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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.
| 38301. |
A block of mass m requires a horizontal force F0 to move it on a horizontal metal plate with constant velocity. The metal plate is folded to make it a right angled horizontal trough. Find the horizontal force F that is needed to move the block with constant velocity along this trough. |
| 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/sqrt2-3056932" style="font-weight:bold;" target="_blank" title="Click to know more about SQRT2">SQRT2</a> F_(<a href="https://interviewquestions.tuteehub.com/tag/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</a>)`</body></html> | |
| 38302. |
Amotorof powerP_0 isusedtodeliverwateratacertain ratethroughagivenhorizontalpipe.Toincreasethe rateof flowof waterthroughthesamepipentimes,the power of the moter is increased to P_1 to P_0 is |
| Answer» <html><body><p>n : <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/>`n^(2) : 1`<br/>`n^(3) : 1`<br/>`n^(4) : 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> :C</body></html> | |
| 38303. |
The piston cylinder arrangement shon contains a diatomic gas at temperature 300 K The crooss sectinal area of the cylinder is 1 m^(2). Initially the height of the piston above th base of the cyclinder is 1 m. The temperature is now raised to 400 K at constatn pressure. Find tge bew jgeauggt if tge oustib avive tge vase if tge cylinder.If the psiton is now brouyght back to its original height without any heat losss, find the new equilibrium temperaturef of the gas. You can leave the answer in fraction. |
| 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/4-311707" style="font-weight:bold;" target="_blank" title="Click to know more about 4">4</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`(<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>)`T_3=400((4)/(3))^0.4 <a href="https://interviewquestions.tuteehub.com/tag/k-527196" style="font-weight:bold;" target="_blank" title="Click to know more about K">K</a>`</body></html> | |
| 38304. |
Assertion : Deep inside a liquid density is more than the density on surface. Reason : Density of liquid increases with increase in depth. |
| Answer» <html><body><p>If both Assertion and <a href="https://interviewquestions.tuteehub.com/tag/reason-620214" style="font-weight:bold;" target="_blank" title="Click to know more about REASON">REASON</a> are correct 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 Assertion.<br/>If both Assertion and Reason 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> but Reason is not the correct explanation of Assertion.<br/>If Assertion is true but Reason 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/>If Assertion is false but Reason is true.</p>Solution :N//A</body></html> | |
| 38305. |
Show that a=(v^2)/r and hence obtain an expression for centripetal force. |
| Answer» <html><body><p></p>Solution :Let `vecr` and `vecr`, be the position vectors and `vecv` velocities 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> when it is it point P and P. By definition, velocity at a point is <a href="https://interviewquestions.tuteehub.com/tag/along-1974109" style="font-weight:bold;" target="_blank" title="Click to know more about ALONG">ALONG</a> the tangent at that point in the direction of the motion. Since the path is <a href="https://interviewquestions.tuteehub.com/tag/circular-916697" style="font-weight:bold;" target="_blank" title="Click to know more about CIRCULAR">CIRCULAR</a>, `vecv` is perpendicular to `vecr` and `vecv` is perpendicular to `vecr` <br/> Therefore, `/_\vecv` is perpendicular to `/_\vecr`. Average acceleration `(/_\vecv)/(/_\t)` is perpendicular to `/_\vecr` <br/> The magnitude of `veca` is, by definition, given by `|veca|=lim_(/_\t to0)(/_\vecv)/(/_\t)` <br/> The <a href="https://interviewquestions.tuteehub.com/tag/triangle-1427233" style="font-weight:bold;" target="_blank" title="Click to know more about TRIANGLE">TRIANGLE</a> formed by the position vectors is similarly to the triangle formed by the velocity vectors. <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/SPH_ABT_PHY_QB_XI_C04_E04_006_S01.png" width="80%"/> <br/> i.e, `(|/_\vecv|)/v=(|/_\vecr|)/r "or"|/_\vecv|=v(|/_\vecr|)r` <br/> Therefore, `|veca|=lim_(|/_\vecv)/(/_\t)=lim_(/_\t to0)(v|/_\vecr|)/(r/_\t)=(vecv)/(r)lim_(/_\t to0)(|/_\vecr|)/(/_\t)` <br/> If `/_\t` is very small, `/_\theta` will also small. The arc PP' is approximately equal to `|/_\vecr|` <br/> i.e., `lim_(/_\t to0)(|/_\vecr|)/(/_\t)=v` Thus, centripetal acceleration `|veca|=v/fv=(v^2)/(r)` and `veca=v/r(dvecr)/(dt)` <br/> The centripetal acceleration is always <a href="https://interviewquestions.tuteehub.com/tag/directed-440230" style="font-weight:bold;" target="_blank" title="Click to know more about DIRECTED">DIRECTED</a> towards the centre. The centripetal force=ma.</body></html> | |
| 38306. |
What are the characteristics of geostationary satellite and polar satellite ? Why polar satellite are called weather satellite -Explain. |
| Answer» <html><body><p></p>Solution :Geostationary satellite are <a href="https://interviewquestions.tuteehub.com/tag/placed-591674" style="font-weight:bold;" target="_blank" title="Click to know more about PLACED">PLACED</a> on the equatorial <a href="https://interviewquestions.tuteehub.com/tag/plane-1155510" style="font-weight:bold;" target="_blank" title="Click to know more about PLANE">PLANE</a> at a height of nearly 36000 km from the earth.s surface. Since the motion of such a satellite is same as that of the <a href="https://interviewquestions.tuteehub.com/tag/diurnal-432074" style="font-weight:bold;" target="_blank" title="Click to know more about DIURNAL">DIURNAL</a> motion of the earth, the satellite seems to be stationary with respect to the earth.s surface. <br/> On the other hand, polar satellite are placed at a height of 800 km, from the surface of the earth, aligned with the polar plane. To revolve around the earth once, a polar satellite takes about 2 hours. <br/> Polar satellite are placed comparatively nearer to the earth.s surface and travel over a vast area while revolving . <a href="https://interviewquestions.tuteehub.com/tag/thus-2307358" style="font-weight:bold;" target="_blank" title="Click to know more about THUS">THUS</a> they are suitable for observing the <a href="https://interviewquestions.tuteehub.com/tag/weather-1450884" style="font-weight:bold;" target="_blank" title="Click to know more about WEATHER">WEATHER</a> and hence, are called .weather satellite..</body></html> | |
