="font-weight:bold;" target="_blank" title="Click to know more about MANOMETER">MANOMETER</a> in the fonn of U-tube is <a href="https://interviewquestions.tuteehub.com/tag/also-373387" style="font-weight:bold;" target="_blank" title="Click to know more about ALSO">ALSO</a> attached between the wide and narrow tubes. The manometer contains a liquid of <a href="https://interviewquestions.tuteehub.com/tag/density-17451" style="font-weight:bold;" target="_blank" title="Click to know more about DENSITY">DENSITY</a> `. rho_m .`<br/>Let `P_1`be the pressure of the fluid at the wider region of the tube A. Let us assume that the fluid of density .`rho` . flows .from the pipe with speed `v_1`and into the narrow region, its .speed increases to `v_2` . <a href="https://interviewquestions.tuteehub.com/tag/according-366619" style="font-weight:bold;" target="_blank" title="Click to know more about ACCORDING">ACCORDING</a> to the Bernoulli.s equation, this increase in speed is accompanied by a decrease in the fluid pressure `P_2`at the narrow region of the tube B. Therefore, the pressure <a href="https://interviewquestions.tuteehub.com/tag/difference-951394" style="font-weight:bold;" target="_blank" title="Click to know more about DIFFERENCE">DIFFERENCE</a> between the tubes A and B is noted by measuring the height difference`(Delta P = P_1 + P_2)`between the surfaces of the manometer liquid. From the equation of continuity, we can say that `Av_1 = av_2`which means that <br/> ` v_2= A/a v_1` <br/> Using bernoulli.s equation, <br/> ` P_1 + rho (v_1^2)/(2) = P_2 + rho (v_2^2)/(2) = P_2 + rho 1/2 (A/a v_1)^&2` <br/> from the above equation , the pressure difference<br/> ` Delta P = P_1 - P_2= rho (v_1^2)/(2) (A^2 - a^2)/(a^2)` <br/> Thus , the speed of flow of liquid at the wide end of the tube A <br/> ` v_1^^2 = (2 (Delta P)a^2)/(rho(A^2 - a^2) ) rArr v_1= sqrt((2(Delta P)a^2)/(rho (A^2 - a^2) )` <br/>The volume of the liquid flowing out per second is `V = A v_1 = A sqrt( (2 (Delta P )a^2)/(p (A^2 -a^2) ) )= a A sqrt( (2(Delta P))/(p(A^2 - a^2) ))`</body></html> | |
| 36290. |
A block of 0.5kg is placed on a horizontal platform. The system is making vertical oscillations about a fixed point with a frequency of 0.5Hz. Find the maximum amplitude of oscillation if the block does not lose contact with the horizontal platform? |
| Answer» <html><body><p><br/></p>Solution :At the highest point i.e., upper extreme <a href="https://interviewquestions.tuteehub.com/tag/position-1159826" style="font-weight:bold;" target="_blank" title="Click to know more about POSITION">POSITION</a>, <a href="https://interviewquestions.tuteehub.com/tag/apparent-363672" style="font-weight:bold;" target="_blank" title="Click to know more about APPARENT">APPARENT</a> weight of the block `= <a href="https://interviewquestions.tuteehub.com/tag/mg-1095425" style="font-weight:bold;" target="_blank" title="Click to know more about MG">MG</a> -m omega^2 A `. When it just does contact, mg`-m omega^2 A = 0, A= g//omega^2`</body></html> | |
| 36291. |
The volume of a sphere is 1.76 cm^3 . The total volume of 24 such spheres with due regard to the significant places is |
| Answer» <html><body><p>`0.42 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 10^2 cm^3` <br/>`<a href="https://interviewquestions.tuteehub.com/tag/43-316498" style="font-weight:bold;" target="_blank" title="Click to know more about 43">43</a> cm^3` <br/>`42.42 cm^3`<br/>`42.2 cm^3`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :D</body></html> | |
| 36292. |
There are three sinusoidal waves A, B and C represented by equations- A rarr Y=A sin kx, B rarr y= A/2 sin 2 kx, C rarr y=A/2 sin 3 kx (a) To get a waveform of nearly the shape given in fig (a) which of the two waves B or C shall be superimposed with wave A? (b) To get a waveform close to that in fig (b) which of the two waves B or C shall be superimposed with A? |
| Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :(a) <a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a> <br/> (b) <a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a> <br/> `(##IJA_PHY_V01_C13_E01_084_A01##)`<br/> `(##IJA_PHY_V01_C13_E01_084_A02##)`</body></html> | |
| 36293. |
