InterviewSolution
Saved Bookmarks
InterviewSolution
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.
| 38001. |
(A) : The value of dimensionless constants or proportionality constants cannot be found by dimensional methods. (B) : The equations containing trigonometrical, exponential and logarithmic functions cannot be analysed by dimensional methods. |
| Answer» <html><body><p>Both 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 true<br/>Both A & B are false<br/>Only A is true<br/>Only B is true</p>Answer :A</body></html> | |
| 38002. |
The potential energy of a particle varies with distance x from a fixed origin as V = (Asqrt(x))/(x+B) where A and B are constants. The dinmensions of ABare |
| Answer» <html><body><p>`M^1L^(5//<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)T^(-2)` <br/>`M^1L^2T^(-2)` <br/>`M^(3//2)<a href="https://interviewquestions.tuteehub.com/tag/l-535906" style="font-weight:bold;" target="_blank" title="Click to know more about L">L</a>^(5//2) T^(-2)` <br/>`M^1L^(7//2)T^(-2)` </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> | |
| 38003. |
A drunkard is walking along a straight road. He takes 5 steps forward and 3 steps backward and so on. Each step is 1 m long and takes 1 s. There is a pit 11m away. The drunkard will fall into the pit after |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/21-293276" style="font-weight:bold;" target="_blank" title="Click to know more about 21">21</a> s <br/><a href="https://interviewquestions.tuteehub.com/tag/29-299821" style="font-weight:bold;" target="_blank" title="Click to know more about 29">29</a> s<br/>31 s<br/>37 s</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> | |
| 38004. |
What is perfectly black body? What is its absorbing power and emissivity ? |
| Answer» <html><body><p></p>Solution :A <a href="https://interviewquestions.tuteehub.com/tag/body-900196" style="font-weight:bold;" target="_blank" title="Click to know more about BODY">BODY</a> which absorbs all the radiation <a href="https://interviewquestions.tuteehub.com/tag/incident-1039786" style="font-weight:bold;" target="_blank" title="Click to know more about INCIDENT">INCIDENT</a> on it is <a href="https://interviewquestions.tuteehub.com/tag/called-907796" style="font-weight:bold;" target="_blank" title="Click to know more about CALLED">CALLED</a> a block boby. The absorbing power and also emissivity of it is unity.</body></html> | |
| 38005. |
An equiconvex lens with radii of curvature of magnitude r each, is put over a liquid layer poured on top of a plane mirror. A small needle with its tip on the principle axis of the lens if moved along the axis intil its inverted real image coincides with the needle itself, Fig. The distance of needle from lens is measured to be a. On removing the liquid layer and repeating the experiment, the distance isn found to be b. Given that two values of distances measured represent that real length values in the two cases, obtain a formula for refractive index of the liquid. . |
| Answer» <html><body><p></p>Solution :Here, combined focal length of <a href="https://interviewquestions.tuteehub.com/tag/glass-14164" style="font-weight:bold;" target="_blank" title="Click to know more about GLASS">GLASS</a> lens and liquid lens, `F = a`, and Focal length of <a href="https://interviewquestions.tuteehub.com/tag/convex-933348" style="font-weight:bold;" target="_blank" title="Click to know more about CONVEX">CONVEX</a> lens, `f_1 = b` If `f_2` is focal length of liquid lens, then <br/> `(1)/(f_1)+(1)/(f_2)=(1)/(F)` <br/> `(1)/(f_2)=(1)/(F)-(1)/(f_1) = ((1)/(a) - (1)/(b))` <br/> The liquid lens is plano <a href="https://interviewquestions.tuteehub.com/tag/concave-927978" style="font-weight:bold;" target="_blank" title="Click to know more about CONCAVE">CONCAVE</a> lens for which `R_1 = -r, R_2 = oo` <br/> From `(1)/(f_2)=(<a href="https://interviewquestions.tuteehub.com/tag/mu-566056" style="font-weight:bold;" target="_blank" title="Click to know more about MU">MU</a> -1) ((1)/(R_1) - (1)/(R_2))` <br/> `((1)/(a) - (1)/(b)) = (mu - 1) ((1)/(-r) - (1)/(- oo))` <br/> `:. (mu - 1) = (r)/(b) - (r)/(a)` <br/> `mu = 1 + (r)/(b) - (r)/(a)`.</body></html> | |
| 38006. |
An object of mass m is raised from the surface of the earth to a height equal to the radius of the earth, that is, taken from a distance R to 2R from the centre of the earth. What is the gain in its potential energy ? |
| Answer» <html><body><p></p>Solution :<img src="https://doubtnut-static.s.llnwi.net/static/physics_images/KPK_AIO_PHY_XI_P1_C08_E04_032_S01.png" width="80%"/><br/>Potential energy of the <a href="https://interviewquestions.tuteehub.com/tag/object-11416" style="font-weight:bold;" target="_blank" title="Click to know more about OBJECT">OBJECT</a> at the surface of the earth = `-(<a href="https://interviewquestions.tuteehub.com/tag/gmm-2090431" style="font-weight:bold;" target="_blank" title="Click to know more about GMM">GMM</a>)/R` <br/> PE of the object at a height <a href="https://interviewquestions.tuteehub.com/tag/equal-446400" style="font-weight:bold;" target="_blank" title="Click to know more about EQUAL">EQUAL</a> to the radiusof the earth `=-(GMm)/(<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/> `:.` Gain in PE of the object <br/> `=(-GMm)/(2R) -(-(GMm)/R)` <br/> `=(-GMm+2GMm)/(2R) = +(GMm)/(2R)`<br/> `= (gR^2xxm)/(2R) = 1/2 mgR "" ( :. g = (GM)/(R^2))`</body></html> | |
| 38007. |
The spring balance A read 2 kg. with ab block m suspended from it. A balance B reads 5 kg. When a beaker with liquid is put on the pan of the balance. The two balances are now so arranged that the hanging mass is inside the liquid in the beaker as shown in fig. In tis situation. |