| 38307. |
Is the work done bynon conservative force is always negative ? Discuss . |
| Answer» <html><body><p></p>Solution :No. If the body does not move <a href="https://interviewquestions.tuteehub.com/tag/even-976335" style="font-weight:bold;" target="_blank" title="Click to know more about EVEN">EVEN</a> if a force acting on the body ,then work is zero due to frictional force and <a href="https://interviewquestions.tuteehub.com/tag/reason-620214" style="font-weight:bold;" target="_blank" title="Click to know more about REASON">REASON</a> for <a href="https://interviewquestions.tuteehub.com/tag/motion-1104108" style="font-weight:bold;" target="_blank" title="Click to know more about MOTION">MOTION</a> is <a href="https://interviewquestions.tuteehub.com/tag/friction-21545" style="font-weight:bold;" target="_blank" title="Click to know more about FRICTION">FRICTION</a> and work done due to the friction force is positive .</body></html> | |
| 38308. |
A satellite is launched into circular orbit of radius R around the earth while a second satellite is launched into an orbit of Radius 1.02 R. The percentage change in the time periods of the two satellite is |
| Answer» <html><body><p>0.7<br/>`1.0`<br/>`1.5`<br/>3</p>Answer :D</body></html> | |
| 38309. |
A uniform disc is rotating at a constantt speed in a vertical plane about a fixed horizontal axis passing through the centre of the disc. A piece of the disc from its rim detaches itself from the disc at the instant when it is at horizontal level with the centre of the disc and moving upward. Then about the fixed axis, the angular speed of the |
| Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/remaining-1184297" style="font-weight:bold;" target="_blank" title="Click to know more about REMAINING">REMAINING</a> <a href="https://interviewquestions.tuteehub.com/tag/disc-955290" style="font-weight:bold;" target="_blank" title="Click to know more about DISC">DISC</a> remains unchanged<br/>remaining disc decreases<br/>remaining disc increases <br/> <a href="https://interviewquestions.tuteehub.com/tag/broken-402937" style="font-weight:bold;" target="_blank" title="Click to know more about BROKEN">BROKEN</a> away <a href="https://interviewquestions.tuteehub.com/tag/piece-1154494" style="font-weight:bold;" target="_blank" title="Click to know more about PIECE">PIECE</a> decrease <a href="https://interviewquestions.tuteehub.com/tag/initially-516121" style="font-weight:bold;" target="_blank" title="Click to know more about INITIALLY">INITIALLY</a> and later</p>Answer :A::D</body></html> | |
| 38310. |
A block of mass m hangs on a vertical spring. Initially the spring is unstreched, it is now allowed to fall from rest. Find (a) the distance the block falls if the block is released slowely, (b) the maximum distance the block falls before it beigns to move up. |
| Answer» <html><body><p></p>Solution :When the block <a href="https://interviewquestions.tuteehub.com/tag/falls-983294" style="font-weight:bold;" target="_blank" title="Click to know more about FALLS">FALLS</a> slowely, it comes to rest at a distance `y_(0)`, which is referred to as the equlibrium <a href="https://interviewquestions.tuteehub.com/tag/position-1159826" style="font-weight:bold;" target="_blank" title="Click to know more about POSITION">POSITION</a>. From, condition of equilibrium, <br/> `sumF_(y)=ky_(0)-<a href="https://interviewquestions.tuteehub.com/tag/mg-1095425" style="font-weight:bold;" target="_blank" title="Click to know more about MG">MG</a>=0 ory_(0)=(mg)/(k)` <br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/AKS_NEO_CAO_PHY_XI_V01_MP2_C06_SLV_039_S01.png" width="80%"/> <br/> (b) When the block is released suddenly, it oscilates about the equilibrium position. Initially the speed of the block <a href="https://interviewquestions.tuteehub.com/tag/increases-1040626" style="font-weight:bold;" target="_blank" title="Click to know more about INCREASES">INCREASES</a> then reaches maximum value and then <a href="https://interviewquestions.tuteehub.com/tag/decreases-946143" style="font-weight:bold;" target="_blank" title="Click to know more about DECREASES">DECREASES</a> to zero at the lowest position. In this situation the block oscillates about the equilibrium position. The block is released from rest, there fore its total mechanical energy initially. <br/> `E_(i)=U_(g)+U_(s)+K=0+0+0` <br/> Final total mechanical energy, `E_(f)=-mgy_(m)+(1)/(2)ky^(2)+0` <br/> From conservation of energy, `E_(i) = E_(f)` <br/> `0=-mgy_(m)+(1)/(2)ky^(2)ory_(m)=(2mg)/(k)=2y_(0)`</body></html> | |
| 38311. |
Hailstones falling vertically with a speed of 10 m s^-1, hit the wind screen (wind screen makes an angle 30^@ with the horizontal) of a moving car and rebound elastically. Find the velocity of the car if the driver finds the hailstones rebound vertically after striking. . |
| Answer» <html><body><p><br/></p>Solution :For the driver to observe, the hailstones move vertically <a href="https://interviewquestions.tuteehub.com/tag/upward-721781" style="font-weight:bold;" target="_blank" title="Click to know more about UPWARD">UPWARD</a> after the <a href="https://interviewquestions.tuteehub.com/tag/elastic-967540" style="font-weight:bold;" target="_blank" title="Click to know more about ELASTIC">ELASTIC</a> collision. <br/> Let the velocity of hailstone w.r.t. car be `vec v_(h,c)` <br/> Then `vec v_h = vec v_(h,c) + vec v_c` <br/> `(vec v_h)_x = (vec v_(h,c) + vec v_c)_x` <br/> `vec v_h = vec v_(h,c) + vec v_c` <br/> `(vec v_h)_x = (vec v_(h,c) + vec v_c)_x` <br/> But `(vec v_h)_x = 0`, since hailstones fall vertically down. <br/> `rArr (vec v_h)_x = -(v cos 30^@) rArr |vec v_c|_x = |vec v_(h,c)|_x` <br/> Now, `(vec v_h)_y = (vec v_(h,c) + vec v_c)_y` <br/> Since `(v_h)_y = -10 ms^-1 , (v_c)_y = 0 rArr -10 = -V sin 30^@ + 0` <br/> `V sin 30^@ = 10 rArr V = <a href="https://interviewquestions.tuteehub.com/tag/20-287209" style="font-weight:bold;" target="_blank" title="Click to know more about 20">20</a> ms^-1` <br/> `(vec v_c)_x = V cos 30^@ = 20 xx (<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/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>))/(2) = 10 sqrt(3) ms^-1` <br/> `vec v_c = 10 sqrt(3) ms^-1`. <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/BMS_V01_C05_E01_070_S01.png" width="80%"/> , <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/BMS_V01_C05_E01_070_S02.png" width="80%"/>.</body></html> | |