A source S and a detector D are placed at a distance d apart. A big cardboard is placed at a distance sqrt2d from the source and the detector as shown in figure. The source emits a wave of wavelength = d/2 which is received by the detector after reflection from the cardboard. It is found to be in phase with the direct wave received from the source. By what minimum distance should the cardboard be shifted away so that the reflected wave becomes out of phase with the direct wave ? |
| Answer» <html><body><p><br/></p>Solution :Here given `lamda=d/<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>` <br/> Initial path <a href="https://interviewquestions.tuteehub.com/tag/difference-951394" style="font-weight:bold;" target="_blank" title="Click to know more about DIFFERENCE">DIFFERENCE</a> is given by <br/> `2sqrt((d/2)^2+2d^2)=(2xx(3d)/2)-d` <br/> If it is now <a href="https://interviewquestions.tuteehub.com/tag/shifted-7301841" style="font-weight:bold;" target="_blank" title="Click to know more about SHIFTED">SHIFTED</a> a distance x then <a href="https://interviewquestions.tuteehub.com/tag/pathdifference-2917693" style="font-weight:bold;" target="_blank" title="Click to know more about PATHDIFFERENCE">PATHDIFFERENCE</a> will be <br/> `=2sqrt((d/2)^2)+(sqrt2d+x)^2-d` <br/> `=2d+d/4` <br/> `rarr (d/2)^2+(sqrt2dx)^2=169/64d^2` <br/> `rarr (sqrt2d+x)^2=d^2(9169-16))/64` <br/> `rarr sqrt2d+x=1.54d` <br/> `rarr x=1.54d-1.414d` <br/> `=0.3d`</body></html> | |
| 36294. |
When a man is standing, rain drops appears to himfalling at 60^(@) from the horizontal from his front side. When he is travelling at 5 kmph on a horizontal road they appear to him falling at 30^(@) from the horizontal from his front side. The actual speed of the rain is (in kmph) |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a><br/>4<br/>5<br/>`5sqrt(3)`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html> | |
| 36295. |
Match the physical quantities given in column I with dimensions given in column II. (Dimension of charge is Q) |
| Answer» <html><body><p>1 - (<a href="https://interviewquestions.tuteehub.com/tag/iv-501699" style="font-weight:bold;" target="_blank" title="Click to know more about IV">IV</a>), 2 - (iv), <a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a> - (v), 4 - (ii)<br/>1 - (v), 2 - (iv), 3 - (ii), 4 - (i)<br/>1 - (vi), 2 - (v), 3 - (i), 4 - (ii)<br/>1 - (v), 2 - (i), 3 - (ii), 4 - (<a href="https://interviewquestions.tuteehub.com/tag/iii-497983" style="font-weight:bold;" target="_blank" title="Click to know more about III">III</a>)</p>Answer :C</body></html> | |
| 36296. |
A wave travels in a medium according to the equation of displacement given by y (x , t) = 0.03 sin {pi (2t - 0.01 x)} where y and x are in metres and t in seconds . The wavelength of the wave is ………. . |
| Answer» <html><body><p>200 m <br/><a href="https://interviewquestions.tuteehub.com/tag/100-263808" style="font-weight:bold;" target="_blank" title="Click to know more about 100">100</a> m <br/>20 m <br/>10 m </p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`<a href="https://interviewquestions.tuteehub.com/tag/lambda-539003" style="font-weight:bold;" target="_blank" title="Click to know more about LAMBDA">LAMBDA</a> = (2pi)/(k) = (2pi)/(0.01 <a href="https://interviewquestions.tuteehub.com/tag/pi-600185" style="font-weight:bold;" target="_blank" title="Click to know more about PI">PI</a>) = 200 m`</body></html> | |
| 36297. |
If momentum (p), area (A) and time (T) are taken to be fundamental quantities, then energy has the dimensional formula. |
| Answer» <html><body><p>`(pA^(-1)T^(1)]`<br/>`[p^(2)AT]`<br/>`[pA^(-(1)/(2))T]`<br/>`[pA^((1)/(2)T)]`<br/></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Given, fundamental quantities are momentum (p), <a href="https://interviewquestions.tuteehub.com/tag/area-13372" style="font-weight:bold;" target="_blank" title="Click to know more about AREA">AREA</a> (A) and time (T). <br/>We can write energy E as <br/> `E <a href="https://interviewquestions.tuteehub.com/tag/prop-607409" style="font-weight:bold;" target="_blank" title="Click to know more about PROP">PROP</a> p^(A)A^(b)T^(c)` <br/> `E= kp^(a)A^(A)T^(c)` <br/> `:. [E]=[p]^(a)[A]^(A)[T]^(c)` <br/> where <a href="https://interviewquestions.tuteehub.com/tag/k-527196" style="font-weight:bold;" target="_blank" title="Click to know more about K">K</a> is <a href="https://interviewquestions.tuteehub.com/tag/dimensionless-954191" style="font-weight:bold;" target="_blank" title="Click to know more about DIMENSIONLESS">DIMENSIONLESS</a> constant of proportionality <br/> Dimensions of `E=[E]=[M^(1)L^(2)T^(-2)] and p]=[M^(1)L^(1)T^(-1)]` <br/> `[A]=[L^(2)]` <br/> `[T]=[T^(1)]` <br/> Putting all the dimensions, we get <br/> `M^(1)L^(2)T^(-2)=[M^(1)L^(1)T^(-1)]^(a)[L^(2)][T]^(c)` <br/> `:. M^(1)L^(2)T^(-2)=M^(a)L^(2b+b)T^(-a+c)` <br/> By comparison of power <br/> `a=1, 2b+a=2` <br/> `:.2b+1=2` <br/> `b=(1)/(2)` <br/> `-a+c=-2` <br/> `c=-2+a=-2+1=-1` <br/> Thus`|E|=[pA^((1)/(2))T^(-1)]`</body></html> | |