| Answer» <html><body><p>The balance A will read read more than <a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a> <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/>The balance <a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a> will read more than5 kg.<br/>The balance A will read less than 2 kg . And B will read more than <a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a> kg.<br/>The balance A and B will reak 2 kg and 5 kg respectively.</p>Answer :B::C</body></html> | |
| 38008. |
Consider the followig statements: Statement-1: Work done by friction is always negative. Statement-2 : If frictional force acts on a body its K.E. may decrease Statement 3: A rigid disc rolls without slipping on a fixed rough horizontal surface with uniform angular velocity.Then the acceleration of lowest point on the disc is zero. Statements-4: Consider an isolated sun planet system in which the planet moves in elliptical path around the fixed sun. Kinetic energy of the planet changes with time but its angular momentum with respect to centre of the sun remains constants. STatement-5: Net electric field inside the material of a conductor must always be zero Then, the number of incorrect statements is |
| Answer» <html><body><p>2<br/>2<br/>3<br/>4</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> | |
| 38009. |
A sky lab of mass 2xx10^3 kg is first launched from the surface of earth in a circular orbit of radius 2R (from the centre of earth ) and then it is shifted from this circular orbit to another circular orbit of radius 3R . Calculate the minimum energy required (a) to place the lab in the first orbit (b) to shift the lab form first orbit to the second orbit . Given , R = 6400 Km and g = 10m//s^2 |
| Answer» <html><body><p></p>Solution :The energy of the sky lab on the <a href="https://interviewquestions.tuteehub.com/tag/surface-1235573" style="font-weight:bold;" target="_blank" title="Click to know more about SURFACE">SURFACE</a> of <a href="https://interviewquestions.tuteehub.com/tag/earth-13129" style="font-weight:bold;" target="_blank" title="Click to know more about EARTH">EARTH</a> <br/> `E_(s)=KE +PE=<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>+(-(GMm)/R)=-(GMm)/R`<br/> And the energy of the sky lab in an orbit of radius r <br/> `E=1/2mv_(0)^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)+[-(GMm)/r]=(-GMm)/(2r)[ " as" v_(0) = sqrt((GM)/r)]`<br/> (a) So the energy required to place the lab from the surface of earth to the orbit of radius 2R, <br/> `E_(1)-E_(s)=-(GMm)/(2(2R))-[-(GMm)/(R)]=3/4(GMm)/R` <br/> i.e `DeltaE=3/4m/RxxgR^(2)=3/4mgR"" ["asg" =(GM)/R^2]`<br/> i.e. `DeltaE=3/4(2xx10^(3)xx10xx6.4xx10^6)=` <br/> `3/4(12.8xx10^(10))=9.6xx10^(10)J` <br/> (b) As for <a href="https://interviewquestions.tuteehub.com/tag/ii-1036832" style="font-weight:bold;" target="_blank" title="Click to know more about II">II</a> orbit `r = 3R , E_(11)=-(GMm)/(2(3R))=-(GMm)/(6R)` <br/> `:. E_(11)-E_(1)=-(GMm)/(6R)-(-(GMm)/(4R))=1/12 (GMm)/R` <br/> But as `g=(Gm//R^2), ` i.e. `GM=gR^2` or <br/> `DeltaE=1/2mgR=1/12(12.8xx10^(10))=1.1xx10^(10)J`</body></html> | |
| 38010. |
The thrust developed by a rocket-motor is given by F= mv + A(P_1 -P_2)where m is the mass of the gas ejected per unit time, v is velocity of the gas, A is area of cross-section of the nozzle, P_1 and P_2are the pressures of the exhaust gas and surrounding atmosphere. The formula is dimensionally |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/correct-409949" style="font-weight:bold;" target="_blank" title="Click to know more about CORRECT">CORRECT</a> <br/><a href="https://interviewquestions.tuteehub.com/tag/wrong-1462035" style="font-weight:bold;" target="_blank" title="Click to know more about WRONG">WRONG</a> <br/>Some <a href="https://interviewquestions.tuteehub.com/tag/time-19467" style="font-weight:bold;" target="_blank" title="Click to know more about TIME">TIME</a> wrong, <a href="https://interviewquestions.tuteehub.com/tag/sometimes-648384" style="font-weight:bold;" target="_blank" title="Click to know more about SOMETIMES">SOMETIMES</a> correct<br/>Data is not adequate</p>Answer :A</body></html> | |
| 38011. |
A horizontal force of 150N produces an acceleration of 2m//s^(2) on a body placed on a horizontal surface. A horizontal force of 200N produces an acceleration of 3m//s^(2). Find the mass of the body and the coefficient of kinetic friction (g=10ms^(-2)) |
| 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/50kg-324195" style="font-weight:bold;" target="_blank" title="Click to know more about 50KG">50KG</a>, 0.1</body></html> | |
| 38012. |
Pressure depends on distance as P =(alpha)/(beta) "exp" (-(alphaz)/(ktheta)) where alpha betaare constants z is distance k is Boltzmann is constant and theta is temperature. The dimension of beta are |
| Answer» <html><body><p>`M^(0)<a href="https://interviewquestions.tuteehub.com/tag/l-535906" style="font-weight:bold;" target="_blank" title="Click to know more about L">L</a>^(0)T^(0)`<br/>`M^(-<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)L^(-1)T^(-1)`<br/>`M^(0)L^(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)T^(0)`<br/>`M^(-1)L^(1)T^(2)`</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> | |
| 38013. |
Give some practical applications of Stoke's law. |