| 38312. |
Give the quantities for which the following are the dimensions: M^(0)L^(0)T^(0) |
| 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/angle-875388" style="font-weight:bold;" target="_blank" title="Click to know more about ANGLE">ANGLE</a> <a href="https://interviewquestions.tuteehub.com/tag/etc-975753" style="font-weight:bold;" target="_blank" title="Click to know more about ETC">ETC</a> <a href="https://interviewquestions.tuteehub.com/tag/without-3281350" style="font-weight:bold;" target="_blank" title="Click to know more about WITHOUT">WITHOUT</a> any <a href="https://interviewquestions.tuteehub.com/tag/dimensions-439808" style="font-weight:bold;" target="_blank" title="Click to know more about DIMENSIONS">DIMENSIONS</a></body></html> | |
| 38313. |
A lift is freely falling from 8^(th) floor and is just about to stop at 4^(th) floor. Taking ground floor as origin and positive direction upwards for displacement, velocity, acceleration. Which one of the following is correct? |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/xlt0-3292391" style="font-weight:bold;" target="_blank" title="Click to know more about XLT0">XLT0</a>, vlt0, <a href="https://interviewquestions.tuteehub.com/tag/agt0-2404772" style="font-weight:bold;" target="_blank" title="Click to know more about AGT0">AGT0</a>` <br/>`<a href="https://interviewquestions.tuteehub.com/tag/xgt0-3292278" style="font-weight:bold;" target="_blank" title="Click to know more about XGT0">XGT0</a>,vlt0,alt0` <br/>`xgt0,vlt0,agt0` <br/>`xgt0,vgt0,alt0` </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> | |
| 38314. |
A man walks on a straight road from his home to a market 2.5km away with a speed of 5km//h . Finding the market closed , heinstantley turns and walks back home with a speed of 7.5km//h. The average speed of the man over the interval of time 0 to 40 nim , is equal to |
| 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> :`45/8km//h`</body></html> | |
| 38315. |
A wheel in uniform motion about an axis passing through its centre and perpendicular to its plane is considered to be in mechanical (translation plus rotational) eqilibrium because no net external force ot torque is required to sustain its motion. However, the particles that constitute the wheel do experience a centripletal acceleration directed towards the centre. How do you reconcile this fact with the wheel being in equilibrium ? How would you set a half-wheel into uniform motion about an axis passing through the centre of mass of the wheel and perpendicular to its plane ? Will you require external forces to sustain the motion ? |
| Answer» <html><body><p></p>Solution :A <a href="https://interviewquestions.tuteehub.com/tag/wheel-736315" style="font-weight:bold;" target="_blank" title="Click to know more about WHEEL">WHEEL</a> is a rigid body. The centripetal <a href="https://interviewquestions.tuteehub.com/tag/accelerations-846663" style="font-weight:bold;" target="_blank" title="Click to know more about ACCELERATIONS">ACCELERATIONS</a> of the particle of the wheel arise due to the internal elastic forces which cancel out in <a href="https://interviewquestions.tuteehub.com/tag/paries-2913851" style="font-weight:bold;" target="_blank" title="Click to know more about PARIES">PARIES</a>. <br/> In a half wheel, the distribution of mass about its centre of mass (through which axis of rotation <a href="https://interviewquestions.tuteehub.com/tag/passes-1148517" style="font-weight:bold;" target="_blank" title="Click to know more about PASSES">PASSES</a>) is not symmertical. Therefore, the direction of angular <a href="https://interviewquestions.tuteehub.com/tag/momentum-1100588" style="font-weight:bold;" target="_blank" title="Click to know more about MOMENTUM">MOMENTUM</a> of the wheel does not coincide with the direction of its angular velocity. Hence, an external torque is required to maintain the motion of the half wheel.</body></html> | |
| 38316. |
Two identical springs each of force constant k are connected in series Calculate the frequency of the system |
| Answer» <html><body><p></p>Solution :For series <a href="https://interviewquestions.tuteehub.com/tag/combination-922669" style="font-weight:bold;" target="_blank" title="Click to know more about COMBINATION">COMBINATION</a>, the <a href="https://interviewquestions.tuteehub.com/tag/effective-966090" style="font-weight:bold;" target="_blank" title="Click to know more about EFFECTIVE">EFFECTIVE</a> force <a href="https://interviewquestions.tuteehub.com/tag/costant-2550125" style="font-weight:bold;" target="_blank" title="Click to know more about COSTANT">COSTANT</a>, <br/> `k_(s)=(kxxk)/(k+k)= (k)/(2)` <br/> `:.` Frequency, <br/> `v_(s)=(1)/(2pi) sqrt((k_(s))/(m))=(1)/(2pi) sqrt((k//2)/(m))` <br/> `=(1)/(2pi) sqrt((k)/(2m)) ""...(1)`</body></html> | |
| 38317. |
Figure shows the P- V diagram of an ideal gas undergoing a change of state from A to B. Four different parts I, II, III and IV as shown in the figure may lead to the same change of state. |