| 36298. |
A cord of negligible mass is wound round the rim of a fly wheel of mass 20 kg and radius 20 cm. A steady pull of 25 N is applied on the cord as shown in Fig. 7.35. The flywheel is mounted on a horizontal axle with frictionless bearings. (a) Compute the angular acceleration of the wheel. (b) Find the work done by the pull, when 2m of the cord is unwound. (c) Find also the kinetic energy of the wheel at this point. Assume that the wheel starts from rest. (d) Compare answers to parts (b) and (c). |
| Answer» <html><body><p></p>Solution :<img src="https://doubtnut-static.s.llnwi.net/static/physics_images/KPK_AIO_PHY_XI_P1_C07_E02_015_S01.png" width="80%"/> <br/> (a) Torque `<a href="https://interviewquestions.tuteehub.com/tag/tau-663951" style="font-weight:bold;" target="_blank" title="Click to know more about TAU">TAU</a>=RFsintheta` <br/> `=0.2xx25xxsin90^(@)` <br/> `tau=5.0Nm[because sin90^(@)=<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>]` <br/> Moment of inertia `I=(MR^(2))/(2)` <br/> `=(<a href="https://interviewquestions.tuteehub.com/tag/20xx-292987" style="font-weight:bold;" target="_blank" title="Click to know more about 20XX">20XX</a>(0.2)^(2))/(2)` <br/> `therefore I=0.4kgm^(2)` <br/> Now `Ialpha=tau` <br/> `therefore alpha=(tau)/(I)=(5)/(0.4)` <br/> `therefore alpha=12.5"rad/s"^(2)""....(1)` <br/> (b) Work <a href="https://interviewquestions.tuteehub.com/tag/done-2591742" style="font-weight:bold;" target="_blank" title="Click to know more about DONE">DONE</a> by the pull unwinding 2m of the cord <br/> `W=Fl` <br/> `=25xx2` <br/> `therefore W=50J` <br/> (c ) Let `omega` be the final <a href="https://interviewquestions.tuteehub.com/tag/angular-11524" style="font-weight:bold;" target="_blank" title="Click to know more about ANGULAR">ANGULAR</a> velocity <br/> The kinetic energy gained `=(1)/(2)Iomega^(2)...(2)` <br/> If wheel gains angular velocity `omega` after starting from rest, then <br/> Now `omega^(2)=omega_(0)^(2)+2alphatheta` <br/> `therefore omega^(2)=2alphatheta""[because omega_(0)=0]...(3)` <br/> Angular displacement <br/> `theta=("length of unwound string")/("radius of wheel")` <br/> `=(2)/(0.2)` <br/> `therefore theta=10` rad `""....(4)` <br/> From eqn. (1), (3) and (4) <br/> `omega^(2)=2alphtheta` <br/> `=2xx12.5xx10` <br/> `therefore omega^(2)=250 ("rad s"^(-1))^(2)` <br/> `therefore` kinetic energy gained <br/> `K.=(1)/(2)Iomega^(2)` <br/> `=(1)/(2)xx0.4xx250` <br/> `therefore K=50J""...(5)` <br/> (d) Equation (2) and (5) are equal, means kinetic energy gained by the wheel = work done by the force. <br/> Here, there is no loss of energy due to friction.</body></html> | |
| 36299. |
If R is the radius of the Earth and g is the acceleration due to gravity on the Earth's surface. Find the mean density of the Earth. |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a><br/>`(R)/(<a href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a>)`<br/><a href="https://interviewquestions.tuteehub.com/tag/2r-300472" style="font-weight:bold;" target="_blank" title="Click to know more about 2R">2R</a><br/>`(R)/(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)`</p><a href="https://interviewquestions.tuteehub.com/tag/answer-15557" style="font-weight:bold;" target="_blank" title="Click to know more about ANSWER">ANSWER</a> :D</body></html> | |
| 36300. |
A body is uniformly rotating about an axis fixed in an interial frame of reference. Let barA be a unit vector along the axis of rotation and barB be the unit vector along the resultant force on a particle P of the body away from the axis. The value of barA.barB is |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a><br/>`-1`<br/>0<br/>none of these</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a></body></html> | |