| Answer» <html><body><p></p>Solution :The viscous force F acting on a spherical body of <a href="https://interviewquestions.tuteehub.com/tag/radius-1176229" style="font-weight:bold;" target="_blank" title="Click to know more about RADIUS">RADIUS</a> r depends directly on: (i) radius (r) of the sphere <br/> (ii) velocity (v) of the sphere and <br/>(iii) coefficient of viscosity n of the liquid <br/>Therefore `F propeta^(x)r^(y)r^(y) = F k eta^(x)v^2 `where k is a dimensionless constant. . Using dimensions, the above equation can be written as<br/> `[MLT^(-2)]=K [ML^(-1)T^(-1) ] ^(z )xx [L]^(y) xx [LT^(-1) ]^(x)` <br/> On solving, we get x = 1, y = 1 and z = 1. Therefore, F = `k eta rv `<br/> Experimentally, Stoke found that the value of `k = 6 pi` <br/> ` f= 6 pietar v `<br/> This relation is <a href="https://interviewquestions.tuteehub.com/tag/known-534098" style="font-weight:bold;" target="_blank" title="Click to know more about KNOWN">KNOWN</a> as Stoke.s law.,<br/>Practical applications of Stoke.s law Since the raindrops are smaller in size and their terminal velocities are small, <a href="https://interviewquestions.tuteehub.com/tag/remain-1184279" style="font-weight:bold;" target="_blank" title="Click to know more about REMAIN">REMAIN</a> suspended in air in the form of clouds. As they grow up in size, their terminal velocities increase and they start falling in the form of rain. This law explains the following: <br/>(a) Floatation of clouds<br/> (b) Larger raindrops hurt us more than the smaller ones <br/> (c) A man <a href="https://interviewquestions.tuteehub.com/tag/coming-922990" style="font-weight:bold;" target="_blank" title="Click to know more about COMING">COMING</a> down with the <a href="https://interviewquestions.tuteehub.com/tag/help-1018089" style="font-weight:bold;" target="_blank" title="Click to know more about HELP">HELP</a> of a parachute acquires constant terminal velocity.</body></html> | |
| 38014. |
A simple pendulum of length 1 m is oscillated at a place where g=9.8m//s^(2). Find the maximum velocity of the bob of the simple pendulum if the amplitude of oscillation is 3 cm. |
| Answer» <html><body><p>`5.219cms^(-<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/>`7.216cms^(-1)`<br/>`6.181cms^(-1)`<br/>`9.391cms^(-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> :D</body></html> | |
| 38015. |
The work done in time t on a body of mass m which is accelerated from rest to a speed v in time t_1 as a function of time t is given by |
| Answer» <html><body><p>`1/2 m (v)/(t_1) t^2`<br/>`m v/(t_1)t^2`<br/>`1/2 ((<a href="https://interviewquestions.tuteehub.com/tag/mv-1082193" style="font-weight:bold;" target="_blank" title="Click to know more about MV">MV</a>)/(t_1))^(2) t^2`<br/>`1/2 m (v^2)/(t_1^2) t^2`</p>Solution :As `v =u + at = 0 + at_1 "" ( :. u = 0)` <br/> `:. A = (v)/(t_1) "" ….(i)` <br/> Force, `F = ma = (mv)/(t_1) ` <br/> <a href="https://interviewquestions.tuteehub.com/tag/distance-116" style="font-weight:bold;" target="_blank" title="Click to know more about DISTANCE">DISTANCE</a> travelled by the body in time t is <br/> `s = <a href="https://interviewquestions.tuteehub.com/tag/ut-718961" style="font-weight:bold;" target="_blank" title="Click to know more about UT">UT</a> + 1/2 at^2 = 0 + 1/2 (vt^2)/(t_1) = (vt^2)/(2t_1) "(Using (i))"` <br/> `:. <a href="https://interviewquestions.tuteehub.com/tag/w-729065" style="font-weight:bold;" target="_blank" title="Click to know more about W">W</a> = F xx s = (mv)/(t_1) (vt^2)/(2t_1) = 1/2 (mv^2 t^2)/(t_1^2)`</body></html> | |
| 38016. |
Under the influence of which force the oscillation of a pendulum gradually dies out? |
| 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/resistive-2248841" style="font-weight:bold;" target="_blank" title="Click to know more about RESISTIVE">RESISTIVE</a> <a href="https://interviewquestions.tuteehub.com/tag/force-22342" style="font-weight:bold;" target="_blank" title="Click to know more about FORCE">FORCE</a></body></html> | |
| 38017. |
The angle of friction between two surface in contact is 60^(@) What is the cofficient of friction between them . |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/sqrt3-3056952" style="font-weight:bold;" target="_blank" title="Click to know more about SQRT3">SQRT3</a>`<br/>`(<a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>)/sqrt3`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>`<br/>`1`</p>Solution :`mu = <a href="https://interviewquestions.tuteehub.com/tag/tan-1238781" style="font-weight:bold;" target="_blank" title="Click to know more about TAN">TAN</a> theta = tan 60^(@) = sqrt3` .</body></html> | |
| 38018. |
Four small objects each of mass m are fixed at the corners of a rectangular wire-frame of negligible mass and of sides a and b (a gt b). If the wire frame is now rotated about an axis passing along the side of length b, then the moment of inertia of the system for this axis of rotation is |
| Answer» <html><body><p>`2ma^2`<br/>`4ma^2`<br/>`2 m (a^2 + b^2)`<br/>`2m (a^2- b^2)`</p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/moment-25786" style="font-weight:bold;" target="_blank" title="Click to know more about MOMENT">MOMENT</a> 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 the system about an <a href="https://interviewquestions.tuteehub.com/tag/axis-889931" style="font-weight:bold;" target="_blank" title="Click to know more about AXIS">AXIS</a> passing along the side b is `I = ma^2 + <a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a> + ma^2 + 0 = 2ma^(2)` <br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/MTG_NEET_GID_PHY_XI_C05_E01_038_S01.png" width="80%"/></body></html> | |
| 38019. |
A cylinder of ideal gas is closed by an 8 kg movable piston of area 60cm^2. The atmospheric pressure is 100 kPa. When the gas is heated from 30^@ C to 100^@ C the piston rises 20 cms. The piston is then fastened in the place and the gas is cooled back to 30^@ C. IfDelta Q_1 is the heat added to gas during heating and Delta Q_2 is the heat lost during cooling, find the difference. |