| Answer» <html><body><p>Change in internal energy is same in IV and III cases, but not in I and II.<br/>Change in internal energy is same in all the four cases.<br/>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> is maximum in case I <br/>Work done is minimum in case II.</p>Solution :Change in internal energy from state A to state B, <br/> `dU_(A to B) = n C_V DeltaT` <br/> `=nC_V(T_B-T_A)` <br/> <a href="https://interviewquestions.tuteehub.com/tag/depends-948548" style="font-weight:bold;" target="_blank" title="Click to know more about DEPENDS">DEPENDS</a> on only the temperature of A and B. <a href="https://interviewquestions.tuteehub.com/tag/hence-484344" style="font-weight:bold;" target="_blank" title="Click to know more about HENCE">HENCE</a>, internal energy is same for all four paths. It depends on the temperature of A and B. Moreover, the work done from state A to state B.`W_(A to B)` = <a href="https://interviewquestions.tuteehub.com/tag/area-13372" style="font-weight:bold;" target="_blank" title="Click to know more about AREA">AREA</a> under the graph `P to V` It is maximum for path I.</body></html> | |
| 38318. |
The angular frequency of a fan increases from 30 rpm to 60 rpm in pi s. A dust particle is present at a distance of 20 cm from axis of rotation. The tangential acceleration of the particle in pi s is |
| Answer» <html><body><p>`0.8 <a href="https://interviewquestions.tuteehub.com/tag/ms-549331" style="font-weight:bold;" target="_blank" title="Click to know more about MS">MS</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/>`0.34 ms^(-2)`<br/>`0.2 ms^(-2)`<br/>`1.2 ms^(-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> :<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> | |
| 38319. |
A horizontal disc rotates freely about a vertical axis through its centre. A ring having the same mass and radius as the disc is now gently placed on the disc After same time, the two rotate with a common angular velocity |
| Answer» <html><body><p>some friction exists between the <a href="https://interviewquestions.tuteehub.com/tag/disc-955290" style="font-weight:bold;" target="_blank" title="Click to know more about DISC">DISC</a> and the rind <br/>the angular momentum of the disc plus ring is <a href="https://interviewquestions.tuteehub.com/tag/conserved-2539138" style="font-weight:bold;" target="_blank" title="Click to know more about CONSERVED">CONSERVED</a> <br/>the <a href="https://interviewquestions.tuteehub.com/tag/final-461168" style="font-weight:bold;" target="_blank" title="Click to know more about FINAL">FINAL</a> common angular angular velocity is `2//3^("<a href="https://interviewquestions.tuteehub.com/tag/rd-613594" style="font-weight:bold;" target="_blank" title="Click to know more about RD">RD</a>")` of the initial angular velocity of the disc <br/>`2//3^("rd")` of the initial kinetic energy <a href="https://interviewquestions.tuteehub.com/tag/changes-913881" style="font-weight:bold;" target="_blank" title="Click to know more about CHANGES">CHANGES</a> to heat </p>Answer :A::B::D</body></html> | |
| 38320. |
A projectile is thrown at an angle of 30^@with a velocity of 10m/s. the change in velocity during the time interval in which it reaches the highest point is |
| Answer» <html><body><p>10m/s<br/>5m/s<br/>`<a href="https://interviewquestions.tuteehub.com/tag/5sqrt3-1900573" style="font-weight:bold;" target="_blank" title="Click to know more about 5SQRT3">5SQRT3</a> m/s ` <br/>`10sqrt3 m/s `</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> | |
| 38321. |
The two ends of a metal rod are maintained at temperatures 100^(@)C and 110^(@)C. The rate of heat flow in the rod is found to be 4.0" Js"^(-1). If the ends are maintained at temperatures 200^(@)C and 210^(@)C, the rate of heat flow will be : |
| Answer» <html><body><p>`16.8" Js"^(-<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/>`8.0" Js"^(-1)`<br/>`4.0" Js"^(-1)`<br/>`44.0" Js"^(-1)`</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> :From `H=kA((T_(1)-T_(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>))`, <br/> (i) `H_(1)kA((110-100)/(L))` <br/> `H_(1)=<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>(kA)/(L)` <br/> (ii) `H_(2)=kA((210-20)/(L))=10(kA)/(L)` <br/> `:.H_(1)=H_(2)` <br/> `:." Js"^(-1)=H_(2)`</body></html> | |
| 38322. |
Is whole of the kinetic energy lost in any perfectly inelastic collision? |
| 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, only that <a href="https://interviewquestions.tuteehub.com/tag/much-2164829" style="font-weight:bold;" target="_blank" title="Click to know more about MUCH">MUCH</a> <a href="https://interviewquestions.tuteehub.com/tag/amount-374803" style="font-weight:bold;" target="_blank" title="Click to know more about AMOUNT">AMOUNT</a> of kinetic <a href="https://interviewquestions.tuteehub.com/tag/energy-15288" style="font-weight:bold;" target="_blank" title="Click to know more about ENERGY">ENERGY</a> is lost as is necessary for the <a href="https://interviewquestions.tuteehub.com/tag/conservation-929810" style="font-weight:bold;" target="_blank" title="Click to know more about CONSERVATION">CONSERVATION</a> of momentum.</body></html> | |
| 38323. |
A ball is projected vertically upwards with a velocity v . It comes back to ground in time t. which v-t graph shows the motion correctly ? |
| Answer» <html><body><p><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/FM_PHY_XI_SP_20_E01_004_O01.png" width="30%"/> <br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/FM_PHY_XI_SP_20_E01_004_O02.png" width="30%"/> <br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/FM_PHY_XI_SP_20_E01_004_O03.png" width="30%"/> <br/><img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/FM_PHY_XI_SP_20_E01_004_O04.png" width="30%"/> </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> | |
| 38324. |
State Newton's law of cooling verify with an experiment. |