| 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> :`32.5cal`</body></html> | |
| 38020. |
(i+i)×(i-j)= |
| Answer» <html><body><p>`-<a href="https://interviewquestions.tuteehub.com/tag/2k-300402" style="font-weight:bold;" target="_blank" title="Click to know more about 2K">2K</a>`<br/>`2k`<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/>`<a href="https://interviewquestions.tuteehub.com/tag/2i-300397" style="font-weight:bold;" target="_blank" title="Click to know more about 2I">2I</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> | |
| 38021. |
If the direction of the torque is inward the paper then the rotation is |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/clockwise-919294" style="font-weight:bold;" target="_blank" title="Click to know more about CLOCKWISE">CLOCKWISE</a> <br/>anti-clockwise <br/><a href="https://interviewquestions.tuteehub.com/tag/straight-633156" style="font-weight:bold;" target="_blank" title="Click to know more about STRAIGHT">STRAIGHT</a> line<br/>random direction</p>Answer :A</body></html> | |
| 38022. |
In the following question, a statement of Assertion is followed by a statement of Reason. Assertion: When a body moves along a circular path, the work done by a centripetal force is zero. Reason: The centripetal force is utilised in moving the body along the circular path and hence the work is done. Choose one of the following statements is correct? |
| Answer» <html><body><p>Both assertion and reason are true and reason is the correct <a href="https://interviewquestions.tuteehub.com/tag/explanation-455162" style="font-weight:bold;" target="_blank" title="Click to know more about EXPLANATION">EXPLANATION</a> of the assertion<br/>Both assertion and reason are true but reason is not correct explanation of the assertion<br/>Assertion is true but reason is false<br/>Both assertion and reason are false </p>Solution :The <a href="https://interviewquestions.tuteehub.com/tag/work-20377" style="font-weight:bold;" target="_blank" title="Click to know more about WORK">WORK</a> done by the centripetal force is zero because it <a href="https://interviewquestions.tuteehub.com/tag/acts-848461" style="font-weight:bold;" target="_blank" title="Click to know more about ACTS">ACTS</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> to the circular path.</body></html> | |
| 38023. |
A train standing in a station-yard, blows a whistle of frequency 400Hz in still air. The wind starts blowing in the direction from the yard to the station with a speed of 10 ms^(-1). What are the frequency wavelength and speed of sound for an observer standing on the station.s platform? Is the situation exactly identical the case when the air is still and the observer runs towards the yard at a speed of 10 ms^(-1) ? The speed of sound in still air can be taken as 340 ms^(-1) |
| Answer» <html><body><p></p>Solution :400Hz, 0.875m, `350ms^(-1)`. No, because in this <a href="https://interviewquestions.tuteehub.com/tag/case-910082" style="font-weight:bold;" target="_blank" title="Click to know more about CASE">CASE</a>, with respect to the <a href="https://interviewquestions.tuteehub.com/tag/medium-1092763" style="font-weight:bold;" target="_blank" title="Click to know more about MEDIUM">MEDIUM</a> both the <a href="https://interviewquestions.tuteehub.com/tag/observer-1127626" style="font-weight:bold;" target="_blank" title="Click to know more about OBSERVER">OBSERVER</a> and the source are in <a href="https://interviewquestions.tuteehub.com/tag/motion-1104108" style="font-weight:bold;" target="_blank" title="Click to know more about MOTION">MOTION</a>.</body></html> | |
| 38024. |
The escape velocity of a body on the earth's surface is v_e. A body is thrown up with a speed of sqrt(5)V_e. Assuming that the sun and planets to not influence the motion of the body, then the velocity of the body at the infinite distance is |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>`<br/>`V_e`<br/>`<a href="https://interviewquestions.tuteehub.com/tag/sqrt-1223129" style="font-weight:bold;" target="_blank" title="Click to know more about SQRT">SQRT</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>)V_e`<br/>`2V_e`</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> | |
| 38025. |
Which of the following is an example of forced vibrations ? |
| Answer» <html><body><p>Vibration of the <a href="https://interviewquestions.tuteehub.com/tag/diaphragm-438991" style="font-weight:bold;" target="_blank" title="Click to know more about DIAPHRAGM">DIAPHRAGM</a> of <a href="https://interviewquestions.tuteehub.com/tag/microphone-1095778" style="font-weight:bold;" target="_blank" title="Click to know more about MICROPHONE">MICROPHONE</a> <br/><a href="https://interviewquestions.tuteehub.com/tag/tuned-7717291" style="font-weight:bold;" target="_blank" title="Click to know more about TUNED">TUNED</a> radio receiver<br/>Vibrating <a href="https://interviewquestions.tuteehub.com/tag/sonometer-1218971" style="font-weight:bold;" target="_blank" title="Click to know more about SONOMETER">SONOMETER</a> wire giving maximum displacement amplitude<br/>Vibrating <a href="https://interviewquestions.tuteehub.com/tag/tuning-1428825" style="font-weight:bold;" target="_blank" title="Click to know more about TUNING">TUNING</a> fork </p>Answer :A</body></html> | |
| 38026. |
From a single source, two wave trains are sent in two different string. The two wave equations are ((area of corss-section and tension of both string are same y_(1)= A sin (w_(1)t-k_(1)x)and y_(2) = 2A sin (w_(1)t -k_(2)z)Suppose u = energy density , P = power trasmitted and I = intensity of the wave, then match the following. |