| 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>></p>Solution :(i) It states that the rate of cooling of a body is directly proportional to the temperature differ between the body and the surroundings. <br/> (ii) Consider a spherical calorimeter of mass m whose outer surface is blackened. It is filled with <a href="https://interviewquestions.tuteehub.com/tag/hot-1029680" style="font-weight:bold;" target="_blank" title="Click to know more about HOT">HOT</a> water of mass m, the calorimeter with thermometer is suspended from a stand. <br/> (iii) The calorimeter & the hot water radiate heat energy to the surrounding. Using a stop clock, the temperature is noted for every 30 second interval of time see the temperature falls by about `20^(@)<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a>`. The readings are tabulated. <br/> If the temperature falls from `T_(1)` to `T_(2)` in t sec, the quantity of heat energy lost by radiation `<a href="https://interviewquestions.tuteehub.com/tag/q-609558" style="font-weight:bold;" target="_blank" title="Click to know more about Q">Q</a>=(ms+m_(1)s_(1))(T_(1)-T_(2))`, where 'S' is the specific heat capacity of the <a href="https://interviewquestions.tuteehub.com/tag/material-1089170" style="font-weight:bold;" target="_blank" title="Click to know more about MATERIAL">MATERIAL</a> of the calorimeter & `S_(1)` - specific heat capacity of water. <br/> Rate of cooling = `("Heat energy lost")/("time "t_(n))` <br/> `therefore (Q)/(E)=((ms+m_(1)s_(1))(T_(1)-T_(2)))/(t)` <br/> Room temperature - `T_(0)` <br/> (v) Average excess temperature of the colorimeter over that of the surroundings = `(T_(1)+T_(2))/(2)-T_(0)` <br/> (vi) Acceleration to Newton's law of cooling `(Q)/(t)prop((T_(1)+T_(2))/(2)-T_(0))` <br/> (vii) Assume the pressure of the gas remains constant during an infinitesimally small outward displacement dy then work done `dW-F.dx=P.A.dx` <br/> `dW=P.dV` <br/> (viii) Total work done by the gas from volume `V_(1)` to `V_(2)` is <br/> `W=underset(V_(1))overset(V_(2))intP.dv` <br/> (ix) But `PV^(gamma)` = constant (k) <br/> `gamma=(C_(P))/(C_(V))` <br/> `therefore W=underset(V_(1))overset(V_(2))intkV^(gamma)dV=k[(V^(1-gamma))/(1-gamma)]_(v_(1))^(v_(2))" "[becauseP=(k)/(V^(gamma))]` <br/> `thereforeW=(k)/(1-gamma)[V_(2)^(1-gamma)-V_(1)^(1-gamma)]` <br/> `W=(1)/(1-gamma)[kV_(2)^(1-gamma)-kV_(1)^(1-gamma)]` <br/> `P_(2)V_(2)^(gamma)=P_(1)V_(2)^(gamma)=k` <br/> (x) Subtract the value of k <br/> `therefore W=(1)/(1-gamma)[P_(2)V_(2)^(gamma),V_(2)^(1-gamma)-P_(1)V_(1)^(gamma),V_(1)^(1-gamma)]` <br/> `W=(1)/(1-gamma)[P_(2)V_(2)-P_(1)V_(1)]` <br/> It `T_(2)` is the final temperature of the gas in adiable expansion, then <br/> `P_(1)V_(1)=RT_(1),P_(2)V_(2)=RT_(2)` <br/> `therefore W=(1)/(1-gamma)[RT_(2)-RT_(1)]` <br/> This is the equation for the work done during adiabating process.</body></html> | |
| 38325. |
If radius of earth becomes half, then find the weight. |
| 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> :`implies` Weight `W = <a href="https://interviewquestions.tuteehub.com/tag/mg-1095425" style="font-weight:bold;" target="_blank" title="Click to know more about MG">MG</a> =(GM_em)/R_e^2` <br/> Since the <a href="https://interviewquestions.tuteehub.com/tag/radius-1176229" style="font-weight:bold;" target="_blank" title="Click to know more about RADIUS">RADIUS</a> becomes half, <br/> `W. = mg. = (GM_(e) m)/((R_e^2)/4)` <br/> `:. W. = 4 ((GM_em)/R_e^2)` <br/> `W. = 4W`</body></html> | |
| 38326. |
A body of mass 1.5 kg is allowed to slide down along a quadrant of a circle from the horizontal position. In reaching to the bottom, Its velocity is 8m/s. The work done in overcoming the friction is 12J. The radius of circle is (g = 10 ms^(-2)) |
| Answer» <html><body><p><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/>3 m<br/>5 m<br/>2 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> | |
| 38327. |
In Millikan's oil drop experiment what is the terminal speed of an unchanged drop of radius 20xx10^-5m and density 1.2xx10^3 kg. m^-3. Take the viscosity of air at the temperature of the experiment to be 1.8xx10^-5Pa. s. How much is the viscous force on the drop at that speed ? Neglect buoyancy of the drop due to air. |
| 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> :`5.8cm. S^-1, 3.9xx10^(-<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/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a>`</body></html> | |
| 38328. |
Two capacitors of capacities (3pm0.09)mu F and (6pm 0.09)muF are connected in parallel. Their equivalent capacity is |
| Answer» <html><body><p>`( 2 <a href="https://interviewquestions.tuteehub.com/tag/pm-1156895" style="font-weight:bold;" target="_blank" title="Click to know more about PM">PM</a> 0.13 ) <a href="https://interviewquestions.tuteehub.com/tag/mu-566056" style="font-weight:bold;" target="_blank" title="Click to know more about MU">MU</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>`<br/>`( <a href="https://interviewquestions.tuteehub.com/tag/9-340408" style="font-weight:bold;" target="_blank" title="Click to know more about 9">9</a> pm 0.18 ) mu F`<br/>`( 2 pm 0.09 ) mu F`<br/>`( 9 pm 0.09) muF`</p>Answer :B</body></html> | |
| 38329. |
Explain surface energy and surface tension. |
| Answer» <html><body><p></p>Solution :An extra energy is associated with surface of liquid , the creation of more surface keeping volume fixed requires additional energy.<br/><img src="https://doubtnut-static.s.llnwi.net/static/physics_images/KPK_AIO_PHY_XI_P2_C10_E01_059_S01.png" width="80%"/><br/>As <a href="https://interviewquestions.tuteehub.com/tag/shown-1206565" style="font-weight:bold;" target="_blank" title="Click to know more about SHOWN">SHOWN</a> in figure (a) a U- shape frame is made from wire , the wire PQ slide without firction over the rides AP and BQ.<br/>This frame dipped and taken out of the soap solution , thin film APQB is formed from the frame.<br/>In figure (a) film is in equilibrium. <br/>In figure (b) it is shown that the film is <a href="https://interviewquestions.tuteehub.com/tag/stretched-7711564" style="font-weight:bold;" target="_blank" title="Click to know more about STRETCHED">STRETCHED</a> at extra distance d.<br/> Since the area of the surface increases ,the system now more energy , means that some work has been done against an internal force.<br/>Let this internal force be F.The work done by the applied force is , <br/> `W=vecF*vecd=Fd` ...(1)<br/>From the <a href="https://interviewquestions.tuteehub.com/tag/conservation-929810" style="font-weight:bold;" target="_blank" title="Click to know more about CONSERVATION">CONSERVATION</a> of energy , this is stored as additional energy in the film.<br/>If the surface energy of the film is S per unit area , the extra area is 2ld. (Because `DeltaA`=length `xx` breadth ld and film has two free surface,so 2ld is present.<br/> The liquid in between , so there are two surfaces and the extra energy is`=(2ld)S`.<br/>`thereforeW=(2ld)S=SDeltaA[because2ld`= increase in area]<br/>`thereforeS=(W)/(DeltaA)`<br/>`thereforeS=(fd)/(2ld)=(F)/(2l)` <br/>This quantity S is the magnitude of surface tension .It is <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 surface energy per unit area of the liquid interface and is also equal to the forceper unit lenght exerted by the <a href="https://interviewquestions.tuteehub.com/tag/fluid-993483" style="font-weight:bold;" target="_blank" title="Click to know more about FLUID">FLUID</a> on the movable bar.</body></html> | |
| 38330. |
The gravitational potential energy of a body at distance r from the centre of earth is U. Its weight at a distance 2r from the centre of earth is |
| Answer» <html><body><p>U/r <br/>U/<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/>U/4r <br/>`U//sqrt(2r)` </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> | |
| 38331. |
……. Is the best conductor of heat. |
| 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/sliver-645857" style="font-weight:bold;" target="_blank" title="Click to know more about SLIVER">SLIVER</a></body></html> | |
| 38332. |
Less accurate of the four options given below |
| Answer» <html><body><p>9.27<br/>41<br/>1.01<br/>`9.00 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> <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/0-251616" style="font-weight:bold;" target="_blank" title="Click to know more about 0">0</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> | |
| 38333. |
A child running a temperature of 101^(@)F is given an antipyrin (i.e. a medicine that lowers fever) which causes an increase in the rate of evaporation of sweet from his body. If the fever is brought down to 98^(@)F in 20 min., what is the average rate of extra evaporation caused, by the drug ? Assume the evaporation mechanism to be the only way by which heat is lost. The mass of the child is 30 kg. The specific heat of human body is approximately the same as that of water and latent heat of evaporation of water at that temperature is about 580cal.g^(-1). |
| Answer» <html><body><p></p>Solution :Here, fall in <a href="https://interviewquestions.tuteehub.com/tag/temp-1240937" style="font-weight:bold;" target="_blank" title="Click to know more about TEMP">TEMP</a>. <br/> = `DeltaT=101-98=3^(@)F=3xx5/(9)""^(@)C=5//3""^(@)C` <br/> Mass of child, m = 30 kg <br/> <a href="https://interviewquestions.tuteehub.com/tag/sp-1219706" style="font-weight:bold;" target="_blank" title="Click to know more about SP">SP</a>. heat of human body = sp. heat of water, <br/> `c=1000cal.kg^(-1)""^(@)C^(-1)` <br/> `therefore` Heat lost by the child, <br/> `DeltaQ=mcDeltaT=30xx1000xx5/3=50000"cals"` <br/> If m. be the mass of water evaporation in 20 min. <br/> then, `m.L=DeltaQorm.=(DeltaQ)/L=50000/580=86.2g` <br/> `therefore` <a href="https://interviewquestions.tuteehub.com/tag/average-13416" style="font-weight:bold;" target="_blank" title="Click to know more about AVERAGE">AVERAGE</a> rate of <a href="https://interviewquestions.tuteehub.com/tag/extra-447191" style="font-weight:bold;" target="_blank" title="Click to know more about EXTRA">EXTRA</a> evaporation <br/> = `86.2/20=4.31g"min"^(-1)`</body></html> | |
| 38334. |
In non-uniform circular motion , particle will have . . . . . And tangential acceleration |
| Answer» <html><body><p>centripetal<br/>centrifugal<br/>both (a) and (<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/><a href="https://interviewquestions.tuteehub.com/tag/neither-1113494" style="font-weight:bold;" target="_blank" title="Click to know more about NEITHER">NEITHER</a> (a) nor (b)</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> | |
| 38335. |
A particle moves on a circle of radius r with centripetal acceleration as function of time as a_e = k^2 rt^2where k is a positive constant. Find the resultant acceleration. |
| Answer» <html><body><p>`kt^2` <br/>`<a href="https://interviewquestions.tuteehub.com/tag/kr-534811" style="font-weight:bold;" target="_blank" title="Click to know more about KR">KR</a>` <br/>`kr <a href="https://interviewquestions.tuteehub.com/tag/sqrt-1223129" style="font-weight:bold;" target="_blank" title="Click to know more about SQRT">SQRT</a>(k^2 t^4 + <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/>`kr sqrt(k^2 r^4 - 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 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> | |
| 38336. |
Law od gravitation is not applicable if (A) Velocity of moving objects are comparable to velocity of light (B) Gravitational field between objects whose masses are greater than the mass of sun. |
| Answer» <html><body><p>A is true, B 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, B is true<br/>Both A & B are true <br/>Both A & B are false</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> | |
| 38337. |
The dimensional equation for magnetic flux is |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</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>)T^(-2)I^(-<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/>`ML^(2)T^(-2)I^(-2)`<br/>`ML^(-2) T^(-2) I^(-1)`<br/>`ML^(-2)T^(-2)I^(-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> :A</body></html> | |
| 38338. |
The potential energy in joules of a particle of mass 1 kg moving in a plane is given by U = 3x + 4y, the position coordinates of the point being x and y, measured in metres. If the particle is initially at rest at (6, 4), then: |
| Answer» <html><body><p>its <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 of magnitude `<a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a> m//s^(2)`<br/>its speed when it <a href="https://interviewquestions.tuteehub.com/tag/crosses-939624" style="font-weight:bold;" target="_blank" title="Click to know more about CROSSES">CROSSES</a> they-axis is `10 m//s`<br/>it crosses the y-axis `(x = 0)` at `y = -4`<br/>it moves in a straight line passing through the origin (0, O) </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::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></body></html> | |
| 38339. |