| Answer» <html><body><p><<a href="https://interviewquestions.tuteehub.com/tag/p-588962" style="font-weight:bold;" target="_blank" title="Click to know more about P">P</a>><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/q-609558" style="font-weight:bold;" target="_blank" title="Click to know more about Q">Q</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>)P, (<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a> )P</body></html> | |
| 38027. |
From a circular disc of radius R and mass 9M, a small disc of radius R/3 is removed as shown in figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc andpassing through O is |
| Answer» <html><body><p>`4 MR^(2)`<br/>`40/9 MR^(2)`<br/>`40 MR^(2)`<br/>`37/9 MR^(2)`</p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/mass-1088425" style="font-weight:bold;" target="_blank" title="Click to know more about MASS">MASS</a> per unit area of disc <br/> `=(9M)/(piR^(2))` <br/> Mass of <a href="https://interviewquestions.tuteehub.com/tag/removed-2986295" style="font-weight:bold;" target="_blank" title="Click to know more about REMOVED">REMOVED</a> portion of disc <br/> `=(9M)/(piR^(2)) xx pi(R/3)^(3) = M` <br/> <a href="https://interviewquestions.tuteehub.com/tag/moment-25786" style="font-weight:bold;" target="_blank" title="Click to know more about MOMENT">MOMENT</a> 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 removed portion about an <a href="https://interviewquestions.tuteehub.com/tag/axis-889931" style="font-weight:bold;" target="_blank" title="Click to know more about AXIS">AXIS</a> passing through center of disc and perpendicular to the plane of disc, using theorem of parallel axes is <br/> `I_(1)=M/2(R/3)^(2)+1/2(MR^(2))=4MR^(2)` <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/NCERT_OBJ_FING_PHY_XI_C07_E01_052_S01.png" width="80%"/></body></html> | |
| 38028. |
Show that for a particle in linear SHM the average kinetic energy over a period of oscillation equals the average potential energy over the same period. |
| Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/hint-486738" style="font-weight:bold;" target="_blank" title="Click to know more about HINT">HINT</a>: Average K.E. `=1/Tint_(<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>)^(T)1/2mv^(2)<a href="https://interviewquestions.tuteehub.com/tag/dt-960413" style="font-weight:bold;" target="_blank" title="Click to know more about DT">DT</a>,` Average P.E. `=1/Tint_(0)^(T)1/2kx^(2)dt`</body></html> | |
| 38029. |
A particle is fired vertically upwards from the surface of earth and reaches a height 6400 km. Find the initial velocity of the particle if R = 6400 km and g at the surface of earth is 10 m//s^(2). |
| Answer» <html><body><p></p>Solution :`(1)/(2) mv^(2) = (mgh)/([1+(h//R)])` <br/> here h = R = 6400 <a href="https://interviewquestions.tuteehub.com/tag/km-1064498" style="font-weight:bold;" target="_blank" title="Click to know more about KM">KM</a> and `<a href="https://interviewquestions.tuteehub.com/tag/g-1003017" style="font-weight:bold;" target="_blank" title="Click to know more about G">G</a> = 10 m//s^(2)` <br/> so `v^(2) = gh`, <br/> i.e., `v = sqrt(10 xx 6400 xx 10^(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)) = <a href="https://interviewquestions.tuteehub.com/tag/8-336412" style="font-weight:bold;" target="_blank" title="Click to know more about 8">8</a> km//s`</body></html> | |
| 38030. |
Two metal spheres of same material and each of radius r are in contact with each other. The gravitational force of attraction between the spheres is proportional to |
| Answer» <html><body><p>`1/<a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</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/>`1/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/>`r^(4)`<br/>`r^(<a href="https://interviewquestions.tuteehub.com/tag/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</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> | |
| 38031. |
A sphere and a cube both made of copper have equal volumes and are blackened. These are heated to same temperature and are allowed to cool under same surroundings. The ratio of their rates of loss of heat is |
| Answer» <html><body><p>`1:1`<br/>`p//6`<br/>`6//p`<br/>`p//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</body></html> | |
| 38032. |
Draw a graph showing the variation of K.E and P.E with respect to displacement for a spring mass system. |
| Answer» <html><body><p></p>Solution :Let us consider the nature of a compressed or extended spring. A compressed or extended spring will transfer its <a href="https://interviewquestions.tuteehub.com/tag/stored-7260220" style="font-weight:bold;" target="_blank" title="Click to know more about STORED">STORED</a> potential <a href="https://interviewquestions.tuteehub.com/tag/energy-15288" style="font-weight:bold;" target="_blank" title="Click to know more about ENERGY">ENERGY</a> into kinetic energy of the mass attached to the spring. The potential energy-displacement graph is shown in the figure. It <a href="https://interviewquestions.tuteehub.com/tag/shows-1206590" style="font-weight:bold;" target="_blank" title="Click to know more about SHOWS">SHOWS</a> variation of potential energy with respect to displacement. <br/> In a frictionless environment, the energy gets transferred from kinetic to potential and potential to kinetic repeatedly in such a way that the total energy of the system remains <a href="https://interviewquestions.tuteehub.com/tag/constant-930172" style="font-weight:bold;" target="_blank" title="Click to know more about CONSTANT">CONSTANT</a>. At the <a href="https://interviewquestions.tuteehub.com/tag/mean-1091459" style="font-weight:bold;" target="_blank" title="Click to know more about MEAN">MEAN</a> position, <br/> `DeltaKE=DeltaU` <br/> i.e., change in kinetic energy = Change in potential energy. <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/PRE_GRG_PHY_XI_V02_C04_E02_099_S01.png" width="80%"/></body></html> | |
| 38033. |
A stone is released from an elevator going up with an acceleration 'a'. The acceleration of the stone after the release is |
| Answer» <html><body><p>a <a href="https://interviewquestions.tuteehub.com/tag/upward-721781" style="font-weight:bold;" target="_blank" title="Click to know more about UPWARD">UPWARD</a><br/>`(g-a)` upward<br/>`(g-a)` downward<br/>g downward</p>Answer :D</body></html> | |
| 38034. |
What is meant by reverberation time? |
| Answer» <html><body><p></p>Solution :The <a href="https://interviewquestions.tuteehub.com/tag/persistence-1151770" style="font-weight:bold;" target="_blank" title="Click to know more about PERSISTENCE">PERSISTENCE</a> of audible <a href="https://interviewquestions.tuteehub.com/tag/sound-648690" style="font-weight:bold;" target="_blank" title="Click to know more about SOUND">SOUND</a> after the <a href="https://interviewquestions.tuteehub.com/tag/source-1219297" style="font-weight:bold;" target="_blank" title="Click to know more about SOURCE">SOURCE</a> has <a href="https://interviewquestions.tuteehub.com/tag/ceased-7669727" style="font-weight:bold;" target="_blank" title="Click to know more about CEASED">CEASED</a> to <a href="https://interviewquestions.tuteehub.com/tag/emit-970204" style="font-weight:bold;" target="_blank" title="Click to know more about EMIT">EMIT</a> sound is called reverberation .</body></html> | |
| 38035. |
A steel wire 0.72 m long has a mass of 5.0 xx 10^(-3) kg. If the wire is under a tension of 60 N. What is the speed of transverse waves on the wire ? |
| Answer» <html><body><p></p>Solution :Mass <a href="https://interviewquestions.tuteehub.com/tag/per-590802" style="font-weight:bold;" target="_blank" title="Click to know more about PER">PER</a> unit length of the wire, <br/> `mu = (5.0 x 10^(-<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>) kg)/(0.72 m)` <br/> `= 6.9 xx 10^(-3) kg m^(-1)` <br/> Tension, T = 60 N <br/> The <a href="https://interviewquestions.tuteehub.com/tag/speed-1221896" style="font-weight:bold;" target="_blank" title="Click to know more about SPEED">SPEED</a> of wave on the wire is <a href="https://interviewquestions.tuteehub.com/tag/given-473447" style="font-weight:bold;" target="_blank" title="Click to know more about GIVEN">GIVEN</a> by <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((60 N)/(6.9 xx 10^(-3) kg m^(-1))) = 93 s^(-1)`</body></html> | |
| 38036. |
The breaking stress of a wire depends upon …………. . |
| Answer» <html><body><p>lengthof a wire <br/>nature of the wire <br/><a href="https://interviewquestions.tuteehub.com/tag/diameter-950836" style="font-weight:bold;" target="_blank" title="Click to know more about DIAMETER">DIAMETER</a> of the wire<br/>shape of the cross <a href="https://interviewquestions.tuteehub.com/tag/section-25668" style="font-weight:bold;" target="_blank" title="Click to know more about SECTION">SECTION</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 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> | |
| 38037. |
(A) : A body is momentarily at rest at the instant it reverses the direction. ( R) : A body cannot have acceleration if its velocity is zero at a given instant of time. |
| Answer» <html><body><p>Both (A) and ( R) are <a href="https://interviewquestions.tuteehub.com/tag/ture-1429019" style="font-weight:bold;" target="_blank" title="Click to know more about TURE">TURE</a> and ( R) is the correct explanation of (A) <br/>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> and ( R) is not the correct explanation of (A)<br/>(A) is true but ( 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/>Both (A) and ( R) are false </p>Answer :C</body></html> | |
| 38038. |
A hiker stands on the edge of cliff 490 m above the ground and throws a stone horizontally with an initial speed of 15 "m s "^(-1) Neglecting airresistance , find the time taken by the stone to reach the ground , and the speed with which it hits the ground . (Take g=9.8 " m s"^(2)). |
| Answer» <html><body><p></p>Solution :We choose the <a href="https://interviewquestions.tuteehub.com/tag/origin-1139399" style="font-weight:bold;" target="_blank" title="Click to know more about ORIGIN">ORIGIN</a> of the x - and y-axis at the <a href="https://interviewquestions.tuteehub.com/tag/edge-965664" style="font-weight:bold;" target="_blank" title="Click to know more about EDGE">EDGE</a> of the cliff and t =0 s at the instant the stone is <a href="https://interviewquestions.tuteehub.com/tag/thrown-7258593" style="font-weight:bold;" target="_blank" title="Click to know more about THROWN">THROWN</a> . Choose the positive direction of x-axis to be along the initial velocity and the positivedirection of axis to be the <a href="https://interviewquestions.tuteehub.com/tag/vertically-3260386" style="font-weight:bold;" target="_blank" title="Click to know more about VERTICALLY">VERTICALLY</a> upward direction The x- , and y- <a href="https://interviewquestions.tuteehub.com/tag/components-926700" style="font-weight:bold;" target="_blank" title="Click to know more about COMPONENTS">COMPONENTS</a> of the motion can be treated independently . Theequations of motion are :<br/>`x(t)=x_(@)+v_(@x)t`<br/>`y(t)=y_(@)+v_(@y)t+(1//2)a_(y)t^(2)`<br/>Here , `x_(@)=y_(@)=0,v_(@y)=0,a_(y)=-r=-9.8 " m s "^(-2),v_(@x)=15 " m s"^(-1)`.<br/>The stone hits the ground when y(t) =- 490 m. <br/>`-490m=-(1//2)(9.8)t^(2)`.<br/>This gives `t=10s`.<br/>The velocity components are `v_(x)=v_(@x)andv_(y)=v_(@y)- gt`<br/>so that when the stonehits the ground :<br/>`v_(@y)=15 " m s"^(-1)`<br/>`v_(@y)=0-9.8xx10=-98 " m s"^(-1)`<br/>Therefore , the speed of the stone is<br/>`sqrt(v_(x)^(2)+v_(y)^(2))=sqrt(15^(2)+98^(2))=99 " m s"^(-1)`</body></html> | |