A disc of mass 50 kg rotates freely about a fixed horizontal axis through its centre. A thin cotton pad fixed to its rim can absorb water. Water drips onto the pad at the rate of 25 grams per sec, falling along a line passing through the horizontal axis. The angular velocity of the disc will get reduced to one fourth of its initial value time ____x10^(3) s. |
| 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/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a></body></html> | |
| 38340. |
Statement I : For a floating body to be in stable equilibrium, its centre of buoyancy must be located above the centre of gravity. Statement II : The torque developed by the weight of the body and the upthrust will restore the body back to its normal position, after the body is disturbed. |
| Answer» <html><body><p>Statement I is true, statement <a href="https://interviewquestions.tuteehub.com/tag/ii-1036832" style="font-weight:bold;" target="_blank" title="Click to know more about II">II</a> is true , statement II is a <a href="https://interviewquestions.tuteehub.com/tag/correct-409949" style="font-weight:bold;" target="_blank" title="Click to know more about CORRECT">CORRECT</a> <a href="https://interviewquestions.tuteehub.com/tag/explanation-455162" style="font-weight:bold;" target="_blank" title="Click to know more about EXPLANATION">EXPLANATION</a> for statement I.<br/>Statement I is true, statement II is true , statement II is not a correct explanation for statement I.<br/>Statement I is true, statement II is false.<br/>Statement I is false, statement II is true.</p>Answer :D</body></html> | |
| 38341. |
Supposeapersonwantsto increasethe efficienyof thereversibleheatenginethatisoperatingbetween100^(@)C and 300^(@)C Hehadtwowaysto increasethe efficiency . ( a)Bydecreasing thecoldreservoirtemperaturefrom 100^(@)C to 50^(@)Candkeepingthe hotreservoirtemperatureof the hotreservoirfrom 300^(@)Cto 350^(@)Cbykeepingthecoldreservoirtemperatureconstnatwhichis thesuitablemethod ? |
| Answer» <html><body><p></p>Solution :Heat engine operates at initial temperature `=100^(@)C+273=373K` <br/> Final temperature `=300^(@)C+273=573 K` <br/> At <a href="https://interviewquestions.tuteehub.com/tag/melting-1093090" style="font-weight:bold;" target="_blank" title="Click to know more about MELTING">MELTING</a> point `=273K` <br/> `"Efficiency "eta=1-(T_(2))/(T_(1))=1-(<a href="https://interviewquestions.tuteehub.com/tag/272-1832893" style="font-weight:bold;" target="_blank" title="Click to know more about 272">272</a>)/(573)=0.3491, eta=34.9%` <br/> (a) By <a href="https://interviewquestions.tuteehub.com/tag/decreasing-946148" style="font-weight:bold;" target="_blank" title="Click to know more about DECREASING">DECREASING</a> the cold <a href="https://interviewquestions.tuteehub.com/tag/reservoir-1186140" style="font-weight:bold;" target="_blank" title="Click to know more about RESERVOIR">RESERVOIR</a>, efficiency <br/> `T_(1)=350^(@)C+273=623K, T_(2)=100^(@)C+273=373K` <br/> `eta=1-(T_(2))/(T_(1))=1-(323)/(573)=0.436` <br/> `eta=43.6%` <br/> (<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>) By increasing the temperature of hot reservoir, efficiency, <br/> `T_(1)=350^(@)C+273=623K, T_(2)=100^(@)C+273=373K` <br/> `eta=1-(T_(2))/(T_(1))=1-(373)/(623)=0.401` <br/> `eta=40.12%` <br/> Method (a) More efficiency than method (b).</body></html> | |
| 38342. |
A ballet dancer spins about a vertical axis at 120 rpm with arms outstretched. With her arms folded, the moment of inertia about the axis of rotation decreases by 40%. Calculate the new rate of rotation. |
| Answer» <html><body><p></p>Solution :`n_(1)=120 rpm , n_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)= ?"" I_(1)omega_(1)=I_(2)omega_(2)`<br/>Let `I_(1)=<a href="https://interviewquestions.tuteehub.com/tag/100-263808" style="font-weight:bold;" target="_blank" title="Click to know more about 100">100</a>` <a href="https://interviewquestions.tuteehub.com/tag/units-1438468" style="font-weight:bold;" target="_blank" title="Click to know more about UNITS">UNITS</a>, then `I_(2)=60` units <br/>`100xx120 = 60xx omega_(2) rArr omega_(2)=<a href="https://interviewquestions.tuteehub.com/tag/200-288914" style="font-weight:bold;" target="_blank" title="Click to know more about 200">200</a>` rpm.</body></html> | |
| 38343. |
The error in the measurement of radius of a drde is 0.6%. Find the percentage error in the calculation of the area of the circle. |
| 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/area-13372" style="font-weight:bold;" target="_blank" title="Click to know more about AREA">AREA</a> `A=pir^2(<a href="https://interviewquestions.tuteehub.com/tag/deltaa-2570644" style="font-weight:bold;" target="_blank" title="Click to know more about DELTAA">DELTAA</a>)/Axx100=2xx(<a href="https://interviewquestions.tuteehub.com/tag/deltar-2053542" style="font-weight:bold;" target="_blank" title="Click to know more about DELTAR">DELTAR</a>)/rxx100=2xx0.6=1.2%`</body></html> | |
| 38344. |
Two rods of different materials having coefficients of thermal expansion alpha_(1) , alpha_(2) and Young's Moduli Y_(1), Y_(2) respectively are fixed between two rigid massive walls. The rods are heated such that they undergo the same increase in temperature. There is no bending of the rods. If alpha_(1) : alpha_(2) = 2 : 3, the thermal stresses developed in the two rods are equal provided Y_(1) : Y_(2) is equal to |
| Answer» <html><body><p>`<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>:3`<br/>`1:1`<br/>`3:2`<br/>`4:9`<br/></p>Solution :Expansion in the rod due to <a href="https://interviewquestions.tuteehub.com/tag/rise-1189674" style="font-weight:bold;" target="_blank" title="Click to know more about RISE">RISE</a> in temp Compression in rod. For the first rod`Deltal_(1)=alpha_(1)l_(1)Deltatheta` <br/> Compression in the rod `Deltal_(1)= (-F)/(A) (l_(1))/(Y_(1))` <br/> or the length of the rod remains unchanged <br/> `alpha_(1)l_(1)<a href="https://interviewquestions.tuteehub.com/tag/delta-947703" style="font-weight:bold;" target="_blank" title="Click to know more about DELTA">DELTA</a> theta= (-Fl_(1))/(AY_(1))` <br/> or <br/> `alpha_(1) Delta theta = (-F)/(AY_(1))--- (i)` <br/> Similarly for second rod <br/> `alpha_(2) Delta theta = (-F)/(AY_(2))--- (ii)` <br/>`(alpha_(1))/(alpha_(2))=(Y_(2))/(Y_(1))or(Y_(1))/(Y_(2))=(alpha_(2))/(alpha_(1))=(3)/(2)`</body></html> | |