| 38039. |
One rod vibrates with frequency 200 Hz. It produces a sound which travels in air with velocity 340 m/s. Find wavelength of this wave |
| Answer» <html><body><p>`1.7cm`<br/>`6.8cm`<br/>`1.7m`<br/>`6.8m` </p>Solution :We <a href="https://interviewquestions.tuteehub.com/tag/know-534065" style="font-weight:bold;" target="_blank" title="Click to know more about KNOW">KNOW</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> = f lamda <a href="https://interviewquestions.tuteehub.com/tag/implies-1037962" style="font-weight:bold;" target="_blank" title="Click to know more about IMPLIES">IMPLIES</a> lamda = (v)/(f)` <br/> `therefore lamda = (340)/(<a href="https://interviewquestions.tuteehub.com/tag/200-288914" style="font-weight:bold;" target="_blank" title="Click to know more about 200">200</a>) =1.7 m`</body></html> | |
| 38040. |
Is it possible to have apparent expansion equal to real expansion? |
| Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/yes-749457" style="font-weight:bold;" target="_blank" title="Click to know more about YES">YES</a>, if the <a href="https://interviewquestions.tuteehub.com/tag/expansion-980011" style="font-weight:bold;" target="_blank" title="Click to know more about EXPANSION">EXPANSION</a> of the container is compensated by introducing a <a href="https://interviewquestions.tuteehub.com/tag/certain-407894" style="font-weight:bold;" target="_blank" title="Click to know more about CERTAIN">CERTAIN</a> amount of <a href="https://interviewquestions.tuteehub.com/tag/mercury-558950" style="font-weight:bold;" target="_blank" title="Click to know more about MERCURY">MERCURY</a> whose expansion is just equal to that of the container used,</body></html> | |
| 38041. |
An air bubble starts rising from the bottom of a lake. Its diameter is 3.6 mm at the bottom and 4 mm at the surface. The depth of the lake is 2.5 m and the temperature at the surface is 40^(@)C. What is the temperatue at the bottom of the lake ? |
| 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> :Let `P_(1), V_(1), T_(1)" and "P_(2), V_(2), T_(2)` are the parameters of the air bubble at the bottom and surface of the lake respectively. <br/> `P_(1)=` <a href="https://interviewquestions.tuteehub.com/tag/atm-364409" style="font-weight:bold;" target="_blank" title="Click to know more about ATM">ATM</a>/<a href="https://interviewquestions.tuteehub.com/tag/pressure-1164240" style="font-weight:bold;" target="_blank" title="Click to know more about PRESSURE">PRESSURE</a> + Pressure of water at the bottom <br/> `""=(76 times 13.6 times <a href="https://interviewquestions.tuteehub.com/tag/980-342886" style="font-weight:bold;" target="_blank" title="Click to know more about 980">980</a>)+(250 times 1 times 980)=1283.6 times 980" dyne "cm^(-2)` <br/> `P_(2)="Atm. pressure"=76 times 13.6 times 980=1033.6 times 980" dyne "cm^(-2)` <br/> `V_(1)=4/3pi(0.36/2)^(3)=0.007776 times <a href="https://interviewquestions.tuteehub.com/tag/pi-600185" style="font-weight:bold;" target="_blank" title="Click to know more about PI">PI</a> cm^(3) rArr V_(2)=4/3pi(0.4/2)^(3)=0.01067 times picm^(3)` <br/> `T_(2)=273+40=313K` <br/> From the equation, `(P_(1)V_(1))/T_(1)=(P_(2)V_(2))/T_(2)` <br/> `rArr T_(1)=P_(1)/P_(2) times V_(1)/V_(2) times T_(2)=(1283.6 times 980)/(1033.6 times 980) times (0.00776pi)/(0.01067pi) times 313=1.242 times 0.7288 times 313=283.3K` <br/> Temperature at the bottom of the lake = 283.3 - 273 = `10.3^(@)C`</body></html> | |
| 38042. |
A bob of mass m is attached to one end of the rod of negligible mass and length r, the other end of which is pivoted freely at a fixed centre O as shown in the figure. What initial speed must be given to the object to reach the top of the circle? (Hint: Use law of conservation of energy). Is this speed less or greater than speed obtained in the section 4.2.9? |
| Answer» <html><body><p></p>Solution :<img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/PRE_GRG_PHY_XI_V02_C04_E01_031_S01.png" width="80%"/> <br/> <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 law of conservation of energy <br/> `(1)/(2)mv_(<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>)^(2)+0=0+mg(2r)` <br/> `v_(0)=sqrt(4gr)ms^(-1)` <br/> This <a href="https://interviewquestions.tuteehub.com/tag/value-1442530" style="font-weight:bold;" target="_blank" title="Click to know more about VALUE">VALUE</a> is less than <br/> `v_(1)=sqrt(5gr)` and <a href="https://interviewquestions.tuteehub.com/tag/greater-476627" style="font-weight:bold;" target="_blank" title="Click to know more about GREATER">GREATER</a> than <br/> `v_(2)=sqrt(gr)`</body></html> | |
| 38043. |
A uniform metre scale of mass 2kg is suspended from one end. If it is displaced through an angle 60^(@) from the vertical, the increase in its potential energy is |
| Answer» <html><body><p>4.9 <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/>9.8 J<br/>`9.8sqrt(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)J`<br/>`4.9(2-sqrt(3))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> | |
| 38044. |
Vessel A is filled with hydrogen while vessel B, whose volume is twice that of A, is filled with the same mass of oxygen at the same temperature. The ratio of the mean kinetic energies of hydrogen and oxygen is |