| 38345. |
A block of mass 15 kg is placed on a long trolley . The coefficient of friction between the block andthe trolley is 0.18. The trolley accelerates from rest with 0.5ms^(-2) for 20s and then moves with uniform velocity. Discuss the motion of theblock as viewed by (a) a stationary observer on theground (b) as observer moving with the trolley. |
| Answer» <html><body><p><br/></p>Solution :`F_(s)=mu_(s)N=mu_(s)mg=0.18xx15xx9.8=26.46N`. Force on block `=F=ma=15xx7.5=7.5N F_(s) <a href="https://interviewquestions.tuteehub.com/tag/gt-1013864" style="font-weight:bold;" target="_blank" title="Click to know more about GT">GT</a> F`.So the block cannot move. (a) The stationary <a href="https://interviewquestions.tuteehub.com/tag/observer-1127626" style="font-weight:bold;" target="_blank" title="Click to know more about OBSERVER">OBSERVER</a> will see the accelerated and the uniformmotion . (b) The observeris in an accelerated <a href="https://interviewquestions.tuteehub.com/tag/frame-998974" style="font-weight:bold;" target="_blank" title="Click to know more about FRAME">FRAME</a>. So laws of motion are not applicable. But during uniform motion, he will see that the block is at <a href="https://interviewquestions.tuteehub.com/tag/rest-11386" style="font-weight:bold;" target="_blank" title="Click to know more about REST">REST</a> w.r.t him.</body></html> | |
| 38346. |
The moment of inertia of a thin uniform rod of mass M and length L about an axis passing through its midpoint and perpendicular to its length is I_(0) . Its moment of inertia about an axis passing through one of its ends and perpendicular to its length is |
| Answer» <html><body><p>`I_(0) +ML^(2)//2`<br/>`I_(0) + ML^(2)//4`<br/>`I_(0)+ <a href="https://interviewquestions.tuteehub.com/tag/2ml-1837952" style="font-weight:bold;" target="_blank" title="Click to know more about 2ML">2ML</a>^(2)`<br/>`I_(0) + ML^(2)`</p>Solution :According to the theorem of parallel axes , the moment of inertia of the <a href="https://interviewquestions.tuteehub.com/tag/thin-707663" style="font-weight:bold;" target="_blank" title="Click to know more about THIN">THIN</a> rod 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> M and length L about an axis passing through one of the ends is <br/> `I = I_(CM) + Md^2` <br/> where `I_(CM)` is the moment of inertia of the given rod about an axis passing through its centre of mass and perpendicular to its length and d is the distance between <a href="https://interviewquestions.tuteehub.com/tag/two-714195" style="font-weight:bold;" target="_blank" title="Click to know more about TWO">TWO</a> parallel axes. <br/> Here `I_(CM) = I_(0) , d = (L)/(2) therefore I = I_(0) + M ((L)/(2))^(2) = I_(0) + (ML^(2))/(4)`</body></html> | |
| 38347. |
From a uniform disk of radius R, a circular hole of radius R/2 is cut out. The centre of the hole is at R/2 from the centre of the original disc. Locate the centre of gravity of the resulting flat body. |
| 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> :at R/6 from the <a href="https://interviewquestions.tuteehub.com/tag/center-11455" style="font-weight:bold;" target="_blank" title="Click to know more about CENTER">CENTER</a> of original disc <a href="https://interviewquestions.tuteehub.com/tag/opposite-1137237" style="font-weight:bold;" target="_blank" title="Click to know more about OPPOSITE">OPPOSITE</a> to the center of <a href="https://interviewquestions.tuteehub.com/tag/cut-942198" style="font-weight:bold;" target="_blank" title="Click to know more about CUT">CUT</a> <a href="https://interviewquestions.tuteehub.com/tag/portion-1159785" style="font-weight:bold;" target="_blank" title="Click to know more about PORTION">PORTION</a></body></html> | |
| 38348. |
A satellite s is moving in an elliptical orbit ,paround the earth. The mass of the satellite is very small compared to the mass of the earth, Then, |
| Answer» <html><body><p>The acceleration of s is always directedtowards the centre 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/>The angular momentum of s about the centreof the earth <a href="https://interviewquestions.tuteehub.com/tag/changes-913881" style="font-weight:bold;" target="_blank" title="Click to know more about CHANGES">CHANGES</a> in direction, but <a href="https://interviewquestions.tuteehub.com/tag/itsmagnitude-2757917" style="font-weight:bold;" target="_blank" title="Click to know more about ITSMAGNITUDE">ITSMAGNITUDE</a> remains constant. <br/>The <a href="https://interviewquestions.tuteehub.com/tag/total-711110" style="font-weight:bold;" target="_blank" title="Click to know more about TOTAL">TOTAL</a> mechanical energy of s variesperiodically with time.<br/>The magnitude of s remains constant</p>Answer :A</body></html> | |
| 38349. |
If the value of 'g' on the surface of the moon is g// 6, the time period of a pendulum on the surface of the moon w.r.t time period on the earth will be |
| Answer» <html><body><p>`<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/6-327005" style="font-weight:bold;" target="_blank" title="Click to know more about 6">6</a>)` <a href="https://interviewquestions.tuteehub.com/tag/times-1420471" style="font-weight:bold;" target="_blank" title="Click to know more about TIMES">TIMES</a><br/>6 times<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>)/(sqrt(6))` times<br/>`(1)/(6)` times</p>Answer :A</body></html> | |
| 38350. |
(A) : The soldiers marching on a suspended bridge are advised to go out of steps. (R): Frequency of marching steps may match with the natural frequency of oscillation of bridge. |
| Answer» <html><body><p>Both 'A' and '<a href="https://interviewquestions.tuteehub.com/tag/r-611811" style="font-weight:bold;" target="_blank" title="Click to know more about R">R</a>' are true and R' 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 'A'<br/>Both 'A' and 'R' are true and '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' is true</p>Answer :A</body></html> | |