| Answer» <html><body><p>`16:1`<br/>`1:8`<br/>`8:1`<br/>`1:1`</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> :D</body></html> | |
| 38045. |
Does it matter if one uses gauge instead of absolute pressures in applying Bernoulli's equation Explain? |
| 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> :No. unless the atmospheric <a href="https://interviewquestions.tuteehub.com/tag/pressures-1164423" style="font-weight:bold;" target="_blank" title="Click to know more about PRESSURES">PRESSURES</a> at the two <a href="https://interviewquestions.tuteehub.com/tag/points-1157347" style="font-weight:bold;" target="_blank" title="Click to know more about POINTS">POINTS</a> where Bernoulli.s equation is applied are significantly <a href="https://interviewquestions.tuteehub.com/tag/different-951434" style="font-weight:bold;" target="_blank" title="Click to know more about DIFFERENT">DIFFERENT</a>.</body></html> | |
| 38046. |
Two identical copper spheres are put in contact. The gravitational force of attraction between them is proportional to r^(x). Find the value of x |
| 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 href="https://interviewquestions.tuteehub.com/tag/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</a></body></html> | |
| 38047. |
The refractive index of a lens material is 1.5 and focal length f. Due to some chemical changes in the material, its refractive index has increased by 2%. The percentage change in its focal length is |
| Answer» <html><body><p>`+4.5%`<br/>`-4.5%`<br/>`+5.67%`<br/>`-5.67%`</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> | |
| 38048. |
Consider two cylindrical rods of identicaly dimensions, one of rubber and the other steel. Both the rods are fixed rigidly at one end to the roof. A mass M is attached to each of the free ends at the centre of the rods. Then |
| Answer» <html><body><p>Both the rods will elongate but there no <a href="https://interviewquestions.tuteehub.com/tag/perceptible-2920990" style="font-weight:bold;" target="_blank" title="Click to know more about PERCEPTIBLE">PERCEPTIBLE</a> change in the shape.<br/>The steel <a href="https://interviewquestions.tuteehub.com/tag/rod-1190628" style="font-weight:bold;" target="_blank" title="Click to know more about ROD">ROD</a> will elongate and change shape but the rubber rod will only elongate.<br/> The steel rod will elongate without any perceptible change in shape, but the rubber rod will elongate and the shape of the bottom <a href="https://interviewquestions.tuteehub.com/tag/edge-965664" style="font-weight:bold;" target="_blank" title="Click to know more about EDGE">EDGE</a> will change to an ellipse.<br/>The steel rod will elongate, without any perceptible change in shape, but the rubber rod will elongate with the shape of the bottom edge <a href="https://interviewquestions.tuteehub.com/tag/tapered-1239219" style="font-weight:bold;" target="_blank" title="Click to know more about TAPERED">TAPERED</a> to a tip at the centre.</p>Answer :C</body></html> | |
| 38049. |
Find the angle between the vectors 2hati+3hatj-6hatk" "and 6hati-3hatj+2hatk |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>^@` <br/>`30^@` <br/>`<a href="https://interviewquestions.tuteehub.com/tag/60-328817" style="font-weight:bold;" target="_blank" title="Click to know more about 60">60</a>^@` <br/>`<a href="https://interviewquestions.tuteehub.com/tag/90-341351" style="font-weight:bold;" target="_blank" title="Click to know more about 90">90</a>^@` </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 angle between the two vector is`90^@ . [ cos theta = (vecA.vecB)/(AB)]`</body></html> | |
| 38050. |
An artificial satellite is describing an equatorial orbit at 1600 km above the surface earth. Calculate its orbital speed and the period of revolution. If satellite is travelling in the same direction as the rotation of the earth (i.e., from west to east), calculate the interval between two successive time at wich it will appear verically overhead to an observer at a fixed point on the equator. Radius of earth = 6400 km. |
| 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> :We <a href="https://interviewquestions.tuteehub.com/tag/know-534065" style="font-weight:bold;" target="_blank" title="Click to know more about KNOW">KNOW</a> that period of the satellite is <br/> `T = (2pi)/(sqrt(GM))r^(3//2) = (2pi)/(sqrt(gR^(2)))r^(3//2)` <br/> Where `r = 6400+1600=8000xx10^(3)m`, <br/> `g = 9.8 m/sec^(2)` and `R = 6400xx10^(3)m` <br/> <a href="https://interviewquestions.tuteehub.com/tag/substituting-1231652" style="font-weight:bold;" target="_blank" title="Click to know more about SUBSTITUTING">SUBSTITUTING</a> values we get <br/> `T=2xx3.14[((8000xx10^(3))^(3))/(9.8xx(6400xx10^(3))^(2))]^(1//2)` <br/> `= 7096` <br/> <a href="https://interviewquestions.tuteehub.com/tag/futher-2664615" style="font-weight:bold;" target="_blank" title="Click to know more about FUTHER">FUTHER</a>, orbital speed, <br/> `v = sqrt((GM)/(r )) = sqrt((gR^(2))/(r ))` <br/> or `v = sqrt(((9.8)/(8000xx10^(3))))xx(6400xx10^(3))` <br/> `= 7083.5 m//s` <br/> Let t be the time interval between two successive moments at which the satellite is overhead to an observer at fixed position on the equator. As both satelite and earth are moving in same direction with angular speed `omega_(S)` and `omega_(E )` respectively, we can write the time of separation as <br/> `t = (2pi)/(omega_(S)-omega_(E ))` <br/> Here `omega_(S)=(2pi)/(7096)` and `omega_(E )=(2pi)/(86400)` <br/> Thus we have `t = (86400xx7096)/(86400-7096)` <br/> `= 7731 s`</body></html> | |